MED PART I OPTIMIZING ENERGY EFFICIENCY AND HEAT … · 10.1 Purpose of heating process water 10.2...

ENERGIETRANSITIE P1/MED I/DRYING/FdG/V3/06-11-2009/PAGE 1

MED PART I

OPTIMIZING ENERGY EFFICIENCY AND HEAT RECOVERY

DRYING SECTION

Model improving Energy efficiency Drying process (MED)

FdG/6th November 2009/Version: V3 Copyright FdG ©2009 All rights reserved.

ROYAL VNP


Preface

By far the largest share of energy use in a paper mill is in the drying sections.

Figures from the Dutch paper industry show that 84% of the steam consumption and

about 50% of the total energy consumption in paper production is in the drying

sections. About 87% of the energy used in these drying sections is in the form of

steam, 8,5% in the form of electricity and 4,5 % is directly used fuel (mainly gas).

The total amount of product water evaporated in the Dutch paper and board mills is

3.800.000 m3 per year. The heat consumption is between 3050 and 7000 MJ/m3

product water evaporation; this is about 400.000.000 m3 natural gas equivalent per

year.

This handout presents an overview of the possibilities to optimize the energy efficiency

and heat recovery of the drying sections in paper and board mills.

For better understanding the energy efficiency of drying processes basic knowledge

about psychrometrics and thermodynamics will be presented first.

Next the principles of drying are pointed out as well as the theoretical background

behind the energy consumption of water evaporation. This is explained with help of the

Mollier chart.

Then several possibilities to optimize the energy efficiency of the drying sections and

the heat recovery are discussed; it will be shown that:

- At higher temperatures energy consumption decreases and quality of heat in

exhaust air for heat recovery (HRC) improves

- The higher the temperature the less the energy consumption decreases

- The higher the temperature the more the quality of heat in exhaust air improves

- At higher temperatures the heat for process water, spray water and

space-heating can (almost) be covered by HRC (no fresh steam!!)

To assess the optimal process conditions and the energy saving possibilities belonging

to these conditions often extensive and time-consuming calculations are required.

To prevent this the MED-software is developed. MED is an abbreviation for Model

improving Energy efficiency Drying process.

MED calculates, for the whole product range (g/m2), the current and optimal conditions,

the possibilities for heat recovery and the energy savings.

This handout provides also the necessary knowledge for understanding MED.

Royal VNP

Postbus 731

2130AS Hoofddorp

www.vnp-online.nl

T 020 6543055

http://www.vnp-online.nl/


Table of contents MED part I

PART I FUNDAMENTALS OF DRYING

1 PSYCHROMETRICS

1.1 Introduction

1.2 Thermo physical properties of humid air

1.3 Relation between pressure, temperature and vapor content

2 THERMO-DYNAMICS

2.1 Introduction

2.2 Gas laws

2.3 Relation between mass, amount of substance and volume

2.4 Calculation mass flow from volume flow and temperature

2.5 Pitot-tube for measuring volume flows

2.6 Calculation enthalpy of humid air

3 FORMULA LIBRARY FOR DRYING PROCESSES

3.1 Overview formulas

3.2 Method of working humid air

3.3 Method of working combustion gas

4 MOLLIER CHART

4.1 Introduction

4.2 Explanation Mollier chart

PART II THERMAL PAPER AND BOARD DRYING

5 DRYING PRINCIPLES

5.1 Product water removal

5.2 Thermal drying

5.3 Drying periods and rates

5.4 Air systems

5.5 Supply air

5.6 Dryer hoods

5.7 Zero pressure level

5.8 Exhaust air

6 ENERGY CONSUMPTION DRYERS

6.1 System boundary heat consumption drying section

6.2 Heat and mass balance drying section

6.3 Heat consumption per kg evaporated product water

6.4 Relationship between exhaust air temperature and heat consumption

PART III HEAT RECOVERY

7 IMPROVING QUALITY OF HEAT IN EXHAUST AIR

7.1 Increasing the exhaust air temperature

7.2 Explanation heat recovery process in Mollier chart


7.3 Improving heat recovery potential multi-cylinder dryers

7.4 Improving heat recovery potential yankee-cylinder dryers

7.5 Demonstration energy-efficiency and heat recovery potential dryers

8 SELECTION OF HEAT EXCHANGERS

8.1 Design and selection

8.2 Indicative k-values for heat-exchangers

9 HEATING SUPPLY AIR

9.1 Purpose heating supply air

9.2 Temperature efficiency air-air heat exchangers

9.3 Calculating heat consumption supply air

10 PROCESSWATER HEATING

10.1 Purpose of heating process water

10.2 Increase dry solids content after press section

10.3 Spontaneous water evaporation process water

10.4 Calculating heat consumption process water

10.5 Simplified diagram for process water

11 SPRAYWATER HEATING

11.1 Purpose of heating spray water

11.2 Calculating heat consumption spray water

12 SPACEHEATING

12.1 Heat losses buildings

12.2 Degree-days and base-temperature

12.3 Capacity heat exchangers steam and heat recovery

12.4 Heat consumption heat exchangers and heat recovery

13. DRYING PROCESS AND HEATRECOVERY IN MOLLIER CHART

13.1 Introduction

13.2 Reference and optimal situation

13.3 Power consumption fans

PART IV UTILITIES FOR DRYING

14 FANS, PUMPS AND COMPRESSORS

14.1 Introduction

14.2 Power consumption pumps


15 ENERGY CONVERSION

15.1 Introduction

15.2 Heat recovery in energy conversions 15.3 Heat savings and electricity production

ANNEXES A COMBUSTION PROCESSES


A1 INTRODUCTION

A2 COMBUSTION EQUATION

A2.1 Components in natural gas

A2.2 Genereal combustion equation

A2.3 Combustion G-gas

B ENERGY CONVERSION

B1 INTRODUCTION

B1.1 Configuration energy conversion

B1.2 Thermal efficiency

B2 GASTURBINES

B2.1 Principle

B2.2 Types of gas-turbines

B2.3 Gas-turbine cycle

B2.4 Gas-turbine efficiency

B3 STEAMTURBINES

B3.1 Principle

B3.2 Types of steam-turbines

B3.3 Steam-turbine cycle

B3.4 Steam-turbine efficiency

B4 STEAMBOILERS

B4.1 Steam-boiler plant

B4.2 Types of steam-boilers

B4.3 Exhaust heat steam-boiler

B4.4 Efficiency steam-boilers

B5 COMBINED HEAT POWER (CHP)

B5.1 Combined heat power plant

B5.2 Combined cycle plant

C CENTRIFUGAL FANS AND PUMPS

C1 Types of centrifugal fans and compressors

C2 Power consumption centrifugal fans and compressors

C3 Relation between pressure and required power

C4 Electricity consumption supply, exhaust and circulation fans

C5 Types of pumps

C6 Head pressure

C7 Power consumption pumps

C8 Pump and system characteristic

D TABLES AND UNITS

D1 Steamtable

D2 Units

D3 Relation between θ, θdew, pvapour and wvapour

D4 Average heat capacities

D5 Mollier charts


1. PSYCHROMETRICS

1.1 Introduction

Psychrometrics or psychrometry is the science of moist air properties and processes.

Psychrometrics describes the physical and thermodynamic properties of gas-vapor

mixtures.

Though the principles of psychrometry apply to any physical system consisting of gas-

vapor mixtures, we consider only mixtures of water vapor and dry air. These

components will be named hereafter vapor and air.

At atmospheric circumstances the mixture consists of a non-condensing component

(air) and a condensing component (vapor); the ratio between these components

depends on the circumstances of the drying process.

In this chapter we will explain how the thermo physical properties of these components

can be determined.

1.2 Thermo physical properties of humid air

There are seven physical properties of humid air applied in drying processes.

As, without elaborate study, the relation between these properties is very hard to

understand these properties are discussed in the next chapters in detail.

In this chapter we will confine to a definition of these properties:

Dry-bulb-temperature (θ in oC)

The dry bulb temperature of an air sample is determined by an ordinary thermometer,

the thermometer’s bulb being dry. The dry-bulb temperature will hereafter be indicated

as temperature.

Wet-bulb-temperature (θwet-bulb in oC)

In order to understand the concept of wet bulb temperature for air – water vapor

mixtures, it is necessary to understand two processes:

- 1st process: large amount of water is contacting an air flow (temperature air after is

lower and humidity is higher; water becomes adiabatic saturation temperature; see

fig.1)

In a well-insulated chamber entering air contacts a spray of circulating liquid water.

The air leaving the chamber is at a higher humidity and lower temperature than the

air that enters.

The evaporation of water into the air results in saturation of the air by converting

part of the enthalpy (sensible heat) of the entering air into latent heat for vaporizing

water.

Or in other words: increase in latent heat = decrease in sensible heat.

Exchange of heat between the air and the water with no loss across the chamber

walls is defined as adiabatic saturation.

The temperature of the water being circulated reaches a steady-state temperature

called adiabatic saturation temperature (θsat).

This θsat is attained when a large amount of water continuously contacts the entering

air in an adiabatic chamber.

The enthalpy balance over the chamber for 1 kg air can be expressed as follows:


(wvapor– w”vapor) * hRo = (cp av. air+ cp av.vapor* wvapor) * (θ –θ”) (1)

Where:

θ = temperature of the air before the insulated chamber in °C

θ” = temperature of the air after the insulated chamber in °C

wvapor = absolute humidity supply air in kg water per kg dry gas (kg/kg)

w”vapor= absolute humidity exhaust air in kg water per kg dry gas (kg/kg)

cp av.air = average humid heat or specific heat of dry air (= 1,01 kJ/(kg.°C))

cp av.vapor = average humid heat or specific heat of water vapor (= 1,84 kJ/(kg.°C))

hRo = latent heat of vaporization of water (= 2501,6 kJ/kg).

Adiabatic chamber

Supply air Exhaust air

θ, wvapour θ", w"vapour

Make-up water

Fig. 1. Adiabatic saturation of a gas in an insulated chamber.

- 2nd process: small amount of water is contacting a large air flow (temperature and

humidity air after is same as before; water becomes adiabatic saturation

temperature; wet-bulb-thermometer process):

When a small amount of water is exposed to a continuous stream of entering gas

under adiabatic conditions, the water temperature decreases to a steady-state non

equilibrium temperature. As the amount of liquid is small and the air flow is big, the

temperature and humidity of the air are not modified, as is the case in the first

process, adiabatic saturation. Rather, the cooling effect of evaporating water into gas

decreases the temperature of the remaining water to what is known as the wet bulb

temperature (Twet).

For an accurate wet-bulb thermometer, “the wet-bulb temperature” and “the adiabatic

saturation temperature” are approximately equal for air-water vapor mixtures at

atmospheric temperature and pressure. In literature the wet-bulb temperature is often

indicated as “thermodynamic wet-bulb temperature”.

The thermodynamic wet-bulb temperature is the minimum temperature which may be

achieved by purely evaporative cooling of a water-wetted ventilated surface.

For a given parcel air at a known pressure and dry-bulb temperature, the

thermodynamic wet-bulb temperature corresponds to unique values of relative

humidity, dew point temperature and other properties.


The wet-bulb-temperature is the same as the dry-bulb-temperature when the air

sample is saturated with water. For air that is less than saturated (<100 % relative

humidity), the wet-bulb temperature is lower than the dry-bulb temperature; and the

dew point temperature is less than the wet-bulb temperature.

In the paper drying process sensible heat from the drying air is converted into latent

heat in the vapor from the paper web. As long as the surface of the paper web is wet by

surface-moist or macro-capillary-moist the wet surface will obtain the wet bulb

temperature.

Cooling of the human body through perspiration is inhibited as the wet-bulb

temperature (and relative humidity) of the surrounding air increases in summer.

It is the thermodynamic wet-bulb temperature that is plotted on psychrometric charts.

In practice, this is the reading of a thermometer whose sensing bulb is covered with a

wet sock evaporating into a rapid stream of the sample air.

Isenthalpic, adiabatic or isothermal ? Isenthalpic process An isenthalpic process is a process that proceeds without any change in enthalpy. A process will be isenthalpic if there is:

- no transfer of heat to or from the surroundings, - no work done on or by the surroundings, and - no change in the kinetic energy of the fluid. In an isenthalpic process: h1 = h2

The throttling process (for instance in pressure reducing valves) is a good example of an isenthalpic process. Adiabatic process

An adiabatic process is a process in which no heat is transferred. This can happen: - if the process happens so quickly that there is no time to transfer heat, or - if the system is very well insulated from its surroundings.

In an adiabatic process: Q = 0

Adiabatic heating occurs in a diesel engine. During the compression stroke adiabatic heating will elevate the temperature sufficiently to ignite the fuel. Another example of an adiabatic process is the drying process in a closed chamber without any heat exchange to the surroundings. An adiabatic process that is reversible is also called an isentropic process. Additionally, an adiabatic process that is irreversible and extracts no work is in an isenthalpic

process. Isothermal process An opposite extreme of the adiabatic process is the isothermal process. This process allows heat transfer with the surroundings, causing the temperature to remain constant.

Dew point temperature (θdew in oC)

The dew point is the temperature to which a given parcel of moist air must be cooled,

at constant barometric pressure, to condense the water vapor in the air into water. The

point indicates, independent of temperature, the mole fraction of water vapor in the air,

and therefore determines the specific humidity (NL: absolute vochtigheid) of the air.

The dew point is a saturation point.


The dew point is associated with relative humidity. A high relative humidity indicates

that the dew point is closer to the current air temperature. Relative humidity of 100%

indicates that the dew point is equal to the current temperature (and the air is

maximally saturated with water). When the dew point stays constant and temperature

increases, relative humidity will decrease.

Relative humidity (φ in -)

For a given parcel of air (=per unit volume) the relative humidity (φ) is the ratio of the

actual mass of the water vapor in the air to the (maximum) mass of the water vapor in

this parcel of air at the same temperature when the air is saturated with water vapor

(the saturation vapor pressure).

The ratio of the actual mass of water vapor and the maximum mass of water vapor is

the same as the actual partial vapor pressure and the maximum vapor pressure.

Absolute humidity (wvapor in g vapor / kg dry air or simply g/kg)

The absolute humidity is also known as specific humidity, moisture content, mixing

ratio, or humidity ratio (NL: absolute vochtigheid).

It is the proportion of mass of water vapor per unit mass of dry air at the given

conditions (θ, θwet-bulb ,θdew , φ etc.).

Specific enthalpy (h in kJ/kg dry air or simply kJ/kg)

The specific enthalpy , also called heat content per unit mass, is the sum of the internal

(heat) energy of the moist air in question, including the heat of the air and water vapor

within. In the approximation of ideal gases, lines of constant enthalpy are about parallel

to lines of constant θwet-bulb .

Specific volume (v in m3/kg dry air or simply m3/kg)

The specific volume, also called inverse density, is the volume per unit mass of the air

sample.

1.3 Relation between pressure, temperature and vapor content

When air is saturated with vapor, the relative humidity (φ) is 100% and the

temperature (θ), the dewpoint (θdew) and wet-bulb temperature (θwet-bulb) are the same.

Saturated air: φ = 100%; θ = θdew = θwet-bulb (2)

Furthermore there is an empirical relation between the dew point (θdew), the partial

vapor pressure (pvapor) and the specific humidity (wvapor).

This empiric relation is shown in table 1.

The atmospheric pressure (po) consists of two partial pressures: the partial pressure of

the dry air (pair) and the partial pressure of the water vapor in the air (pvapor):

po = pair + pvapor (see paragraph 2.1: gas laws).

The relation between the saturated partial pressure p”vapor and θ is shown in the

equations (3a) and (3b) and explained in box 1:

"θ237,3

"θx17,27exp.x0,611p

"

D

(3a)


In chapter 2 (about thermo-dynamics) it will be shown that the partial vapor pressure

(pvapor) is a very important humid air condition.

For most calculations in the dryer section pvapor must be known.

Table 1 Empirical relations between θ, θdew, pvapor and wvapor

Dew

poin

t

Absolu

te

hum

idity

Part

ial vapor

pre

ssure

Rela

tive

hum

idity

Tem

pera

ture

Wet

bulb

tem

pera

ture

Enth

alp

y

Dew

poin

t

Absolu

te

hum

idity

Part

ial vapor

pre

ssure

Rela

tive

hum

idity

Tem

pera

ture

Wet

bulb

tem

pera

ture

Enth

alp

y

θdew wvapor pvapor φ θ θwet.bulb h θdew wvapor pvapor φ θ θwet.bulb h

°C g/kg kPa % °C °C J/g °C g/kg kPa % °C °C J/g

1,0 4,1 0,7 100 1,0 1,0 11 51,0 92,6 13,0 100 51,0 51,0 2922,0 4,4 0,7 100 2,0 2,0 13 52,0 98,0 13,6 100 52,0 52,0 3073,0 4,8 0,8 100 3,0 3,0 15 53,0 103,7 14,3 100 53,0 53,0 3234,0 5,1 0,8 100 4,0 4,0 17 54,0 109,8 15,0 100 54,0 54,0 3405,0 5,5 0,9 100 5,0 5,0 19 55,0 116,1 15,7 100 55,0 55,0 3586,0 5,9 0,9 100 6,0 6,0 21 56,0 123,0 16,5 100 56,0 56,0 3777,0 6,3 1,0 100 7,0 7,0 23 57,0 130,2 17,3 100 57,0 57,0 3978,0 6,7 1,1 100 8,0 8,0 25 58,0 137,9 18,2 100 58,0 58,0 4189,0 7,2 1,1 100 9,0 9,0 27 59,0 146,0 19,0 100 59,0 59,0 441

10,0 7,7 1,2 100 10,0 10,0 30 60,0 154,7 19,9 100 60,0 60,0 46511,0 8,3 1,3 100 11,0 11,0 32 61,0 163,9 20,9 100 61,0 61,0 49012,0 8,8 1,4 100 12,0 12,0 34 62,0 173,8 21,8 100 62,0 62,0 51713,0 9,5 1,5 100 13,0 13,0 37 63,0 184,3 22,9 100 63,0 63,0 54614,0 10,1 1,6 100 14,0 14,0 40 64,0 195,4 23,9 100 64,0 64,0 57615,0 10,8 1,7 100 15,0 15,0 42 65,0 207,3 25,0 100 65,0 65,0 60916,0 11,5 1,8 100 16,0 16,0 45 66,0 220,2 26,2 100 66,0 66,0 64417,0 12,3 1,9 100 17,0 17,0 48 67,0 234,0 27,3 100 67,0 67,0 68218,0 13,1 2,1 100 18,0 18,0 51 68,0 248,7 28,6 100 68,0 68,0 72219,0 14,0 2,2 100 19,0 19,0 55 69,0 264,4 29,8 100 69,0 69,0 76520,0 14,9 2,3 100 20,0 20,0 58 70,0 281,5 31,2 100 70,0 70,0 81121,0 15,9 2,5 100 21,0 21,0 62 71,0 299,8 32,5 100 71,0 71,0 86122,0 16,9 2,6 100 22,0 22,0 65 72,0 319,7 34,0 100 72,0 72,0 91523,0 18,0 2,8 100 23,0 23,0 69 73,0 341,3 35,4 100 73,0 73,0 97324,0 19,1 3,0 100 24,0 24,0 73 74,0 364,7 37,0 100 74,0 74,0 1.03725,0 20,3 3,2 100 25,0 25,0 77 75,0 390,2 38,6 100 75,0 75,0 1.10626,0 21,6 3,4 100 26,0 26,0 81 76,0 417,9 40,2 100 76,0 76,0 1.18127,0 23,0 3,6 100 27,0 27,0 86 77,0 448,2 41,9 100 77,0 77,0 1.26228,0 24,4 3,8 100 28,0 28,0 91 78,0 480,5 43,7 100 78,0 78,0 1.35029,0 26,0 4,0 100 29,0 29,0 96 79,0 518,6 45,5 100 79,0 79,0 1.45230,0 27,6 4,2 100 30,0 30,0 101 80,0 559,3 47,4 100 80,0 80,0 1.56231,0 29,3 4,5 100 31,0 31,0 106 81,0 604,7 49,3 100 81,0 81,0 1.68532,0 31,1 4,8 100 32,0 32,0 112 82,0 655,7 51,4 100 82,0 82,0 1.82233,0 32,9 5,0 100 33,0 33,0 118 83,0 713,2 53,4 100 83,0 83,0 1.97734,0 35,0 5,3 100 34,0 34,0 124 84,0 777,8 55,6 100 84,0 84,0 2.15135,0 37,1 5,6 100 35,0 35,0 130 85,0 852,4 57,8 100 85,0 85,0 2.35236,0 39,3 5,9 100 36,0 36,0 137 86,0 937,7 60,1 100 86,0 86,0 2.58137,0 41,7 6,3 100 37,0 37,0 144 87,0 1.036,4 62,5 100 87,0 87,0 2.84638,0 44,1 6,6 100 38,0 38,0 152 88,0 1.152,4 65,0 100 88,0 88,0 3.15839,0 46,8 7,0 100 39,0 39,0 160 89,0 1.290,7 67,5 100 89,0 89,0 3.53040,0 49,5 7,4 100 40,0 40,0 168 90,0 1.458,7 70,1 100 90,0 90,0 3.98241,0 52,5 7,8 100 41,0 41,0 177 91,0 1.665,9 72,8 100 91,0 91,0 4.53842,0 55,6 8,2 100 42,0 42,0 186 92,0 1.927,2 75,6 100 92,0 92,0 5.24043,0 58,8 8,6 100 43,0 43,0 195 93,0 2.268,3 78,5 100 93,0 93,0 6.15744,0 62,3 9,5 100 44,0 44,0 205 94,0 2.731,7 81,5 100 94,0 94,0 7.40145,0 65,9 9,6 100 45,0 45,0 216 95,0 3.396,1 84,6 100 95,0 95,0 9.18546,0 69,8 10,1 100 46,0 46,0 227 96,0 4.426,7 87,7 100 96,0 96,0 11.95347,0 73,9 10,6 100 47,0 47,0 239 97,0 6.248,4 91,0 100 97,0 97,0 16.84448,0 78,1 11,2 100 48,0 48,0 251 98,0 10.303,1 94,3 100 98,0 98,0 27.73149,0 82,7 11,7 100 49,0 49,0 264 99,0 27.145,9 97,8 100 99,0 99,0 72.95350,0 87,5 12,3 100 50,0 50,0 277 100,0 - 101,4 100 100,0 100,0 -


and

ln (1,639 x "

Dp ) = "3,237

"27,17

x (3b)

Box 1 Relation pressure, temperature and vapor content

60 isotherm p"vapour

θ"

50 Vapour

40

30

20

Water

10

7

0

1 5 10 15 20

Vapour pressure kPa

Tem

pera

ture

oC

Formula for the relation between the saturated vapor pressure (p”vapor) and the

saturated temperature (θ”); valid for temperatures between 0 and 100 oC:

"θ237,3

"θx17,27exp.x0,611p

"

vapor

(3a)

ln (1,639 x "

vaporp ) = "3,237

"27,17

x (3b)

Question 1:

What happens when vapor (pvapor = 1 kPa, θ =60 oC), in a parcel void of air, is

compressed ?

Answer:

When water vapor is compressed slowly (isothermal process) the rise of the pressure

will be limited. At a certain pressure the vapor will start to condense. Further

isothermal compression will result in more condensation of vapor till all vapor is

condensed.

The pressure stays constant during the condensation process. This socalled

saturated pressure depends on the temperature of the system

Question 2:

How much vapor can be absorbed in air when at atmospheric pressure the

temperature is more than 100 oC ?

Question 3:

The temperature of the humid air is 70 oC and the partial vapor pressure is 20 kPA.

What is the dew point temperature ?


2 THERMO-DYNAMICS

2.1 Introduction

Thermodynamics is the study of the conversion of energy into work and heat and its

relation to variables such as temperature, volume and pressure. Historically, thermo-

dynamics developed out of need to increase the efficiency of early steam engines.

In our calculations thermodynamic processes are defined as systems that move from

state 1 to state 2, where the state number is denoted by subscript.

The relationships between the pressure, volume and temperature of a sample of gas are

set down in gas laws.

2.2 Gas laws

The early gas laws were developed at the end of the eighteenth century, when

scientists began to realize that relationships between the pressure, volume and

temperature of a sample of gas could be obtained which could be applied to all gases.

Gases behave in a similar way over a wide variety of conditions because to a good

approximation they all have molecules which are widely spaced.

The earlier gas laws (Boyle, Gay-Lussac etc) are now considered as special cases of the

Ideal gas equation or General gas law.

For all gas laws the basic assumption is that the gas is an “ideal gas”. This means that

both molecular size and intermolecular attractions are neglected.

The ideal gas law is most accurate for mono atomic gases at high temperatures and low

pressures. The neglect of molecular size becomes less important for larger volumes,

i.e., for lower pressures. The relative importance of intermolecular attractions

diminishes with increasing temperatures.

Boyle’s law

Boyle’s Law shows that for an ideal process at constant temperature the product of

pressure and volume is always constant.

p.V =constant (5)

Where:

p = the pressure of the gas, N/m2

V = the volume of the gas, m3

The product is: N/m2 x m3 = Nm = J

When air is compressed and the conditions before compression are p1,V1 and after p2,V2

the equation will be:

p1 x V1 = p2 x V2

In case of compression this equation is only correct when during compression heat is

removed because the temperature should be constant.

Gay-Lussac’s law

Gay-Lussac’s Law shows that for an ideal process at constant volume the quotient of

pressure and absolute temperature is always constant.

p/T =constant (6)


Where:

p = the pressure of the gas, N/m2

T = absolute temperature, K

Note: 0 oC = 273 K; 20 oC = (20 +273) K

Avogadro’s Law

Avogadro’s Law states that the volume occupied by an ideal gas is proportional to the

amount of moles (or molecules) present in the container.

This gives rise to the molar volume of a gas, which at standard conditions for

temperature and pressure is 22,4 m3.

Note: standard conditions for temperature and pressure are 273 K (= 0 oC) and

101,325 kPa (=101.325 Pa = 1,01325 bar = 1 atm.)

Dalton’s Law

Dalton’s Law states that the the pressure of a mixture of gases simply is the sum of the

partial pressures of the individual components.

Dalton’s Law is as follows:

ptotal = p1 + p2 + p3 + ……. Pn (7)

Where:

ptotal = total pressure of the gas mixture

p…. = partial pressure of each component in the mixture

For drying processes at standard pressure conditions Dalton’s Law is as follows:

po = pair + pvapor

Where:

po = 101,325 kPa (“standard” total pressure of the atmosphere)

pair = partial pressure of the dry air component in the mixture

pvapor = partial pressure of the water vapor component in the mixture

Dalton’s law is not exactly followed by real gases. Those deviations are considerably

large at high pressures. In such conditions, the volume occupied by the molecules can

become significant compared to the free space between them.

Combined and ideal gas laws

The combined gas law or general gas equation (NL: Algemene gaswet) is formed by the

combination of the laws of Boyle, Gay-Lussac and Avogadro.

The law shows that the state of an amount of gas is determined by its pressure, volume

and temperature.

The modern form of the equation is:

p.V = n.R.T (kPa.m3=(kN/m2).m3=kNm=kJ) (8)

Where:

p = the absolute pressure of the gas, in kPa

V = the volume of the gas, in m3

n = amount of substance in kmol

R = the universal gas law constant of 8,3145 m3·kPa/(kmol·K)

T = the absolute temperature of the gas, in K


This law has the following important consequences:

- If temperature and pressure are kept constant, then the volume of the gas is

directly proportional to the number of molecules of gas.

- If the temperature and volume remain constant, then the pressure of the gas

changes is directly proportional to the number of molecules of gas present.

- If the number of gas molecules and the temperature remain constant, then the

pressure is inversely proportional to the volume.

- If the temperature changes and the number of gas molecules are kept constant,

then either pressure or volume (or both) will change in direct proportion to the

temperature.

The law is the equation of state of a hypothetical ideal gas.

It is a good approximation to the behaviour of many gases under many conditions,

although it has several limitations.

As the amount of substance could be given in mass instead of moles, sometimes an

alternative form of the ideal gas law is useful:

m = n x M (9)

Where:

m = mass in kg

n = number of moles

M = molar mass in kg/kmol

2.3 Relation between mass, amount of substance and volume

This relation is shown in box 2.

In box 3 an exercise for this relation is depicted.

2.4 Calculation mass flow from volume flow and temperature

During field measurements usually volume flows and temperatures are laid down.

In box 4 is shown how, with help of the gas laws, the mass of these flows is calculated.

To calculate the mass of humid air the vapor content of the flow should be known.

Calculations for determining the vapor content of a flow are shown in chapter 3.


Box 2 Relation between mass, amount of substance and volume

Name Symbol Unit Explanation

Mass m kg

Amount of substance n kmol

Volume V m3

Normal volume Vo m3o p = 101,325 kPa; θ = 0 oC

Molar volume VM m3o VM = 22,4 m3

o

Molar mass M kg/kmol See explanation

Pressure p kPa 100 kPa = 1 bar

Normal pressure po kPa po = 101,325 kPa

Temperature θ oC

Temperature T K T = (θ + 273) K

Density ρ kg/m3o ρ = M/VM

Universal gas constant R kJ/(kmol.K) R = 8,3145 kJ/(kmol.K)

see explanation

Explanation:

Molar mass

Molar mass = atomic or molecular weight x molar mass constant

Molar mass constant = 1 kg/kmol

Atomic weight: H=1, C=12, N=14, O=16

Molar mass:

Water vapor H2O: M = (2 x1 +16) x 1 = 18 kg/kmol (= 22,4 m3o)

Carbon-di-oxide CO2: M = (12 + 2 x 16) x 1 = 44 kg/kmol (= 22,4 m3o)

Nitrogen N2: M = (2 x 14) x 1 = 28 kg/kmol (= 22,4 m3o)

Oxigen O2: M = (2 x 16) x 1 = 32 kg/kmol (= 22,4 m3o)

Example: 1 kmol O2 = 22,4 m3o = 32 kg; ρ = (32/22,4=) 1,43 kg/m3

o

Universal gas constant

Calculated for 1 kmol gas:

po = standard pressure (= 101,325 kPa);

VM = molar volume (= 22,4 m3

o /kmol);

To = 273,15 K (= 0oC);

R = universal gas constant.

R = po x VM / To (kJ/(kmol.K) (12a)

(kPa x m3o/kmol/K = kN/m2 x m3/kmol/K = kN.m/(kmol.K)= kJ/(kmol.K)

R = (101,325 x 22,4) / 273,15 = 8,3145 kJ/(kmol.K) (12b)


Box 2 Relation between mass, amount of substance and volume

Name Symbol Unit Explanation

Mass m kg

Amount of substance n kmol

Volume V m3

Normal volume Vo m3o p = 101,325 kPa, θ = 0 oC

Molar volume VM m3o VM = 22,4 m3

o

Molar mass M kg/kmol See explanation

Pressure p kPA 100 kPa = 1 bar

Normal pressure po kPa po = 101,325 kPa

Temperature θ oC

Temperature T K T = (θ + 273) K

Density ρ kg/m3o ρ = M/VM

Universal gas constant R kJ/(kmol·

K)

R =8,3145 kJ/(kmol·K)

See explanation

Explanation:

Molar mass

Molar mass = atomic or molecular weight x molar mass constant

Molar mass constant = 1 kg/kmol

Atomic weight: H=1, C=12, N=14, O=16

Molar mass:

Water vapor H2O: M = (2 x1 +16) x 1 = 18 kg/kmol (= 22,4 m3o)

Carbon-di-oxide CO2: M = (12 + 2 x 16) x 1 = 44 kg/kmol (= 22,4 m3o)

Nitrogen N2: M = (2 x 14) x 1 = 28 kg/kmol (= 22,4 m3o)

Oxigen O2: M = (2 x 16) x 1 = 32 kg/kmol (= 22,4 m3o)

Example: 1 kmol O2 = 22,4 m3o = 32 kg; ρ = (32/22,4=) 1,43 kg/m3

o

Universal gas constant

Calculated for 1 kmol gas:

po = standard pressure (= 101,325 kPa);

VM = molar volume (= 22,4 m3

o /kmol);

To = 273,15 K (= 0oC);

R = universal gas constant.

R = po x VM / To (kJ/(kmol.K) (12a)

(kPa x m3o/kmol/K = kN/m2 x m3/kmol/K = kN.m/(kmol.K)= kJ/(kmol.K)

R = (101,325 x 22,4) / 273,15 = 8,3145 kJ/(kmol.K) (12b)

Box 3 Exercise 1: Relation between mass, amount substance & volume

Given:

In box 2 the relation between mass, amount of substance and volume is shown.

Question:

Complete the table in this box and answer the questions.

Answer:

Gas M(kg/kmol) VM(m3o / kmol) ρ(kg/m3

o)

Water vapor H20

Air

(79%N2&21% O2)

Methane CH4

Propane C3H8

Butane

C4H10

2x1+16 = 18

0,79x2x14+0,21x

2x16 = 28,84

1x12+4x1 = 16

3x12+8x1 = 44

4x12+10x1 = 58

22,4

22,4

22,4

22,4

22,4

18/22,4=0,804

28,84:22,4=1,2875

16:22,4 = 0,714

44:22,4 = 1,964

58:22,4 = 2,589

The combustion equation for methane (CH4) and oxygen (O2) is:

1 CH4 + 2 O2 1 CO2 + 2H2O

How many kg water vapor will be formed during combustion of 1 m3o methane and

how much combustion air is required ?

1 kmolCH4 + 2 kmol O2 1 kmol CO2 + 2 kmol H2O

22,4m3CH4+44,8m3O2 22,4m3CO2+44,8m3H2O

16 kg CH4+64 kg O2 44 kg CO2+36 kg H2O

Vapor formed during the combustion of 1 m3o methane:

36kg H2O:22,4m3CH4=1,61kgH2O/m3CH4

Air required for the combustion of 1 m3o methane:

2m3O2/m3CH4x

zuurstofm

luchtm3

3

21,0

1= 9,52m3 air/m3CH4

Why is it not allowed to park your LPG-car in a “closed” building ?


Box 4 Calculation mass flows from volume flow and temp

(1) Universal gas constant for 1 kmol gas:

K).kJ/(kmol8,314273,15

22,4x101,325

T

V.pR

o

mo (12)

(2) General gas law (amount of substance = n kmol gas):

kJ/KT

V.pR.n (8)

(3) Mass m (kg) of n (kmol) with molar mass M (kg/kmol):

m = n . M. (9)

(4) Formula (2) substituted in formula (3):

kg/sT

V.p

R

Mm (13a)

(5) Air flow with atmospheric conditions: only V and T should be measured

kg/sT

V. 101,325

8,314

29m (13b)

(6) Dalton:

po = pG + pD kPa (7)

For drying processes:

po = atmospheric pressure (101,235 kPa)

pair = partial pressure non-condensable components (dry air)

pvapor = partial pressure condensable components (water vapor)

(7) Mass vapor flow (mvapor):

T

V.p

8,314

18m

vapour

vapour (13c)

(8) Mass dry air flow (mair):

T

V.p

8,314

28,84m air

air (13d)

The question is how to calculate pvapor en pair


2.5 Pitot-tube for measurering volume flows

A pitot tube is a pressure meter that measures head according Bernoulli ‘s law:

h = v2 / 2g

Where:

h = head in m

v = gas velocity in m/s

g = gravity (9,81 m/s2)

The pitot-tube measures ∆p over the connections:

∆p (N/m2) = ρ (kg/m3) x g (m/s2) x h (m)

Where:

∆p = pressure difference over the connections in N/m2

ρ = specific density in kg/m3

Note: unit Newton (N) is kg.m/s2

Substitution of (formule 1) in (formule 2):

v = (2 * ∆p / ρ) )^0,5

Substitution of formule 19 from table 3:

v =(2*∆p/((28,84*pdry air+18*pvapor)/(8,314*(273,15+θdry bulb))))^0,5 (17)

2.6 Calculation enthalpy of humid air

In thermodynamics the enthalpy (denoted as H, or specific enthalpy denoted as h) is

a thermodynamic property used to calculate the heat transfer during a process taking

place under constant pressure (isobaric process).

Enthalpy H is an arbitrary concept but the enthalpy change ΔH is more useful because it

is equal to the change in the internal energy of the system, plus the work that the

system has done on its surroundings.

The specific enthalpy may be defined by:

h = u + p.V (10)

Where:

h = specific enthalpy (kJ/kg)

u = internal energy (kJ/kg)

p = pressure (kPa)

V = volume in (m3)

For gasses h = 0 when the temperature (θ) of that gas is 0 oC.

Humid air is a mixture of dry air and water vapor.

The enthalpy of humid air includes the enthalpy of the dry air (= sensible heat) and

the enthalpy of the vapor in the air (= latent heat)

Specific enthalpy of humid air is defined as the total enthalpy of the mixture of dry air

and vapor per kg dry air:

h = cp av.airx θ + wvapor x (hRo+cp av.vapor x θ) (kJ/kg (dry!) air) (11)

Where:

h = specific enthalpy of humid air (kJ/kg dry air)


cp av.air = average specific heat of dry air (kJ/(kg.oC); see table 2

θ = temperature (oC)

wvapor = absolute humidity supply air in kg water per kg dry gas (kg/kg)

hRo = heat of evaporation (= 2501,6 kJ/kg)

cp av.vapor = average specific heat of vapor (kJ/(kg.oC); see table 2

We distinguish specific heat capacity at constant pressure (cp) and average specific heat

capacity at constant pressure (cp av). Both are heat capacities for constant pressures.

The specific heat capacity at constant pressure (cp) is not a fixed constant and varies

somewhat depending on the temperature. For air, for instance, this “true” specific heat

capacity is 1,01 at 0 oC and about 1,30 at 2000 oC.

To simplify calculations the average specific heat capacity (cp av) is introduced.

By definition cp av represents the average quantity of heat needed to raise 1 kg of gas

through 1 oC in the range between 0 and θ oC. For air, for instance, this average

specific heat capacity is 1,01 at 0 oC and about 1,16 for the temperature range 0 –

2000 oC.

In box 5 is shown how the enthalpy of humid air is calculated. Also some average

specific heat capacities of some gasses are graphically shown.

Table 2 Average specific heat

'Cp.av. in kJ/(kg.oC) of components in air or flue gas

temp

range °CCO2 Ar O2 N2 H2O Dry air

0 0,819 0,519 0,917 1,041 1,83 1,006

0 - 100 0,870 0,519 0,923 1,041 1,84 1,009

0 - 200 0,915 0,519 0,935 1,045 1,87 1,012

0 - 300 0,953 0,519 0,950 1,048 1,89 1,019

0 - 400 0,989 0,519 0,964 1,055 1,92 1,028

0 - 500 1,021 0,519 0,978 1,065 1,95 1,038

0 - 600 1,046 0,519 0,993 1,075 1,98 1,048

0 - 700 1,072 0,519 1,005 1,087 2,01 1,061

0 - 800 1,092 0,519 1,017 1,098 2,04 1,071

0 - 900 1,112 0,519 1,029 1,109 2,07 1,080

0 - 1000 1,129 0,519 1,037 1,118 2,11 1,090

0 - 1100 1,144 0,519 1,045 1,129 2,14 1,100

0 - 1200 1,158 0,519 1,052 1,139 2,17 1,109

0 - 1300 1,170 0,519 1,060 1,145 2,20 1,118

0 - 1400 1,181 0,519 1,066 1,152 2,23 1,125

0 - 1500 1,192 0,519 1,072 1,160 2,26 1,132

0 - 1600 1,203 0,519 1,078 1,168 2,29 1,139

0 - 1800 1,219 0,519 1,090 1,182 2,34 1,153

0 - 2000 1,234 0,519 1,101 1,192 2,39 1,162


Box 5 Calculation enthalpy of humid air

Average specific heat gasses and water vapor

(H2O*=water vapor as ideal gas en H2O** water vapor is nota n ideal gas)

Dry gas (sensible heat) Water vapor (latent heat)

Average specific heat dry gas in kJ/(kg.K)

absolute humidity in kg water per kg dry gas (kg/kg)

Heat of evaporation water at 0oC (=2501,6 kJ/kg)

Average specific heat water vapor

h = cp av.air . θ + wvapor . (hRo + cp av.vapor . θ ) kJ/kg (dry) gas (11)

Avera

ge s

pecific

heat

dry

gas in kJ/

(kg.K

)

Temperature θ in oC


3 FORMULA LIBRARY FOR DRYING PROCESSES

3.1 Overview formulas

In box 6 formulas for relative and absolute humidity are derived from the definitions

and formulas in the chapters psychrometrics and thermodynamics.

In table 3 a “Formula library for calculating physical properties of drying air” is depicted.

The formulas are derived from the formulas in chapter 2 and 3. The polynom formulas,

derived from table 1, are not depicted in the table.

As an example the formula for deriving θdew from wvapor shows as follows: θdew=ALS(W(vapor)>4427;1,3169*LN(W(vapor))+85,626;ALS(W(vapor)>1928;4,4688* LN(W(vapor))+ 58,565;ALS(W(vapor)>1153;7,4765*LN(W(vapor))+35,491;ALS(W(vapor)>605;10,644*LN(W(vapor))+13,081;ALS(W(vapor)>265;14,287*LN(W(vapor))-10,328; ALS(W(vapor)> 155; 0,00000000000137* W(vapor)^6-0,0000000019017* W(vapor)^5+ 0,00000109041874* W(vapor)^4-0,00032957623934* W(vapor)^3 +0,05507945136945* W(vapor)^2-4,70562681251105* W(vapor)+215,243754649645; ALS(W(vapor)>38; -0,00000000000476* W(vapor)^6 + 0,0000000033553* W(vapor)^5 - 0,00000099856273* W(vapor)^4 + 0,00016432944202* W(vapor)^3 – 0,01669988421594* W(vapor)^2 + 1,18909495672254* W(vapor) + 7,17981604351348;-0,00000011228793*W(vapor)^6 + 0,00001583312096* W(vapor)^5 – 0,00091311105523* W(vapor)^4 + 0,02800484300993* W(vapor)^3 – 0,50553648368732* W(vapor)^2 + 6,14499909516863* W(vapor) – 17,3718810425193)))))))

3.2 Method of working humid air

In table 4 the method of working for calculating air conditions and mixing flows is shown.

Table 4 Method of working calculating humid air conditions and mixing flow

a b c d e f g h i j k l m n o p q r s t u v w

Calculating flow Formulas for calculations air with watervapour

- h b/Ф A ∆p k v V po θdew θdry bulb wvapour h φ pvapour"θwet bulb pdry air mdry air mvapour mwet air H cpm dry air cpm vapour

- m m m2 Pa - m/s m3/s kPa oC oC g/kg kJ/kg % kPa °C kPa kg/s kg/s kg/s kJ/s kJ/(kg.K)kJ/(kg.K)

Step 1: dimensions of the ducts, air velocities, dry- and wet-blub temperatures are measured are measured

1 1,68 3,92 3,14 101,3 47 31,4 1,01 1,84

2 0,66 0,66 2,15 101,3 128 63,4 1,01 1,84

Step 2: copy the corresponding formulas from the library and paste them at the right place

1 1,68 3,92 6,59 3,14 20,7 101,3 26,7 47 22 106 32,3 3,5 31,4 97,8 21,9 0,49 22,4 2312 1,01 1,84

2 0,66 0,66 0,44 2,15 1,00 2,30 1,00 101,3 58,8 128 143 521 7,2 18,9 63,4 82,5 0,7 0,1 0,8 372 1,01 1,84

Step 3: flows 1 and 2 are mixed; flow 3 is sum mdry air, sum mvapour and sum H

1 21,9 0,5 2312

2 0,7 0,1 372

3 22,6 0,6 2684

Step 4: calculate h = H/mdry air and wvapour = mvapour*1000/mdry air

1

2

3 26,1 119 22,6 0,6 2684

Step 5: copy the corresponding formulas from the library and paste them at the right place

1

2 *3 101,3 29,1 50,6 26,1 119 31,7 4,02 33,4 97,3 22,6 0,6 23,2 2684 1,01 1,84

Values in white cells are input data !!; * The validity of the formula for wet bulb temperature is limited, check mollierdiagram

Specific

heat

vapour

Mass f

low

dry a

ir

Mass f

low

vapour

Mass f

low

wet

air

Enth

alp

y

Specific

heat

dry a

ir

Enth

alp

y

Rela

tive h

um

idity

Partial vap.

Pressure

Wet

bulb

tem

peratu

re

*

Partial dry a

ir

pressure

Flo

w n

um

ber

Hig

ht

duct

Wid

th/dia

met

er d

uct

Cross-section

duct

pressure-

diffe

rence

pitot-

tube

calibr.-

facto

r

pitot-

tube

gas-velo

city

Volu

me f

low

Tota

l

pressure

Dew

poin

t

Dry b

ulb

tem

peratu

re

Absolu

te

hum

idity


3.3 Method of working combustion gas

In table 4 of chapter 3 the method of working for calculating air conditions and mixing

flows is shown.

These flows are a mixture of only two components: dry air and vapor. Although dry air

consists mainly of two components, N2 and O2, its composition will not change. In

drying processes only the amount of vapor (H2O) per kg dry air will change.

In fact the calculations for humid air only relate to the components dry air and vapor.

Only these components need to be considered in calculations on drying processes.

Calculations on combustion processes are more complicated: CH4 + 2O2 -- 2H2O + CO2

This means that during combustion the chemical reactions will change the composition

of the gas; the amount of O2 will decrease and the amounts of O2 and H2O will increase.

For calculations in drying processes when combustion products are involved four

components need to be considered: N2, O2, H2O and CO2.

In annex A the method of working for combustion gases is explained.

Box 6 Formulas for relative and absolute humidity

Relative humidity (φ in -)

For a given parcel of air the relative humidity (φ) is the ratio of the actual mass of the

water vapor in the air (mvapor in kg) to the (maximum) mass of the water vapor

(m”vapor in kg) in the same parcel of air at the same temperature when the air is

saturated with water vapor.

Note that relatve humidity is per unit volume.

In “Papermaking Part 2, Drying” the relative humidity is defined per unit mass as φ =

mvapor / mdry air; this is not correct.

φ = (mvapor / m”vapor) x 100 % (14)

With formula (13c) can be derived :

φ = (pvapor / p”vapor) x 100 % (15)

When θ is measured p”vapor can be laid down with formula (3a),

When φ is measured pvapor can be calculated.

Absolute humidity

The absolute humidity is also known as specific humidity, moisture content, mixing

ratio, or humidity ratio.

Absolute humidity is the proportion of mass of water vapor per unit mass of dry air

(wvapor in kg/kg (dry!) air) at the given conditions (θ, θwet-bulb ,θdew , φ etc.).

wvapor = (mvapor / mdry air) kg vapor/kg dry air (16)

Formula (17) is derived from formulas (13c), (13d) and (15)

wvapor = ''

vapouro

''

vapour

dryair

vapour

pp

p

M

M

kg vapor / kg drygas (17)


a b c d e f g h i j k l m n o p q r s t u v w

Calculating flow Formulas for calculations air with watervapour

= value to be calculated from and

Values in white cells are input data !!

Validity formula wet bulb temp. is limited, check mollierdiagram

- h b/Ф A ∆p k v V po θdew θdry bulb wvapour h φ pvapour"

θwet bulb pdry air mdry airmvapour mwet air H cpm dry air cpm vapour Mdry air=28,84 kg/kmol; Mvapour=18,0 kg/kmol; T= 273,15 K (=0oC); R=8,314 kJ/(kmol.K)

- m m m2 Pa - m/s m3/s kPa oC oC g/kg kJ/kg % kPa °C kPa kg/s kg/s kg/s kJ/s kJ/(kg.K)kJ/(kg.K)

1 58,43 18,51 θdew = ln((1,639*pvapour)*(237,3+θdew)/17,27)

2 58,42 141,3 θdew (kPa) = polynom based on wvapour

3 58,42 79,9 141,3 455 θ = (h - 0,001*wvapour*(2501,6+1,85*θ))/1,01

4 θsupply air out = 0,01*ηθ*(θexhaust air in-θsupply air in) +θsupply air in

5 58,42 139,6 wvapour(g/kg)=1000*(18/28,84)*((0,611*exp(1)^((17,27*θdew))))/(237,3+θdew))/(101,325-(0,611*exp((17,27*θdew))))/(237,3+θdew))))))

6 151,4 455 air is saturated with vapour !! wvapour(g/kg)= polynom based on h when air is saturated with vapour

7 80,0 141,3 455 h (J/g) = 1,01*θ + 0,001*wvapour*(2501,6+1,85*θ)

8 58,42 80,0 39,0 φ (%) = ((0,611*exp(1)^((17,27*θdew)/(237,3+θdew)) )) / (((0,611*exp(1)^(17,27*θ)/(237,3+θ)))))*100

9 58,42 18,52 pvapour'" (kPa) = 0,611*exp(1)^((17,27*θdew)/(237,3+θdew))

10 141,3 18,71 pvapour '" (kPa) = wvapour *(101,325-pvapour)/0,624; condition: pvapour < 0,5*po (in this example po = 101,325 kPa)

11 455 19,57 air saturated with vapour !! pvapour "(kPa) = polynoom based on h when air is saturated with vapour

12 58,42 18,51 pvapour "(kPa) = polynom based on θdew

13 58,42 141,3 18,52 pvapour "(kPa) = polynom based on wvapour

14 58,62 80,0 18,67 60,14 pvapour (kPa) = 0,611*exp(1)^((17,27*θ)/(237,3+θ))

15 58,62 80,0 141,2 455 18,67 60,14 θwetbulb = ((0,00066*101,325)*θ+(4098*pvapour/(θdew +237,3)^2*θdew))/((0,00066*101,325)+4098*pvapour/(θdew+237,3)^2)

16 58,62 80,0 141,2 455 18,67 60,00 pvapour (kPa) = 0,611*exp(1)^((17,27*θwet bulb/(237,3+θwet bulb) - 101,325*(θ(dry bulb) - θwet bulb)*0,000660*(1+0,00115*θwet bulb)

17 58,62 80,0 141,3 455 18,67 60,00 pvapour (kPa) = 0,6112*exp(1)^((17,67*θwet bulb/(243,5+θwet bulb) - 101,325*(θ(dry bulb) - θwet bulb)*0,000660*(1+0,00115*θwet bulb)

18 58,62 80,0 141,3 455 39,3 18,67 60,00 θdew = ln((1,639*(0,6112*exp(1)^((17,67*θwet bulb/(243,5+θwet bulb) - 101,325*(θ(dry bulb) - θwet bulb)*0,000660*(1+0,00115*θwet bulb)))*(237,3+θdew)/17,27)

19 3,00 1,00 3,00 15,0 1,00 5,69 17,07 101,3 58,62 80,0 141,3 455 39,3 18,70 60,00 82,6 ## 1,96 15,8 6305 1,010 1,840 mdry air = (28,84/8,314)*((pdry air*V)/(273,15+θ(dry bulb));mvapour = (18/8,314)*((pvapour*V)/(273,15+θ(dry bulb))

20 101,3 58,62 80,0 141,3 455 39,3 18,70 60,00 82,6 ## 1,96 15,8 6305 1,010 1,840

21 3,00 1,00 3,00 15,0 1,00 5,69 17,07 101,3 80,0 18,67 60,00 82,7 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

22 3,00 1,00 3,00 15,0 1,00 0,54 101,3 58,62 80,0 18,70 82,6 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

23 3,00 1,00 3,00 3,00 9,00 101,3 58,62 80,0 141,3 455 39,3 18,70 60,00 82,6 7,3 1,03 8,3 3323 1,010 1,840 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

24 3,00 1,00 3,00 3,00 9,00 101,3 58,62 80,0 141,2 455 39,3 18,70 60,14 82,6 7,3 1,03 8,3 3323 1,010 1,840 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

Different volume flows to be combined to one common flow

25a 3,00 1,00 3,0 15,0 1,00 5,69 17,07 101,3 58,62 80,0 141,3 455 39,3 18,70 60,00 82,6 13,9 1,96 15,8 6305 1,010 1,840 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

25b 1,00 0,8 15,0 1,00 5,69 4,47 101,3 58,42 80,0 139,6 451 39,0 18,52 59,96 82,8 3,6 0,51 4,1 1640 1,010 1,840 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

25c 3,00 1,00 3,0 15,0 1,00 5,69 17,07 101,3 58,62 80,0 141,3 455 39,3 18,70 60,00 82,6 13,9 1,96 15,8 6305 1,010 1,840 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

25d 3,00 1,00 3,0 15,0 1,00 5,69 17,07 101,3 58,42 80,0 139,6 451 39,0 18,52 59,96 82,8 13,9 1,94 15,8 6256 1,010 1,840 v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k (formule 5)

25 58,58 80,0 140,6 453 39,3 18,64 60,10 45,2 6,4 51,6 20505 1,010 1,840 Calculate Σ mdry air, mvapour and H; subsequently h and wvapour; for remaining values use common formula's.

Enth

alp

y

Rela

tive h

um

idity

Partial vap.

Pressure

Wet

bulb

tem

peratu

re

*

Partial dry a

ir

pressure

Mass f

low

dry a

ir

Mass f

low

vapour

Mass f

low

wet

air

Enth

alp

y

Specific

heat

dry a

ir

Specific

heat

vapour

Absolu

te

hum

idity

Form

ula

num

ber

Channel

heig

ht

Channel

wid

th/ d

iam

.

press.d

iffe

r.

pitot

tube

drukverschil

pitotb

uis

kalibr.-

facto

r

pitotb

uis

velo

city

(m

ole

ntj

e)

Volu

me f

low

Tota

l

pressure

Dew

poin

t

Dry b

ulb

tem

peratu

re

Table 3 Formula library for calculating physical properties of drying air


4 MOLLIER CHART

4.1 Introduction

To understand the relation between the properties of humid air in processes,

psychro-metric charts are developed.

The versatility of the psychrometric chart lies in the fact that by knowing three

independent properties of some humid air (one of which is the barometric pressure),

the other properties can be determined. Changes in state, such as when two air

streams mix, can be modeled easily.

The most welkwown psychrometric chart is the Mollier diagram. The barometric air

pressure for the diagram described will be 101,325 kPa.

To determine the thermophysical properties of a mixture at least two of the six

independent properties must be known (θ, θwet-bulb, θdew, φ, h and v).

4.2 Explanation Mollier chart

The Mollier chart is explained in eight steps in box 7.1 u/i 7.8.

For each step the applied formulas from table 3 are depicted in the box concerned.

Exercise 2 is depicted in box 7.6.


6,1 absolute humidity 136 (wvapour in g/kg dryair) 144

Mollier chart: h and wvapour

The Mollier-chart is a roadmap for humid air with total pressure 101,325 kPa

Humid air = 1 kg dry air + vapour

x-axis = absolute humidity of humid air (wvapour in g vapour / kg dry air)

y-axis = enthalpy of humid air (h in kJ / kg dry air)

Vapour means that the relative humidity φ =< 100 %

(increasing the vapour content when φ =100 % will result in condensation of vapour)

So one of the borders will be the constant line for φ =< 100 %

Box 7.1. Mollier chart for humid air (h and wvapour)

Ab

s.h

um

idit

y

Ab

so

lute

hu

mid

ity

(w

vap

ou

rin

g

/k

g d

ry a

ir)

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y


1,2 2,5 (pvap in kPa) partial vapour pressure

430 °C

408 °C

59 °C

58 °C

21 °C

i1

10 °C

7,7 16 absolute humidity 138 (wvapour in g/kg dryair) 146

Mollier chart: pvapour and θdew

For air saturated with water vapour there are empirical relations between θdew, wvapour, & pvapour

From the table some values from these relations are depicted in the Mollier-chart

The values can also be calculated by the following formula's:

wvapour(g/kg) =1000*(18/28,84)*((0,611*exp(1)^((17,27*θdew))))/(237,3+θdew))/

(101,325-(0,611*exp((17,27*θdew))))/(237,3+θdew))))))

pvapour" (kPa) = 0,611*exp(1)^((17,27*θdew)/(237,3+θdew))

θdew = ln((1,639*pvapour)*(237,3+θdew)/17,27)

Box 7.2. Mollier chart for humid air (pvapour and θdew)

18,2 19,0

The total pressure (po) of the humid air is the sum of the partial pressure of dry air (pdry air) and the

partial pressure of the vapour (pvapour); po = pdry air + pvapour (kPa)

The partial pressure of the vapour (pvapour) depends on the amount of vapour in the humid air

(wvapour) and the latter depends on the temperature at saturation (θdew) of the humid air.

Temperature

Ab

s.h

um

idit

y

(g

/kg

dry

air

)

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

Pa

rtia

l va

po

ur

pre

ssu

re (

kP

a)


1,2 2,5 (pvap in kPa) partial vapour pressure

430 °C

408 °C

Heat capacity J/(g.K)

temp

rang

vapo

ur

dry

air

0 1,83 1,01

0 - 100 1,84 1,01

0 - 200 1,87 1,01

0 - 300 1,89 1,02

0 - 400 1,92 1,03

0 - 500 1,95 1,04

0 - 600 1,98 1,05

0 - 700 2,01 1,06

0 - 800 2,04 1,07

80 °C 0 - 900 2,07 1,08

0 - 1000 2,11 1,09

59 °C A 0 - 1100 2,14 1,10 B

C D 0 - 1200 2,17 1,11

58 °C 0 - 1300 2,20 1,12

21 °C 0 - 1400 2,23 1,13

0 - 1500 2,26 1,13

0 - 1600 2,29 1,14

0 - 1800 2,34 1,15

10 °C 0 - 2000 2,39 1,16

7,7 16 absolute humidity 138 (wvapour in g/kg dryair) 146

Mollier chart: θ and φ

The temp.(θ) can be derived from the relation between enthalpy (h), abs.humidity (wvapour) and temp.(θ):

h (J/g) = cp av.dry air (J/(g.K)) * θ (oC)+ 0,001 * wvapour (g/kg) * (hRo (J/g) + cp av.vapour * θ)

cp av. = average heat capacity (see insert)

R = heat of evaporation; for watervapour compared to 0 oC: R = 2501,6 J/g

The rel. humidity (φ) in A is the actual mvapour divided by the maximum mvapour at the same temperature:

mvapour =mass of vapour present in a space

m"vapour =maximum possible mass of vapour present in the same space at the same temperature

φ = mvapour /m"vapour * 100 %

When V = volume-flow (m3/s) and T = absolute temperature (K):

mvapour = (18/8,314)*pvapour * V/T (kg/s); also:

φ = pvapour /p"vapour * 100 %

Example: In point A θdry bulb=59 oC, pvapour=2,5 kPA; max. mass at 59 oC is at B: p"vapour=19,0 kPa

φ = 2,5/19,0 *100% = 13,2 %

Box 7.3. Mollier chart for humid air (θ and φ)

18,2 19,0

Temperature

Temperature

Temperature

Temperature

Ab

so

lute

hu

mid

ity

(g

/k

g d

ry a

ir)

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

Pa

rtia

l va

po

ur

pre

ssu

re (

kP

a)


1 (pvap in kPa) 18 partial vapour pressure 19 20

430 °Ci4

408 °C

i3

80 °C A

59 °C u5=60,5oC

58 °Cθwetbulb in A

21 °C

10 °C

i1= 6,1 absolute humidity u4=136 (wvapour in g/kg dryair) u2= 144

Mollier chart: θwet bulb and difference with θdew

pvapour (and subsequently φ) can be calculated from:

pvapour (kPa) = 0,611*exp(1)^((17,27*θwet bulb/(237,3+θwet bulb) - 101,325*(θ(dry bulb) - θwet bulb)*0,000660*(1+0,00115*θwet bulb)

Note: in point A the wet bulb temperature is 60,5 oC and the dew point is 59 oC

Box 7.4. Mollier chart for humid air ( θwet bulb and difference with θdew)

Temperature

Ab

so

lute

hu

mid

ity (

w v

ap

ou

r) =

1

44

g/

kg

Temperature

Temperature

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

The (thermodynamic) wet-bulb temperature (θwetbulb) is the temperature a volume of air

will have if cooled adiabatically to saturation by evaporation of water into it, all latent

heat being supplied by the volume of air.

Wet-bulb temperature is measured using a thermometer that has its bulb wrapped in

cloth, called a sock, that is kept wet with water.

When φ <100%, water evaporates from the bulb which cools the bulb below ambient

temperature.

Compare: the temperature of drying air that passes over wet paper web in a dryer and

becomes saturated is the wet-bulb temperature.

De

w p

oin

t in

A

(θ

de

w)

= 5

9 o

C



430 °Ci4

408 °C

i3

80 °C A

59 °C u5=60,5oC

58 °Cθwetbulb in A

21 °C

10 °C

i1= 6,1 absolute humidity u4=136 (wvapour in g/kg dryair) u2= 144

Mollier chart: density ρ in kg/m3; not depicted in this chart

ρ is an indication for the air-duct-dimensions and is used for pitot-tube calculations

A pitot tube measures the head (h); with (h in m) the air velocity (v in m/s) can be calculated.

h (m) = v2 (m/s) / 2g (m/s2) (1)

The pressure difference (∆p) between the connections on the pitot tube is measured:

∆p (N/m2) = ρ (kg/m3) x g (m/s2) x h (m) (2) (Attention: unit newton (N) = kg.m/s2 )

v (m/s) = (2 * ∆p (Pa) / ρ (kg/m3) )^0,5 (3)

ρ (kg/m3) = mwet (kg) / V (m3) (4)

v (m/s) = (2 * ∆p (Pa) / (mwet/V) (kg/m3) )^0,5 *k

mwet = mdry air + mvapour

mdry air = (28,84/8,314)*((pdry air*V)/(273,15+θ(dry bulb));

mvapour = (18/8,314)*((pvapour*V)/(273,15+θ(dry bulb))

v (m/s) = (2 * ∆p (Pa) / ((28,84*pdry air + 18*pvapour)/(8,314*(273,15+θdry bulb))))^0,5 * k

k= calibration factor

Box 7.5. Mollier chart for humid air ( density ρ in kg/m3 of humid air)

Temperature

Ab

so

lute

hu

mid

ity (

w v

ap

ou

r) =

1

44

g/

kg

Temperature

Temperature

Ab

s.h

um

idit

y

Ab

s.h

um

idit

y

De

w p

oin

t in

A

(θ

dew

) =

59

oC



430 °C

408 °C

80 °C A

59 °C θwet bulb A

58 °Cθdew A=59

oC 60,5

oC

21 °C

B10 °C


Mollier chart: draw the following processes in the Mollier chart !

1) Air with conditions as depicted in B is heated in a steamheater

2) Air with conditions as depicted in B is heated in a natural gas heater

3) How is the drying process depicted in the Mollier chart

4) Air with conditions as depicted in A is cooled in the heatrecovery

5) Supply- and exhaust air conditions of a dryer are B and A; what is the heat consumption

Box 7.6. Mollier chart for humid air ( processes in the Mollier chart)

Dewpoint A

Temperature A

Pa

rtia

l va

po

ur

pre

ssu

re in

A 1

9 k

Pa

Ab

so

lute

hu

mid

ity i

n A

14

4 g

/k

g d

ry a

ir



430 °C i4

408 °C i3

80 °C u1

59 °C u2 θwet bulb u1

u3 θdew u1=59oC 60,5

oC

58 °C u4

21 °C i2 Real process in multi cylinder dryer

(isothermal, adiabatic or isenthalpic ?)

10 °C i1


Moisture increase per kg dry air: ∆w = wu1-wi1 = 138 g moisture / kg dry air

Required air per kg PWE: 1000 g : ∆w g/kg = 7,2 kg air / kg PWE

Cons.with HRC supplyair: 7,2 x (hi3 - hi2) = 2920 kJ/kg PWE

Mollier chart: heat input process water

(1 kg PWE * 4,2 kJ/(kg.oC) * 40oC) kJ/kg PWE : ((1000/(143-6)) kg air/kg PWE = 23 kJ/kg

Box 7.7. Mollier chart for humid air (heat input process water)

"Real process in multi cylinder dryer":

From i1 till i4 air is heated by cylinders and (pocket ventilation)air heaters from 10oC to "430"oC.

This pocket ventilation air is preheated (i1-i2) by heat recovery from the exhaust air (u1-u3).

It is assumed that during heating the moisture content of the air will not change:

the process follows the line of constant moisture content wvap (= 6,1 g moisture/kg).

In realty the process will be as depicted on the insert on the Mollier chart

"Heat contribution process water"

The entalpie in the Mollier chart is based on 0oC

As the process-water temperature is higher than 0oC between i3-i4 no heating is required.

When the process water temperature is f.i. 40 oC the heat between i3 and i4 can be calculated as follows:

Dewpoint u1A

Temperature u1

Pa

rtia

l va

po

ur

pre

ssu

re in

u1

19

kP

a

Ab

s.h

um

idit

y i

n u

1 1

44

g/

kg

dry

air

He

ati

ng



430 °C i4

PWE = productwater evaporation

θwb = wet bulbtemperature exhaust air

408 °C i3 φ = relative humidity (%)

ŋθ = temp.-efficiency air heatexchanger

ŋθ = 15 %

80 °C u1 Exhaust air

59 °C u2 θwet bulb u1

u3 60,5oC

58 °C u4

21 °C i2

Supply air

10 °C i1


Moisture increase per kg dry air: ∆w = wu1-wi1 = 138 g moisture / kg dry air


Cons.with HRC supplyair: 7,2 x (hi3 - hi2) = 2920 kJ/kg PWE

HRC when cooling exhaust air till 58 oC is 12 % of heat consumption

Consumption dryer - savings processwater is 2564 kJ/kg PWE

Mollier chart: heat recovery

Temperature-efficiency:

ŋθ = (θi2 -θi1) /(θu1 - θi1) ***

ŋθ = temperature efficiency of the heat exchanger


*** Heat capacity of the exhaust air flow => supply air flow

Box 7.8. Mollier chart for humid air (heat recovery)

Temperature efficiency values are supplied by the supplier of the equipment and are employed to

calculate the amount of heat transferred in air to air heat recovery equipment

Dewpoint u1A

Temperature u1

Pa

rtia

l va

po

ur

pre

ssu

re in

u1

19

kP

a

Ab

s.h

um

idit

y i

n u

1 1

44

g/

kg

dry

air

He

ati

ng


5 DRYING PRINCIPLES

5.1 Product water removal

After the stock preparation the stock flow enters the head box of the paper machine

with a consistency between about 0,2 and 1,3 %. After drainage on the wire or forming

section the web consistency increases to 15 till 26 %. After the press section

consistency is called dry solids content (ds). The dry solids contents after the press

section is between 40 and 55 %.

After the press section the web enters the drying section where the remaining water

thermally is removed till the dry solids content is between 91 and 95 %.

The product water removed in the drying section is:

PWE = (100/dsin – 100/dsout) x Pbone-dry (kg water/s or ton water/h) (18)

Where:

PWE = product water evaporation in kg/s or ton/h

dsin = dry solids content of the web entering the drying section in %

dsout = dry solids content of the paper or board leaving the drying section in %

P bone-dry = the mass of the product dried in the drying section in kgbone-dry /s or ton/h

Some dryers consist of a pre-dryer and an after-dryer; in between is a size-press, a

coater or a glue-machine for applying respectively starch, coating or glues:

Starch

In the case of starch, typically, size press additions involve re-wetting the pre-dried

paper on flooded rollers with a starch solution so as to soak the starch into the paper.

Once soaked, the treated paper is dried again, thereby providing paper of increased

strength. In the production of paper for corrugated board from recovered paper,

approximately 3,5% of the end weight of the product is starch. Currently, starch is

typically added in a solution with only 8% dry matter content. This means that for every

100.000 ton of paper produced, 3500 ton of starch is added in a solution of 40.250 ton

of water, that needs to be evaporated in the after-dryer section.

Coating

Coatings can constitute a large share of the total end weight of the paper (more than

30%). Coatings are added to the paper in much higher consistency than starch, typical

values are between 60 and 70 % dry matter content. This means that for every

100.000 ton of paper produced, 30.000 ton of coating is added in a solution of 14.118

ton of water.

Glues

Glues are used to laminate board with a layer of paper. The amount of paper added to

the board can be a large share of the total end weight of the product, but the glue is

only about 1,5% of the final board mass (without paper additions). Glue is typically

added to the board in a solution of 30% dry matter content. This means that for every

100.000 ton of board produced, 1500 ton of glue is added in a solution of 3500 ton of

water.

In all cases the product is rewetted after the pre-dryer.

In table 5 is shown the increase in productwater evaporation (PWE) in the after-dryer at

different additive solutions.


When, for instance, the consistency of the starch-solution for the size-press is

increased from 8 till 12 % (see column e in table 5) the PWE in the after-dryer will

decrease from 9,5 till 6,0 ton/h (see column i).

Table 5 Waterevaporation in after-dryer at different additives solutions *

a b c d e f g h i j

Input

pre

-dry

er

Dry

solids c

onte

nt

befo

re p

re-d

ryer

Dry

solids c

onte

nt

aft

er

pre

-dry

er

Pro

duct

wate

r evap-

ora

tion p

re-d

ryer

Additiv

es s

olu

tion

(to s

ize p

ress o

r

coate

r)

Input

aft

er-

dry

er

(fib

res +

additiv

es)

Dry

solids c

onte

nt

befo

re a

fter-

dry

er

Dry

solids c

onte

nt

aft

er

aft

er-

dry

er

Pro

duct

wate

r evap-

ora

tion a

fter-

dry

er

Pro

duct

wate

r evap-

ora

ted in p

re-

dry

ers

I&

II a

nd

aft

er-

dry

er

Pdryer dsin dsout PWE - Pdryer dsin dsout PWE PWE

t bone

dry/h% %

t PWE

/h

%-bone

dry

t bone

dry/h% %

t PWE

/h

t PWE

/h

22,2 46,3 95,0 24,6 8,0 23,1 68,3 95,0 9,5 34,1

22,2 46,3 95,0 24,6 12,0 23,1 76,1 95,0 6,0 30,7

22,2 46,3 95,0 24,6 20,0 23,1 83,7 95,0 3,3 27,9

22,2 46,3 95,0 24,6 30,0 23,1 88,1 95,0 1,9 26,5

22,2 46,3 95,0 24,6 40,0 23,1 90,5 95,0 1,2 25,8

22,2 46,3 95,0 24,6 60,0 23,1 93,0 95,0 0,5 25,1

22,2 46,3 95,0 24,6 90,0 23,1 94,8 95,0 0,0 24,7

22,2 46,3 95,0 24,6 20,3 23,1 83,9 95,0 3,2 27,8 ‘* Final product 90 g airdry/m2, ds=95%; product to after-dryer (0,95*90=) 85,5 g bone-dry/m2; additives = 3,6% bone-dry; product to pre-dryer ((100-3,6)/100)*85,5 = 82,4 g bone-dry/m2

5.2 Thermal drying

Drying methods

In the Netherlands are three types of thermal drying methods:

- Multi-cylinder drying for paper and boards

- Yankee cylinder drying for tissues

- Infrared drying for drying after surface sizing and coating

The temperatures in these dryers are up to 100 oC for multi-cylinder dryers, 400 oC for

yankee-dryers and 900 oC for infrared dryers.

In thermal drying of a product, we can distinguish 3 steps:

- the separation of the liquid from the product,

- the transition of the liquid from a fluid to a gaseous phase and

- the removal of the produced vapor.

The manner and temperature at which these steps occur largely determine the energy-

efficiency of the drying process.

For the removal of the produced vapor convective type drying and boiling type drying

are distinguished.

Convective type drying

(Translation: DE: verdunsten, NL: verstrooien)

During convective drying the vapor is removed by a (molecular) diffusion process.


Molecular diffusion, often called simply diffusion, is a net transport of molecules from a

region of higher concentration to one of lower concentration by random molecular

motion. The result of diffusion during convective drying processes is a gradual mixing of

water vapor and dry air molecules. Basically diffusion is the movement of molecules

from an area of high concentration to a lower area.

In the multi-cylinder dryers in the paper industry, heat transfer is mainly conductive

(from the cylinder wall) and for a neglectable part convective (from the heated pocket

air).

Vapor transfer is convective.

The evaporated water diffuses in the air surrounding the cylinders. This air needs to be

replaced all the time in order to maintain a difference in vapor concentration between

the vapor at the surface of the paper web and the vapor in the air. The difference in

vapor concentration is proportional to the water vapor pressure at the paper web

surface and the partial water vapor pressure in the air, which also determines the speed

of diffusion. The speed of diffusion decreases towards zero if the moisture content of

the product would be in balance with the relative moisture content of the surrounding

air.

The Stefan equation expresses evaporation from the web surface to the surrounding

air:

mevaporation = β x R/(Mvapor x T) x (pwebsurface – pvapor) (19)

Where :

mevaporation = drying rate

β = mass transfer coefficient

R = universal gas constant = 8,3145 kJ/(kmol.K)

Mvapor = molar mass water vapor = 18 kg/kmol

T = temperature in K

pwebsurface = partial vapor pressure on the web surface in kPa

pvapor = partial vapor pressure of the vapor in the surrounding air in kPa

As long as the surface of the paper web is wet by surface-moist or macro-capillary-

moist, the wet surface will obtain the wet bulb temperature (θwet-bulb). In this case the

partial vapor pressure on the web surface (pwebsurface) is derived from this wet bulb

temperature (θwet-bulb).

An important feature of the diffusion process is:

Convective type drying : pvapor < po (19a)

Where :

po = atmospheric pressure (101,325 kPa)

During convective type drying air transports the vapor from the dryer to outside.

Boiling type drying

During boiling type drying under atmospheric conditions the temperature of the water

vapor is higher than 100 oC. This results in:

Boiling type drying : pvapor => po (19b)

There is no longer diffusion the vapor flows free in the surrounding air like the boiling

process in a singing teakettle.


No air is needed to transport the vapor from the dryer to outside; the vapor just flows

freely to the surrounding air.

Examples of drying processes with boiling type drying air:

- generally processes with temperatures > 100 oC

- infrared dryers

- yankee dryers

- vacuum dryers

5.3 Drying periods and rates

As shown in the example in figure 2a, the drying process has typically three phases:

- the heat up period

- the constant rate period

- and the falling rate period

During the heat up period the temperature of the web increases tot the the level of

adiabatic saturation. The length of this period is typically 4 to 6 heated cylinders.

During the constant rate period the temperature of the web is typically 65 – 75 oC.

As long as there the surface is saturated with moist during this period, the temperature

of the surface of the web will be theoretically the wet bulb temperature. Due to

conductive heat transfer it will be somewhat higher.

The heat transfer rate from the cylinder to the web determines the rate of

evaporation.

The constant rate period ends when all free water has been removed from the voids

between the fibers. Moisture content is then critical moisture content.

0,50 g/g

70oC

55oC

40oC

15 kg/(m2.h)

0,05 g/g

Temperature

Drying rate

Time

Heat up period Falling rate periodConstant rate period

Figure 2a Drying periods (typical values)


The falling rate period starts at the critical moisture content. The uniform water film

has disappeared and the web is partly dry. The evaporating surface is decreasing;

this results in more resistance between the web and the cylinder. As a result the

heat conductivity decreases and the temperature of the web starts to rise.

The values for drying rate depuicted in figure 1 are typically for multi cylinder

dryers. Drying rates in Yankees are much higher. In figure 2b indicative average

drying rates for multi-cylinder and yankee dryers are shown.

28

22

Multi-cylinder dryers

14

12

120 200

Steam temperature oC

170

140

Air impingement dryers

(Yankee hood etc)

Jet speed > 100 m/s

25

120 500

Jet temperature oC

Dry

ing r

ate

kg/(

m2.h

)D

ryin

g r

ate

kg/(

m2.h

)

Fig. 2b Indicative drying rates for multi-cylinder and yankee dryers

(values for illustration only; not for calculations)

Drying rates decrease considerably when dryness of paper increases. The average

drying rate reflects the average evaporation from wet to dry.

5.4 Air systems

The dryer section ventilation sytem consists essentially of the following parts:

- Supply air system (including distribution elements, pocket ventilators, blow boxes,

air knives, impulse blowing sytems etc)

- Hood and basement enclosures

- Exhaust air sytems

- Heat recovery equipment


5.5 Supply air

Some decades ago the supply air to the drying section was primarily distributed through

the basement. Now the aim is to provide as much dry air as close as possible to the

web to promote evaporation. To obtain this correct distribution of the the supply air is

necessary. Correct distribution will improve both drying rate and sheet uniformity.

The air humidity (wvapor) close to the sheet, which has a fixed relation with the partial

vapor pressure (pvapor), should be at a reasonable level to create sufficient driving force

for evaporation.

The driving force (pwebsurface – pvapor) can be derived from formula (19):

mevaporation = β x R/(Mvapor x T) x (pwebsurface – pvapor) .

The supply air is distributed mainly by blow boxes. Pocket ventilation is one of the

concepts to improve the drying rate and sheet uniformity. The supply air provided by

pocket ventilation to areas where evaporation occurs should be as dry as possible.

A suitable supply air temperature is 90 – 95 oC. Testing showed that higher

temperatures resulted in only negligible increase in evaporation rates while specific heat

consumption increased.

In multi-cylinder dryers the web is heated by conductive heat transfer from the steam

heated cylinders. The heat transfer from the pocket air is negligible.

In yankee dryers, however, air speeds of more than 100 m/s are often maintained. In

this case the heat transfer has an impingement character and will be considerably.

To restrict cold and moist air from the press section from entering the hood and to

restrict hot humid drying air from the hood entering the machine room sometimes

air knives are build in the hood.

5.6 Dryer hoods

The removal of the produced vapor in the drying section of paper mills can be done in

three ways:

No real hood

This means that the ceiling of the production location functions as a hood. In order to

avoid condensation (droplets) the production space is ventilated.

Because of labor conditions, the temperature in summer can not be above 30-40 ºC and

the relative humidity will be below 75 %.

In winter the incoming air should be heated till about 25 oC.

Because of low drying temperatures, the amount of drying air to be removed is high.

This results in poor energy efficiencies.

Open hood

This means that a hood is placed about 2 meters above the machine floor. The incoming

air is first pre-heated by the exhaust air. The temperature of the exhaust air is about

45-50 ºC. By installing an open hood the exhaust air temperature can be raised without

increasing the hall temperature.

Because of low drying temperatures, energy efficiency is better but still poor.

Labor conditions were improved by these hoods.

Closed hood

Here the drying sections of the paper machine are completely separated from the

production space. The ventilation air can be controlled and is supplied to those places

where evaporation is highest. Because of high drying temperatures, energy efficiency is


much better than in the former two options. We distinguish low, medium and high

humidity hoods.

Table 6 shows typical operating values for the different types of closed hoods for

multicylinder and yankee dryers.

The webtemperature at entering the dryer is in all cases supposed to be 35 oC.

Table 6 Typical operating values for different hood concepts

Multi cylinder dryer Yankee-dryer

Unit

Open hood

Low humidit closed hood

Medium humidit closed hood

High humidity

closed hood

Medium humiditclosed hood

High humidit closed hood

Exhaust air:

θ

φ

θdew

wvapor

kg air / kg PWE

oC

% oC

g/kg

kg/kg

55,0

40,0

37,1

41

28,3

80,0

33,0

54,9

114

9,2

85,0

38,0

62,2

174

6,0

90,0

38,0

66,6

225

4,6

320

0,12

53,8

107

9,9

320

0,54

88,6

1215

0,8

Supply air

wvapor

θ before HRC

ŋ θ HRC

θ after HRC

θ after HE steam

msupply/mexhaust

g/kg oC

% oC oC

%

7

10

-

-

-

-

7

10

50

45

95

60

7

10

55

51

95

75

7

10

60

54

95

90

7

10

50

165

-

80

7

10

60

196

-

90

Heat consumption

Without HRC supply air

With HRC supply air

Savings (in relation to

(preceding column)

J/gPWE

J/gPWE

%

3757

-

0

3161

2832

24,6

2967

2716

4,1

2894

2688

1,0

6069

4506

0

3205

3048

32

Heatrecovery

θ before HRC

θ after HRC

Recovery rate

oC oC

%

-

-

-

80

60

7

85

60

18

90

60

36

320

60

51

320

60

93

Note that, when medium humidity hoods of multi-cylinder dryers are replaced by

high humidity hoods, the heat savings will only be 1%. But at the same time, after

cooling the exhaust air till 60 oC, the the heat-recovery rate (= recovered heat / heat

consumption), increases from 18 to 36 % !!

Conclusion: by increasing the temperature of the exhaust air, not only the energy use

in the drying section decreases, also particularly the heat recovery potential increases

with increasing temperatures.

Other advantages of high humidity hoods are smaller air ducts and heat recovery

equipment.

There will also be savings on electricity consumption.


5.7 Zero pressure level

In dryer enclosures, where air temperature and humidity or both are higher than

outside, a difference in density will exist between the air in the enclosure and the air

outside.

The value for this density (ρ in kg/m3) can be derived from the formulas:

ρ = mwet / V ( kg/m3) (21)

Where:

ρ = density of the humid air in kg/m3

mwet = mair + mvapor for volume V m3

mair and mvapor can be derived from:

T

V.p

8,314

18m

vapour

vapour (13c)

T

V.p

8,314

29m

air

air (13d)

The difference in densities causes the so-called chimney effect. Air of low density in the

upper parts of the enclosure tries to escape while air of higher density tries to enter

through the lower parts of the enclosure.

The upper part of the enclosure has a slight excess pressure while in the lower part

there is a slight under pressure compared to the outside. The level where the in- and

outside pressures are equal is the zero pressure level.

This level can be adjusted by balancing the amounts of supply and exhaust air. More

supply air will result in a falling zero level; more exhaust air will result in a rising zero

level.

The zero level is adjusted to where the largest openings are; these are the sheet

entrance and exit openings.

5.8 Exhaust air

To remove the exhaust air in a controlled way from the hood the air passes a false

ceiling or a plenum which is provided with slide dampers along the length of the

machine.

To compensate different evaporation rates these slide dampers can be separately

adjusted.

Normally adjustment is only performed at start-up of the machine after rebuild of the

dryer section.

A better solution to control the humidity of the exhaust air is installing speed controlled

supply and exhaust air fans.

Especially in dryers for a wide product range the exhaust air conditions will vary

depending on the product weight and machine speed. In this case installing speed

controlled fans will result in substantial heat savings.


6 ENERGY CONSUMPTION DRYERS

6.1 System boundary heat consumption drying section

Figure 3 shows a block diagram for defining the borderline for heat consumption of the

dryer section. The depicted situation is just an example for explaining what belongs and

what belongs not to the heat consumption of the dryer.

Example of calculating heat consumption:

- Enthalpy steam supply + 5510 MJ

- Condensate to heat-exchanger processwater - 495 MJ

- Flash steam to steambox - 321 MJ

- Condensate to steamboiler - 416 MJ

- Heat consumption drying section + 4278 MJ

The heat consumption is caused by:

- Heat consumption dryer section I + 1939 MJ

- Heat consumption dryer section II + 1939 MJ

- Heating river water + 387 MJ

- Flash on condensate tank dryer + 13 MJ

- Heat consumption drying section + 4278 MJ

Boundary balance heat consumption drying-sections

Dryer-section I

Dryer-section II

Flashtank

Condensor

steam2000 kg5500 MJ6 bar,159 oC

condensate930 kg, 670 J/g +

blowthrough steam70 kg, 2755 J/g

6 bar,159 oC

condensate930 kg, 670 J/g +

blowthrough steam70 kg, 2755 J/g

6 bar,159 oC Flashtank

condensate882 kg, 561 J/g

3 bar,133 oC

HP Flash steam118 kg, 2723 J/g

3 bar,133 oC

Flashtank

condensate828 kg, 429 J/g 1,1 bar,103 oC

LP Flash steam172 kg, 2678 J/g

1,1 bar,103 oC

cooling waterto river

∆Q= 387 MJ

consumptiondrying section I

Q= 1939 MJ

consumptiondrying section II

Q= 1939 MJ

steam

1000kg2755 J/g

6 bar,159 oC

steam

1000kg2755 J/g

6 bar,159 oC

condensate172 kg, 429 J/g 1,1 bar,103 oC

Condensatetank

flash losscondens.tank1,0 bar,100 oC

∆Q= 13 MJ

condensate to boiler995 kg, 418 J/g, 1,0 bar, 100 oC,

416 MJ

condensate to heat exchanger

space-heating495 MJ

HP Flash steamto steam-boxes

321 MJ

Heat supply to drying sectionsSteam to drying section I 2755 MJSteam to drying section II 2755 MJ +Steam to drying sections 5510 MJ

Heat consumption drying sectionsDrying section I 2755 MJDrying section II 2755 MJ Cooling water from river 387 MJ Flash steam condensate tank 13 MJ +Total heat consumption 4278 MJ

Heat delivery to other consumersHeat-exchanger space-heating 495 MJSteamboxes 321 MJ Condensate to boiler 416 MJ +Total delivery other consumers 1232

Figure 3 Defining system boundary heat consumption drying section


Expressing the heat consumption in tons steam does not give the right information for

heat consumption. Example from practice:

Example heat consumption expressed in ton steam:

Mill A: Steam supply to drying section 1 ton (10 bar, 190 oC, >> h= 2806 MJ/ton) Condensate to boilerhouse 1 ton ( 1 bar, 95 oC, >> h= 399 MJ/ton)

Heat consumption 0 ton ∆h= 2407 MJ/ton

Mill B: Steam supply to drying section 1 ton (10 bar, 190 oC, >> h= 2806 MJ/ton) Condensate to boilerhouse 1 ton ( 8 bar, 170 oC, >> h= 720 MJ/ton)

Heat consumption 0 ton ∆h = 2086 MJ/ton

Though the steam supply is in both cases 1 ton; the “steam consumption” is 0 ton and

the heat consumption in mill A is 15% higher than in mill B.

6.2 Heat and mass balance drying section

In figure 4 an example of a heat and mass balance of drying section is shown. In this

figure also the symbols and units applied in the following chapters are explained.

Heat and mass transfer

Drying section

Paper web in

θpr.in

mpr.in = m.pr.ds. in + mpr.H2O.in

Supply airheater

Product WaterEvaporationθevap; PWE= m.pr.H2O. in - mpr.H2O.out

Paper web out

θpr.out

mpr.out = m.pr.ds. out + mpr.H2O.out

Exhaust air

θexh. ; θdew.exh.; θwet-b.exh.; φexh.;mair.exh.; mvapour.exh.; wvapour.exh.; hexh

Supply air

θsupply; θdew.supply.; θwet-b.supply;

φsupply;mair.supply; mvapour.supply; wvapour.supply; hsupply

Infiltration air

Exfiltration air

Steam supply

Hsteam

Condensate Hcond.

Heat lossesRadiation & convection,

f lash, condensor, leaks

Symbols:

θ = temperature in oC

θdew. = dew point in oC

θwet-b.= wet-bulb temp.in oC

φ = relative humidity

mair = mass dry air in kg

mvapour = mass vapour in kg

wvapour = absolute humidity

in g vapour/ kg dry air

H = enthalpy in kJ

h = specific enthalpy in kJ/kg

PWE = product water evaporation

mpr.= mass wet product m.pr.ds.= mass dry product

mpr.H2O = mass water in product

Figure 4 Heat and mass balance drying section

On the basis of this figure the heat consumption of the in- and outgoing flows for a

typical contemporary drying section are calculated in table 7.

The heat consumption for the average drying section can be roughly divided as follows:

a) Heating of the paper web (2%)


dry material & final moisture

b) Evaporating water (75 %)

- heating of water

- water evaporation

- superheating vapor

- heat of sorption

c) Heating of the supply air (19%)

- heating supply air including vapor

- heating of the air infiltrating in the enclosure of the dryer

d) Convective and radiation heat losses (2%)

e) Other losses (2 %)

- flash steam, leaks, vacuum pumps etc.

The depicted percentages are typical values (state of the art about 1995); the situation

in the mill can be different. In the next chapters energy use for a), d) and e) is not

taken into account. This means that about 94% of the heat consumption of the dryer is

covered.

For multi-cylinder dryers the heat consumption is covered by the difference of

enthalpies of steam supply and condensate:

In general, the heat needed for drying paper is dependent on:

a) the amount of water that needs to be removed in the drying sections

b) the efficiency of water evaporation

c) the amount of heat that is recovered from the exhaust air

6.3 Heat consumption per kg evaporated product water

In box 7.1 u/i 7.8 the Mollier chart is explained.

In box 7.6 an example is given for calculating the heat consumption for evaporating 1

kg product water (MJ/kg PWE).

In excercise 3 you will make this calculations yourself.

A summary of the formulas based on the Mollier chart will be given.

There is no infiltration and exfiltration air so:

mdrying = msupply = mexh

Where:

mdrying = mass drying air in kg/s

msupply = mass supply air in kg/s

mexh = mass exhaust air in kg/s

Main heat consumers dryer section (enthalpies):

- Exhaust air - supply air = 94 %

- Paper web out - paper web in = 2 %

- Heat and other losses = 4 %

- Steam - condensate = 100 %


Table 7. Calculation heat consumption drying section without heatrecovery (see fig. 4 for mass en heat balance)

Nr Heat consumer Formula Typical values closed hoods kJ/kgPWE *

1

1a

1b

Heating product (mpr.in)

from θPr.in tot θPr.out

- dry matter

- final moisture

QPr.ds=mPr.ds . cPr. (θPr.out–θPr.in)

QPr.H2O.out=mPr.H2O.out.cW.(θPr.out– θPr.in)

θPr.in = 45oC, θPr.out = 80oC, cPr = 1,34 J/(g.K),

cW = 4,2 J/(g.K), mPr.H2O.out= mPr.ds.(1/ds-1);ds=0,93

50

10

2

2a

2b

2c

2d

Productwater evaporation (PWE) at θevap

- heating productwater from θPr.i to θevap

- evaporating productwater at θevap

- superheating vapor from θevap to θexh

- bound water (hS)

QHW = PWE . cW . (θevap – θPr.in)

QEW = PWE . (hRO – 2,4. θevap)

QSV = PWE .cvapor . (θexh- θevap)

QBW = PWE. hS

PWE = m Pr.H2O.in – mPr.H2O.out , cvapor = 1,8 J/(g.K),

hRo = 2502 J/(g.K), θevap = 80oC, θexh = 85oC

150

2310

10

0-20

3

3a

3b

Heating supply air

- heating supply air from θsupply to θexh

- heating vapor supp.air (θsupply to θexh)

Qsa = mair.supply . cair . (θexh – θsupply)

Qsv=mair.supply.wvapor.cvapor.(θexh– θsupply)

cair = 1,01 J/(g.K), θsupply = 10oC

wvapor.supply = 0,010 kg vapor/kg (dry) supply air

384

4

4

4a

4b

Heating infiltration air and air for heating

machine room (in winter)

- heating infiltration air from θinfiltr to θexh

- heating room air from θsupply to θroom

Qinf.air = mair.infiltr . cair . (θexh – θinfiltr)

Qroom.air=mair.room . cair . (θroom – θsupply)

θroom = 20oC (average machine room temperature)

mair.infiltr = depends on air balance

mair.room= about 25 kg air for each kg PWE

Year average θsupply = 10oC and year average θroom

during heating season is 16oC

100

150

5

5a

Radiation and convection losses

to surroundings

Qrad.conv = Σ (.A.Δθ)

= 0,3 – 0,5 W/m2. oC, A(m2), θ(oC)

40

6

6a

Other losses

(flash, condensor, leaks etc)

0,3 – 0,5 W/m2 oC

45

7

Total (excl. machine room)

3110


The amount of drying air for 1 kg (= 1000 g) product water evaporation is:

mdryingPWE =(1000 gPWE/kgPWE)/(wvapor.exh- wvapor.supply) kg air/kgPWE (22)

Where:

mdryingPWE = kg drying air/ kg PWE

wvapor.exh = absolute humidity g vapor / kg (dry) air

wvapor.supply = absolute humidity g vapor / kg (dry) air

When the temperature of the process water is 0 oC (θ=0 > h=0), the heat consumption

for 1 kg (= 1000 g) product water evaporation is:

QPWE,0o

C = mdryingPWE x (hexh - hsupply) kJ / kg PWE (23)

Where:


h.exh = enthalpy of the exhaust air in kJ / kg (dry) air

hsupply = enthalpy of the supply air in kJ / kg (dry) air

Remark: the drying process is an adiabatic process; when θ = 0 oC the adiabatic line in

the Mollier chart is the same as the isenthalpic line.

When the temperature of the process water is θ oC, the enthalpy of the process water

will be:

Hprocess.θo

C = cw x θ process kJ / kg PWE (24)

Where:

Hprocess.θoC = enthalpy of the processwater in kJ / kg PWE

cw = specific heat water (=4,2) kJ/(kg.K)

θprocess = temperature process water oC

In the Mollier chart specific enthalpy (h) and abspolute humidity (wvapor) are always per

kg drying air:

∆hprocess.θo

C = Hprocess.θo

C / mdryingPWE kJ / kg (dry) air (25)

When the temperature of the process water is θ oC (θ >0 oC »»» h >0 kJ/kg), the heat

consumption for 1 kg (= 1000 g) product water evaporation is:

QPWE,θo

C = mdryingPWE x {(hexh - hsupply) - ∆hprocess.θo

C } kJ / kg PWE (26)

Where:


h.exh = enthalpy of the exhaust air in kJ / kg (dry) air

hsupply = enthalpy of the supply air in kJ / kg (dry) air

6.4 Relationship between exhaust air temperature and heat consumption

In order to define the optimal drying conditions (i.e. with lowest heat consumption) the

relation between the temperature of the exhaust air and the heat consumption per kg

water evaporation is calculated with help of the formulas in the previous chapters. This


relation is shown in figure 5. Also different dew point temperatures, relative humidities

and drying air temperatures are indicated. The conditions of the supply (outside) air for

this figure are set at 10°C and 80% relative humidity.

Figure 5 Indicative heat consumption ideal dryer

(values for illustration only, not for calculations;

Based on “Trockner und Trocknungsverfahren, Kast, 1956)

The figure shows that from a point of view of heat consumption optimal exhaust air

conditions are: temperature 100 oC and relative humidity (φ) 100%.

This point, however, is located on the pure vapor line where moisture content of the air

is 100% and drying will be impossible. In order to keep the drying process at sufficient

speed, the relative humidity of the air should be around 40%. If the line of 40%

moisture content is followed, it can be seen that the energy use per kg evaporated

water decreases with increasing temperature. This may seem contradictory since higher

temperatures demand a higher energy use. However, the capacity of air to hold

moisture increases exponentially with increasing temperatures, meaning that much less


air is needed to evaporate a kg of water. Therefore, although the temperature

increases, the amount of air that needs to be heated decreases which results in a

decrease in total energy consumption. Given the above, in theory, the optimal drying

condition would be a relative moisture content of around 40% with as high temperature

as possible (i.e. just above the pure vapor line).

Apart from the process constraint (i.e. φ is about 40 %) there is also a construction

limitation. Current best available dryers have dew points of maximum 70°C while many

paper mills with closed hoods have dryers with operating dew points only around 55°C

or even lower. A transition from a dew point of 55°C to a dew point of 70° C would decrease energy use per kg of water evaporation with more than 8%.

The relative humidity in the pockets of the drying section are mostly between 50 and

60% (see figure 6).

As a result of short circuit flows the relative humidity of the exhaust air is considerably

lower. Typical differences between these values vary in practice from 10 till 30%. As

these differences have a high impact on the heat consumption of the dryer, the 10%-

value may be considered as good practice and the 30%-value as bad practice.

Figure 6 Example of relative humidity in pockets of cylinder dryers

Conditions for optimizing exhaust air conditions and heat consumption:

- process constraint:

relative humidity drying air (φ)

(φ lays down the “drying rate”)

- construction constraint:

dew point in the hood (θdew)

(θdew lays down the vapor content of the exhaust air)


The relation between the exhaust air conditions temperature and dewpoint and the heat

consumption are shown in table 8 for relative humidities in exhaust air of 20% and

40%.

The table shows that for a given relative humidity of the exhaust air the heat

consumption will decrease at higher temperatures.

Table 8 Relative humidity and temperature exhaust air and heat consumption

φ= 20% φ= 40%

Dew

poin

t

Tem

pera

ture

Absolu

te

hum

idity

Heat

consum

ption

Savin

gs

Dew

poin

t

Tem

pera

ture

Absolu

te

hum

idity

Heat

consum

ption

Savin

gs

oC oC g/kg kJ oC oC g/kg kJ

24,1 54 19 6.069 36,2 54 39 3.952

28,9 60 26 5.233 14% 41,5 60 53 3.692 7%

33,7 66 34 4.669 46,8 66 72 3.490

38,5 72 45 4.262 52,0 72 97 3.330

42,5 77 57 4.000 56,4 77 124 3.222

47,2 83 74 3.752 61,6 83 168 3.115

52,0 89 97 3.556 5% 66,8 89 229 3.027 3%

56,7 95 126 3.399 72,1 95 317 2.956

61,4 101 165 3.271 77,2 101 450 2.897

66,0 107 218 3.166 82,4 107 669 2.848

70,7 113 290 3.080 87,6 113 1.084 2.807

Exhaust air conditions I Exhaust air conditions II

However at higher temperatures the decrease will be less than at lower temperatures.

The table shows that for a relative humidity of the exhaust air of 20%:

- Increasing the temperature 6 oC from 54 till 60 oC will decrease consumption 14 %

- Increasing the temperature 6 oC from 83 till 89 oC will decrease consumption 5 %

For exhaust air with a relative hunisity of 40% these values are only 7 and 3%.

Example good practice:

Relative humidity between cylinders = 50%

Relative humidity in exhaust pipe = 40%

Difference = 10%

Example bad practice:

Relative humidity between cylinders = 50%

Relative humidity in exhaust pipe = 30%

Difference = 20%


See also “Demonstration drying temperatures, relative humidity and energy

consumption” of the excel file “Demonstration” of MED for a demonstration of the

influence of the actual and optimal drying temperature and relative humidity on the

energy consumption.

Statements:

- At higher temperatures heat consumption decreases

(and quality of heat in exhaust air for heat recovery (HRC) improves)

- The higher the temperature the less the energy consumption decreases

(However the more the quality of the exhaust air improves)

- The higher the relative humidity the less the energy consumption decreases

(However the more the quality of the exhaust air improves)

- The higher the temperature the more the quality of heat in exhaust

air improves

- At higher temperatures the heat for process water, spray water and space-

heating can (almost) be covered by HRC (no fresh steam!!)


Exercise 3 (page 1). Relation heat consumption and exhaust air temperature

Givens: Ideal dryer: there are no heatlosses to the surroundings and no leakages.

The dryer is shown in figure 4 and below.

Product (P):

Temperature product in θPr.in = 0oC

Temperature product out θPr.out = θexh

Mass dry product = 0 kg

Product water evaporation =mpr.H2O.in–mpr.H2O.out=1kg

Supply air (supply):

Temperature θsupply = 10oC

Relative humidity φsupply = 80%

Mass dry air mair.supply = ? kg

Absolute humidity (g vapor/kg dry air) wvapor.supply = ? g/kg

Specific enthalpy (air + vapor/kg dry air) hsupply = ? kJ/kg

Exhaust air (exh):

Temperature ………………………………….……………………… θu =see questions

relative humidity …………………………………………….. φu =see questions

Heat of evaporation, evaporation-temperature and specific heat:

Heat of evaporation at 0oC hRo =2.501,6 kJ/kg

Evaporation-temperature θevap = θexh

Heat of evaporation at θexh oC see steam table in annex

Specific heat water cwater = 4,2 kJ/kg . K.

Specific heat vapor cvapor = 1,8 kJ/kg . K.

Specific heat dry solids product cPr.ds = 1,34 kJ/kg. K.

Specific heat dry air cair = 1,01 kJ/kg . K.

Ideal dryer Supply airheater

Product WaterEvaporationθevap; PWE = m.pr.H2O. in - mpr.H2O.out

Paper web out

θpr.out

mpr.out = m.pr.ds. out + mpr.H2O.out

Exhaust air

θexh. ; θdew.exh.; θwet-b.exh.; φexh.;mair.exh.; mvapour.exh.; wvapour.exh.; hexh

Supply air

θsupply; θdew.supply.; θwet-b.supply;

φsupply;mair.supply; mvapour.supply; wvapour.supply; hsupply

Steam supply

Hsteam

Condensate

Hcond.

Heat consumption Q = Hsteam - H cond

Paper web in

θpr.in

mpr.in = m.pr.ds. in + mpr.H2O.in


Exercise 3 (page 2). Relation heat consumption and exhaust temperature

Questions :

a) Along which constant line in the Mollier chart is the drying process ?

b) Calculate heat consumption if φexh = 20% at θexh = 40, 60, 80 and 95oC.

Write the results in the table on page 3 of this exercise.

c) Ditto if φ = 50%

d) Ditto if φ = 80%

e) Determine the dewpoint at the exhaust air conditions.

f) Draw the results on page 4 of this exercise.

g) At question c) you calculated the heat consumption (Q) for the evaporation of 1

kg water when φexh = 50% and θexh = 40, 60, 80 and 95oC on the basis of the

difference of the enthalpy of the exhaust and supply air.

This consumption can be divided in four parts and after that it can also be

calculated for each part:

- Heating air (Qair.supply)

- Heating vapor in the air (Qvapor.supply)

- Evaporating the product water (QPWE.evap)

- Heating the product water (QPWE.water)

Calculate the consumption for each part and total (Qtotal).

Write the result in the table on page 3 of this exercise.

Compare the results with those you calculated before with the Mollier chart.

Answers :

a) Heat for the vapor in the supply air may be neglected.

Note that the temperature of the ingoing product water is 00C,

In this case the drying process is along the constant line for …………… .

b) u/i f)

The table on page 3 of this exercise can be completed as follows:

- wvapor.supplyi, hsupply, wvapor.exh, hexh and θdew.exh ; see Mollier-chart

- mexh =1 kg PWE : (wvapor.exh – wvapor.supply) kg vapor/kg dry air (=kg air/kg PWE)

- Q = mexh . (hexh – hsupply) kJ/kg PWE.

g) Calculating the heat consumption for each part:

given are φexh = 50% and θexh = 40, 60, 80 and 95oC; calculate p”vapor from: (θ

= θ”)

p”vapor = 0,611 x exp. (17,27 x θ”) : (237,3 + θ”) ;

Calculate wvapor.exh from: (Mvapor = 2x1+16=18 kg; Mair =

0,21x2x16+0,79x2x14=28,84 kg)

wvapor =

vapouro

vapour

air

vapour

pp

p

M

M

kg vapor/kg (dry) air

msupply= mexh = 1 kg PWE : (wvapor.exh – wvapor.supply) kg (dry) air/kg PWE

Qair.supply = msupply . cair . (θexh – θsupply) kJ/kg PWE

Qvapor.supply= msupply .wvapor.supply .cvapor . (θexh - θsupply) kJ/kg PWE

QPWE.evap = hRo – 2,4 . θu * kJ/kg PWE

QPWE.water = cWater . (θexh – 0) kJ/kg PWE

Qtotal = Qair.supply + Qvapor.suppl + QPWE.evap + QPWE.water kJ/kg PWE

Remarks:

* Approximating formula; see also steamtable

∆hair.supply = cair . (θexh - θsupply) kJ/kg (dry) air

∆hair.vapor = cvapor . (θexh - θsupply) kJ/kg vapor

∆hsupply.total = cair.(θexh-θsupply)+ wvapor.supply .(θexh-θsupply) kJ/kg (dry) air

∆hPWE.water = cWater . (θexh – 0) kJ/kg PWE

∆hPWE.evap = hRo – 2,4 . θu kJ/kg PWE ∆hPWE.total = cWater . (θexh – 0) + (hRo – 2,4 . θu) kJ/kg PWE


Exercise 3 (page 3) Relation heat consumption and exhaust temperature

Results calculations heat consumption for 1 kg product water (PWE)

Supply air (θsupply = 10oC; φsupply = 80%)

Temperature θsupply oC

Relative humidity φsupply %

Absolute humidity wvapor.supply g/kg

Enthalpy hsupply kJ/kg

10

80

6

25

10

80

6

25

10

80

6

25

10

80

6

25

Exhaust air ( φexh = 20%)

Temperature θexh oC

Absolute humidity wvapor.exh g/kg

Mass mex kg/kgPWE

Enthalpy hexh kJ/kg

Heat consumption Q kJ/kg H2O

Dew point θdew.exh oC

40

9

333

64

13.00

13

60

26

50

128

5.150

29

80

69

15,9

260

3.736

46

95

133

7,9

450

3.357

57,5




Mass mexh kg/kgPWE

Enthalpy hexh kJ/kg



40

24

56

101

4.222

27,6

60

68

16,1

238

3.435

45,8

80

190

5,43

584

3.038

63,8

95

446

2,27

1.289

2.873

77,2

Heating supply air Q kJ/kg H2O

Heating vapor in supply air.Q kJ/kg H2O

Evaporating product water.Q kJ/kg H2O

Heating product water. Q kJ/kg H2O

Total. Q kJ/kg H2O

1.683

18

2.406

168

4.275

815

9

2.358

252

3.433

384

4

2.310

336

3.034

195

2

2.274

399

2.870




Mass mexh kg/kgPWE

Enthalpy hexh kJ/kg



40

38,5

31

139

3.508

35,9

60

116

9,1

364

3.082

55,3

80

372

2,73

1.066

2.844

74,6

95

1.250

0,80

3.444

2.748

89,1


Exercise 3 (page 4) Relation heat consumption and exhaust air temperature

Heat consumption in kJ / kg PWE

12.000

11.000

10.000

9.000

8.000

7.000

6.000

5.000

4.000

3.000

2.000

30 40 50 60 70 80 90 100

Temperature exhaust air °C

Results calculations heat consumption in kJ / kg PWE

40%

%

80%

50%

0%

%

20%


7 IMPROVING QUALITY OF HEAT IN EXHAUST AIR

7.1 Increasing the exhaust air temperature

By increasing the temperature of the exhaust air, not only the heat consumption in the

drying section decreases, also the heat recovery potential increases with increasing

temperatures.

When the exhaust air temperature is increased and the relative humidity of this exhaust

air is not changed we may assume that the drying rate will not change.

Table 8 showed that when the exhaust temperature and the relative humidity in the

current situation are already rather high the heat savings will be small.

In this chapter we shall demonstrate that, though the heat savings will be small in

these cases, the quality of the exhaust heat will improve sharply and as a result of this

the heat recovery potential rises sharply.

In this chapter we shall demonstrate the increase of the heat recovery potential as well

for multi-cylinder dryers as for yankeedryers.

Hence we will discuss the possibilities to apply the recovered heat for:

- Pre-heating supply air

- Heating process water (head box water)

- Heating spray water (cleaning wires, felts etc)

- Space heating machine room

- Space heating other spaces

In exercise 6 in chapter 13 you can calculate the heat recovery potential in your

current situation and the exhaust heat quality improvement in your optimal situation.

7.2 Explanation heat recovery process in Mollier chart

In figures 7MC(R), 7MC(O) 7YC(R) and 7YC(O) Mollier-charts for heat recovery

processes for exhaust air from multi-cylinder and yankee dryers are depicted.

“MC” indicates multi cylinder, YC yankee cylinder, R is reference situation and O is

optimal situation.

These processes are described in paragraphs 7.3 and 7.4.

In this paragraph we shall give an explanation; the values in this explanation refer to

figure 7MC(R).

i1 – i4 = Heating drying air

Drying air is heated from 10oC to “430”oC. This happens mainly indirect by the cylinders

via the web (conductive) and direct by heating air for pocket ventilation. Air for pocket

ventilation is preheated (i1-i2) by heat recovery from the exhaust air (u1-u3). It is

assumed that during heating the moisture content of the air will not change. This

means that the heating process follows the line of constant moisture content wvap ( 6,1

g moisture/kg is the moisture content of the supply air). As a result of the heat in the

process-water the temperature of this water will be higher than 0oC; this means that

between i3-i4 no heating is required. When the process water temperature is for

instance 40 oC the distance between i3 and i4 can be calculated as follows:


(1 kg PWE * 4,2 kJ/(kg.oC) * 40oC) kJ/kg PWE : ((1000/(144-6)) kg air/kg PWE = 23

kJ/kg

In realty the process in multi-cylinder dryers will not reach a temperature of 430 oC.

The real process will be as depicted below. The heating-line i1-i4 is is supposed to be

the sum of all blue lines (in the same way the drying-line i4-u1 is supposed to be the

sum of all red lines). In this way a process that looks isothermal can be split up in a

combination of many small processes of constant moisture en constant enthalpy.

Real process in multi cylinder dryer

(isothermal, adiabatic or isenthalpic ?)

i4-u1 = Drying

Though the drying process is an adiabatic process, it is assumed to be isenthalpic so it

will follow the line of constant enthalpy. This is only about correct when the

temperature of the product is 0 oC; as we corrected already before for this temperature

the final result will be correct. During drying the temperature of the air decreases

(sensible heat) and at the same time the moisture content of the air increases (latent

heat). The sum of sensible heat and latent heat does not change; this means that the

enthalpy stays constant during the drying process. In u1 the temperature of the

exhaust air is 80oC, the dew point 59 oC, the enthalpy 463 kJ/kg and the relative humidity φ =40%.

u1-u3 = Heat recovery (HRC) from exhaust air (u1-u3) to supply air (i1-i2).

Assuming supply air (i1) is 10 oC, exhaust air (u1) is 80 oC and temperature efficiency

of the heat exchanger is 45%. The preheating of the supply air (i1-i2) can be calculated

as follows: 0,45 x (80-10) = 31,5 oC and θi2 = 31,5+10=41,5 oC.

The Mollier-chart shows that 32,2 kJ/kg is transferred from the exhaust air to the

supply air.

u3-u4 = Heat recovery (HRC) from exhaust air to process- & spraying water

Heat transfer between u3 and u4 is about (431–414=) 17,2 kJ/kg exhaust air.

Heat recover capacity

The amount of heat transferred between u1 and u4 per 1 kg product-water-evaporation

(PWE) is:

(1000 g PWE : (144-6)g PWE/kg air) * (463-414) kJ/kg air = 357 MJ/t PWE.

This is 12% of the energy consumption.

7.3 Improving heat recovery potential multi-cylinder dryers


In figures 7MCc and 7MCo the reference and optimal situation for the heat recovery

potentials are depicted in Mollier charts.

In both situations the temperature of the supply air is 10 oC and the relative humidity is

80%.

The realative humidity of the exhaust air is in both cases 40%; so we may assume that

the drying rate will stay about the same. In the reference situation the exhaust air

temperature is 80 oC and in the optimal situation this temperature is 92,6 oC.c The dew

point under these circumstances will be respectively 59 oC and 70 oC.

In both situations the temperature efficiency of the heat-exchanger for pre-heating the

supply air is 45% (see chapter 8 for explanation).

The fuelprice is in both situations 45,50 €/MWh and the steamboiler efficiency is 90%

In both situations the exhaust air is cooled down in the heat recovery till 58 oC; below

this temperature there are no applications for further heat recovery available.

Comparing the heat recovery potentials for both situations shows that there is a big

difference in heat recovery capacity. In case all exhaust heat with a temperature above

58 oC can be utilised for heat recovery (HRC) the amount of recovered heat in the

current and optimal situation will be respectively 12 and 56% of the heat consumption.

7.4 Improving heat recovery potential yankee-cylinder dryers

In figures 7YCc and 7YCo the reference and optimal situation for the heat recovery

potentials are depicted in Mollier charts.

In both situations the temperature of the supply air is 10 oC and the relative humidity is

80%.

The relative humidity of the exhaust air is in the current situation 0,20% and in the

optimal situation 1,00%; so we may assume that the drying rate will stay about the

same. In the current situation the exhaust air temperature is 300 oC and in the optimal

situation this temperature is also 300 oC. The dew point under these circumstances will

be respectively 58,8 oC and 88,3 oC.

In the reference situation the temperature efficiency of the heat-exchanger for pre-

heating the supply air is 30% and in the optimal situation 60%. (see chapter 9 for

explanation).

The fuel price is in both situations 45,50 €/MWh and the steam boiler efficiency is 90%

In both situations the exhaust air is cooled down in the heat recovery till 58 oC; below

this temperature there are no applications for further heat recovery available.

Comparing the heat recovery potentials for both situations show that there is a big

difference in heat recovery capacity. In case all exhaust heat with a temperature above

58 oC can be utilised for heat recovery (HRC) the amount of recovered heat in the

current and optimal situation will be respectively 47 and 95% of the heat consumption.

7.5 Demonstration energy-efficiency and heat recovery potential

The excel file “Demonstration” of the MED-program contains a demonstration program

for simulating multi-cylinder dryers and yankee-dryers.

For instance fill in:


“Demo MC(R)”: for exhaust air from dryer 83 oC and 40%, for exhaust air after HRC 60

oC, for process water temperature 40 oC and for the temperature efficiency of the

supply air heat recovery 50%

“Demo MC(O)”: for exhaust air from dryer 89 oC and 40%, for exhaust air after HRC 60

oC, for process water temperature 40 oC and for the temperature efficiency of the

supply air heat recovery 50%

The figures show that energy consumption is reduced from 2717 to 2678 J/g PWE

(PWE = Product Water Evaporation).

In case all exhaust heat with a temperature above 60 oC can be utilised for heat

recovery (HRC) the amount of recovered heat in the actual and future situation will be

respectively 15 and 38 % of the energy consumption.


1,0 18,2 partial vapour pressure (pvap in kPa) 19,0 20,4

430 °C i4



φ = relative humidity (%)


ŋθ = 45% 15

408 °C i3 23

40

440437

463

Exhaust air

80 °C U1

U2 u5 θwb=

59,0 °C 60,5

u3

40,0 431 3

2,2

58,0 °C u4

80

41,5 °C i2

100

57,7

414 17,2

49

357

Supply air i1

10 °C

25,5 32,2

12

6,1 136 absolute humidity (wvap in g/kg air) 144

=i1 =u4 =u2

(Fuelprice 45,50 €/MWh)

Moisture increase per kg dry air: ∆w = wu1-wi1 = 138 g moisture / kg air (ŋst.boiler 90 % )


Cons.with HRC supplyair: 7,23 x (hi3 - hi2) = 2765 kJ/kg PWE Cost: 38,83 €/ t PWE

HRC when cooling exhaust air till 58 °C is 12 % of heat consumption

Consumption dryer - savings processwater = 2409 kJ/kg PWE Cost: 33,82 €/"t PWE"

Dewpoint exhaust air

Temperature exhaust air

Figure 7MC(R) Example of the reference Multi Cylinder drying



787 °C i4





ŋθ = 45%

727 °C i3 63

55

772801

835

Exhaust air

92,6 °C u1

u2 u5 θwb=

70,0 °C 71,0

u3

40797 3

8

58,0 °C u4

80

47,2 °C i2

100

63

414

383

421

1551

Supply air (i1)

10,0 °C

25,5 38 56

6,1 136 absolute humidity (wvap in g /kg air) 278

i1 u4 u2




Cons. with HRC supplyair: 3,68 x (hi3 - hi2) = 2612 kJ/kg PWE Cost: 35,94 €/ t PWE

HRC when cooling exh. air till 58,0 °C is 56 % of heat consumption




Figure 7MC(O) Example of the optmal Multi Cylinder drying process



694 °C i4





ŋθ = 30% 15

672 °C i3 23

40

716432

739

Exhaust air

300 °C U1

U2 u5 θwb=

58,8 °C 75,8

u3

0,2

0650 8

8,8

58,0 °C u4

80

97 °C i2

100

114,4

414 236

325

2377

Supply air i1

10 °C

25,5 88,8

47

6,1 136 absolute humidity (wvap in g/kg air) 143

=i1 =u4 =u2




Cons.with HRC supplyair: 7,32 x (hi3 - hi2) = 4403 kJ/kg PWE Cost: 61,83 €/ t PWE

HRC when cooling exhaust air till 58 °C is 47 % of heat consumption




Figure 7YC(R) Example of the reference Yankee Cylinder drying process



3.694 °C i4





ŋθ = 60%

3.507 °C i3 195

40

3.669

3.195

3.864

Exhaust air

300 °C u1

u2 u5 θwb=

88,3 °C 93,7

u3

1

3.686 178

58,0 °C u4

80

184 °C i2

100

203

414

3.2

72

3.4

50

2974

Supply air (i1)

10,0 °C

25,5 178

95

6,1 136 absolute humidity (wvap in g /kg air) 1166

i1 u4 u2




Cons. with HRC supplyair: 0,86 x (hi3 - hi2) = 2988 kJ/kg PWE Cost: 41,12 €/ t PWE

HRC when cooling exh. air till 58,0 °C is 95 % of heat consumption




Figure 7YC(O) Example of the optimal Yankee Cylinder drying process


8 SELECTION OF HEAT EXCHANGERS

8.1 Design and selection

Sequence and design of heat exchangers should be determined in consultation with

the designer of this equipment. Special attention should be paid to anti fouling

measures.

Figure 8 (example from file “Calculations” MED) shows possible configurations of the

heat exchangers in the heat recovery.

Exhaust heat can be applied, for instance, as follows:

- preheating supply air (air for pocket ventilation)

- keeping process water at the desired temperature without applying fresh steam

- heating spray-water for cleaning wires and felts; if necessary in combination with

the exhaust gasses of the steam-boilers

- heating the machine room by the glycol-system;

Selection heat exchangers:

- closed heat exchanger

- open heat exchanger

- half open heat exchanger

Criterion for “low temperature” applications of heat exchangers:

- counter-current flow or counter-current/cross flow

- temperature difference between “water in” and “air out”

- temperature-efficiency air/air heat exchangers

Selection of heat exchangers

“Direct contact” heat exchangers, like scrubbers, perform very good heat transfer and

are also preferred from a point of view of investments. The absorption of air in the

process water may be a disadvantage.

High quality exhaust heat means that the vapor content is high. This results in

condensing heat exchangers which have even higher heat transfer coefficients.

In direct counter-current flow heat exchangers the temperature of the outgoing water

flow may reach the temperature of the outgoing exhaust air.

8.2 Indicative k-values for heat-exchangers

Indicative k-values for the total heat transfer in W/(m2.K) are as follows:

- air/air typical value: 8 ( 5 - 35)

- water/air typical value: 12 (10 - 70)

- steam/air typical value: 14 (10 - 80)

- water/water typical value: 325 (300 - 1200)

- steam/water typical value: 1050 (1000-4000)

(= steam-condenser with thin cupper pipes)

- exhaust gas boiler (15-50)


Heat consumption in GJt or MJt mdry.air mvapor θ θdew Q

Steam Cyl.+HE I&II 29,5 GJt / h 24,5 1,5 44 44 5,0

Steam Cyl.+HE I&II 8,2 MW

Prod.w.evap.(PWE) 10,8 t PWE / h HRC IV 40,0 °C 100 m3/h

Net consumption 2721 MJ / t PWE glycol

Heat consumption in ton steam/h 33,9 °C HE V 65 °C

Steam pressure 4,9 bar ∆Q=-0,68 MW ∆Q= 3,46 MW 2,778

Steam temperature 160,0 oC Steam/glycol

Steam enthalpy 2770 kJ / kg 24,5 1,3 41 41 4,3

Condens.pressure 1,0 bar kg/s kg/s °C °C MW

Condens.temperature 89,8 oC

Condens. enthalpy 377 kJ/kg HRC III 13 °C 120 m3/h

Ton steam / hour 12,3 t / h spraywater

Ton steam / t PWE 1,14 t / t PWE 49,3 °C HE IV 50 °C

∆Q= 5,08 MW ∆Q= 0,10 MW

Gas 0,27 €/m3o ŋboiler 100 % Steam/sprayw.

Heatcosts 8,45 €/GJt 24,5 3,1 56,3 56,3 9,4

Dryers

Gross consumption 25,68 €/t PWE HRC II 45,0 °C 2403 m3/h

HRC for supply air 2,70 €/t PWE processwater

Net consumption 22,98 €/t PWE 45,0 °C HE III 45,5 °C

Processwater ∆Q= 0,00 MW ∆Q= 1,32 MW

Gross consumption 3,7 €/t PWE Steam/processw.

HRC in dryer 0,00 €/t PWE 24,5 3,07 56,3 56,3 9,4

Net consumption 3,7 €/t PWE

Spraywater

Gross consumption 14,52 €/t PWE

HRC in dryer 14,25 €/t PWE

Net consumption 0,27 €/t PWE Supply air mdry.air mvapor θ θdew Q

Spaceheating machine hall 24,5 0,1 10 6,0 0,6

Gross consumption 2,58 €/t PWE HRC I kg/s kg/s °C °C MW

HRC in dryer -1,77 €/t PWE supply air

Net consumption 4,35 €/t PWE ∆Q= 0,96 MW

Spaceheating other

Gross consumption 2,65 €/t PWE 24,5 0,1 48,5 6,0 1,6

HRC in boilerhouse 2,64 €/t PWE

Net consumption 0,01 €/t PWE HE II

Total net consumption Condens./air HRC = heat recovery

Gross 49,14 €/t PWE HE I HE = heat exchanger

HRC 17,82 €/t PWE Steam/air PWE=ProductWaterEvap

Net 31,31 €/t PWE ∆Q= 1,2 MW

24,5 0,1 95 6,0 2,7

24,5 3,15 80 57 10,3

Exfiltration air Infiltration air 0,0 0,0 10 6,0 -

95,0 5,8 112

24,5 3,15 80 57 10,3 24,5 0,1 95,0 6,0 2,7

Paper in (temp. 45 oC) PM1 Pre-dryer PWE= 3,01 kg/s

v 1.000 m/min dm 66,6 %

width 5,00 m

bonedry 54,9 g/m2 Condensate out

bonedry 16,47 t/h Steam cylinders in

dry-m. 46,3 % ∆Q= 7,0 MW

dry a

ir

vapor

tem

peratu

re

dew

poin

t

heat

Ex

ha

ust

air

Figure 8 Configuration heat recovery for exhaust air drying section

(from file “Calculations MED”)


9 HEATING SUPPLY AIR

9.1 Purpose heating supply air

An option to recover the heat of the exhaust air is to pre-heat the supply air of the

drying section first and subsequently heat the process water, spray-water and glycol for

space heating.

As already stated in paragraph 5.5 correct distribution of the the supply air is

necessary. Correct distribution will improve both drying rate and sheet uniformity.

In multi-cylinder dryers the web is heated by conductive heat transfer from the steam

heated cylinders. Pocket ventilation is one of the concepts to improve the drying rate

and sheet uniformity. Because air velocities are low the convective heat transfer from

the pocket air is negligible.

A suitable supply air temperature is 90 – 95 oC; about the exhaust air temperature.

Testing showed only negligible increase in evaporation rates while specific heat

consumption increased.

In yankee dryers, however, air speeds of more than 100 m/s are often maintained. In

this case the heat transfer has an impingement character and will be considerably.

Figure 2b shows that at 500 oC and air velocities of more than 100 m/s drying rates of

up to 150 kg/(m2.h) are possible.

Pre-heating supply air with exhaust air is already common practice. For economic

reasons the temperature efficiencies of the heat exchangers are seldom higher than

60%.

9.2 Temperature-efficiency air-air heat exchangers

The efficiency of heat recovery from the exhaust air to preheat the incoming air

depends on the efficiency of the (air-air) heat exchanger.

Temperature efficiency values are employed to calculate the amount of heat transferred

in air to air heat recovery equipment.

For definition temperature efficiency see figure 9.

9.3 Calculating heat consumption supply air

In table 9 is shown how the heat recovery from the exhaust air and the additional

steam heating is calculated.

The data in the white cells are obtained from field measurements:

- Mass supply air to heatrecovery HRC I as percentage of the drying air mass (column b) - Mass exhaust air to heatrecovery HRC I as percentage of the drying air mass (column l)

- Temperatures and dewpoints supply air and exhaust air for the whole product range

- Temperature efficiency of the heat exchanger

- Temperature supply air after steamheater

The data in the coloured cells are calculated with the formulas from the formula library

in table 3.


Figure 9 Temperature efficiency heat exchanger

Table 9 Calculations heatrecovery and steam heating supply air

a b c d e f g h i j k l m n o p q r s t u v w x y

Supply air after HRC I Exhaust air after HRC I

Mass s

upply

air

to

HR

C in %

of

mdry

Supply

air

consum

ption

Dew

poin

t

Tem

pera

ture

Abs.

hum

idity

Enth

alp

y

Dew

poin

t

Tem

pera

ture

Abs.

hum

idity

Enth

alp

y

Mass e

xh.

Air

to

HR

C in %

of

mdry

Exhaust

air

consum

ption

Dew

poin

t

Tem

pera

ture

Abs.

hum

idity

Enth

alp

y

Dew

poin

t

Tem

pera

ture

Abs.

hum

idity

Enth

alp

y

Tem

p.e

ffic

iency

heat

exchanger

Heat

reception in

heat

recovery

HR

C I

Tem

p.

aft

er

ste

am

-heate

r

Heat

reception in

ste

am

heat

exchanger

%msupply ∆mi2 θdew θ wvapour h θdew θ wvapour h % mexh ∆mu3 θdew θ wvapour h θdew θ wvapour h ηθ ∆hiI θi3 ∆HHE

% kg/s oC oC g/kg J/g oC oC g/kg J/g % kg/s oC oC g/kg J/g oC oC g/kg J/g % MW °C MW

60 80 26,6 6,7 10,0 6,1 25 6,7 46,0 6,1 62,3 90 29,9 52,0 82,0 97 340 52,0 54,5 97 307 50 0,98 95 1,33

80 80 33,2 6,7 10,0 6,1 25 6,7 46,5 6,1 62,8 90 37,4 53,0 83,0 103 356 53,0 55,3 103 323 50 1,24 95 1,65

90 80 32,4 6,7 10,0 6,1 25 6,7 47,0 6,1 63,3 90 36,5 54,0 84,0 109 373 54,0 56,2 109 340 50 1,23 95 1,59

100 80 30,5 6,7 10,0 6,1 25 6,7 47,5 6,1 63,8 90 34,3 55,0 85,0 115 391 55,0 57,1 115 357 50 1,17 95 1,48

110 80 28,7 6,7 10,0 6,1 25 6,7 48,0 6,1 64,3 90 32,3 56,0 86,0 122 410 56,0 58,0 122 376 50 1,11 95 1,38

120 80 27,0 6,7 10,0 6,1 25 6,7 48,5 6,1 64,8 90 30,4 57,0 87,0 129 431 57,0 58,9 129 396 50 1,06 95 1,28

130 80 25,3 6,7 10,0 6,1 25 6,7 49,0 6,1 65,3 90 28,5 58,0 88,0 136 452 58,0 59,9 136 417 50 1,01 95 1,19

91,8 80 29,2 6,7 10,0 6,1 25 6,7 47,3 6,1 63,6 90 32,8 54,8 84,7 113 386 54,8 56,8 113 353 50 1,11 95 1,42

Exhaust air

Pro

duct

code

Heat recovery supply air

g air

dry

/m2

Supply air before HRC I

Supply air

Exhaust air before HRC I

ηθ = temperature efficiency of the heat exchanger

m = mass (kg)

h = enthalpy of the humid air (kJ/kg dry air)

wvapor = moisture content (g/kg dry air)


c = specific heat (kJ/(kg.oC))

When: mexh . cexh < msupply . csupply

supply1exh1exh

supply1supply2supplysupply

.c.m

c.m

exh

(27)

When: mexh . cexh => msupply . csupply

supply1exh1

supply1supply2

supply1exh1supplysupply

supply1supply2supplysupply

)(.c.m

)(c.m

(28)

Exhaust air in (heat delivery)

mexh, θexh1, wvapor.exh1,

hexh1, cexh

Exhaust air out

mexh, θexh2, wvapor.exh2,

hexh2, cexh

Supply air in (heat reception)

msupply, θsupply1, wvapor.supply1, hsupply1, csupply

Supply air out

msupply, θsupply2, wvapor.supply2,

hsupply2, csupply

Heat

exchan

ger air - air


10 PROCESSWATER HEATING

10.1 Purpose of heating processwater

The purpose of heating processwater is to obtain a higher dry matter content of the

web after the press section. This will decrease the amount of water to be evaporated in

the drying section and also result in a lower heat consumption in the drying section.

Further heating process water will improve the runnability in many mills.

There are many advantages and disadvantages:

Advantages

- Steam savings due to higher dry matter content after press section in some cases

- Lower specific energy consumption for pulping and vacuum

- Improved felt- and wire cleaning

- Decreased “picking” on cylinders dryer

- Lower biological activity in process water circulation loop (θ >50 oC; to prevent

legionella pneumophila θ >55 oC)

- Lower specific water consumption

Disadvantages

- Higher temperatures waste water and higher pipe temperatures

- In case of direct contact heat exchangers (=scrubbers): after-scrubbing necessary,

extra maintenance demisters, air in process water

- Heat losses by spontaneous water-evaporation in wire and press section

10.2 Increase dry solids content after press section

The most important advantage is the increase of the dry solids content of the web after

the press section.

Heating the process water will lower energy use since an increased process water

temperature results in a lower viscosity of this water. The lower viscosity leads to

increased drain velocity in the wire and press section which results in some mills in a

higher dry solids content after the press section.

Calculations show that for each percent increase in the dry solids content, the amount

of water to be evaporated decreases with 4%. As a rule of thumb, a temperature

increase of 10°C yields an increase of approximately 1% in dry solids content of the

web after the press section (Cutshall and Hudspeth, 1987) However the precise

increase depends on several conditions (Patterson and Iwasama, 1999, Szikla, 1992

and Maloney et al., 1998).

See example in exercise 4.

The following assumptions based on several of the above mentioned sources:

- Process-water from 40 oC to 50 oC, increase d.s. 1,5%, (steam saving 6 %)

- Process-water from 50 oC to 60 oC, increase d.s. 1,3%, (steam saving 5,2%)



After the stock preparation the stock flow enters the headbox of the papermachine with

a consistency between about 0,2 and 1,3 %. After drainage on the wire or forming

section the web consistency increases tot 15 till 26 %. After the press section

consistency is called dry solids content (ds). The dry solids contents after the press

section is between 40 and 55 %.

After the press section the web enters the drying section where the remaining water

thermally is removed till the dry solids content is between 91 and 95 %.


10.3 Spontaneous water evaporation process water

The most important disadvantage of process water heating is the sharp increase of the

spontaneous water evaporation in the wire and press section.

This heat loss should be compensated in order to keep the process water at a constant

high temperature.

The spontaneous water evaporation is a linear function of the saturated partial vapor

pressure and the relation between both follows from the formula:

pvap = 0,611 exp ((17,27 x θ”)/(237,3 + θ”) (29)

Figure 10 depicts an example of spontaneous water evaporation (Qevap.wire) for different

process water temperatures.

Exercise 4 Example heat savings after increasing ds-contents after press section

The product water removed in the drying section can be calculated with formula

(18):

PWE = (100/dsin – 100/dsout) x Pbone-dry (kg water/s or ton water/h)

Given:

dsout = 92% (dry solids content of the paper leaving the drying section)

P bone-dry = 1 ton/h (mass of the product dried in the drying section)

dsin 50 oC = 48,0% (dry solids content of the web entering the drying section θ=50 oC)

dsin 60 oC = 49,3% (dry solids content of the web entering the drying section θ=60 oC)

Question:

Calculate PWE (product water evaporation in ton/h) at dsin 50 oC=48,0% and dsin 60

oC=49,3%

Answer:

PWE at dsin 50 oC=48,0% = (100/48,0 – 92/100) x 1 = 0,996 t PWE/h

PWE at dsin 60 oC=49,3% = (100/49,3 – 92/100) x 1 = 0,945 t PWE/h

A ds- increase of 1,3% results in a PWE-decrease of 5,1% this results in a heat

saving of (5,1/1,3=) about 4% per 1% ds increase in the press section.


Figure 10 Indication spontaneous water evaporation in wire & press section

10.4 Calculating heat consumption processwater

In order to maintain a constant process water temperature, the energy losses need to

be compensated. In table 10 is shown how the heat recovery from the exhaust air and

the additional steam heating is calculated.

The data in the white cells are obtained from field measurements.

The data in the coloured cells in columns t u/i ae are calculated with the formulas from

the formula library in table 3.

The other data are calculated as follows:

∆Hgross = Wweb.width x Qevap.wire x 0,001 x (hRo-2,4xθ) /3600 (s/h) (30a)

∆Hint.pump = mprocess x 0,001 x ∆Hpump

∆Hint.refiner= Pdryer x 0,001 x ∆Hrefiner (30b)

∆Hnet = ∆Hgross - ∆Hint.pump - ∆Hint.refiner

θin(gross) = {∆Hgross x 3600(s/h) + mprocess x 4,2 x θout }/{mprocess x 4,2} (30c)

θin(net) = {∆Hnet x 3600(s/h) + mprocess x 4,2 x θout }/{mprocess x 4,2 } (30d)

Stock temperature oC

Sponta

neous w

ate

r evapora

tion o

n w

ire &

pre

ss

section in k

g w

ate

r /

(h .

m w

eb w

idth

)


Where:

- Wweb.width = paper web width in m (column c)

- Qevap.wire = spontaneous water evaporation in wire & press section in kg/(h.m)

- θ = temperature process water after wire in oC; column n

- hRo = heat of evaporation (kJ/kg evaporated water); see steam table

- mprocess = process water flow in m3/h (column g)

- Pdryer = wire production in tbone.dry/h (column b)

- dsheadbox = % dry solids content in head box (column d)

- R = % retention on wire section (column e)

- f = - multiply factor for calculating process water (column f)

- ∆Hrefiner = electricity consumption refiners based on ton bone dry wire production

in kWh/tonbone.dry production (column j)

- ∆Hpump = electricity consumption head box pumps based on process water flow

in kWh/m3 process water (column k)

- θout = temperature processwater after wire section in oC (column n)

- ∆Hgross = heat losses on wire section caused by spontaneous water

evaporation in MW (column i)

- ∆Hint.pump = heat delivery by electricity consumption head box pumps based

on process water flow in MW (column m); is part of total consumption

- ∆Hint.refiner = heat delivery by electricity consumption refiners based on ton bone dry

wire production in MW (column l); is part of total consumption

- ∆Hnet = net heat losses in wire & press section to be compensated by

heatrecovery or steam heating in MW (column n)

- θin(gross) = (virtual) temperature processwater before wire section

calculated on base of gross heat loss in oC (column p)

- θin(net) = real temperature processwater before wire section

calculated on base of net heat loss in oC (column q)

When there is a lot of refiner heat the process water temperature may rise above the

desired temperature of θout = 50oC. When the temperature becomes for instance 60 oC

the increase in the spontaneous water evaporation can be calculated from the formula

(29) by calculating pvap at 50 and 60 oC : pvap.50oC and pvap60oC.

Formula (29): pvap = 0,611 exp ((17,27 x θ”)/(237,3 + θ”)

When the spontaneous water evaporation at 50 oC is 1000 kg/((h.m) it will be at 60

oC: Qevap.wire.60oC = Qevap.wire.50oC x (pvap.60oC : pvap50oC)

Heat calculations can be simplified by applying the following specific units:

Amount process water expressed in kg process water per kg PWE:

mprocess.water/PWE = mprocess.water / PWE ton process water/ton PWE

Amount spray water expressed in kg spray water per kg PWE:

msrray.water/PWE = mspray.water / PWE ton spray water/ton PWE


Table 10 Calculations heatrecovery and steam heating process water (MED)

a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae ad

50 oC is 1.000kg/(h

.m)D

ry solids content

in head box

Retentio

n

Multip

ly-factor for

calc

. processw

ater

Process w

ater

(head box w

ater)

Water evap. in

w

ire &

press sectio

n (kg/(h.m

wir

e w

idth)

Gross lo

ss by spontane-

ous w

ater evaporatio

n

in w

ire &

press sectio

n

Ele

ctr.cons. refin

ers per

ton bone dry dryer in

Ele

ctr.cons. pum

ps per

ton processw

ater

Internal heat productio

n

processw

. by refin

ers

Internal heat productio

n

in processw

. by pum

ps

Net lo

ss by spontaneous

water evaporatio

n in

wir

e &

press sectio

n

Tem

perature after w

ire

Tem

perature before

wir

e (based on ∆

Hgross )

Tem

p.before w

ire(based

on ∆

Hnet i.s.o. ∆

Hgross)

Processw

.expr. in

g

pro

cess w

ater

/g PW

E

Part of m

process c

overed

by steam

Part of m

process c

overed

by h

eatrec. d

ryer

Mass exh. air

to H

RC

II

in %

of m

dry

Exhaust air

consum

ptio

n

Dew

poin

t

Tem

perature

Absolu

te hum

idity

Enthalp

y

Dew

poin

t

Tem

perature

Absolu

te hum

idity

Enthalp

y

Heat receptio

n in

heat

recovery H

RC

II

Heat receptio

n in

steam

heat exchanger

- PdryerWweb

ds(hb) R f mproc WWE ∆Hgross ∆Hrefin ∆Hpump ∆Hint.refiners ∆Hint.pump ∆Hnet θout θin θin mproc - -%

mexh

∆mu4 θdew θ wvapour h θdew θ wvapour h ∆HHRCII ∆Hsteam

g air

dry

/m2

t

bone

dry/h

m % % - m3/hkg/

(h.m)MW

kWh

/tbonedr

kWh/

tproc.w.

MW MW MW oC oC oCg/g

PWE

0/100

%

0/100

%% kg/s °C °C g/kg J/g °C °C g/kg J/g MW MW

1.000 50,0

60 16,5 5,00 0,70 83,0 1,05 2.958 1.000 3,29 90 0,07 1,48 0,21 1,60 50,0 50,95 50,46 132 100 0 90 69,9 50,2 58,5 88 289 50,2 58,6 88 289 0,00 1,60

80 22,0 5,00 0,75 84,0 1,05 3.636 1.000 3,29 90 0,07 1,98 0,25 1,06 50,0 50,78 50,25 122 100 0 90 86,1 51,2 57,0 93 301 51,2 57,0 93 301 0,00 1,06

90 22,9 5,00 0,80 85,0 1,05 3.503 1.000 3,29 90 0,07 2,06 0,25 0,98 50,0 50,80 50,24 113 100 0 90 85,1 52,2 57,7 99 316 52,2 57,7 99 316 0,00 0,98

100 22,9 5,00 0,85 86,0 1,05 3.267 1.000 3,29 90 0,07 2,06 0,23 0,99 50,0 50,86 50,26 106 100 0 90 79,1 53,2 58,5 105 332 53,2 58,5 105 332 0,00 0,99

110 23,0 5,00 0,90 87,0 1,05 3.052 1.000 3,29 90 0,07 2,07 0,21 1,01 50,0 50,92 50,28 99 100 0 90 75,1 54,2 59,2 111 349 54,2 59,2 111 349 0,00 1,01

120 23,1 5,00 0,95 88,0 1,05 2.871 1.000 3,29 90 0,07 2,08 0,20 1,01 50,0 50,98 50,30 93 100 0 90 70,0 55,2 60,0 117 367 55,2 60,0 117 367 0,00 1,01

130 23,0 5,00 1,00 89,0 1,05 2.691 1.000 3,29 90 0,07 2,07 0,19 1,03 50,0 51,05 50,33 88 100 0 90 67,0 56,0 60,7 123 381 56,0 60,7 123 381 0,00 1,03

91,8 21,6 5,00 0,830 85,83 1,05 3.155 1.000 3,29 90 0,07 1,94 0,22 1,12 50,0 50,89 50,31 108 100 0 90 76,3 52,9 58,6 103 328 52,9 58,6 103 328 0,00 1,12

Paper w

eb w

idth

Quantity processwaterLosses in wire-&press-section

(evaporation loss at

Heat sources for

processwater

Exhaust air

(from HRC I pre-& after dryers)

Exhaust air (to HRC III

pre-& after dryers)

Heat rec.

proc. water

Product code

Input dryer


Heating headbox water 1289 m3/h

On the processwater side all heatexchangers HRC II are connected in parallel

The processwater flow is proportional to the waterevaporation in each drying section

1289

mwater4= 358 kg/s mwater5= 358 kg/s

dilution θwater4= 55,0 oC HRC II θwater5= 57,0 oC Att. temperature headboxwater out

Steamconsumption 0,00 MW water Hwater4= 82,7 MW 2,95 Hwater5= 85,6 MW Maximum θwater5 = about 60,6 oC

p(bar) = 4,85 temp. (oC) =160 tank Check with designer heat exchanger

steam (J/g) = 2770 cond.(J/g)= 237

Steamconsumption 0,00 t/h Exhaust air MW Exhaust air

363 kg stock/s mG= 20,5 kg/s mG= 20,5 kg/s

1074 kg water/s θ= 60,7 oC θ= 65,6 oC

251 MW θdew out= 60,7 oC θdew in= 65,6 oC

55,7 oC

mbone dry 1= 4,15 kg stock/s Spontaneous waterevaporation

mwater 1= 111,0 kg water/s 1,9 kg water/s

Refiner dry matter 1= 3,60 % dm Headbox pump (incl. refiners) wire- 4,42 MW

θ1= 55,0 oC 1,47 MW section Retention: press-

Hwater1= 25,6 MW Thin stock 2578 m3/h head 83 % section To dryer-section

Thick stock fwater 1= 26,78 5,05 mbone dry 2= 5,05 kg stock/s box mbo.dry3= 3,99 kg stock/s

716 kg water/s mwater 2= 716,0 kg water/s mwater 3= 4,7 kg water/s

0,7 % dm dry matter 2 = 0,70 % dm dry s. 3 = 46 % dm

55,7 oC θ2= 56,5 oC θ3= 55,0oC

mbone dry 5= 0,90 kg stock/s 168 MW Hwater2= 169,8 MW Hwater3= 1,08 MW

mwater 5= 605,0 kg water/s

dry matter 5 = 0,149 % dm

θ5= 55,0 oC

Hwater5= 139,8 MW

fwater 5= 670,84

mbone dry 6= 0,16 kg stock/s mbone dry 4= 1,06 kg stock/s

mwater 6= 104,5 kg water/s mwater 4= 709,5 kg water/s

dry matter 6 = 0,15 % dm dry matter 4 = 0,15 % dm

θ6= 55,0 oC θ4= 55,0 oC

Hwater6= 24,1 MW Hwater4= 163,9 MW

To thick stock preparation

Inputdata 376,0339 Inputdata Inputdata

Final productweight 90 g a.d./m2 Dry matter input drying section 46,0 % Exhaust air HE in & out in out

Basisweight 86 g b.d./m2 Dry matter output drying section 97 % Dry mass exhaust air 20,5 20,5 kg/s

Input drying section 14,4 t b.d./h Evaporation in wire &press section 1.350 kg/(h.m) Temperature exhaust air (θ) 65,6 60,7 oC

Input drying section 4,0 kg b.d./s Wire width (WW) 5,00 m Dew point exhaust air (θdew) 65,6 60,7 oC

Dry matter content thick stock 3,60 % Evaporation heat (EH) 2.357 kJ/kg Heattransfer in heatexchanger HE 2,95 MW

Dry matter content thin stock 0,70 % Powercons. headbox pump and refiners 1,47 MW Mass head box water 716,0 kg/s

Retention wire section 83 % Temperature processwater after wire 55 oC Mass HRCwater/mass head box water *** 0,500 -

Multiply-factor for calc. processwater 1,05 - Steam consumption heating processwater 0,00 MW θdew in (exh.air) - θwater5 out HE=minimal 5,0 oC

Maximum mass water to heat-exchanger 716 kg/s Steampressure and steamtemperature 4,85 160 oC *** Evaporation in wire- & press-section also changes

Figure 11 Simplified diagram for process water (MED)


10.5 Simplified diagram for process water loop

Figure 11 shows a simplified diagram for the process water loop of the paper machine.

There are two reasons why there is always an excess of water present in the paper

machine loop. First, the paper machine loop is continuously fed with fresh water, used for

spraying and cleaning wires and felts. Secondly, the incoming water content of the stock

is higher than the water content in the sheet after the press section.

This excess water is sent as make-up water to the stock preparation, thus following the

counter current principle.

By replacing the input values in the white cells of figure 11 heat and electricity savings

can be simulated:

- Keeping the refreshing by spray water as low as possible (for instance < 5 m3 per ton

wire production). In chapter 11 the heating of spray water will be discussed.

- By replacing the 1,0%-value in the white cell in figure 11 by 3,6% the amount of

excess water that is sent back to the stock preparation will reduce from 1640 m3/ to

376 m3/h.

- Keeping the dry matter content of the thin stock as high as possible. By replacing the

0,7%-value in the white cell in figure 11 by 1,2% the amount of head box water in

the machine loop will reduce from 2578 m3/ to 1496 m3/h.

- Keeping the multiplier f = “mass HRC water / mass head box water” as low as

possible:

f = 1,00 > mass HRC water = 2578 m3/h; water is heated from 55,0 to 56,0 oC



At lower water flows temperature differences increase; this is less favorable for the

efficiency of the heating process.

Electricity savings will be discussed in chapter 14.


11 SPRAYWATER HEATING

11.1 Purpose of heating spray water

Spray water is used for cleaning wires, felts etc. For good cleaning results the

temperature of the spray water should be at least equal to the temperature of the

process water.

Further heating of the spray water will improve the runnability.

A second function of the spray water is refreshment of the machine water loop.

11.2 Calculating heat consumption spray water

Because most fresh water is first used for cooling purposes the temperature will be 5 till

10 oC higher than the source and will be about 22 oC.

In table 11 is shown how the heat recovery from the exhaust air and the additional

steam heating is calculated.

The data in the white cells are obtained from field measurements.

The data in the coloured cells in columns “t” u/i “ae” are calculated with the formulas

from the formula library in table 3.

The temperature difference between the dew point of the exhaust air and the spray

water after the heat exchanger is set at 7 oC.


Table 11 Calculations heatrecovery and steam heating spray water (MED)

a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab

Spraywater; θ=55,0oC; ∆θ exh. HRC & spr.water =7,0

oC

From HRC dryers

- mspray θin θout Qspray - θin θout - θin θout - θin θout

%

mexh

∆mu4 θdew θ wvapour h θdew θ wvapour h mw2 QsprayHRC Qspray BH Qspray ST

- m3/h oC oC MW on/off oC oC on/off oC oC on/off oC oC % kg/s °C °C g/kg J/g °C °C g/kg J/gg w/g

PWEMW MW MW

60 130 22 55,0 5,01 on 22,0 42,3 off 42,3 42,3 on 42,3 55,0 90 69,9 50,2 58,6 88 289 47,4 47,2 75 243 5,79 3,26 0,00 1,74

80 130 22 55,0 5,01 on 22,0 43,9 off 43,9 43,9 on 43,9 55,0 90 86,1 51,2 57,0 93 301 48,8 48,7 82 261 4,36 3,41 0,00 1,59

90 130 22 55,0 5,01 on 22,0 45,0 off 45,0 45,0 on 45,0 55,0 90 85,1 52,2 57,7 99 316 49,8 49,7 86 274 4,20 3,57 0,00 1,44

100 130 22 55,0 5,01 on 22,0 45,9 off 45,9 45,9 on 45,9 55,0 90 79,1 53,2 58,5 105 332 50,6 50,4 90 285 4,20 3,72 0,00 1,29

110 130 22 55,0 5,01 on 22,0 46,9 off 46,9 46,9 on 46,9 55,0 90 75,1 54,2 59,2 111 349 51,4 51,3 95 297 4,21 3,87 0,00 1,14

120 130 22 55,0 5,01 on 22,0 47,8 off 47,8 47,8 on 47,8 55,0 90 70,0 55,2 60,0 117 367 52,2 52,0 99 309 4,21 4,02 0,00 0,99

130 130 22 55,0 5,01 on 22,0 48,6 off 48,6 48,6 on 48,6 55,0 90 67,0 56,0 60,7 123 381 52,8 52,7 103 320 4,23 4,15 0,00 0,86

92 130,0 22,0 55,0 5,01 on 22,0 45,5 off 45,5 45,5 on 45,5 55,0 90 76,3 52,9 58,6 103 328 50,2 50,0 88 280 4,46 3,67 0,00 1,34

Heat covered

by boilerhouse

Mass w

ater to H

RC

III

Heat sources spraywater

Heat covered

by steam

Dew

poin

t

Tem

perature

Absolu

te hum

idity

Enthalp

y

Exhaust air

(to HRC IV pre-& after dryers)

Heat covered

by H

RC

III dryer

Mass exh. air

to H

RC

III

in %

of m

dry

Exhaust air

consum

ptio

n

Exhaust air

(from HRC II pre-& after dryers)

Tem

p. after steam

heater

(H

E IV

in

fig

.1-S

um

m)

Dew

poin

t

Tem

perature

Absolu

te hum

idity

Enthalp

y

Tem

p. sprayw

ater after H

RC

III (fig

.1-S

um

m)

Heat recovery from

H

R

boilerhouse [o

n] or [o

ff]

Tem

p. before H

E3 in

boilerhouse (see fig

.3.1H

)

Tem

p. after H

E3 in

boilerhouse

(see fig

.3.1H

)

Heatin

g by steam

[o

n] or [o

ff]

Tem

p. before steam

heater

(H

E IV

in

fig

.1-S

um

m)

Code

Required heat spraywater From HR boilerhouse From steam

Sprayw

ater quantity

(cle

anin

g w

ire and felts)

Tem

perature sprayw

ater

before heatin

g

Tem

perature sprayw

ater

after heatin

g

Requir

ed heat for

heatin

g sprayw

ater

Heat rec. H

RC

III dryin

g

sectio

ns [o

n] or [o

ff]

Tem

p. sprayw

ater before H

RC

III (fig

.1-S

um

m)


12 SPACEHEATING MACHINE ROOM AND OTHER SPACES

12.1 Heat losses buildings

The amount of heat required to maintain the desired conditions in a building depends

on the required amount of ventilation, the insulation of the building and the outside air

temperature. The amount of ventilation depends on the desired conditions in the

machine room or other building. The insulation depends on the thermal resistance (R-

value) or the reciprocal overall heat transfer coefficient (U-value).

The R-value is a measure of thermal resistance of a building construction:

R = ∆T / qarea m²·K/W (31)

Where:

R = thermal resistance of a construction (building) in m²·K/W

∆T = the temperature difference over the construction

qarea = heat flow per unit area in W/m2

The bigger the number, the better the building insulation's effectiveness.

The U-value, more correctly called the overall heat transfer coefficient, is the reciprocal

of the R-value:

U = 1 /R W/(m²·K) (32)

Increasing the thickness of an insulating layer increases the thermal resistance.

Perfect wall insulation only eliminates conduction through the insulation but leaves

unaffected the conductive heat loss through such materials as glass windows as well as

heat losses from air exchange.

12.2 Degree days (Dgd) and base temperature

Degree-days

Unlike heating process water or spray water, there is no “fixed” temperature difference

for calculating heat consumption for space heating.

During the space heating season, the outside temperature varies in the Netherlands,

between about + 20 oC and -10 oC.

This means, when space heating starts at an outside temperature of 20 oC, the

temperature difference might be between (20-20=) 0 oC and (20 - - 10=) 30 oC.

To simplify calculations a measure of heating or cooling was introduced: degree day

(Dgd).

Distinction is made between “heating degree days” (Dgdheating) and “cooling degree

days” (Dgdcooling).

Dgdheating reflect the demand for energy needed to heat a building or home. Dgd’s are

derived from daily temperature observations. The heating requirements for a given

building are considered to be directly proportional to the number of Dgd’s.

The cooling degree day' (Dgdcooling), reflects the amount of energy used to cool a

building.

A zero degree-day is when both heating and cooling consumption is at a minimum.

Totalized degree days for a certain period may be used to monitor the heating costs of

buildings.

http://en.wikipedia.org/wiki/Thermal_resistance

http://en.wikipedia.org/wiki/Building_insulation

http://en.wikipedia.org/wiki/Heat_transfer_coefficient#Overall_heat_transfer_coefficient

http://en.wikipedia.org/wiki/Heat

http://en.wikipedia.org/wiki/Demand

http://en.wikipedia.org/wiki/Energy

http://en.wikipedia.org/wiki/Temperature


Table 12 Calculations space heating machine room

a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab

Spaceheating: Room PM 1 θgly.out.HRC IV -θdew.before.HRC IV = 7,0oC Glycol heating machine room

Heatrec. [on] or [off] on 2

Base temperature θbase 20oC 3

Degree-days at θbase 2399 dgd/a 4

HR temperature θHR oC 5

Degree-days at θHR dgd/a 6

Lowest temp. outdoor θlow -10oC 7

Degree-days at θlow 0 dgd/a 8

Required heat per Dgd q 8,0 GJ/dgd 9

Req. heating cap. (Pgly.max) 2,78 MW 10

Temp. glycol before heating 35oC 11

Temp. glycol after heating 60 oC 12

Glycolwater flow 100 m3/h 13 PUnet θdew mglycol θinθgly.outH

RC

θout.b.houseθout.ste.h. Pgly.max Pgly.HR Pgly.steam θbase θHRC θlow - θHRC θHRC θHRC q Qglycol Qgly HRC Qgly steam

∆ θglyco & θexhaus t air / flue 7,0oC 14 h/a °C m3/h

oC oC oC oC MW MW MW oC oC oC dgd/a dgd/a dgd/a dgd/aGJ/dgd GJ/a GJ/a GJ/a

Av. cap. HRC/HE (Pgly.HR) MW 15

Av. cap. steam-h. (Pgly.steam) MW 16 60 1100 45,7 100 35,0 38,7 38,7 60,0 2,78 0,41 2,37 20 15,6 -10,0 455 2281 1511 286 8,0 3641 1349 2292

Degree-days at Pgly.max- dgd/a 17 80 900 48,1 100 35,0 41,1 41,1 60,0 2,78 0,68 2,10 20 12,7 -10,0 372 1569 1039 161 8,0 2979 1689 1290

Degree-days at Pgly.HR - dgd/a 18 90 850 49,1 100 35,0 42,1 42,1 60,0 2,78 0,79 1,99 20 11,5 -10,0 352 1287 852 125 8,0 2813 1814 999

Heatdelivery HRC/HE 12.339 GJ/a 19 100 800 49,9 100 35,0 42,9 42,9 60,0 2,78 0,87 1,91 20 10,6 -10,0 331 1094 725 100 8,0 2648 1848 800

20 110 750 50,7 100 35,0 43,7 43,7 60,0 2,78 0,96 1,81 20 9,6 -10,0 310 901 597 77 8,0 2482 1865 617

Heat delivery steam-heater 6.857 GJ/a 21 120 700 51,4 100 35,0 44,4 44,4 60,0 2,78 1,05 1,73 20 8,7 -10,0 290 738 488 59 8,0 2317 1845 472

22 130 700 52,1 100 35,0 45,1 45,1 60,0 2,78 1,12 1,65 20 7,9 -10,0 290 608 402 49 8,0 2317 1928 388

Total heat consumption 19.196 GJ/a 23 20 8,0

24 92 5800 49,2 100 35,0 42,2 42,2 60,0 2,78 0,81 1,97 20 11,3 -10,0 2399 - 1295 857 8,0 19196 12339 6857

Heat

covere

d b

y H

RC

IV

dry

er

Required heating

capacity

Outdoor

temperaturesDegree days (Dgd)

Dew

poin

t exhaust

air

befo

re H

RC

IV

Gly

colw

ate

r flow

(heating m

achin

e h

all)

Tem

pera

ture

gly

colw

ate

r

befo

re h

eating

Tota

l re

quir

ed c

apacity (

HR

C

IV,

HE b

oilerh

.&ste

am

heate

r)

Req.

heat

per

degre

e-d

ay

Degre

e-d

ays a

t θ

HR

C

(wit

h c

orr

. fo

r pro

d.

days

)

Degre

e-d

ays a

t θ

HR

C

(wit

h c

orr

. fo

r pro

d.

ho

urs

)

θH

R is low

est

tem

p.

heatr

ec.

is s

uff

icie

nt;

here

aft

er

addit.

ste

am

heating is r

equir

ed

Low

est

tem

pera

ture

outs

ide

Degre

e-d

ays a

t θ

HR

C

(without

corr

. fo

r pro

d.

hours

)

Heat consumptionMass flow and temp. glycol

Pro

duction h

ours

PM

(=re

el pro

duction)

Depends

product

Depends

product

Code

Heat

covere

d b

y s

team

Tem

p.

gly

col aft

er

ste

am

heate

r

Base t

em

pera

ture

Degre

e-d

ays a

t θbase f

or

the w

hole

pro

duct

range

Tem

pera

ture

gly

colw

ate

r

aft

er

HR

C I

V (

fig.1

H-S

um

m)

Capacity c

overe

d

by H

RC

IV

dry

er

Real to

tal heatc

onsum

ption

(fro

m H

RC

IV

, boilerh

ouse &

ste

am

heate

r)

Tem

p.

gly

col aft

er

boilerh

ouse (

HE3 in f

ig.3

H)

Capacity c

overe

d

by s

team

g air

dry

/m2

Maximum required heating capacity (Pgly.max; table 12 cell d10)

Pgly.max =(θbase - θlow)oC x q GJ/Dgd x1000 MJ/GJ/(24h x 3600s/h) MW (33)Pgly.max =(20 – (-10))oC x 8,0 GJ/Dgd x1000 MJ/GJ/(24h x 3600s/h) =2,78 MW

Available heating capacity heat recovery dryer (Pgly.HR; table 12 cell d15) Pgly.HR1=(θbase - θHR)oC x q GJ/Dgd x 1000 MJ/GJ/(24h x 3600s/h) MW (34)

Pgly.HR1=(20 - θHR)oC x 8 GJ/Dgd x 1000 MJ/GJ/(24h x 3600s/h)= 0,09x(20-θHR)MW

θHR is the lowest temp. HRC is sufficient; hereafter additional steam heating is required

How to calculate θHR ??θHR depends on the heat that can be extracted economically from the exhaust air:

minimum ∆ between θdew.exhaust.air and θglycol.after.HRC (45,7 – 7=) 38,7 oC (cell l16).

Heat extracted by the glycol water from the exhaust air heat:Pgly.HR2 =(θgly.outHRC-θgly.inHRC )oC x mgly m

3/h x (1/3600) x sh MJ/(oC.m3) MW

Pgly.HR2 = (38,7-35)oC x 100 m3/h x(1h/3600s) x 4,0 MJ/(oC.m3) = 0,41 MWBecause Pgly.HR2 =Pgly.HR1 temperature θHR (=15,6oC) can be calculated

Additional heating capacity steam heater (Pgly.steam; table 12 cell d16 > q16)

Pgly.steam=(θgly.outSteam- θgly.outHRC)oC x mgly m3/h x(1/3600)x sh MJ/(oC.m3) MW

Pgly.steam=(60 – 38,7) x 100 x (1/3600) x 4 = 2,37 MW

Degree days (Dgd; table 12 cells d4, d8 etc)Base temperature θbase = 20oC (cell r16), based on 8760 h/a > 3624 Dgd.

Dgd during production (5800/8760) x 3624 Dgd = 2399 Dgd

60 g air.dry/m2: 1100 h/a (cell h16) > Dgd=(1100/5800) x 2399 = 455 Dgd (cell u16).At θHRC= 15,6oC (cell s16) there 2281 Dgd / 365 d (cell v16)

Total production hours: (5800/8760) x 2281 Dgd = 1511 Dgd (cell w16)60 g air.dry/m2 > 1100h/a(cell h16) > Dgd=(1100/5800)x1511= 286 Dgd (cell x16)

Heat delivery HRC IV and heat consumption steam heater

Q = d Dgd/a x q GJ/DgdQglycol = (455 – 0) Dgd x 8 GJ/Dgd = 3641 GJ/a

Qgly.HRC = (455 – 286) Dgd x 8 GJ/Dgd = 1349 GJ/aQglysteam = (286 – 0) Dgd x 8 GJ/Dgd = 2292 GJ


30 oC 1330 Dgd 1833 Dgd

(30--10)x45,8=

Heatdelivery sun, motors, people etc.

1 day

20oC (20--10)x45,8=

20oC=θbase 455 Dgd 1375 Dgd

Dgd=1 day x (20-16)oC= 4 Dgd

Pgly.HR Heat delivery by heat recovery

0,41 16oC Actual heat delivery (168 Dgd x 8 GJ/Dgd =) 1.349 GJ/a

MW * Available heat delivery (204 Dgd x 8 GJ/Dgd =) 1.631 GJ/a

"HRC-SURFACE"= Max. Dgd HRC= Capacity utilization factor (f) = actual heat delivery : available heat delivery

Pgly.max Actual Dgd HRC= (1375-1171=) Capacity utilization factor (f) = 0,83

2,78 (455-286=)168 Dgd 204 Dgd

MW

15,6 oC = θHR 286 Dgd 1171 Dgd

(15,6--10)x45,8=)

"STEAM-SURFACE" Heat delivery by steam heaters

Pgly.s team Actual Dgd steam = Max. Dgd steam= Actual heat delivery (286 Dgd x 8 GJ/Dgd =) 2.292 GJ/a

2,37 (=286-0=) 286 Dgd (1171-0=) Available heat delivery (1171 Dgd x 8 GJ/Dgd =) 9.369 GJ/a

MW * 1171 Dgd Capacity utilization factor (f) = actual heat delivery : available heat delivery

Capacity utilization factor (f) = 0,24

-10 oC = θlow 0 Dgd 0 Dgd

1-jul 1-feb 30-jun

Days = production hours / 24 (= 1100/24=) 45,8 d/a

(Total surface = (30 -(-10))oC x 45,8 d/a = 1.833 Dgd)

* Pgly.max = (θbase - θlow) oC x (q) GJ/dgd x 1000 MJ/GJ / (24 x3600) s/d = ….MW* Pgly.HR = (θbase - θHR) oC x (q) GJ/dgd x 1000 MJ/GJ / (24 x3600) s/d = ….MW

* Pgly.steam = (θHR - θlow) oC x (q) GJ/dgd x 1000 MJ/GJ / (24 x3600) s/d = ….MW

Heat consumption q = 8,0 GJ/Dgd

Example for product 60 g air dry / m2; degree-days and capacities heat exchangers, heat recovery & heat consumption

Figure 12 Calculations space heating machine room


Base-temperature

Dgd’s are defined relative to a base temperature.

The base-temperature is the outdoor temperature above which a building needs no

heating.

For historical reasons Dgd’s are often made available with base temperatures of 18°C

or 15°C. These base temperatures are approximately appropriate for a good proportion

of buildings. For machine rooms and other buildings in paper mills, however, these

base-temperatures are often not appropriate for the following reasons:

- The temperature the building is heated to is deviating from normal

- Large heat and moist production by production equipment

- Heavy insulation of the building

- Solar radiation

- The amount of wind outside

- Individuals' opinions about what constitutes a comfortable indoor temperature.

An approximation method for calculating Dgd is to take the average temperature on

any given day, and subtract it from the base temperature. If the value is less than or

equal to zero, that day has zero Dgd. But if the value is positive, that number

represents the number of Dgd on that day.

Dgd can be added over periods of time to provide a rough estimate of seasonal heating

requirements. In the course of a year, for example, the number of Dgd for Amsterdam

is around 3000 whereas that for a place in Alaska is over 18.000. Thus, one can say

that, for a given building, six times the energy would be required to heat the building in

Alaska.

Example:

For a typical Amsterdam winter day (= 24h), the temperature is during 8 hours 2 oC,

during 12 hours 6 oC and during the other 4 hours 8 oC.

The average temperature during this day (= 24 hours) is ((8x2+12x6+4x8)/24=) 5 oC.

When the base temperature is 18 oC the number of degree-days (Dgd) for that day is

18- 5 = 8 Dgd. A month of thirty similar days might accumulate Dgd (=30x8=) 240

Dgd.

A year (including summer average temperatures above 18 oC) might accumulate an

annual Dgd = 3200.

12.3 Capacities heat exchangers steam and heat recovery

On the basis of the examples depicted in table 12 and figure 12 the method of working

for calculating the required capacity for steam heaters and heat recovery equipment

will be demonstrated. As far as the product range is involved the example will be based

on 60 g air dry / m2 (table 12 cell g16).

Maximum required heating capacity (Pgly.max; table 12 cell d10)

Pgly.max=(θbase- θlow)oC * q GJ/Dgd *1000 MJ/GJ/(24h*3600s/h) MW (33)

[ table 12: d10 = (d3 – d7) x d9 x 1000 /(24 x 3600) ]

Pgly.max =(20 – (-10))oC x 8,0 GJ/Dgd x1000 MJ/GJ/(24h x 3600s/h) =2,78 MW

Available heating capacity heat recovery dryer (Pgly.HR; table 12 cell d15)

Pgly.HR1=(θbase-θHR)oC * q GJ/Dgd * 1000 MJ/GJ/(24h * 3600s/h) MW (34)


[ table 12: d15 = (d3 – d5) x d9 x 1000 /(24 x 3600) ]

Pgly.HR1=(20 - θHR)oC x 8 GJ/Dgd x 1000 MJ/GJ/(24h x 3600s/h)= 0,09x(20-θHR)MW

θHR is the lowest temperature the heatrecovery is sufficient; hereafter additional steam

heating is required.

The question is how to calculate θHR;

To do this we need to know how much heat can be recovered in HRC IV.

This differs for each product. As an example the method of working will be explained for

the weight 60 g air dry / m2.

The heat recover capacity depends on the amount of heat that can be extracted

economically from the exhaust air. Economically means that there should be a

minimum temperature difference between the (dew point) temperature of the exhaust

air before HRC IV (=45,7 oC, cell i16) and the temperature of the glycol after HRC IV. In

the example this temperature is set at 7 oC (cell n1) which results in a glycol

temperature after HRC IV of (45,7 – 7=) 38,7 oC (cell l16).

Now the heat extracted by the glycol water from the exhaust air heat can be calculated:

Pgly.HR2 =(θgly.outHRC-θgly.inHRC )oC*mgly m

3/h*(1/3600)*sh MJ/(oC.m3) MW (35)

[ table 12: p16 = (l16 – k16) x j16 x 4,0 MJ/(oC.m3) ] Pgly.HR2 = (38,7-35)oC x 100 m3/h x(1h/3600s) x 4,0 MJ/(oC.m3) = 0,41 MW

sh = specific heat (for 10% glycol solution 4,0 MJ/(oC.m3))

Because Pgly.HR2 =Pgly.HR1 temperature θHR (=15,6oC) can be calculated with formula

(34).

Additional heating capacity steam heater (Pgly.steam; table 12 cell d16 > q16)

Pgly.steam=(θgly.outSteam- θgly.outHRC)oC*mgly m

3/h*(1/3600)*sh MJ/(oC.m3) MW (36)

[ table 12: q16 = (n16 – l16) x j16 x (1/3600) x 4,0 ) ]

Pgly.steam=(60 – 38,7) x 100 x (1/3600) x 4 = 2,37 MW

12.4 Heat consumption heat exchangers and heat recovery

On the basis of the examples depicted in table 12 and figure 12 the method of working

for calculating degree-days and the heat consumption will be demonstrated. As far as

the product range is involved the example will be based on 60 g air dry / m2 (table 12

cell h16).

Degree days (Dgd; table 12 cells d4, d8 etc)

Are calculated in MED and/or can be obtained from www.kwa.nl:

At base temperature θbase = 20oC (cell r16) there are, based on 365 d/a (=8760 h/a),

3624 Dgd.

Though there are 8760 hours in a calendar year, there are only 5800 h/a for production

(cell h24); the difference between calendar and production hours are left aside in our

calculations.

This result in (5800/8760) x 3624 Dgd = 2399 Dgd (cell u24)for the whole product

range.

http://www.kwa.nl/


For the weight 60 g air dry / m2 the number of production hours is 1100 h/a (cell h16);

the number of degree days is: (1100/5800) x 2399 = 455 Dgd (cell u16).

At the lowest temperature heat recovery is still sufficient for heating θHRC= 15,6oC (cell

s16) there 2281 Dgd / 365 d (cell v16)

For total production hours there are: (5800/8760) x 2281 Dgd = 1511 Dgd (cell

w16)for the whole product range.

For the weight 60 g air dry / m2 the number of production hours is 1100 h/a (cell h16);

the number of degree days is: (1100/5800) x 1511 = 286 Dgd (cell x16).

Heat delivery HRC IV and heat consumption steam heater

Q = d Dgd/a x q GJ/Dgd (37)

Where:

Q = Heat delivery or heat consumption in GJ/a

d = degree days in Dgd/a

q = heat consumption in GJ/Dgd

Total heat consumption for space heating machine room:

Qglycol = (455 – 0) Dgd x 8 GJ/Dgd = 3641 GJ/a

Total heat delivered by heat recovery HRC IV:

Qgly.HRC = (455 – 286) Dgd x 8 GJ/Dgd = 1349 GJ/a

Total heat consumption for space heating machine room:

Qglysteam = (286 – 0) Dgd x 8 GJ/Dgd = 2292 GJ/a

How to calculate the heat consumption per degree day

See figure 12.

Base temperature = 20 oC

Average outdoor temperature during 24 hours (=1 day) = 16 oC

Number of degree days = (20 – 16)oC x 1 day = 4 Dgd

Heat consumption during that day = 200 m3o natural gas x 31,65 MJ/m3

o= 6330 MJ/d

Heat consumption per degree day = 6330 MJ/d : 4 Dgd/d = 1582 MJ/Dgd


13 DRYING PROCESS AND HEATRECOVERY IN MOLLIER CHART

13.1 Introduction

In chapter 7 was shown how by increasing the temperature of the exhaust air, not only

the heat consumption in the drying section decreases, but also the heat recovery

potential.

It was demonstrated that when the exhaust temperature and the relative humidity are

already rather high the heat savings are small, but the quality of the exhaust heat will

improve sharply and as a result of this the heat recovery potential rises sharply.

When the exhaust air temperature is increased and the relative humidity of this exhaust

air is not changed we may assume that the drying rate will not change.

In chapters 8 u/i 12 possibilities and constraints for applying recovered heat for pre-

heating supply air, heating process water, heating spray water and heating glycol for

space heating

were discussed.

In this chapter the knowledge we gained will be applied on a case from practice.

13.2 Reference and optimal situation

Introduction

In exercise 6 two situations are compared.

The first situation, the reference situation, is an example for calculations to be carried

out in the optimal situation.

The exercise should be carried out in the tables and pictures on next pages.

For this task make use of the Mollier-chart 300oC, 190 g/kg; this chart is included as an

attachment (note: unit for h is Wh/kg).

Specify the entire process in the Mollier-chart.

Specify how the efficiency of the dryer can be increased.

Explain in a callout how far the process water can be heated in the heat recovery if no

additional steam injection is applied.


Exercise 5 & 6 (page 1)

Comparing reference and optimal situation

Given:

See MED-tables 1.1 u/i 1.3 and MED-table 4 on next pages.

Question:

Calculate the values in the yellow cells in tables 1.1 u/i 1.3

Remark: the letters above the columns are the same as in the tables of MED IV

Answer:

PRODUCTION

Input drying section:

Reference situation:

Pdryer = 91,6 g bone.dry/m2 x 1200 m/min x 5m x 10-6x 60min/h = 33,0 t bone.dry/h

Optimal situation:

Pdryer = 91,6 g bone.dry/m2 x1200 m/min x 5m x 10-6x 60min/h = 33,0 t bone.dry/h

Product water evaporation:

PWE = (100/dsin – 100/dsout) x Pbone-dry kg water/s or ton water/h (formula 18)


PWE = (100/50 – 100/95) x 33,0 = 31,2 ton water/h

Optimal situation:

PWE = (100/50 – 100/95) x 33,0 = 31,2 ton water/h

PROCESWATER (= HEAD BOX WATER)

Process water (head box water):


mprocess = (33,0 t bone.dry/h / (85,0/100)) x ((100-0,80)/0,80) x 1,00 = 4810 t/h

Optimal situation:

mprocess = (33,0 t bone.dry/h / (85,0/100)) x ((100-0,80)/0,80) x 1,00 = 4810 t/h

For calculating heat losses in the wire and press section the following symbols are

used:

Wweb.width = paper web width in m

Qevap.wire = spontaneous water evaporation in wire & press section in kg/(h.m)

hRo = heat of evaporation (kJ/kg evaporated water); see steam table

mprocess = process water flow in m3/h

Pdryer = wire production in tbone.dry/h

dsheadbox = % dry solids content in head box

R = % retention on wire section

f = multiply factor for calculating process water

θout = temperature process water after wire section in oC

∆Hint.pump = heat delivery by electricity consumption head box pumps based on

process water flow in MW

∆Hint.refiner = heat delivery by electricity consumption refiners based on

ton bone dry wire production in MW

∆Hnet = net heat losses in wire & press section to be compensated by

heat recovery or steam heating in MW

θin = temperature process water before wire section in oC


MED: Table 1.1 Calculations Product Water Evaporation (PWE) for different paper grades

b c d e f g h i j k l m n o p q r s t u v

Production Spraywater ∆θ between exh. HRC & spr.water = 7,0oC

Required heat spraywater HR sources heating sprayw.

- PUnet vnom Wweb Pdryer dsin dsuit PWE mspray θin θout Pspray Qspray Scos t θin θout θout mproces PsprayHRC Pspray STScos t add.

g bd

/m2 h/a m/min m t bd/h % %t PWE

/hm3/h oC oC MW GJ/a €/a oC oC oC

g/g

PWEMW MW €/a

Reference situation

91,6 7.000 1.200 5,00 33,0 50,0 95,0 31,2 100,0 20,0 55,0 4,08 102.900 874.169 20,0 49,3 55,0 3,20 3,42 0,66 142.216

Optimal situation

91,6 7.000 1.200 5,00 33,0 50,0 95,0 31,2 100,0 20,0 55,0 4,08 102.900 874.169 20,0 52,6 55,0 3,20 3,81 0,27 58.742

MED: Table 1.2 Required heat from Heat ReCovery (HRC) and/or (additional) steam for heating processwater


Quantity processwater Heat losses in wire-&press-section HR sources processwater

Pro

duct

to d

ryin

g-s

ection

Dry

solids c

onte

nt

in h

ead b

ox

Rete

ntion

Multip

ly-f

acto

r fo

r

calc

ula

ting p

rocessw

ate

r

Pro

cess w

ate

r

(head b

ox w

ate

r)

Wate

r evapora

tion in

wir

e &

pre

ss s

ection

Gro

ss loss w

ate

r

evapora

tion in w

ire a

ns

pre

ss s

ection

Ele

ctr

.cons.

refiners

per

t

bone d

ry d

ryer

in

Ele

ctr

.consum

ption p

um

ps

per

ton p

rocessw

ate

r

Inte

rnal heat

pro

duction in

pro

cessw

ate

r by r

efiners

Inte

rnal heat

pro

duction in

pro

cessw

ate

r by p

um

ps

Net

loss w

ate

r evap.

in

wir

e a

ns p

ress s

ection

Tem

pera

ture

aft

er

wir

e

Tem

pera

ture

befo

re w

ire

(based o

n ∆

Hgro

ss )

Tem

p.

befo

re w

ire (

based

on ∆

Hnet i.s

.o.

∆H

gro

ss)

Net

loss (

=ste

am

) w

ate

r

evap.

wir

e &

pre

ss s

ection

Cost

ste

am

if

no H

RC

Part

of

mpro

cess covere

d

by s

team

Part

of

mpro

cess covere

d b

y

heatr

ecovery

dry

ers

(HRC)

Pro

cessw

.expre

ssed in

g p

rocessw

ate

r/g P

WE

Part

of

mpro

cess covere

d

by h

eat

recovery

(H

RC)

- ds(hb) R f mpro. WWE ∆Hgros ∆Hrefi ∆Hpu ∆Hint.refiners ∆Hint.pump ∆Hnet θout θin θin ∆Hnet Scos t - - mproces mproHRC

g bd

/m2% % - t/h

kg/

(h.m)MW

kWh

/tbd

kWh/

tpr.w

MW MW MW oC oC oC GJ/a €/a0/100

%

0/100

%

g/g

PWE

g/g

PWE

Reference situation

91,6 0,80 85,0 1,00 4.810 1.400 4,58 5 0,07 0,16 0,34 4,08 55,0 55,82 55,73 102.878 873.984 100 0 154 0

Optimal situation

91,6 0,80 85,0 1,00 4.810 1.400 4,58 5 0,07 0,16 0,34 4,08 55,0 55,82 55,73 102.878 873.984 0 100 154 154

MED: Table 1.3 Heating spraywater and supply air and exhaust air conditions drying section


Energy costs

Net heating value (NHV)31,65 31,65 MJ/m3o

Electricity price 0,085 0,085 €/kWh

CO2-emiss.rights 22,00 22,00 €/t CO2

Gas.price(commodity) 0,30 0,30 €/m3o

Gas.price(transport) 0,01 0,01 €/m3o

CO2-emiss.rights 0,04 0,04 €/m3o

Natural gas price 0,35 0,35 €/m3o

Adjustm.gasprice*1) 76,9 76,9 %

Adjusted gas price 0,27 0,27 €/m3o

Steam - feedwater 2.543 2.543 MJ/t

- θdew θ wvapor h pvapor ηθ θsupply msuppl θdew θ wvapor h pvapor mexh Efficiency boiler 100 100 %

g bd

/m2oC oC g/kg J/g kPa % oC % oC oC g/kg J/g kPa %

Reference situation

91,6 6,0 10,0 6 25 0,9 55,0 95 100 57,0 80,0 129 422 17,3 100

Optimal situation

91,6 6,0 10,0 6 25 0,9 55,0 95 100 66,0 90,0 218 671 25,9 100,0

Explanation: White cells are givens; yellow cells are to be calculated.

*1) adjustment gas price due to

electr.production loss in steamturbine

Tem

pera

ture

supply

air

aft

er

ste

am

heat

exchanger

mass s

upply

air

/

mass d

ryin

g a

ir

Dew

poin

t exhaust

air

Tem

pera

ture

exhaust

air

Absolu

te h

um

idity e

xhaust

air

Spra

yw

ate

r expr.

in g

spra

yw

./g P

WE

Requir

.heat

covere

d b

y

HR

C I

II d

ryer

Requi.heat

covere

d b

y

ste

am

Additio

nal ste

am

costs

Pro

duct

to d

ryin

g-s

ection

Drying-section Refer

ence

Opti

malUnits

Supply air Exhaust air

Dew

poin

t supply

air

Tem

pera

ture

supply

air

Abs.

hum

idity s

upply

air

Enth

alp

y s

upply

air

Part

ial vapour

pre

ssure

supply

air

Tem

pera

ture

eff

iecie

ncy

heat

exchanger

supply

- /

exh.

air

Enth

alp

y e

xhaust

air

Part

ial vapour

pre

ssure

exhaust

air

mass e

xhaust

air

/

mass d

ryin

g a

ir

Product Production & water evap. (PWE) Temperatures

Pro

duct

to d

ryin

g-

section

Pro

d.

hours

PM

(=

reel

pro

duction)

Machin

e s

peed

Paper

web w

idth

Input

dry

er

(fib

res o

nly

)

Dry

solids c

onte

nt

befo

re d

ryer

Dry

solids c

onte

nt

aft

er

dry

ing-s

ection

Pro

duct

wate

r evap-

ora

tion in d

ryer

Spra

yw

. Q

uantity

(c

leanin

g w

ire &

felts)

Tem

p.

spra

yw

ate

r

befo

re h

eating

Tem

p.

spra

yw

ate

r

aft

er

heating

Requir

ed h

eat

for

heating s

pra

yw

ate

r

Requir

ed h

eat

for

heating s

pra

yw

ate

r

Cost

ste

am

if

no H

RC

III

for

heating s

pra

yw

.

Tem

p.

spra

yw

ate

r

befo

re H

RCII

I

Tem

p.

spra

yw

ate

r

aft

er

HRC I

II

Tem

p.

spra

yw

ate

r

aft

er

ste

am

heate

r


Exercise 6 (page 2)

Gross heat loss by spontaneous water evaporation in wire and press section:

∆Hgross = Wweb.width x Qevap.wire x 0,001 x (hRo-2,4xθ) /3600 (s/h) MW (formula 30a)


Spontaneous water evaporation:

θout = 55 oC >> 1400 kg/(h.m), derived from figure 10.

∆Hgross = 5 m x 1400 kg/(h.m) x 0,001 t/kg x (2501,6-2,66x55)/3600s/h= 4,58 MW

Optimal situation:

Spontaneous water evaporation: θout = 55 oC >>

pvap.55oC(1) = 0,611 exp ((17,27 x 55)/(237,3 + 55) = 15,75 kPa(1) (formula 24)

pvap.55oC(2) = 0,611 exp ((17,27 x 55)/(237,3 + 55) = 15,75 kPa(2) (formula 24)

When the spontaneous water evaporation at 55oC(1) is 1400 kg/((h.m) it will be at

55oC(2):

Qevap.wire.55oC(2) = Qevap.wire.55oC(1) x (pvap.55oC(2) : pvap55oC(1)) (formula 29)

Qevap.wire.55oC = 1400 x (15,75/15,75) = 1400 kg/(h.m)

∆Hgross =5 m x 1400 kg/(h.m) x 0,001t/kg x (2501,6-2,66x55)/3600s/h=4,58 MW

Heat delivery by electricity consumption head box pumps (∆Hint.pump ):

∆Hint.pump = mprocess x 0,001 x ∆Hpump MW


∆Hint.pump = 4810 t proc.w./h x 0,001 MW/kW x 0,07 kWh/tproc.water = 0,34 MW

Optimal situation:

∆Hint.pump = 4810 t proc.w./h x 0,001 MW/kW x 0,07 kWh/tproc.water = 0,34 MW

Electricity consumption refiners based on ton bone dry wire production (∆Hrefiner ):

∆Hint.refiner= Pdryer x 0,001 x ∆Hrefiner MW (formula 30b)


∆Hint.refiner= 33,0 t bone.dry/h x 0,001 MW/kW x 5 kWh/ t bone.dry = 0,16 MW

Optimal situation:

∆Hint.refiner= 33,0 t bone.dry/h x 0,001 MW/kW x 5 kWh/ t bone.dry = 0,16 MW

Net heat losses in wire & press section to be compensated by HRC or steam heating

∆Hnet = ∆Hgross - ∆Hint.pump - ∆Hint.refiner


∆Hnet = 4,58 – 0,34 – 0,16 = 4,08 MW

Optimal situation:

∆Hnet = 4,58 – 0,34 – 0,16 = 4,08 MW

Virtual temperature process water before wire section calculated on base of gross

heat loss in wire and press section (θin(gross) ):

θin(gross) ={∆Hgross x3600(s/h)+mprocess x 4,2 x θout}/{mprocess x 4,2}oC (formula 30c)


θin(gross) = {4,58 x 3600 + 4810 x 4,2 x 55 }/{4810 x 4,2} = 55,82 oC

Optimal situation:

θin(gross) = {4,58 x 3600 + 4810 x 4,2 x 55 }/{4810 x 4,2} = 55,82 oC

Real temperature process water before wire section calculated on base of net heat

loss:

θin(net) ={∆Hnet x3600 (s/h)xmprocessx4,2 x θout }/{mprocessx 4,2 }oC (formula 30d)


Exercise 6 (page 3)


θin(net) = {4,08 x 3600 + 4810 x 4,2 x 55 }/{4810 x 4,2} = 55,73 oC

Optimal situation:

θin(net) = {4,08 x 3600 + 4810 x 4,2 x 55 }/{4810 x 4,2} = 55,73 oC

Amount process water expressed in kg process water per kg PWE:

mprocess.water/PWE = mprocess.water / PWE ton process water/ton PWE


mprocess.water/PWE = 4810 t/h / 31,2 t/h = 154 g process water/ g PWE

Optimal situation:

mprocess.water/PWE = 4810 t/h / 31,2 t/h = 154 g process water/ g PWE

Net heat loss (= additional steam consumption) by spontaneous water evaporation in

wire and press section:

∆Hnet.GJ/a = ∆Hnet.MW x 3600 s/h x PUnet h/a x 0,001 GJ/MJ GJ/a


∆Hnet.GJ/a = 4,08 MW x 3600 s/h x 7000 h/a x 0,001 GJ/MJ = 102.878 GJ/a

Optimal situation:

∆Hnet.GJ/a = 4,08 MW x 3600 s/h x 7000 h/a x 0,001 GJ/MJ = 102.878 GJ/a

Cost steam for compensating net heat loss if no HRC is available:

Scost = [{∆Hnet.MW x 3600 s/h x PUnet h/a x (100/ŋboiler)} / Hi MJ/m3o] x sgas €/ m3

o


Scost = [{4,08 x 3600 x 7000 x (100/100)} / 31,65] x 0,27 = 102.878 €/a

Optimal situation:

Scost = [{4,08 x 3600 x 7000 x (100/100)} / 31,65] x 0,27 = 102.878 €/a

For more accurate calculations:

The process water is heated by steam injection:

Steampressure 6 bar, saturated; hsteam = 2.755 MJ/t (see steam table)

Steam injection is live steam, this means the condensate is lost at θprocess.water and is

substituted by make-up water of 10 oC.

Heat for producing 1 ton steam: h∆steam = 2755 – 10 x 4,2 = 2713 MJ

msteam = ∆Hnet.GJ/a x 1000 MJ/GJ / h∆steam ;

Reference situation:msteam =102.878/a x 1000MJ/GJ / 2713 MJ=87.920 t steam /a

Optimal situation:msteam =102.878GJ/a x 1000MJ/GJ / 2713 MJ=87.920 steam /a

SPRAY-WATER

Heat consumption spray water:

The required heat for heating spray water (Pspray):

Pspray = mspray m3/h x (1/3600) h/s x (θout – θin)

oC x 4,2 MJ/(m3.oC)


Pspray = 100 m3/h x (1/3600) h/s x (50 – 20) oC x 4,2 MJ/(m3.oC) = 4,08 MW

Optimal situation:

Pspray = 100 m3/h x (1/3600) h/s x (56 – 20) oC x 4,2 MJ/(m3.oC) = 4,08 MW

Note:


Exercise 6 (page 4)

The temperature difference between the spray water coming from the HRC and the dew

point of the exhaust air to the HRC is set at 7 oC. This temperature difference

determines to a large extent the investment for the heat exchanger.

For method for calculating consumption per year and steam costs see process water.

DRYING SECTION

In the calculations the processes move from (for instance) state 1 to state 2, where the

state number is denoted by subscript; the applied subscripts are shown in MED: figure

2 on pages 89 and 90 .

The yellow marked formulas are applied in the answers:

pvapor (kPa) = 0,611*exp(1)^((17,27*θdew)/(237,3+θdew)) (3a)

wvapor = (Mvapour/Mdry.air) * (((p”vapour / ((po/φ) – p”vapour))) (17)

Substitute formula (3a) in formula (17); note: if θ = θdew then φ =100 %

wvapor (g/kg) = 1000 * (18/28,84) * ((0,611*exp(1)^((17,27*θdew))))/(237,3+θdew)) /

(101,325-(0,611*exp((17,27*θdew))))/(237,3+θdew)))))) (17der)

φ (%) = (pvapour / p”vapour) * 100 % (15)

Substitute (3a) in (15) and see definition in box 6 on page 23 and box 1 on page 11:

φ (%) = ((0,611*exp(1)^((17,27*θdew)/(237,3+θdew)) )) /

(((0,611*exp(1)^(17,27*θ)/(237,3+θ))))) * 100 (15)

h = cp av.air . θ + wvapour . (hRo + cp av.vapour . θ ) kJ/kg (dry) gas (11)

For cp av.air . and cp av.vapour see table 2 on page 20; hRo = 2501,6 kJ/kg

ln (1,639 * p”vapour) = (17,27 * θ”) / (237,3 + θ”) (3b)

When the dew point belonging to the temperature θ is θdew:

θdew = ln((1,639*pvapour)*(237,3+θdew)/17,27) (3bder)

θwet.bulb = ((0,00066*101,325)*R12+(4098*V12/(Q12+237,3)^2*Q12))/((0,00066*101,325)+

4098*V12/(Q12+237,3)^2) (table 3) Note: Results of the wet bulb calculation should always be checked !!!

Supply air conditions drying section

Given: θDi1 = 6 oC and θi1 = 10 oC;


pDi1 = 0,611 * 2,71828 ^ (17,27*6)/(237,3+6) = 0,935 kPa

wi1 = 1000 * (18/28,84) * ((0,611*exp(1)^((17,27*6))))/(237,3+6)) /

(101,325-(0,611*exp((17,27*6))))/(237,3+6)))))) = 5,81 g/kg

φi1 = ((0,611*exp(1)^((17,27*6)/(237,3+6)) )) /

(((0,611*exp(1)^(17,27*10)/(237,3+10))))) * 100 = 76,15 %

hi1 =1,009 * 10 + 5,81 * (2501,6 + 1,84 * 10 ) = 24,8 kJ/kg (dry) gas θwet.bulb = 8,05 oC

Optimal situation:

pDi1 = 0,935 kPa

wwi1 = 5,81 g/kg

φi1 = 76,15 %

hi1 = 24,73 kJ/kg (dry) gas

θnbi1 = 8,05 oC

Exhaust air conditions drying section


Given: θdu1 = 57 oC and θu1 = 80 oC;


MED: Table 4 Calculations energy consumption in the reference and optimal situation

A C D G H I J K L Q R S T U V AA AB AC AD AE AM AV

Supply air Exhaust air after drying section Consumption between i1 & u1

Dew

poin

t

Tem

pera

ture

Absolu

te h

um

idity

Enth

alp

y

Rela

tive h

um

idity

Part

ial vapour

pre

ssure

Dew

poin

t

Tem

pera

ture

Absolu

te h

um

idity

Enth

alp

y

Rela

tive h

um

idity

Part

ial vapour

pre

ssure

Dry

ing a

ir c

ons.

(g a

ir/g

PW

E)

Dry

ing a

ir c

ons.

(kg a

ir /

s)

Mois

ture

reception

(g m

ois

ture

/g P

WE)

Heat

consum

ption

(J h

eat

/g P

WE)

Heat

consum

ption

(MW

)

- PUnet PWE θdi1 θi1 wi1 hi1 φi1 pDi1 θdu1 θu1 wu1 hu1 φu1 pD1 mair/PWE mair/s ∆wvapour∆hPWE ∆hs hu2 ∆hu2

g bd

/m2h/a

gPWE/

s°C °C g/kg J/g % kPa °C °C g/kg J/g % kPa

g/g

PWEkg/s

g/g

PWE

J/g

PWEMW J/g

J/g

PWE

91,6 7.000 8.676 6,0 10,0 5,8 24,8 76,4 0,9 57,0 80,0 129 422 36,4 17,3 8,14 70,58 1,000 2.999 26,02 393 234

91,6 7.000 8.676 6,0 10,0 5,8 24,8 76,4 0,9 66,0 90,0 218 671 37,1 25,9 4,72 40,97 1,000 2.822 24,48 637 161

BU BV BW BX BY BZ CA CF CG CH CI CJ CN CO CP CQ CR CS CT CU CZ DA DB DC DD

Exhaust air after HRC I, this is in u3 Heat deliv. between u1 & u3 Supply air after HRC I Heat recept. between i1 & i2

Mass s

upply

air

aft

er

HRCI

in %

of

mu

Dew

poin

t

aft

er

HRC I

Tem

pera

ture

aft

er

HRC I

Absolu

te h

um

idity

aft

er

HRC I

Enth

alp

y

aft

er

HRC I

Rela

tive h

um

idity

aft

er

HRC I

Part

. vapour

pre

ss.

aft

er

HRC I

Dry

ing a

ir c

ons.

(g a

ir/g

PW

E)

Dry

ing a

ir

consum

ption

Mois

ture

delivery

(g m

ois

ture

/g P

WE)

Heat

delivery

(J h

eat/

g P

WE)

Heat

delivery

(MW

)

Tem

p.

eff

icie

ncy

heat-

exchanger

Mass s

upply

-air

to H

RC in %

of

mu

Dew

poin

t

aft

er

HRC I

Tem

pera

ture

aft

er

HRC I

Absolu

te h

um

idity

aft

er

HRC I

Enth

alp

y

aft

er

HRC I

Rela

tive h

um

idity

aft

er

HRC I

Part

ial vap.

pre

ss.

aft

er

HRC I

Supply

air

cons.

(g s

upply

air

/gPW

E)

Supply

air

cons.

(kg s

upply

air

/s)

Mois

ture

reception

(g m

ois

ture

/g P

WE)

Heat

reception

(J h

eat

/g P

WE)

Heat

reception

(MW

)

%mu θdu3 θu3 wu3 hu3 φu3 pDu3 ∆mu3 ∆mu3 ∆wu3 ∆huI ∆huI ηθ i%mu θdi2 θi2 wi2 hi2 φ i2 pDi2 ∆mi2 ∆mi2 ∆wi2 ∆hiI ∆hiI

% °C °C g/kg J/g % kPag/g

PWEkg/s

g/g

PWE

J/g

PWEMW % % °C °C g/kg J/g % kPa

g/g

PWEkg/s

g/g

PWE

J/g

PWEMW

100 56,3 56,3 125 382 100 16,7 8,14 70,6 0,03 320 2,77 55 100 6,0 48,5 5,8 64,1 8,2 0,936 8,1 70,6 0,00 320 2,77

100 65,6 65,6 214 626 100 25,6 4,72 41,0 0,02 212 1,84 55 100 6,0 54,0 5,8 69,7 6,2 0,936 4,7 41,0 0,00 212 1,84

6th 7th 9th 8th 5th 4th 1st 2nd 3rd

DS DT DU DV DW DX DY ED EE EF EG EH EL EM EN EO EP EQ

Exhaust air after HRC II, this is in u3' Heat deliv. between u3&u3' Heat reception between u3 & u3'

Mass s

upply

air

aft

er

HRCII

in %

of

mu

Dew

poin

t

aft

er

HRC I

I

Tem

p.

aft

er

HRC I

I

Absolu

te h

um

idity

aft

er

HRC I

I

Enth

alp

y a

fter

HRC I

I

Rela

tive h

um

idity

aft

er

HRC I

I

Part

. vap.

pre

ss.

aft

er

HRC I

I

Dry

ing a

ir c

ons.

(g a

ir/g

PW

E)

Dry

ing a

ir c

ons.

(kg a

ir/s

)

Mois

ture

delivery

(g m

ois

t./g

PW

E)

Heat

delivery

(J h

eat/

g P

WE)

Heat

delivery

(MW

)

Mass w

ate

r to

HRC I

I

Mass w

ate

r to

HRC I

I

Tem

p.

befo

re

HRC I

I

Tem

p.

aft

er

HRC I

I

Heat

reception

in H

RC I

I

Heat

reception

in H

RC I

I

% mu θdu3 ' θu3 ' wu3 ' hu3 ' φu3 ' pD3 ' ∆mu3 ' ∆mu3 ' ∆wu3 ' ∆huII ∆huII mw1 mw1 θwi1 θwu1 ∆hwII ∆hwII


PWEkg/s

g/g

PWE

J/g

PWEMW

g

w/gkg/s °C °C

J/g

PWEMW

100 56,3 56,3 125 382 100 16,7 8,14 70,6 0,00 0 0,00 0 0 55,0 55,7 0 0,00

100 62,4 62,4 177 527 100 22,2 4,72 41,0 0,17 471 4,08 154 1.336 55,0 55,7 471 4,08

6th 7th 9th 8th 5th 4th 3rd

ET EU EV EW EX EY EZ FE FF FG FH FI FM FN FO FP FQ FR

Exhaust air after HRC III, this is in u3'' Heat delivery between u3' & u3''Heat reception between u3' & u3''

Mass s

upply

air

aft

er

HRC I

II in %

of

mu

Dew

poin

t

aft

er

HRC I

II

Tem

pera

ture

aft

er

HRC I

II

Abs.

hum

idity

aft

er

HRC I

II

Enth

alp

y

aft

er

HRC I

II

Rela

tive h

um

idity

aft

er

HRC I

II

Part

ial vapour

pre

ss.

aft

er

HRC I

II

Dry

ing a

ir c

ons.

(g a

ir/g

PW

E)

Dry

ing a

ir c

ons.

(kg a

ir/s

)

Mois

ture

delivery

(g m

ois

ture

/g P

WE)

Heat

delivery

(J h

eat/

g P

WE)

Heat

delivery

(MW

)

Mass w

ate

r

to H

RC I

II

Mass w

ate

r

to H

RC I

II

Tem

pera

ture

befo

re H

RC I

II

Tem

pera

ture

aft

er

HRC I

II

Heat

reception

in H

RC I

II

Heat

reception

in H

RC I

II

%mu θdu3 '' θu3 '' wu3 '' hu3 '' φu3 '' pDu3 '' mu3 '' mu3 '' ∆wu3 '' ∆huIII ∆huIII mw2 mw2 θwi2 θwu2 ∆hwIII ∆hwIII


PWEkg/s

g/g

PWE

J/g

PWEMW

g w/g

PWEkg/s °C °C

J/g

PWEMW

100 53,7 53,7 108 334 100 14,8 8,14 70,6 0,14 394 3,42 3,20 27,8 20,0 49,3 394 3,42

100 58,7 58,7 144 434 100 18,8 4,72 41,0 0,16 439 3,81 3,20 27,8 20,0 52,6 439 3,81


FU FV FW FX FY FZ GA GF GG GH GI GJ GN GO GP GQ GR GS

Exhaust air after HRC IV, this is in u4 Heat del. between u3'' & u4' Heat reception between u3'' & u4

Mass s

upply

air

aft

er

HRCIV

in %

of

mu

Dew

poin

t aft

er

HRC

IV

Tem

pera

ture

aft

er

HRC I

V

Abs.

hum

idity a

fter

HRC I

V

Enth

alp

y a

fter

HRC

IV

Rela

tive h

um

idity

aft

er

HRC I

V

Part

ial vapour

pre

ss.

aft

er

HRC I

V

Dry

ing a

ir

consum

ption

Dry

ing a

ir

consum

ption

Mois

ture

delivery

(g m

ois

ture

/g P

WE)

Heat

delivery

(J h

eat/

g P

WE)

Heat

delivery

(MW

)

Mass w

ate

r to

HRC

IV

Mass w

ate

r to

HRC

IV

Tem

pera

ture

befo

re

HRC I

V

Tem

pera

ture

aft

er

HRC I

V

Heat

reception in

HRC I

V

Heat

reception in

HRC I

V

% mu θdu4 θu4 wu4 hu4 φu4 pDu4 mu4 mu4 ∆wu4 ∆huIV ∆hu IV mw3 mw3 θwi3 θwu3 ∆hwIV ∆hwIV


PWEkg/s

g/g

PWE

J/g

PWEMW

g w/g

PWEkg/s °C °C

J/g

PWEMW

100 52,3 52,3 100 311 100 13,8 8,14 70,6 0,07 187 1,62 4,00 34,7 35,0 46,7 187 1,62

100 56,0 56,0 123 377 100 16,5 4,72 41,0 0,10 268 2,32 4,00 34,7 35,0 51,7 268 2,32


Code

Pro

duction h

ours

PM

(=re

el pro

duction)

Enth

alp

y e

xh.a

ir

when θ

= θ

dew

HRC I (pre-heating supply air)

HRC II (heating process water)

HRC III (heating spraywater)

HRC IV (heating glycol)

Pro

ductw

ate

r

evapora

tion

Heat

delivery

betw

een θ

& θ

dew

Drying section


0,9 13,8 (pD) partial vapour pressure (kPa) 17,3 18,8

391 °C i4

364 °C i3 28

584 °C i3

393

422

393

temperature (θ in oC)

382 Exhaust air (u1)

80 °C U1

U2 θnb=

57,0 °C 58,8

56 °C

u3' u3

56,3 °C

36 u3''

53,7 °C382 3

9 u3'' 58,7 °C

52,3 °C u4

76 0

88

49 °C i2

100 334 100

64,164,1

311 48

901

Supply air 23

10 °C i1

24,8 39 88

30

5,8 100 absolute humidity (wD in g/kg air) 125,2 128,7

=i1 =u4 u3 =u2

Moisture increase per kg dry air: ∆w = wu1-wi1 = 123 g moisture / kg air


Consumpt. without HRC: 8,14 x (hi3 - hi1) = 2999 kJ/kg PWE

HRC when cooling exhaustair till 52,3 °C is 30 % of heat consumption

Attention: for HRC is a minimal temperature difference required between both flows !!!!!

"% of mu" = mass percentage of the exhaust air from the dryer

Heat receiving flow Heat delivering flow Heat recovery

heating of quantity unit from oC to oC cooling of % of mufrom oC to oC path J/g PWE GJ/a €/a

HRC I supply air 100 % of mu 10,0 48,5 exhaust air 100 80,0 56,3 u1 - u3 320 69.900 593.826

HRC II processwater 4.810 m3/h 55,0 55,7 exhaust air 100 56,3 56,3 u3 - u3' 0 0 0

HRC III spraywater 100 m3/h 20,0 49,3 exhaust air 100 56,3 53,7 u3' - u3'' 394 86.159 731.952

Share HRC in heating process water 0% Share steam in heating process water 100% u1 - u3'' 901 156.060 1.325.778

Share HRC in heating spray water 84% Share steam in heating spray water % 16%

Heat consumption

Heat consumption without HRC (process water is 0 oC) = g air/g PWE x (hi4 - hi1) J / g air = i1 - i4 3.230 706.210 5.999.483

Heat consump. without HRC (process water is 55,7oC) = g air/g PWE x (hi3 - hi1) J / g air = i1 - i3 2.999 655.705 5.570.429

Heat consumpt. incl. HRC I (process water is 55,7oC) = g air/g PWV x (hi3-hi2eq) J/g lucht = i2 - i3 2.679 585.805 4.976.603

Heat consump.incl. HRC I,II&III (processw. is 55,7 oC) = g air/g PWE*((hi3-hi1)-0,01mu(hu1-hu3''))J/g air=i - u 2.285 499.646 4.244.651

MED: Figure 2 Results reference process (exh. air: 80,0 °C; rh = 36 %)

(p

D) p

arti

al va

po

ur p

re

ssu

re

(k

Pa

)

(w

D) a

bso

lute

hu

mid

ity

g/

kg


0,9 16,5 partial vapour pressure (pD in kPa) 25,9 27,7

631 °C i4

584 °C i3 49

622

671

637

Exhaust air (u1)626 Exhaust air (u1)

90 °C temperature U1

U2 θnb=

66,0 °C 67,3

66 °C 62,4 °C u3 45

u3'

u3''

53,7 °C 37 u3'' 58,7 °C 527 J/g

56,0 °C u4

76434 1

00

238

54 °C i2 %mu=

100100

69,769,7

377 93

1389

Supply air 57

10 °C i1 J/g HRC III

25 4

5

238

49

5,8 (w) 123 absolute humidity (wD in g/kg air) 213,7 217,6

=i1 =u4 u3 =u2

Moisture increase per kg dry air: ∆w = wu1-wi1 = 212 g moisture / kg air

Required air per kg PWE: 1000 g: ∆w g/kg = 4,72 kg air / kg PWE

Consumpt. without HRC 4,72 x (hi3 - hi1) = 2822 kJ/kg PWE

HRC when cooling exhaustair till 56,0 °C is 49 % of heat consumption

Attention: for HRC is a minimal temperature difference required between both flows !!!!!

"% of mu" = mass percentage of the exhaust air from the dryer

Heat receiving flow Heat delivering flow Heat recovery

heating of quantity unit from oC to oC cooling of % of mu from oC to oC path J/g PWE GJ/a €/a

HRC I supply air 100 % of mu 10,0 54,0 exhaust air 100 90,0 65,6 u1 - u3 212 46.373 393.956

HRC II processwater 4.810 m3/h 55,0 55,7 exhaust air 100 65,6 62,4 u3 - u3' 471 102.878 873.984

HRC III spraywater 100 m3/h 20,0 52,6 exhaust air 100 62,4 58,7 u3' - u3'' 439 95.985 815.427

Aandeel WTW in opwarmen proceswater % 100% Aandeel stoom in opwarmen proceswater % 0% u1 - u3'' 1.389 245.237 2.083.367

Aandeel WTW in opwarmen sproeiwater % 93% Aandeel stoom in opwarmen sproeiwater % 7%

Heat consumption

Heat consumption without HRC (process water is 0 oC) = g air/g PWE x (hi4 - hi1) J / g air = i1 - i4 3.053 667.458 5.670.268

Heat consump.without HRC (process water is 55,7 oC) = g air/g PWE x (hi3 - hi1) J / g air = i1 - i3 2.822 616.953 5.241.214

Heat consumpt. incl. HRC I (processwater is 55,7 oC) = g air/g PWV x (hi3-hi2eq) J/g lucht = i2 - i3 2.610 570.580 4.847.257

Heat consump.incl.HRC I,II&III (processw. is 55,7 oC) = g air/g PWE*((hi3-hi1)-0,01mu(hu1-hu3''))J/g air= i - u 1.700 371.716 3.157.847

MED: Figure 2 Results optimal process (exh. air: 90,0 °C; rh = 37 %)

pa

rti

al va

po

ur p

re

ssu

re

(p

D i

n k

Pa

)

(w

D) a

bso

lute

hu

mid

ity (

g/

kg

)


Exercise 6 (page 5)

pDu1 = 0,611 * 2,71828 ^ (17,27*57)/(237,3+57) = 17,3 kPa

wu1 = 1000 * (18/28,84) * ((0,611*exp(1)^((17,27*57))))/(237,3+57)) /

(101,325-(0,611*exp((17,27*57))))/(237,3+57)))))) = 129 g/kg

φu1 = ((0,611*exp(1)^((17,27*57)/(237,3+57)) )) /

(((0,611*exp(1)^(17,27*80)/(237,3+80))))) * 100 = 36,4 %

hu1 =1,009 * 80 + 5,81 * (2501,6 + 1,84 * 80 ) = 422 kJ/kg (dry) gas

θnbu1 = 58,8 oC

Optimal situation:

Given: θDu1 = 66 oC and θu1 = 90 oC;

pDu1 = 25,9 kPa

wu1 = 218 g/kg

φu1 = 37,1 %

hu1 = 671 kJ/kg (dry) gas

θnbu1 = 67,28 oC

Consumption drying section

The yellow marked formulas are applied in the answers:

Drying air consumption:

mair/PWE =1000 /(wvap.exh. –wvap.supp.) g air/g PWE

m.air/s = (PWE) g PWE/s * (mair) g air/g PWE * 0,001 kg/s

Moisture reception:

∆wvapour= mair/PWE * (wvap.exh – wvap.supp) * 0,001 g/g PWE

Heat consumption without heat recovery air/air:

∆hPWE = mair/PWE * (hi3 – hi1) J/g PWE; for hi3 and hi1 see Fig.2-Mol

∆hsec = (PWE) g PWE/s * ∆hPWE * 0,000001 MW



mair/PWE = ∆mu = 1000 /(128,7 – 5,8) = 8,14 g air/g PWE

mair/s = mu = 8676 * 8,14 * 0,001 = 70,58 kg/s

Moisture reception:

∆wvapour= ∆wu = 8,14 * (128,7 – 5,8) * 0,001 = 1,000 g/g PWE


∆hPWE = ∆hu = 8,14 * (393 – 24,8) = 2999 J/g PWE; for hi3 and hi1 see Fig.2-Mol

Remark: exhaust air h = 422 J/g (see page 4) and process-water is 55 oC (see page 2); heat already in process-water per g drying air: 55oC * 4,2 J/(oC.g) / 8,14 g air/g PWE = 28,4 J/g air; this means that the air should be heated till 422 – 28,4 = 393 J/g air

∆hsec =hu = 8676 * 2999 * 0,000001 = 26,02 MW

Optimal situation:


mair/PWE =∆mu = 1000 /(218 – 5,8) = 4,72 g air/g PWE

mair/s = mu = 8676 * 4,72 * 0,001 = 40,97 kg/s

Moisture reception:

∆wvapour=∆wu = 4,72 * (218 – 5,8) * 0,001 = 1,000 g/g PWE


∆hPWE = ∆hu = 4,72 * (622 – 24,8) = 2822 J/g PWE; for hi3 and hi1 see Fig.2-Mol


Exercise 6 (page 6)

Remark:

exhaust air h = 671 J/g (see page 5) and process-water is 55 oC (see page 2); heat already in process-water per g drying air: 55oC * 4,2J/(oC.g) / 4,72 g air/g PWE = 48,9 J/g air; this means that the air should be heated till 671 – 49 = 622 J/g air

∆hsec = hu = 8676 * 2822 * 0,000001 = 24,48 MW

Calculating exhaust air temperatures (θu3, θu3’, θu3” and θu4) after respectively HRC I,

II, III and IV:

The temperature after each HRC depends on the amount of heat-exchange (∆hu) in

J/g PWE or in MW:


HRC I: heat-exchange exhaust air – supply air:

1st : ŋθ = (θi2 - θi1) / (θu1 - θi1) * 100% (see figure 9 in chapter 9.3)

θi2 = 55 * 0,01 * (80 – 10) + 10 = 48,5 oC

2nd: h = cp av.air . θ + wvapour . (hRo + cp av.vapour . θ ) kJ/kg (dry) gas

hi2 = 1,01*48,5 + 0,001*5,8*(2501,6 + 1,84*48,5) = 64,1 J/g

3rd: ∆hiI = ∆mi2 (g air/g PWE)* (hi2 – hi1) (J/g air) (= J/g PWE)

∆hiI = 8,14 * (64,2 – 24,8) = 320 J/g PWE

∆hiI = (PWE) g PWE/s * (∆hiI) J/g PWE * 0,000001 MW

∆hiI = 8676 g PWE/s * 320 J/g PWE * 0,000001 = 2,77 MW

4th: Heat reception = heat delivery; ∆hiI = ∆huI

The enthalpy of the exhaust air after HRC I (hu3) is the enthalpy of the

exhaust air after the dryer (hu1) minus ∆huI

For calculating the temperature of the exhaust air after HRC I it is necessary

to know whether this temperature (θu3) is above or below the dewpoint

(θdu2); to do this we can also check if hu3’ is above or below hu2.

First the partial vapour pressure pDu3 should be calculated:

5th: pDu3 = if (hu3 => hu2; pd1; if (hu3>3158; …………(multi-stage polynomial)

pDu3 = 16,7 kPa

After pDu3 is calculated the dew-point θdu3 can be calculated:

6th: θdu3 = ((ln (1,639 * pDu3)*(237,3 + θdu3)) / 17,27)

θdu3 = ((ln (1,639 * 16,7)*(237,3 + θdu3)) / 17,27) = 56,3 oC

7th: θu3 = if (hu3 <= hu2; θdu3; wu3*0,001*(2501,6+1,84*θu3) / 1,01

θu3 = 56,3 oC

8th: hu3 = hu1 - ∆huI / ∆mu3

hu3 = 422 – 320 / 8,14 = 382 J/g

9th: wu3 = if (hu3 => hu2; wu1; if (hu3>3979; …………(multi-stage polynomial)

wu3 = 125,2 g vapour/kg dry air

HRC II: heat-exchange exhaust air – process water:

1st : θwi1= 55,0 oC ( temperature process water before HRC II)

2nd: θwu1= 55,7 oC (temperature process water after HRC II)

3rd u/ i 9th: see HRC I

HRC III: heat-exchange exhaust air – spray water:

1st : θwi2 = 20,0 oC ( temperature spray water before HRC III)

2nd: θwu2= 49,3 oC (temperature spray water after HRC III)

3rd u/ i 9th: see HRC II


Exercise 6 (page 7)

HRC IV: heat-exchange exhaust air – glycol:

1st : θwi3 = 35,0 oC ( temperature glycol before HRC IV)

2nd: θwu3 = 46,7 oC (temperature glycol after HRC IV)

3rd u/ i 9th: see HRC II

Optimal situation:

HRC I: heat-exchange exhaust air – supply air:

1st : θi2 = 54,0 oC

2nd: hi2 = 69,7 J/g

3rd: ∆hiI = 212 J/g PWE

∆hiI = 1,84 MW

4th: ∆hiI = ∆huI = 212 J/g PWE

5th: pDu3 = 22,2 kPa

6th: θdu3 = 62,4 oC

7th: θu3 = 62,4 oC

8th: hu3 = 527 J/g

9th: wu3 = 177 g vapour/kg dry air

Remark: difference temperatures exhaust air and water (or glycol)


Heating spray water in HRC III:

θdu3’ = 56,3 oC (dew point after HRC II; see: MED table 4 column DT)

∆θ = 7,0 oC (temp.diff. exhaust HRC II and spray-water out;MED table1.1, col.r)

θwu2 = 49,3 oC (temp. spray-water after HRC III; see: MED table 4 column FP)

Heating glycol in HRC IV:

θdu3” = 53,7 oC (dew point after HRC III; see: MED table 4 column EU)

∆θ = 7,0 oC (temp.diff. exhaust HRC III and glycol out; MED table 12 cell 14)

θwu3 = 52,6 oC (temp. glycol after HRC IV; see: MED table 4 column GQ)

Optimal situation:

Heating spray water in HRC III:

θdu3’ = 62,4 oC (dew point after HRC II; see: MED table 4 column DT)

∆θ = 7,0 oC (temp.diff. exhaust HRC II and spray-water out;MED table1.1, col.r)

θwu2 = 55,7 oC (temp. spray-water after HRC III; see: MED table 4 column FP)

Heating glycol in HRC IV:

θdu3” = 58,7 oC (dew point after HRC III; see: MED table 4 column EU)

∆θ = 7,0 oC (temp.diff. exhaust HRC III and glycol out; MED table 12 cell 14)

θwu3 = 51,7 oC (temp. glycol after HRC IV; see: MED table 4 column GQ)


PART IV UTILITIES FOR DRYING

14 FANS, PUMPS AND COMPRESSORS

14.1 Introduction

In Annex C an elaborate description is given for calculating electricity consumption for

fans, compressors an pumps.

Here only a summary of this annex will be depicted.

14.2 Power consumption pumps

Total head pumps

Head is a concept that relates the energy in a fluid to the height of an equivalent static

column of that fluid. Head is expressed in m of height.

The static head of a pump is the maximum height (pressure) it can deliver. The

capability of the pump can be read from its Q-H curve (flow vs. height).

Head is equal to the fluid's energy per unit weight.

Head is useful in specifying centrifugal pumps because their pumping characteristics

tend to be independent of the fluid's density.

There are four types of head used to calculate the total head in and out of a pump:

elevation, pressure, velocity and resistance head. The equation is shown in formula (39)

in box 1.

With formula (40) height can be converted in kPa (=kN/m2) and, as the difference

between pressure gauges on the discharge and suction side of the pump is equal to the

total head, formula (42) can be derived from formula’s (39) and (41).

The resistance loss in pipes and tubes due to friction depends on the flow velocity, pipe

or duct length, pipe or duct diameter, and a friction factor based on the roughness of

the pipe or duct, and whether the flow us turbulent or laminar – the Reynolds Number

of the flow. The resistance loss is divided in major loss due to friction and minor loss

due to change of velocity in bends, valves and similar. The latter are also expressed in

equivalent friction losses.

The resistance loss in a tube or duct due to friction can be expressed as:

ploss.resistance = λ* (l / dh) * (½ * ρ * vpipe2 ) (44)

Where:

ploss.resistance = pressure loss (kPa)

λ = friction coefficient

l = length of duct or pipe (m)

dh = hydraulic diameter (m)

vpipe = flow velocity in pipe or duct (m/s)

For an existing pump system λ, l, and dh are constant (Cpipe):

Cpipe = λ* (l / dh) (45)

Substituting formula (44) in (45):

ploss.resistance = Cpipe * (½ * ρ * vpipe2 ) (46)

Power consumption pumps

The power required to drive a pump is defined by formula (43) in box 1.


Box 1 Head and required power pumps

Explanation symbols and units

p = pressure in N/m2 (=Pa)

(1 bar = 105N/m2; 1 kN = 1 kg.m/s2)

ρ = fluid density in kg/m3

(ρwater = 1000 kg/m3)

g = acceleration due to gravity

(g = 9,81 m/s2)

Pinput = required input power in kW

H = energy head added to the flow in m

Q = flow rate in m3/s or m3/h

ŋpump = total efficiency pump (as decimal)

ŋmotor = efficiency electric motor (as decimal)

Total head (Htot in m):

Htot = (h1 – h2) + resistance

2

2

2

121 H2g

vv

gρ

p p

(39)

Elevation head pressure head velocity head resistance head

Static head (independent of flow) Dynamic head (depending on flow)

Resistance head is friction loss in pipes, ducts, bends etc.

Conversion of pressure (p in N/m2) in head (m):

p (N/m2) = ρ (kg/m3) x g (m/s2) x h (m) (40)

(Note: N = kg.m/s2)

Difference between outlet and inlet pressure pump (= pman):

pman = poutlet – pinlet (41)

Total head (pman = ρ.g.Htot = poutlet – pinlet in kN/m2)

pman = ρ.g.Hgeo + (p1 – p2) + ½ . ρ . )v(v2

2

2

1 + Cpipe .(½. ρ. vpipe2) (42)

Required power electric motor when Q in m3 water/s and p in kPa

Pinput =)ηx(η

}kPa)p(px /s{(Q)m

pompmotor

inletoutlet

3 kW (43)

h1

h2

v2

v1

p1

p2

poutlet

outle

pintlet

outle

Hg

eo =

h

1 -

h2

pressure gauges are on the same height level


Pump efficiency (ηpump) is defined as the ratio of the power imparted on the fluid by the

pump in relation to the power supplied to drive the pump.

Efficiency is a function of the discharge and therefore also operating head; this

means its value is not fixed for a given pump. One important part of system design

involves matching the pipeline headloss-flow characteristic with the appropriate pump

which will operate at or close to the point of maximum efficiency.

Pump efficiencies tend to decline over time due to wear.

The energy usage is determined by multiplying the power requirement by the length of

time the pump is operating.

Pinput =)ηx(η

}kPa)p(px /s{(Q)m

pompmotor

inletoutlet

3 kW (43)

In this formula poutlet – pinlet is the pressure difference between the discharge and

suction gauge on the pump; we will call this difference “manometric head’’:


This head is equal to the total head (pman = ρ.g.Htot = poutlet – pinlet in kN/m2):


2

2

1 + Cpipe .(½. ρ. vpipe2) (42)


Power consumption centrifugal fans and compressors

In thermo dynamics two compression processes are distinguished: isothermal

compression and isentropic compression.

Isothermal compression assumes that the compressed gas remains at a constant

temperature throughout the compression process. This is only possible when during

compression heat is removed. Compressors that utilize inter-stage cooling between

compression stages come closest to achieving isothermal compression.

Isentropic compression assumes that no heat is removed from the gas during the

compression, and all supplied work is added to the internal energy of the gas, resulting

in increases of temperature and pressure. Theoretical temperature rise is:

Toutlet = Tinlet x (poutlet/pinlet)(k-1)/k

(47)

Where:

Toutlet = outlet temperature in K

Tintlet = inlet temperature in K

poutlet = outlet pressure in kPa

pinlet = inlet pressure in kPa

k = cp/cv, (= ratio of specific heats, approximately 1,4 for air and 1,3 for water vapor)

Isentropic (also called adiabatic) compression is more closely real life.

Isothermal compression takes less work than isentropic (adiabatic) compression:


Pisentropic=)ηx(η

}1)p

p{(x

1k

kx (kPa)px /s)V(m

compressormotor

k

1k

inlet

outlet

inlet

3

kW (48)

Pisothermic= Wk)ηx(η

p

p lnx (kPa) px /s)(mV

compressormotor

inlet

outlet

inlet

3

(49)

Where:



ηmotor = efficiency electric motor in decimal

ηcompr. = efficiency compressor or fan in decimal

V = flow rate in m3/s (at inlet pressure)

Pisentropic = required power for isentropic compression in kW

Pisothermic = required power for isothermic compression in kW

k = cp/cv, (= ratio of specific heats, appr. 1,4 for air and 1,3 for water vapor)

Total head fans

For fans also four kinds of head can be distinguished.

There are however important differences between the transport fluids:

- the density of water in pumps is 1000 kg/m3 and for air in fans this is about 1,3

kg/m3 at standard conditions (0 oC and 101,325 kPa)

- water is incompressible and air is compressible

In box 1 is derived that for pumps the required power: P(kW)= Q (m3/s) x pman (kPa).

In this formula the required power for liquid pumps depends on the difference between

the inlet and outlet pressure pman is:


pman = ρ.g.Hgeo + (pa – pe) + ½ρ. )v(v 2

e

2

a + Cpipe .(½. ρ. vpipe2) (42)

pman =elevation head + pressure head + velocity head + resistance head

The required power for gas compressors and fans for air depends on the pressure ratio:

“pman” = pinlet x (k/(k-1)) x (((poutlet/pinlet)^((k-1)/k))-1)

(51)

When the suction and discharge pipes of the fans are at the same level (Hgeo=0) and

the air velocities are also the same (va = ve) then the elevation head (ρ.g.Hgeo) and the

velocity head (½ρ. )v(v 2

e

2

a ) can be neglected. In these cases the power consumption

depends on the pressure head and the resistance head.

In yankee hoods pressure head is a result of the pressure drop in the plenum where air

velocities of more than 100 m/s are created. Also blow boxes in multi-cylinder dryers

may

need pressure head.

Resistance head


The pressure loss ducts due to friction depends on the flow velocity, pipe or duct length,

pipe or duct diameter, and a friction factor based on the roughness of the pipe or duct,

and whether the flow is turbulent or laminar – the Reynolds Number of the flow-.

The pressure loss in a tube or duct due to friction can be calculated with formula’s (44,

45 & 46).

Electrticity consumption supply, exhaust and circulation fans

When optimizing the energy efficiency of drying processes very often the dew point of

the exhaust air in the optimal situation will be higher than in the reference situation.

When this happens the amount of air will decrease and the absolute vapour content of

this air (g vapour / kg dry air) will increase. This may result in substantial electricity

savings for supply and exhaust air fans of the dryer.

Usually the electricity consumption in the reference situation is known.

In table C1 of Annex C the formula’s for centrifugal fans are applied on supply air,

exhaust air and circulation fans for multi cylinder and yankee dryers for as well the

reference as the optimal situation.

The known data of the reference situation are applied to try to predict the electric

consumption in the optimal situation.


15 ENERGY CONVERSION

15.1 Introduction

In Annex B an elaborate description is given for equipment, heat recovery possibilities

and efficiencies of the energy conversion. In Annex A the method of working is

described when combustion processes are involved.

Calculations on unit operations with combustion processes are very extensive and to

make the right choice for heat recovery extensive calculations are necessary.

Software for these calculations is provided in MED II and IV. In the next chapter a

sample of these calculations will be shown.

15.2 Heat recovery in the energy conversion

Additional heat recovery for auxiliary processes, such as heating spray water for

cleaning wires, can be obtained from the rest heat of the energy conversion.

As an example the configuration of two of these conversions with heat recovery is

shown in figures 13 and 14.

In figure 13 spray water (spw) is heated by boiler feed water (bfw) in heat exchanger

(HE3). The heat exchange is 5,04 MW. As a result the bfw is cooled down 33,3 oC

(from 105 oC to 71,7 oC). As a result the flue gasses will normally also be cooled 33,3 oC (from 99 oC to 65,7 oC) in the economizer. For economic reasons however there

should be a minimum temperature difference between the flue and the bfw of 15 oC.

The saving is only 1,97 MW.

In figure 14 spray water (spw) is heated by flue gasses (flue) in heat exchanger (HE3).

As a result the flue is cooled down 37 oC (from 107 oC to 70 oC). The heat exchange is

5,04 MW. The saving is also 5,04 MW.

Although the investment for water-water heat exchangers are lower than those for

water-gas heat exchangers, the latter gives better results. Generally high air factors in

th foregoing combustion process will give better results.

In table 13, with help of MED-software some situations are calculated and compared.

Table 13 Comparison heat recovery from feed water and flue gas

a b c d e f g h i j k l m n o

1 Configuration Consumption and production Heat recovery Savings

2

3

4

5

6

7 - - -oC oC MW MW MW MW MW MW MW MW MW MW

8

8 SBno heat

recovery1,15 150 92,28 0 0 85,9 5,04 0 0 5,04 97,32 0

9 SBboiler feed

water1,15 117 96,14 0 0 85,9 5,04 5,04 0 0 96,14 1,18

10 SB flue gas 1,15 150 55 92,28 0 0 85,9 5,04 0 5,04 0 92,28 5,04

11 SBno heat

recovery1,84 103 92,24 0 0 85,9 5,04 0 0 5,04 97,28 0

12 SBboiler feed

water1,84 89* 95,27 0 0 85,9 5,04 5,04 0 0 96,62 0,66

13 SB flue gas 1,84 101 46 92,24 0 0 85,9 5,04 0 5,04 0 92,24 5,04

14 GT+SB+STno heat

recovery1,84 137 158,39 42,13 12,8 85,9 5,04 0 0 5,04 163,43 0

15 GT+SB+STboiler feed

water1,84 104 159,84 42,13 12,8 85,9 5,04 5,04 0 0 159,84 3,59

16 GT+SB+ST flue gas 1,84 101 52 158,39 42,13 12,8 85,9 5,04 0 5,04 0 158,39 5,04

16 GT+SB+STno heat

recovery2,65 101 158,39 42,13 12,8 85,9 5,04 0 0 5,04 163,43 0

18 GT+SB+STboiler feed

water2,65 86,7* 161,46 42,13 12,8 85,9 5,04 5,04 0 0 161,46 1,97

19 GT+SB+ST flue gas 2,65 101 70 158,39 42,13 12,8 85,9 5,04 0 5,04 0 158,39 5,04

** Heating spraywater: 40 kg/s from 25 oC to 55 oC (= 5,04 MW) * Correction due to: θflue => θboilerfeedwater + 15 oC

*** SB = steam boiler; GT = gasturbine + generator; ST = steam turbine + generator

energ

y c

onvers

ion

***

tem

p.

flue a

fter

heat

exchanger

spra

y w

ate

r

heat

recovery

fro

m

air

facto

r

ste

am

pro

duction

ste

am

boiler

Savin

gs

tem

p.

flue a

fter

boiler

gas c

onsum

ption

energ

y c

onvers

ion

(SB

+G

T+

cofiri

ng)

ele

ctr

icity

pro

duction g

as

tubin

e

ele

ctr

icity

pro

duction g

as

tubin

e

gas c

onsum

ption

heating s

pra

y w

ate

r

requir

ed h

eat

for

spra

y w

ate

r **

heat

recovery

fro

m

boiler

feed w

ate

r

heat

recovery

fro

m

flue b

oiler

tota

l gas

consum

ption


15.3 Heat savings and electricity production

Heat savings on the drying section may result in less steam supply to the steam-

turbine. If the amount of steam is below design specifications the flow in the turbine will

not be optimal and there will be more internal losses. This results in an increase of the

entropy of the outlet steam (=”back pressure” steam). Figure B7 of annex B shows

that, at a certain back pressure, this will result in an increase in enthalpy and

temperature. As a result the enthalpy drop and thus the electricity production shall

decrease. To calculate cost savings an adjusted gas-price is applied. See table 14.

Note that there is no energy loss: the sum of the enthalpy drop in the steam turbine

and the steam to the production processes shall not change.

Table 14 Calculating steam and gas-consumption and adjustment gas-price

REFERENCE SITUATION OPTIMAL SITUATION Difference

Net production hours 7.000 h/a Net production hours 7.000 h/a 0 h/a

Steam consumption production process according to MPI-Pems Steam consumption production process according to MPI-Pems

Heat delivery steam (=steam - condensate) 2,392 GJ/t Heat delivery steam (=steam - condensate) 2,392 GJ/t

Consumer GJt/a MW ton/a ton/h Consumer GJt/a MW ton/a ton/h

Processwater heating 24.947 0,99 10.428 1,5 Processwater heating 0- 0,00- 0- 0,0- 24.947 GJt/a

Spraywater heating 2.449 0,10 1.023 0,1 Spraywater heating 145 0,01 61 0,0 2.304 GJt/a

Spaceheating mach.hall 1.368 0,05 572 0,1 Spaceheating mach.hall 876 0,03 366 0,1 493 GJt/a

Pre-dryer (H2O-evap.) 247.902 9,84 103.620 14,8 Pre-dryer (H2O-evap.) 219.915 8,73 91.921 13,1 27.988 GJt/a

Pre-dryer (losses) 22.807 0,91 9.533 1,4 Pre-dryer (losses) 14.294 0,57 5.975 0,9 8.513 GJt/a

Spaceheating other 114 0,00 48 0,0 Spaceheating other 118 0,00 49 0,0 -4 GJt/a

Other 10.000 0,40 4.180 0,6 Other 10.000 0,40 4.180 0,6 0 GJt/a

Balance 42 0,00 17 0,0 Balance 77 0,00 32 0,0 -36 GJt/a

Cons. energy conversion319.449 12,68 133.525 19,1 Cons. energy conversion255.245 10,13 106.689 15,2 64.205 GJt/a

Steam to and from steamturbine Steam to and from steamturbine

Pressure steam before steamturbine 81 bar Pressure steam before steamturbine 81 bar 0 bar

Temp. steam before steamturbine 510 oC Temperature steam before steamturbine 510 oC 0 oC

Steam pressure after steam turbine 5 bar Steam pressure after steam turbine 5 bar 0 bar

Temp. steam after steamturbine 220 oC Temperature steam after steamturbine 225 oC -5 oC

Unit prices natural gas and electricity Unit prices natural gas and electricity

Total natural gas 0,35 €/m3o Total natural gas 0,35 €/m3o 0,00 €/m3o

Yield electricity 85 €/MWh Yield electricity 85 €/MWh 0 €/MWh

Consumption natural gas and yield electricity Consumption natural gas and yield electricity

Gas consumption gasturbine 2.196 m30/h Gas consumption gasturbine 2.196 m30/h 0 m30/h

Gas cons. cofiring flue-gas boiler 590 m30/h Gas consumption cofiring flue-gas boiler 230 m30/h 360 m30/h

Total gas consumption 2.786 m30/h Total gas consumption 2.426 m30/h 360 m30/h

Total gas consumption 19.502.000 m30/a Total gas consumption 16.982.000 m30/a 2.520.000 m30/a

Total cost natural gas 6.814.678 €/a Total cost natural gas 5.934.102 €/a 880.576 €/a

Electricity production gasturbine 5,78 MW Electricity production gasturbine 5,78 MW 0,00 MW

Electricity production steamturbine 2,56 MW Electricity production steamturbine 2,00 MW 0,56 MW

Total electricity production 8,34 MW Total electricity production 7,78 MW 0,56 MW

Total electricity production 58.396 MWh/a Total electricity production 54.454 MWh/a 3.943 MWh/a

Yield electricity 4.963.694 €/a Yield electricity 4.628.556 €/a 335.138 €/a

Balance natural gas and electricity 1.850.984 €/a Balance natural gas and electricity 1.305.545 €/a 545.438 €/a

Adjustment gasprice in % in case of production loss electricity due to heat savings

Adjustment gasprice in €/GJt 8,50 €/GJt 8,50 €/GJt

Adjustment gasprice in €/m3o 0,27 €/m3o 0,27 €/m3o

Adjustment gasprice in % 76,95 % 76,95 %

a b c d e f g h i j k l m n

Explanation:

(1) Steam savings (m10) 64.205 GJt/a (5) Adjusted gasprice (m28) 8,50 €/GJt

(2) Total savings natural gas (m21) 880.576 €/a (6) 1 GJt = (1000/31,65=) 31,6 m3o/GJt

(3) Total yield electricity (m26) 335.138- €/a (7) Adjusted gasprice (m29) 0,27 €/m3o

(4) Total net savings (m27) 545.438 €/a (8) Adjusted gasprice (m30) 76,9 %


φ mdry mvapor θ θdew Q Gas- ŋ gear & generator 95 % Loss 2,11 MJ/s mdry mvapor θ θdew Q

Comb. air 80 147,0 0,89 10 6,74 3,75 tur- dry mass H2O temp. dewtemp

bine ŋ shaft-gasturbine 30 % Electricity 42,13 MW kg/s kg/s oC oC MJ/s

Natural gas mdry mvapor θ θdew Q mdry mvapor θ θdew Q

4,000 m30/s 3,33 5,610 - - 140,4 Oxygen in dry flue gas15,11 % O2 3,37 = λ 144,7 6,50 500 38,5 99,94 Symbols used in figures:

126,71 MW kg/s kg/soC oC MJ/s mdry = dry mass gas (flue, air etc)

Natural gas mdry mvapor θ θdew Q mvapor = vapour

1,097 m30/s Co-firing 0,9 1,539 683 42,3 38,5 m = total mass gas (=mG +mW)

34,75 MW kg/s kg/s oC oC MJ/s θ = temperature

161,46 MW (total gas) θdew = dewpoint

mdry mvapor θ θdew Q Q = enthalpy or heat

Flue losses Without HE2/3 144,1 8,04 99,0 42,3 36,0 Specific heat water 4,20 J/(g.K)

% O2 = 13,46 With HE2/3 144,1 8,05 86,7 42,3 34,1

λ = 2,65 Difference 0,0 0 33,3 0,0 1,97

kg/s kg/s oC oC MJ/s

Correction exhaust gas temperature= 21 oC Steam

(∆Qflue =< ∆Qboiler.feed.water; θflue - θb.feedw =>15oC) boiler

Spraywater 25,0 oC 40,0 kg/s 4,20 MW ŋexhaust

70,0 Losses 0,64 kJ/s

Spraywater 55,0 oC 40,0 kg/s 9,24 MW % ŋgear&gen 95 %

∆=

5,04 ŋco-firing

MW 90,2 Steam

36,0 kg/s HE3 36,0 kg/s % tur-

Deaerator: ∆θ= 10oC; >10oC 441 kJ/kg 441,0 kJ/kg ŋtotal bine Electricity 12,8 MW

105 oC 105 oC 75,4 27,0 kg/s

15,89 MW 15,89 MW % 500 oC

Make-up water 4,2 kg/s ∆= 36,0 kg/s 36,0 kg/s 91,6 MW

(incl. injection water) 12 oC 0,00 De- 441 kJ/kg 301,2 kJ/kg 104,39 Steam

50 kJ/kg Con- 35,1 kg/s MW 35,1 kg/s aer 36,0 kg/s 105,0oC 71,7

oC MW 33,7 5,0 bar 220oC

0,21 MW den- 89,4oC 89,4

oC ator 105oC 15,89 MW 10,85 MW steam 500 27,0 kg/s 2898 kJ/kg

kg make-up water/ sate- 376 kJ/kg HE2 376 kJ/kg 441 kJ/kg 33,7 kg/s 80 78,2 MW

kg steam to process = tank 13,18 MW 13,18 MW 15,9 MW ∆HE3 5,04 MW 71,7oC 3398

12 % ∆θ= 33,3 oC 301 kJ/kg 114,5

10,15 MW

6,7 kg/s Bypass steam

30,9 kg/s 0,96 kg/s 500oC

100oC Steam 184 oC 80 bar

420 kJ/kg to de-aerator 4,85 bar Cooling water 3398 kJ/kg Bypass / total =

12,97 MW Pro- 2824 kJ/kg 2,33 kg/s 22,9 MW 20 %

duction steam to process 2,72 MW 71,7 oC

30,9 kg/s Condens to boilerhouse proces 35,1 kg/s 36,0 kg/s 301 kJ/kg

100 oC 184oC 184

oC 0,70 MW

420 kJ/kg 4,85 bar 4,85 bar Steam

12,97 MW 2824 kJ/kg 2824 kJ/kg cooler 33,7 kg/s

∆MW= 99,1 MW 101,8 MW 2998 kJ/kg

4,2 kg/s 85,9 101,1 MW Nett production hours 7.218 h/a

100oC Condensate losses Check pinch &approach temperatures steamboiler !! Gas(all in) price (GHV=) 35,10 MJ/m3

o) 0,25 €/m3o

420 kJ/kg Check temperature differences and other results Extra HRC from flue gasses steamboiler GJt/a

1,77 MW Minimum ∆θ fluegas boilerout & boilerfeedwater is approx. 20 oC Gross cost heating spraywater in HE 3 €/a

Boiler efficiencies are based on gross heating values (GHV) Extra HRC from flue gasses steamboiler €/a

All costs and savings are based on co-firing Net cost heating spraywater in HE 3 €/a

Figure 13 Gasturbine + exhaust gas boiler with cofiring + steamturbine + heatrecovery from boiler feed water

51.256

1.033.997

404.681

629.316


Waterinjection GT (for extra G-gas, electricity and heat see chart) 0 kg/m3o 0,00 kg/s mG mW θ θd Q

105,0 oC dry mass H2O temp. dewtemp

Loss 2,11 MJ/s 441,0 kJ/kg kg/s kg/s oC oC MJ/s

φ in % mG mW θ θd MW ŋ gear & generator 95 % 0,00 MW

Comb. air 80 144,4 0,88 10 6,74 3,69 Gas- Symbols used in figures:

kg/s kg/s oC oC MJ/s tur- Electricity 42,13 MW mG = dry mass gas (flue, air etc)

bine ŋ shaft-gasturbine 30 % mW = vapour

Natural gas mG mW θ θd MW mG mW θ θd MW Steam m = total mass gas (=mG +mW)

4,000 m30/s 3,33 5,610 - - 140,4 15,00 % O2 3,311 = λ 142,1 6,49 509 38,8 99,87 boiler θ = temperature

126,71 MW kg/s kg/s oC oC MJ/s θd = dewpoint

Q = enthalpy or heat

Specific heat water 4,20 J/(g.K)

Natural gas mG mW θ θd MW

1,000 m30/s Co-firing 0,83 1,403 678 42,3 35,1

31,68 MW kg/s kg/s oC oC MJ/s

158,39 MW (total gas)

Flue losses mG mW θ θd MW ∆MW= mG mW θ θd MW

% O2 = 13,47 141,5 7,89 70,0 42,3 30,55 5,04 141,5 7,89 100,7 42,3 35,59 Losses 0,6 kJ/s

λ = 2,65 ŋexhaus t ŋgear&gen 95 %

∆mW= 50,2

0,000 % Steam Electricity 12,8 MW

m3 H2O/h 26,9 kg/s tur-

Spraywater 40,00 kg/s (total spraywater = 40,0 kg/s; see also fig ….) ŋco-firing 500 oC bine 26,9 kg/s

25,0 oC 4,20 MW HE3 90,2 91,4 kW 220 oC

% Steam 5 bar

Spraywater 40,00 kg/s 2898 kJ/kg

55,0 oC 9,24 MW ŋtotal 77,9 MW

73,6

%

Make-up water 4,2 kg/s Con- De- 99,4

(incl. injection water) 12 oC den- aer MW

50 kJ/kg 35,1 kg/s ator 36,1 kg/s 33,6 kg/s steam 33,6 6,7 kg/s Bypass steam to process

0,00 MW 89,4 oC 105 oC 105 oC 500 500 oC

sate- 376 kJ/kg 441 kJ/kg 441 kJ/kg 80 80 bar

Condensate 30,9 kg/s tank 13,2 MW 15,9 MW 14,8 MW 3398 3398 kJ/kg Bypass/total= 20 %

100 oC 114,2 22,8 MW

420 kJ/kg

0,00 MW 0,96 kg/s 2,46 kg/s

Steam to de-aerator 184 oC Cooling water 105,0 oC

4,85 bar 441 kJ/kg

2824 kJ/kg 1,09 MW

2,72 MW

Pro-

duction 36,1 kg/s

Condensate to boilerhouse 30,9 kg/s proces 35,1 kg/s 184 oC

100 oC 420 kJ/kg 12,98 MW 184 oC Steam to process 4,85 bar Steam 33,6 kg/s

4,85 bar 2824 kJ/kg cooler 2998 kJ/kg

Condens. losses (incl.conversion) 4,2 kg/s 2824 kJ/kg 101,9 MW 100,8 MW

100 oC 420 kJ/kg 1,77 MW ∆MW= 99,1 MW Check pinch &approach temperatures steamboiler !!

85,9 Check temperature differences and other results Nett production hours 7.218 h/a

kg make-up water / kg steam to process12 % Minimum ∆θ de-aerator out & in is 10 oC Gas(all in) price (GHV=) 35,10 MJ/m3o) 0,25 €/m3

o

Make-up water for compensation production losses Minimum ∆θ fluegas boilerout & boilerfeedwater is approx. 20 oC Extra HRC from flue gasses steamboiler GJt/a

4,2 kg/s 12 oC 50,4 kJ/kg 0,21 MW Boiler efficiencies are based on gross heating values (GHV) Gross cost heating spraywater in HE 3 €/a

All costs and savings are based on additional firing Extra HRC from flue gasses steamboiler €/a

Figure 14 Gasturbine + exhaust gas boiler with cofiring + steamturbine + heatrecovery in flue Net cost heating spraywater in HE 3 - €/a

130.963

1.033.997

1.033.997


ANNEX A COMBUSTION PROCESSES

For more information about “Combustion processes see handout “Themadag

verbetering energie-efficiency processen met (vochtige) verbrandingsgassen” published by Royal VNP on 20th august 2006.

A1 INTRODUCTION

In chapter 3 “Formula library for drying processes” an overview of formulas for

calculations on humid air for drying processes is given.

In table 4 of chapter 3 the method of working for calculating air conditions and mixing

flows is shown.

These flows are a mixture of only two components: dry air and vapor. Although dry air

consists mainly of two components, N2 and O2, its composition will not change. In

drying processes only the amount of vapor (H2O) per kg dry air will change.

In fact the calculations for humid air only relate to the components dry air and vapor.

Only these components need to be considered in calculations on drying processes.

Calculations on combustion processes are more complicated: CH4 + 2O2 -- 2H2O + CO2

This means that during combustion the chemical reactions will change the composition

of the gas; the amount of O2 will decrease and the amounts of O2 and H2O will increase.

For calculations in drying processes when combustion products are involved four

components need to be considered: N2, O2, H2O and CO2.

A2 COMBUSTION EQUATION

A2.1 Components in natural gas

In table A1 the some components and some of het properties that may be in natural

gas are shown.

In the last column the composition of “Groningen” natural gas (G-gas) is shown; the

caloric values for G-gas are: GHV = 35,17 MJ/m30 and NHV = 31,67 MJ/m3

0.

A2.2 General combustion equation

Stoichiometric combustion

Stoichiometric is the ideal combustion process where fuel is burned completely with no

excess of air..

Process heating equipment is rarely run that way. Combustion in boilers and process

equipment usually incorporates a modest amount of excess air to burn the fuel

completely. This amount is about 4% for boilers and about 200 till 300 % for gas-

turbines.

An insufficient amount of combustion air may result in unburned fuel, soot, smoke, and

carbon monoxide exhausts from the boiler. This results in heat transfer surface fouling,

pollution, lower combustion efficiency, flame instability and a potential for explosion. To

avoid inefficient and unsafe conditions boilers normally operate at an excess air level.

The chemical equation for stoichiometric combustion of methane (CH4) with oxygen can

be expressed as:

CH4 + 2*O2 --> CO2 + 2*H2O


Table A1 Components, molar mass, -volume and caloric values natural gas

Air contains 21% O2 and 79% N2 :

CH4 + 2*(O2 + (0,79/0,21)N2) --> CO2 + 2*H2O + 2*(0,79/0,21)N2

Air factor

If more air is supplied some of the air will not be involved in the reaction.

The additional air is termed excess air, when air factor λ=2 the theoretical air is 200%

and the excess air is 100%.

The chemical equation for methane burned with excess air can be expressed as:

CH4 + 2*λ(O2 + (0,79/0,21)N2) --> CO2 + 2H2O + 2*λ(0,79/0,21)N2

General combustion equation

s

O2L

222

O2L

nm HLx

λ)

4

n(mO)

4

n(mOH

2

nm.COL

x

λ)

4

n(mHC

(A1)

component

formula molar

mass

(M)

kg/kmol

molar

volume

(VM)

m3o/kmol

Hs =

Gross Caloric

Value

(GHV)

Hi =

Net Caloric

Value

(NHV)

Compo-

sition

“Gron-

ingen”

gas

MJ/kmol MJ/m30 MJ/kmol MJ/m3

0 %

methane

ethene

ethane

propene

propane

1-butane

butane

dimethylpropane

hydrogen sulphide

methanol

carbon monoxide

carbon dioxide

hydrogen

nitrogen

oxygen

water vapor

CH4

C2H4

C2H6

C3H6

C3H8

C4H8

C4H10

C5H12

H2S

CH3OH

CO

CO2

H2

N2

O2

H2O

16,04

28,05

30,07

42,08

44,10

56,11

58,12

72,15

34,08

32,04

28,01

44,01

2,02

28,01

32,00

18,01

22,36

22,24

22,19

21,99

21,93

21,61

21,52

21,29

22,19

21,08

22,40

22,25

22,44

22,40

22,40

21,63

890

1411

1560

2058

2220

2717

2877

3517

563

764

283

-

286

-

-

-

39,82

63,43

70,30

93,61

101,23

125,77

133,69

165,20

25,36

36,25

12,63

-

12,74

-

-

-

802

1323

1428

1926

2044

2541

2657

3253

519

676

283

-

242

-

-

-

35,88

59,48

64,35

87,61

93,21

117,62

123,47

152,80

23,38

32,07

12,63

-

10,78

-

-

-

81,29

2,87

0,38

0,15

0,04

0,89

14,32

0,01

Fuel Combustion air

Combustion products

Consumed oxygen

Combustion air


Where:

m = the number of C-atoms

n = the number of H-atoms

λ = the air factor

xO2L = mole fraction oxygen in air; xO2L = 0,21

xN2L = mole fraction nitrogen in air; xN2L = 0,79

L = 1 mole combustion air; L = xO2L + xN2L; L = 0,79 N2 + 0,21 O2

Hs = Gross heating value (GHV); for G-gas Hs= 35,17 MJ/m30

Hi = Net heating value (NHV); for G-gas Hi = 31,65 MJ/m30

Mole fraction x (also called amount fraction) expresses the composition of a mixture.

The mole fraction of each component i is defined as its amount of substance ni divided

by the total amount of substance in the system, n.

Usually fuel consists of more than one component. In this case the combustion equation

should be applied to each component and subsequently the results have to be

multiplied by the mole fraction of each component. In table A2 this is shown for the

usual components in natural gas.

By applying formulas (A1), (A2) and (A3) calculations are strongly simplified.

Caloric values (ISO 6976)

Gross caloric value (GHV; Hs):

The amount of heat evolved by the complete combustion of a unit volume of gas

(normally V= 1 m30 at 0 oC and 101,325 kPa) with air at a constant pressure of

101,325 kPa and a constant temperature TH (normally 25 oC) when the combustion

products have once more cooled to the starting conditions and whereby the water

produced by combustion is assumed to be completely condensed.

Net caloric value (NHV; Hi):

The amount of heat evolved by the complete combustion of a unit volume of gas

(normally V= 1 m30 at 0 oC and 101,325 kPa) with air at a constant pressure of

101,325 kPa and a constant temperature TH (normally 25 oC) when the combustion

products have once more cooled to the starting conditions and whereby the water

produced by combustion is assumed to remain as vapor.

A2.3 Combustion Gronings natural gas (G-gas)

To simplify calculations the formulas for G-gas are derived from table A1 and A2.

If not otherwise specified all values are calculated for 1 m30 G-gas (1 m3

0 =1 m3 at 0 oC

and 101,325 kPa)

Fuel

For composition G-gas see table A1.

MB = 18,637 kg/kmol G-gas

VMB = 22,363 m3/kmol G-gas

VGB = amount G-gas in 3

om

mB = 0,827 kg/3

om

cpmB = 1,8 kJ/kg.K (heat capacity)

hB = 36 kJ/kg (specific enthalpy at 20oC)


Table A2 Combustion equation for natural gas

xcl . (CH4 + 2

LLO

xx

2

1 CO2 + 2 H2O – 2 O2 + 2 LLO

xx

2

+ 0 + 0 )

xc2 . (C2H6 + 3,5

LLO

xx

2

2 CO2 + 3 H2O – 3,5 O2 + 3,5 LLO

xx

2

+ 0 + 0 )

xc3 . (C3H8 + 5

LLO

xx

2

3 CO2 + 4 H2O – 5 O2 + 5 LLO

xx

2

+ 0 + 0 )

xc4 . (C4H10 + 6,5

LLO

xx

2

4 CO2 + 5 H2O – 6,5 O2 + 6,5 LLO

xx

2

+ 0 + 0 )

xc5 . (C5H12 + 8

LLO

xx

2

5 CO2 + 6 H2O – 8 O2 + 8 LLO

xx

2

+ 0 + 0 )

xCO2B . (CO2 + 0 0 + 0 - 0 + 0 + CO2 + 0 )

xN2B . (N2 + 0 0 + 0 - 0 + 0 + N2 + 0 )

xH2OB . (H2O+ 0 0 + 0 - 0 + 0 + 0 + H2O)

_____________+_________+

___+_____+_________+______________+_____+_____+

(xCB+xNB+xH2OB)+ A

LLO

xx

2

B + C - A + A LLO

xx

2

+ xNB + xH2OB

fuel combustion air chem. chem. comb. combustion air non (see table 1) (see formula (A1)) CO2 H2O O2 (see formula (A1) combustibles Explanation:

xcl = mole fraction CH4 in fuel xc2 = mole fraction C2H6 in fuel xc3 = mole fraction C3H8 in fuel xc4 = mole fraction C4H10 in fuel

xc5 = mole fraction C5H12 in fuel xCO2B = mole fraction CO2 in fuel xN2B = mole fraction N2 in fuel

xH2OB = mole fraction H2O-vapor in fuel xCB = xcl + xc2 + xc3 + xc4 + xc5l xNB = xCO2B + xN2B xO2L = mole fraction O2 in combustion air; xO2L = 0,21 xN2L = mole fraction N2 in combustion air; xN2L = 0,79

XL = combustion air; XL= xO2L + xN2L λ = air factor A = 2 xcl + 3,5 xc2 + 5 xc3 + 6,5 xc4 + 8 xc5

(A2)

B = xcl + 2 xc2 + 3 xc3 + 4 xc4 + 5 xc5

(A3)

C = 2 xcl + 3 xc2 + 4 xc3 + 5 xc4 + 6 xc5

(A4)


HB = 29,8 kJ/3

om (enthalpy per 3

om at 20oC)

Hi = 31. 650 kJ/3

om (Net Heating Value)

Hs = 35.170 kJ/3

om (Gross Heating Value)

Combustion air:

Composition : xN2L= 0,7809 ; xO2L= 0,2095 ; xArL= 0,0093 ; xCO2L= 0,0003

Assumption : rel.humidity φ=50% and θ=20oC (w=6,2 g H2O/kg dry air).

Molar mass combustion air:

ML = 29,18 kg/kmol dry air (=molar mass wet air)

MGL = 29,00 kg/kmol dry air (=molar mass dry air)

MDL = 18 kg/kmol vapor (=molar mass vapor)

Volume combustion air for 1 m30 G-gas:

VL = 8,465.λ m3 wet combustion air

VGL = 8,381.λ m3 dry combustion air

VDL = 0,084.λ m3 vapor in combustion air

mL = 10,918.λ kg wet combustion air

mGL = 10,850.λ kg dry combustion air

mDL = 0,067.λ kg vapor in combustion air

A = 1,76 m3 O2 (=required oxygen)

wDL = 0,0062 kg vapor/kg dry combustion air

pDL = 1,003 kPa (partial vapor pressure in air)

θdL = 7oC dew point vapor in air

cpmD = 1,80 kJ/kg K (average specific heat vapor)

cpmGL = 1,01 kJ/kg K (average specific heat dry air)

hL = 36,6 kJ/kg (specific enthalpy air at 20oC)

HL = 397,1 .λ. kJ (enthalpy combustion air at 20oC)

Flue gas from G-gas

Molar mass flue gas:

MV= (30,61.λ+1,84) : (λ-0,086) kg/kmol dry flue gas (=molar mass wet flue gas)

MGV= (29,00.λ-1,53) : (λ-0,086) kg/kmol dry flue gas (=molar mass dry flue gas)

MDV= 18 kg/kmol vapor (=molar mass vapor)

Volume flue gas from 1 m30 G-gas:

VV= 8,465.λ + 1,02 m30 wet flue gas

VGV= 8,381.λ – 0,72 m30 dry flue gas

VDV= 0,084.λ + 1,74 m30 vapor in flue gas

Mass flue gas from 1 m30 G-gas:

mV= 10,91.λ + 0,83 kg wet flue gas

mGV= 10,84.λ – 0,57 kg dry flue gas

mDV= 0,07 λ + 1,40 kg vapor in flue gas

Mole fraction O2 (xO2) in dry flue gas and air factor (λ ):

xO2= (λ-1) : (4,776.λ-0,407)


λ = (1-0,407.xO2) : (1-4,776.xO2)

Vapor per kg dry flue gas:

wDV= (0,07.λ+1,40) : (10,84.λ-0,57) kg H2O/kg dry flue gas

Partial pressure vapor:

pvapor= (1,01.λ+20,83) : (λ+0,12) kPa

Dew point temperature flue gas:

θdV=(234: (((17,1 : ln (pDV : 0,61))-1)) oC

If flue gases are cooled below the dewpoint, for instance θ”, then p”vapor can be

calculated with formula (3a) and subsequently w”DV with formula (13). The amount

condensed vapor is: Δ wDV = mGV. (wDV-w”DV) kg.

Enthalpy flue gas:

HV= 35.200 + 397,1.λ kJ/ m30 G-gas (=HB+HL+HS)

hV= (35.200 + 397.1.λ) : (10,84.λ-0,57) kJ/kg dry flue gas

(hV = 3450 kJ/kg when λ=1 and 660 kJ/kg when λ=5)

Heat capacity:

Cpm= see table 2.

A2.4 Calculations combustion processes Yankee

Figure A1.1 and A1.2 are examples of MED calculations on yankee dryers.

In figure A1 2 the oxygen in the combustion air to the gas-turbine is consumed in three

steps: first in the gas-turbine, then in the yankee hood and finally for co-firing in the

flue gas boiler.


mG 2,1 kg/s Gasconsumption = 1668 m30/h mG 1,90 kg/s mG 1,9 kg/s mG 11,4 kg/s

mW 0,03 kg/s MJ/t PWE (based on GHV & steam from co-firing) mW 0,027 kg/s mW 3,32 kg/s mW 0,49 kg/s

θ 10oC MJ/t PWE (based on NHV & steam from co-firing) θ 10,0

oC θ 262oC θ 51

oC

θd 6,0oC θd 6,0

oC θd 91,3oC θd 37,7

oC

P 0,05 MW 100% 90% P 0,05 MW P 10,2 MW P 1,9 MW

Combustion and infiltration (supply) air Combustion air

Infiltration air= 10% P 0,01 MW ŋθ= 55% Spraywater

θ 169oC HE 0,31 MW Q= 39 m3/h 13

oC HE

Combustion and infiltration (supply) air P 0,36 MW P 1,6 MW 48oC

mG 2,1 kg/s mG 1,9 kg/s

mW 0,03 kg/s mG rec 40,3 kg/s mW 3,32 kg/s

θ 153oC Recirculation multiplier: mW 69,93 kg/s θ 300

oC

θd 6oC mG rec / mG exh = 20 θ 300

oC θd 91,3oC mG 11,4 kg/s

P 0,36 MW θd 91,3oC P 10,5 MW mW 0,49 kg/s

PWE = Product Water Evaporation P 221 MW θ 180oC

Nat.gas (co-firing 2) R1= 2955 kJ/kg Exfiltration 5% θd 37,7oC

Q 604 m30/h mG 42 kg/s PWE= 3,25 kg H2O/s mG 42,3 kg/s P 3,5 MW ŋcofiring GHV =

Q 0,17 m30/s mW 70,2 kg/s Yankee θin = 35

oC mW 73,42 kg/s 90 %

GHV 35,10 MJ/m3o mG 2,0 kg/s θ 292

oC hood θout = 300oC θ 300

oC mG exh 2,0 kg/s ŋflue-gas GHV =

P 5,9 MW mW 0,25 kg/s θd 90,9oC Q= 9599 kJ/s θd 91,3

oC mW 3,50 kg/s Steam 2842 J/g 54 %

θ 1912oC P 227 MW P 232,1 MW θ 300

oC 17 bar 220oC

θd 56,2oC θd 91,3

oC

P 6,3 MW P 11,1 MW P= 4,52 MW

Natural gas O2 3,0 % w= 1736 g/kg P= GJt

Q 1000 m30/h λ 1,15 - h= 5488 kJ/kg 1,8 kg/s 6,6 t/h

Q 0,28 m30/s

GHV 35,10 MJ/m3o Heat by steam Feedwater 376 J/g

P 9,75 MW 90oC 1,0 bar

50 % * Attention:

Combustion air 4800 kJ/s O2 % based on dry flue gas = Steamboiler

θ 15oC O2 % based on wet flue gas = mG 11,4 kg/s

r.h. 60 % mW 0,49 kg/s

θd 7,3oC mbone dry=10,0 t/h mbone dry=10 t/h θ 528

oC

P 0,36 MW dsin 45 % dsuit 95 % θd 37,7oC

Flue-gas P 8,0 MW

ŋ shaft-gasturbine 27,0 % mG 11,4 kg/s O2 15,5 % *

ŋ gear & generator 95,1 % mW 0,46 kg/s λ 3,59 -

θ 485oC

Natural gas (co-firing 1) 2,63 MW Gross heating value (GHV) 35,10 MJ/m3o θd 36,8

oC

Q 64 m30/h Net heating value (NHV) 31,68 MJ/m3

o P 7,4 MW

Q 0,02 m30/s Loss0,13 MW ŋs teamboiler cofiring based onNHV = 100 % O2 15,8 %

GHV 35,10 MJ/m3o λ 3,82 -

P 0,62 MW

Figure A1.1 Separated direct utilisation natural gas in gasturbine and in Yankee hood in future situation

mG 11,3 kg/s

Consumption = m30/h mW 3,94 kg/s

Savings = θ 100oC

recirculation θd 73,5oC

mG rec 227 kg/s P 11,7 MW

Recirculation multiplier: mW 76,8 kg/s 11,7

mG 11,3 kg/s mG rec / mG exh = 20 θ 310oC Spraywater

mW 0,59 kg/s θd 73,1oC Q= 39,2 m3/h 13

oC HE

θ 686oC P 308 MW P 1,6 MW 48

oC

θd 41,1oC

P 10,60 MW

10,6

PWE = Product Water Evaporation mG 11,3 kg/s

Natural gas (co-firing 1) R1= 2955 kJ/kg mW 3,94 kg/s

Q= 333 m30/h mG 238 kg/s PWE= 3,25 kg H2O/s mG 238 kg/s θ 180

oC

Q= 0,09 m30/s mW 77,4 kg/s Yankee θin = 35

oC mW 80,7 kg/s θd 73,5oC

GHV 35,10 MJ/m3o θ 323

oC hood θout = 310oC θ 310

oC P 13,3 MW ŋcofiring GHV =

P= 3,25 MW θd 72,5oC Q= 9599 kJ/s θd 73,1

oC 90 %

P 319 MW P 323 MW "ŋflue-gas GHV="

Natural gas Steam 2842 J/g 31 %

Q= 1000 m30/h 17 bar 220

oC

Q= 0,28 m30/s Fluegas

GHV 35,10 MJ/m3o mG 11,4 kg/s P= 4,52 MW

P= 9,75 MW mW 0,46 kg/s P= GJt

θ 480oC Heat by steam 1,8 kg/s 6,6 t/h

Combustion air θd 36,8oC

θ 15oC P 7,35 MW 50 % Feedwater 376 J/g

r.h. 60 % O2 15,8 % 4.800 kJ/s 90oC 1,0 bar

θd 7,3oC λ 3,816 -

P= 0,36 MW Steamboiler

mG exh 11,3 kg/s mG 11,3 kg/s

mbone dry=10,0 t/h mbone dry=10 t/h mW 3,84 kg/s mW 3,94 kg/s

dsin 45 % dsuit 95 % θ 310oC θ 421

oC

θd 73,1oC θd 73,5

oC

ŋ shaft-gasturbine 27,0 % P 15,4 MW P 17,9 MW

2,63 ŋ gear & generator 95,1 % O2 14,0 % O2 12,7 % *

Natural gas (co-firing 2) MW λ 2,86 - λ 2,41 -

Q= 252 m30/h Gross heating value (GHV)= 35,10 MJ/m3

o * Attention:

Q= 0,07 m30/s Net heating value (NHV)= 31,68 MJ/m3

o O2 % based on dry flue gas = 12,7 %

GHV 35,10 MJ/m3o Loss 0,1 MW ŋs teamboiler cofiring based onNHV = 100 % O2 % based on wet flue gas = 8,2 %

P= 2,46 MW

Figure A1.2 Direct utilisation exhaust heat gasturbine in Yankee hood in future situation

15,5%

14,5%

1585

131.947

5%

131.947

3450

3113

Burner

Yankee-

cylinder

Gas-turbine

Gene-rator

Bur-ner

Burner

Yankee-

cylinder

Gas-turbine

Gene-rator

Bur-ner


ANNEX B ENERGY CONVERSION

For more information about “Energy Conversion” see handout “Themadag verbetering

energie-efficiency energie conversies” published by Royal VNP on 18th august 2006.

B1 INTRODUCTION

B1.1 Configuration energy conversion

In the Dutch paper and board mills three kinds of energy conversion are distinguished:

- Gas turbine + steam boiler with or without cofiring + back pressure steam turbine

with or without condensing part

- Gas turbine + (flue gas) steam boiler with or without cofiring

- Steam boiler

The configuration of these conversions is shown in figure B1. Normally the heat

exchangers HE1, HE2 and HE3n will not be installed at the same time.

ENERGY CONVERSION

Combustion air Gas- Loss

tur-

bine Electricity

Natural gas Flue gas

Natural gas Co-firing

Flue losses Flue gas

HE1

Process water in Steam

boiler

Process water out

Process water in

Process water out Losses

Steam

HE2 tur-

Boiler feed water steam bine Electricity

Steam

Make-up water

De-

Con- HE3 aer

den- ator

sate-

tank

Bypass steam

PRODUCTION PROCESSES

Steam Cooling water

Condens to boilerhouse to de-aerator

Pro- Steam

duction

proces steam to process Steam

cooler

Condensate losses

Figure B1 Configuration energy conversion


B1.2 Thermal efficiency

Thermal efficiency (ŋth) is a dimensionless performance measure of a thermal device.

It can be defined as follows:

Thermal efficiency = the extent to which a resource is used in a thermal device for the

intended purpose.

Examples thermal devices:

- gas-turbines

- steam-turbines

- steam-boilers

- heat-exchangers

- energy conversion

Examples of resources:

- fuel (for gas-turbines, steam-boilers etc)

- steam (for steam-turbines, heat-exchangers etc)

- exhaust heat (heat recovery equipment, heat exchangers)

Examples of intended purposes:

- steam boiler: transform boiler feed water into steam

- gas-turbine-generator set: (conversion of fuel heat in electricity)

- steam-turbine-generator: (conversion of (enthalpy drop) steam in electricity)

Example of an energy conversion process in a gas fired steam power plant:

1. Chemical energy in the gas (=resource) is converted to thermal energy

2. Thermal energy converted to kinetic energy in steam

3. Kinetic energy converted to mechanical energy in the turbine

4. Mechanical energy of the turbine converted to electrical energy (=intended purpose)

Thermal efficiency can best be described as the ratio between “What you attained in the

conversion process” and “What you had to pay fot that”.

Examples:

In a steam boiler steam is generated from boiler-feed water by heat from fuel:

- “What you attained in the conversion process” = m kg boiler-feed water with

enthalpy hb.f.w. (kJ/kg) converted into steam with enthalpy hsteam (kJ/kg).

- “What you had to pay fot that”= the heat content of the fuel.

thatfor pay to had youWhat

process conversion the in attained youWhat

th ` (B1)

B1.3 Energy monitoring and targeting

Definition

Energy monitoring and targeting is primarily a management technique that uses energy

information as a basis to eliminate waste, reduce and control current level of energy

use and improve the existing operating procedures.

It builds on the principle “you can’t manage what you don’t measure”. It essentially

combines the principles of energy use and statistics.

http://en.wikipedia.org/wiki/Dimensionless_quantity

http://en.wikipedia.org/wiki/Boiler

http://en.wikipedia.org/wiki/Chemical_energy

http://en.wikipedia.org/wiki/Thermal_energy

http://en.wikipedia.org/wiki/Steam

http://en.wikipedia.org/wiki/Turbine


While, monitoring is essentially aimed at establishing the existing pattern of energy

consumption, targeting is the identification of energy consumption level which is

desirable as a management goal to work towards energy conservation.

Monitoring and Targeting is a management technique in which all plant and building

utilities such as fuel, steam, compressed air, water, effluent, and electricity are

managed as controllable resources in the same way that raw materials, finished product

inventory, building occupancy, personnel and capital are managed.

It involves a systematic, disciplined division of the facility into Energy Cost Centers.

The utilities used in each centre are closely monitored, and the energy used is

compared with production volume or any other suitable measure of operation.

Once this information is available on a regular basis, targets can be set, variances can

be spotted and interpreted, and remedial actions can be taken and implemented.

2 GAS TURBINES

B2.1 Principle

A gas turbine is a rotary engine that extracts energy from a flow of combustion gas. It

has an upstream compressor coupled to a downstream turbine, and a combustion

chamber in-between.

Energy is added to the gas stream in the combustor, where air is mixed with fuel and

ignited. Combustion increases the temperature, velocity and volume of the gas flow.

This is directed through a nozzle over the turbine's blades, spinning the turbine and

powering the compressor. In figure B2 is a single shaft and a double shaft machine

depicted.

In Dutch paper and board mills gas turbines are applied in Combined Heat and Power

(CHP) plants for co-generation of power and steam. In these plants the exhaust heat of

the gas turbine is recovered in flue gas steam boilers. To make the plant more flexible

the boiler is provided with cofiring.

B2.2 Types of gas turbines

Aeroderivatives and jet engines

Airbreathing jet engines are gas turbines optimized to produce thrust from the exhaust

gases, or from ducted fans connected to the gas turbines.

Aeroderivatives are also used in electrical power generation due to their ability to

startup, shut down, and handle load changes more quickly than industrial machines.

The GE LM2500 and LM6000 are two common models of this type of machine.

Single shaft turbojet engine (GE type J85)

(axial-flow from left to right: compressor, combustion chambers and turbine)

http://en.wikipedia.org/wiki/Rotary

http://en.wikipedia.org/wiki/Engine


http://en.wikipedia.org/wiki/Combustion

http://en.wikipedia.org/wiki/Compressor

http://en.wikipedia.org/wiki/Turbine


http://en.wikipedia.org/wiki/Combustor

http://en.wikipedia.org/wiki/Earth%27s_atmosphere

http://en.wikipedia.org/wiki/Fuel

http://en.wikipedia.org/wiki/Ignition_system

http://en.wikipedia.org/wiki/Temperature

http://en.wikipedia.org/wiki/Velocity

http://en.wikipedia.org/wiki/Volume

http://en.wikipedia.org/wiki/Nozzle

http://en.wikipedia.org/wiki/Combined_heat_and_power

http://en.wikipedia.org/wiki/Jet_engine

http://en.wikipedia.org/wiki/Ducted_fan

http://en.wikipedia.org/wiki/LM2500

http://en.wikipedia.org/wiki/General_Electric_LM6000

http://en.wikipedia.org/wiki/File:J85_ge_17a_turbojet_engine.jpg


Industrial gas turbines for electrical generation

Industrial gas turbines differ from aeroderivative in that the frames, bearings, and

blading is of heavier construction. They can be particularly efficient—up to 60%—when

waste heat from the gas turbine is recovered by a heat recovery steam generator to

power a conventional steam turbine in a combined cycle configuration.

They can also be run in a cogeneration configuration: the exhaust is used for steam

production in a flue gas boiler. A cogeneration configuration can be over 90% efficient.

The turbines are provided with a reduction gear box to obtain a speed of 3,000 rpm to

match the AC power grid frequency.

As a general rule, the smaller the engine the higher the rotation rate of the shaft(s)

needs to be to maintain top speed. Turbine blade top speed determines the maximum

pressure that can be gained,this produces the maximum power possible independent of

the size of the engine. Jet engines operate around 10,000 rpm and micro turbines

around 100,000 rpm.

Gas generator

Combustionchamber

Combus-

tion air

Compres

-sed air

Air com-

pressorGas

turbine

Natural

gas

Hot

gasses

Exhaust

gasses

Gener-

ator

Single shaft gasturbine and electr. generator (pmc = constant); power down >>>>> air factor up >>>>>>> efficiency down

Combustionchamber

Combus-

tion air

Compres

-sed air

Air com-

pressor

Pow

er

tur

bine

Natural

gas

Hot

gasses

Exhaust

gasses

Double shaft gasturbine for pumps & electr. generator (pmc = variable);HP-rotor + nozzles + LP-rotor (= power turbine)

Figure B2 Single and double shaft gas turbine

To match the AC power grid frequency gasturbines always have a fixed speed of 3,000

rpm after the gear box. For efficiency reasons the load should normally always be 100

%. At lower loads the compressed amount of air will stay the same resulting in an

increase of the air factor. Higher air factors will decrease the efficiency of the

gasturbine.

http://en.wikipedia.org/wiki/Combined_cycle

http://en.wikipedia.org/wiki/Cogeneration

http://en.wikipedia.org/wiki/Revolutions_per_minute

http://en.wikipedia.org/wiki/Alternating_current

http://en.wikipedia.org/wiki/Power_grid

http://en.wikipedia.org/wiki/Jet_engine

http://en.wikipedia.org/w/index.php?title=Micro_turbine&action=edit&redlink=1

http://en.wikipedia.org/wiki/Alternating_current

http://en.wikipedia.org/wiki/Power_grid

http://en.wikipedia.org/wiki/Revolutions_per_minute


B2.3 Gas turbine cycle

Gas turbines are described thermodynamically by the Brayton cycle, in which air is

compressed isentropically, combustion occurs at constant pressure, and expansion over

the turbine occurs isentropically back to the starting pressure. The process is shown in

a p-V-diagram in figure B3.

In practice, friction and turbulence cause:

- non-isentropic compression: for a given overall pressure ratio, the compressor

delivery temperature is higher than ideal.

- non-isentropic expansion: although the turbine temperature drop necessary to

drive the compressor is unaffected, the associated pressure ratio is greater, which

decreases the expansion available to provide useful work.

- pressure losses in the air intake, combustor and exhaust: reduces the expansion

available to provide useful work.

B2.4 Gas turbine efficiency

When transforming thermal energy into mechanical energy, the thermal efficiency of a

heat engine is the percentage of heat energy that is transformed into work:

ŋth = Wout / Win = 1 – (Qout / Qin) (B3)

The second law of thermodynamics puts a fundamental limit on the thermal efficiency of

heat engines (gas- and steam- tubines): even an ideal frictionless engine cannot

convert 100% of its input heat into work. The limiting factors are the temperature at

which the heat enters the engine (Tin in K; for gasturbines called TIT) and the

temperature of the environment into which the engine exhausts its waste heat To in K):

ŋth.carnot = (Tin - To) / Tin (B4)

Where:

ŋth.carnot = the highest achievable thermal efficiency

Tin = turbine inlet temperature (called: TIT) in K

To = the ambient temperature where the turbine is located in K

This limiting value is called the Carnot cycle efficiency because it is the efficiency of an

unattainable ideal reversible engine cycle called the Carnot cycle.

When, for instance, TIT is 1450 oC and the ambient temperature is 20 oC the carnot-

efficiency is 83 %. In practice this will be only about 46%, 37 %-points short of the

Carnot value.

Since To is fixed by the environment, the only way to increase the efficiency of an

engine is to increase Tin, the operating temperature of the engine. For this reason the

operating temperatures of engines have increased greatly over the long term.

As with all cyclic heat engines, higher combustion temperature means greater

efficiency. The limiting factor is the ability of the materials that make up the engine to

withstand heat and pressure. Considerable engineering goes into keeping the turbine

parts cool.

http://en.wikipedia.org/wiki/Thermodynamics

http://en.wikipedia.org/wiki/Brayton_cycle

http://en.wikipedia.org/wiki/Isentropic

http://en.wikipedia.org/wiki/Combustion


http://en.wikipedia.org/wiki/Mechanical_energy

http://en.wikipedia.org/wiki/Heat_engine

http://en.wikipedia.org/wiki/Work_(thermodynamics)

http://en.wikipedia.org/wiki/Second_law_of_thermodynamics

http://en.wikipedia.org/wiki/Reversible_process_(thermodynamics)

http://en.wikipedia.org/wiki/Carnot_cycle


http://en.wikipedia.org/wiki/Fuel_efficiency


Q1

1 – 2 isentrop

p 2 3 2 – 3 isobar

3 – 4 isentrop

4 – 1 isobar

tip loss

Otto & Diesel process

1 4

Q2

V p-V-diagram Brayton process

ηth = thisfor fuel from dtransformeheat

gasturbine in energy mechanical in dtransformeheat

ηth =1

2

1

21

Q

Q1

Q

QQ

=1-

)T(T.cm

)T(T.cm

23p

14p

= 1 -

1T

T.T

1

41 :

1T

T.T

2

32

1

2k

1-k

1

2

T

T

p

p

(a)

4

3k

1-k

4

3

T

T

p

p

(b)

(a) and (b) gives:4

3

1

2

p

p

p

p (= ) so : T2 :T1 = T3 : T4 of: T4 : T1 = T3 : T2

(T4 – T1) : (T3 – T2) = T1: T2 = T4 : T3

ηth = 1 - 2

1

T

T of ηth = 1 –

3

4

T

T of ηth = 1 -

k

1-k

1

Figure B3 Thermal efficiency Brayton process

An important property of a gasturbine is the pressure ratio (α); the relation between α

and Tin for the carnot efficiency is shown in figure B3.

η= 1 -

k

1-k

1

α= 15 η= 53,9% α= 24 η= 59,7% (B5)

In practice the efficiencies will be only about 2/3 of these values.


B3 STEAM TURBINES

B3.1 Principle

A steam turbine is a mechanical device that extracts thermal energy from pressurized

steam, and converts it into rotary motion.

The steam turbine is a form of heat engine that derives much of its improvement in

thermodynamic efficiency through the use of multiple stages in the expansion of the

steam, which results in a closer approach to the ideal reversible process.

B3.2 Types of steam turbines

Depending on the steam supply and exhaust conditions the following types are

distinguished:

- Backpressure or noncondensing turbines

These turbines are most widely used for process steam applications in paper and

board plants. The exhaust pressure is controlled by a regulating valve to suit the

needs of the process.

- Condensing turbines

In paper and board mills these turbines are applied in combination with

backpressure turbines. The combination makes it possible to respond on fluctuations

on the steam demand. When for instance there is a paper web broke the load of the

condensing turbine will be increased by the dropped demand of the paper machine.

The vapor content of the exhaust steam of these turbines is typically 95% at a

pressure of about 0,05 bar in the condenser.

- Reheat turbines

These are used almost exclusively in electrical power plants. In a reheat turbine,

steam flow exits from a high pressure section of the turbine and is returned to the

boiler where additional superheat is added. The steam then goes back into an

intermediate pressure section of the turbine and continues its expansion.

- Extracting type turbines

These are common in paper and board mills. In an extracting type turbine, steam is

released from various stages of the turbine, and used for process needs.

The interior of a turbine comprises several sets of blades, or “buckets” as they are more

commonly referred to. One set of stationary blades is connected to the casing and one

set of rotating blades is connected to the shaft. The sets intermesh with certain

minimum clearances, with the size and configuration of sets varying to efficiently exploit

the expansion of steam at each stage.

To maximize turbine efficiency the steam is expanded in a number of stages. These

stages are characterized by how the energy is extracted from them. Depending on the

way how the energy is extracted from the stages the following types are distinguished:

- Impulse turbines

- Reaction turbines

The difference between both turbines is shown in figure B4.

Most steam turbines use a mixture of the reaction and impulse designs: each stage

behaves as either one or the other, but the overall turbine uses both. Typically, higher

pressure sections are impulse type and lower pressure stages are reaction type.

The turbines used for electric power generation are most often directly coupled to their

generators. As the generators must rotate at constant synchronous speeds according to

the frequency of the electric power system, the most common speeds are 3000 rpm for

50 Hz systems.




http://en.wikipedia.org/wiki/Thermodynamic_efficiency

http://en.wikipedia.org/wiki/Reversible_process_(thermodynamics)


Figure B4 Difference between an impulse and a reaction turbine (Wikipedia)

B3.3 Steam turbine cycle

Steam turbines are described thermodynamically by the Rankine cycle.

The process is shown in the flowchart in figure B7 and in a p-V-diagram in figure B5:

- Water is pumped to high pressure (5-1)

- High pressure water is heated in a boiler at constant pressure and becomes

saturated steam (1-2)

- The saturated steam is heated in a superheater at constant pressure and becomes

superheated steam (2-3)

- Super heated steam expands in the steam turbine generating power; the

temperature and pressure of the steam decrease and at the end some condensation

may occur. The process shown in figure B5 is for an ideal steam turbine. An ideal

steam turbine is considered to be an isentropic process in which the entropy of the

steam entering the turbine is equal to the entropy of the steam leaving the turbine.

No steam turbine is truly “isentropic”; the entropy will increase during the process.

Typical isentropic efficiencies are ranging from 20%-90% based on the application

of the turbine.

- The wet steam (about 5% moisture) is condensed in the condenser at a constant

pressure and becomes water (4-5). The pressure and temperature of the condenser

is fixed by the temperature of the cooling coils as the fluid is undergoing a phase-

change.

http://en.wikipedia.org/wiki/Thermodynamics

http://en.wikipedia.org/wiki/Isentropic_process


Q1

p1 1. 1'. 2. 3.

p2

5. 4.

Q2

V1 V2

water wet steam superheated steam

saturated water saturated steamProcess:

1-2-3 isobararic3-4 isentropic

4-5 isobaric5-1 isochoric

pre

ssure

volume

The thermo dynamic efficiency is defined as:

ηth = 1

21

Q

QQ

Where:

Q1 = heat supplied in boiler (including super heater)

= h3 – h1

Q2 = heat removed in condensor

= h4 – h5

W = work delivered by isentropic expansion

= Q1 – Q2

Also:

ηth = )H(h

)h(h)h(h

13

5413

The difference between h5 and h1, the work required by the pump, is only about 1%

and can be neglected:

ηth = 13

43

hh

hh

(B6)

or in words:

ηth = steam by suppliedheat

expansion isentropic during drop enthalpy (B7)

Figure B5 Thermodynamic efficiency Rankine process


B3.4 Steam turbine efficiency

As already described for gasturbines in chapter B2.4 there is a a fundamental limit on

the thermal efficiency for heat engines.

The limiting factors for steam turbines are the temperature of the steam that enters the

turbine and the temperature of the steam after the turbine. The latter depends on the

cooling in the condenser.

As with all cyclic heat engines higher supply steam temperature means greater

efficiency. The limiting factor is the ability of the materials that make up the engine to

withstand heat and pressure.

Higher supply steam temperatures result in higher enthalpies (h3) of this steam.

For efficiency reasons steam boilers for steam-turbines need to be provided with

superheaters.

To gain a clear insight in the possibilities of improving the efficiency of a steamturbine

a, so called, Mollier h,s diagram (h=enthalpy and s=entropy) can be applied. An

example of this diagram, also called Mollier chart for steam, is shown in figure B6. For

calculations charts on A2-size are commercially available.

Figure B6 Mollier chart for steam

In figure B7 the Rankine process is shown in the h-s-diagram.

Efficiencies can be easily defined by this diagram:

Isentropic Efficiency

The isentropic efficiency for an expansion process is defined as:

ŋisentropic = (B6)

real enthalpy drop after real expansion . enthalpy drop after isentropic expansion


http://en.wikipedia.org/wiki/Fuel_efficiency


ŋisentropic = (B7)

The difference between numerator and denominator are the internal losses in the

turbine.

Rankine efficiency

The Rankine efficiency is defined as:

ŋRankine = (B8)

ŋRankine = (B9)

The difference between numerator and denominator is the Rankine loss in the turbine.

Thermal efficiency

The thermal efficiency is defined as:

ŋThermal = (B10)

ŋThermal = (B11)

The difference between numerator and denominator is the condenser loss, this is the

sum of the Rankine loss and the internal losses in the turbine.

h3

h4"

h4

h1

s1 s2

isotherm (superheated steam)enthalpy superheated steam

enthalpy condensate

enth

alp

y

enthalpy steam after turbine

entropy

internal losses

condenser lossRankine loss

enthalpy steam after ideal turbine

Figure B7 Efficiencies steamturbine in h-s-digram

h3 - h4” . h3 - h4

enthalpy drop after isentropic expansion enthalpy drop after condenser

h3 - h4 . h3 - h1

enthalpy drop after real expansion enthalpy drop after condenser

h3 - h4” .

h3 - h1


Exercise B1: Steam turbine processes in h-s- diagram

Given:

Steam supplyvoer steam turbine 80 bar 590 oC.

The process steam is extracted at 15 bar, the temperature shows to be 408oC.

(in order to make this steam usable for the process, the steam should be cooled first to

around 220oC)

Questions:

(a) What is the isentropic efficiencyrate of return (ηisent)?

What part of the enthalpy of the steam supply is used for electricity generation

(Δhhelectr), and what is then the enthalpy after the process (hproces)?

Calculate the electricity production (E) when the efficiency of the gear box and

the generator (ηgear) is together 96%.

(b) As (a) when the steam is extracted at 3,5 bar and 270oC.

(c) As (a) when all steam is discharged in the condenser, this steam has a pressure

of 0,05 bar and contains 5% moisture.

(d) What is in the efficiency of the electricity generation in case (c) (ηE) when the

steam in the condenser is cooled to 50oC, the boiler efficiency is 97%.

Answer h3 = 3615 J/g

3265 J/g

3010 J/g

h4"

2435 J/g

h4

h1 210 J/g

s1 s2

590 oC

enthalpy condensate

enth

alp

y

enthalpy steam after turbine

entropy

condenser lossRankine loss

408 oC

270 oC

50 oC

Answer

ηisent

(%)

Δhhelectr

(kJ/kg)

hproces

(kJ/kg)

ηE

(%)

E

(Wh/kg)

(a)

(b)

(c)

66

70

80

350

605

1180

3265

3010

n.v.t.

96

96

96

93,4

161,3

314,7

(d) (1180 x 0,96) : ((3615 – 50 x 4,2) : 0,97) = 32,3 %


B4 STEAM BOILERS

B4.1 Steamboiler plant

In figure B8 the main components of a steam-boiler and the auxiliary equipment in the

boiler-house are depicted in a block diagram.

A steam-boiler always consists of an evaporator and combustion equipment.

Additionally the boiler may be provided with an economizer, a super-heater and an air

pre-heater. The boiler feed-water is treated and de-aerated.

BOILER HOUSE STEAM

BOILER

Boiler feed water

Make-up water

Con- De-

HE3 aer

den- ator

sate-

tank

Pro-

duction Steam Cooling water

Condens to boilerhouse proces to de-aerator

Steam

steam to process

cooler

Condensate lossesBurner

Super heater

Evaporator

Economizer

Air pre-heater

Fuel

CombustionairFlue

Figure B8 Boilerhouse equipment

The main components are:

De-aerator

De-aerator is a device that is used for the removal of air and other dissolved gases from

the boiler feed water. In particular, dissolved oxygen in boiler feed water will cause

serious corrosion damage in steam systems by attaching to the walls of metal piping

and other metallic equipment and forming oxides (rust). It also combines with any

dissolved carbon dioxide to form carbonic acid that causes further corrosion.

Air pre-heater

An air pre-heater a device designed to heat for example combustion air. The purpose of

the air pre-heater is to recover the heat from the boiler flue gas which increases the

thermal efficiency of the boiler by reducing the useful heat lost in the flue gas.

Economizers

Economizers are devices, fitted in a boiler chimney, which use the waste heat in the flue

gases from the boiler to preheat the cold boiler feed water.

Superheater


A superheater is a device used to convert saturated steam into dry steam used for

power generation or processes. There are three types of superheaters namely: radiant,

convection, and separately fired.

A radiant superheater is placed directly in the combustion chamber.

A convection superheater is located in the path of the hot gases.

A separately fired superheater is totally separated from the boiler.

Heat exchanger HE3 (see figure B8)

In the de-aerator the boiler feed-water is heated to minimal 105 oC. As the temperature

difference between the flue after the economizer and this feed-water is in practice

about 15 oC the temperature of the flue will be at least 120 oC. When the water in the

condensate tank is at least 10 oC colder than 105 oC heat can be transferred by HE3

from after the de-aerator to before the de-aerator. This will result in lower steam

consumption of the de-aerator and lower feed water temperature to the economizer and

thus in lower flue temperatures.

B4.2 Types of steamboilers

Depending on the fluid that through the pipes two types of boilers are distinguished:

fire-tube boilers and water-tube boilers.

Fire-tube boilers

In this type of boiler, the gases of combustion pass through tubes that are surrounded

by water. Figure B9 illustrates a cutaway view of a fire-tube boiler.

Figure B9 Fire-tube boiler (source: dieselduck.com)

The heat energy from the gases passes through the sides of the tubes by thermal

conduction, heating the water and ultimately creating steam.

Water-tube boilers


In this type of boiler, the water circulates through tubes that are heated externally by

fire. Water-tube boilers operate on the principle of forced or natural water circulation.

Figure B10 illustrates a cross section view of a water-tube boiler with natural

circulation of the water. These boilers consist basically of a steam drum and a water

drum connected by a bank of generating tubes. The two drums are also connected by a

row of water tubes, which forms a water-cooled sidewall opposite the tube bank.

Figure B10 Water-tube boiler with natural circulation (Source: tpub.com)

Figure B11 shows the principle of forced circulation.

Water tube boilers are used for high-pressure boilers in power stations.

The small water content in these boilers result in rapid response to load change

The flexible design possibilities make easy control of the temperature of the

superheated steam possible.

B4.3 Exhaust heat steam boiler

Flue gas steam boilers (also called: waste heat or exhaust heat boilers) are an

important component in combined heat power (CHP) plants.

In flue gas boilers heat from the hot exhaust gas stream from the gas-turbine is

recovered. The steam can be used in production processes or used to drive a steam

turbine.

Flue gas boilers consist of three major components: the evaporator, superheater

and economizer. Super heaters are necessary when the flue gas boiler delivers steam to

a steam-turbine.

Flue gas boilers can be categorized by number of ways such as direction of exhaust

gases flow or number of pressure levels.

Based on the flow of exhaust gases, they are categorized into vertical and horizontal

types. In horizontal types exhaust gas flows horizontally over vertical tubes whereas in

vertical types exhaust gas flow vertically over horizontal tubes.

Based on pressure levels, they can be categorized into single pressure and multi

pressure. Single pressure boilers have only one steam drum and steam is generated at

single pressure level whereas multi pressure boilers employ two or more steam drums.


http://en.wikipedia.org/wiki/Steam_turbine


http://en.wikipedia.org/wiki/Steam_drum


Because of the relatively low temperature of the gas turbine exhaust, compared to the

burner flame in a conventional boiler, a much greater boiler heat transfer area is

required for a given heat load.

Figure B11 Water-tube boiler with forced circulation (Source:Spirax)

B4.4 Efficiency Steam boiler

Total steam-boiler efficiency

The total steam boiler efficiency (ŋsteamboiler) indicates the overall efficiency of the boiler

inclusive flue gas losses, radiation and convection losses and boiler blow down losses.

The total steam boiler efficiency (ŋsteamboiler) is also known as “fuel to fluid efficiency”.

The total steam boiler efficiency (ŋsteamboiler) is defined as follows.

Attained in the conversion process: generating steam from boiler feedwater

Paid for this: fuel

LHV.V

)hh(.m

naturalgas

waterboilerfeedsteamsteam

rsteamboile

Where:

msteam = mass steam produced from boiler feed water in kg/s (or ton/h)

hsteam = enthalpy steam in kJ/kg (or MJ/ton)

hboilerfeedwater = enthalpy boiler feed water in kJ/kg (or MJ/ton)

Vnatural gas= natural gas consumption in m30/s (or m3

o/h)

LHV = lower heating value natural gas in MJ/m30 (G-gas = 31,65 MJ/m3

0)

Losses steam boiler conversion process (= numerator – denominator):

- Flue loss

Heat from gas-turbine


This is the difference (hflue) between the enthalpies of the flue gas and the

combustion air. A measure for the ratio flue loss (∆hflue) and the fuelconsumption

(hfuel) is the combustion efficiency (ŋcomb):

ŋcomb = (hfuel - hflue) / hfuel

- Convection and radiation losses

The external surface of an operating steam boiler is hotter than its surroundings

and therefore loses heat by both radiation and convection.

The radiation and convection loss is proportional to the external surface area.

- Boiler blow down

Steam boilers should be blown down daily to maintain recommended dissolved

solids levels and to remove sludge and sediment.

Co-firing efficiency

Temperature limits at the gas turbine inlet force the turbine to use excess air, above

the optimal stoichiometric ratio to burn the fuel.

The flue gases of the gas-turbine contain between about 12 and 15 % oxygen. Co-firing

makes use of this oxygen and no extra combustion air is required. This means that the

mass exhaust gas will not increase. If the exhaust gas temperature also does not

increase then the flue losses will stay the same. Normally the exhaust gas temperature

does not increase. Even more often this temperature will decrease; this is caused by

the increased feed water flow that will cool the exhaust gas further.

In practice co-firing efficiencies are between 97 and 105 %.

http://en.wikipedia.org/wiki/Stoichiometric


B5 COMBINED HEAT AND POWER PLANTS (CHP)

B5.1 Combined heat and power (CHP) plant

Figure B12 shows a CHP plant in paper and board mills which is composed of a gas-

turbine and a flue gas boiler.

COMBINED HEAT POWER

Combustion air Gas- Loss

tur-

bine Electricity

Natural gas Flue gas

Natural gas Co-firing

Flue losses Flue gas

HE1

Process water in Steam

boiler

Process water out

Boiler feed water steam

Make-up water

De-

Con- HE3 aer

den- ator

sate-

tank


Steam Cooling water

Condens to boilerhouse to de-aerator

Pro- Steam

duction

proces steam to process Steam

cooler

Condensate losses

Figure B12 Combined Heat Power plant

The hot exhaust gases from a gas turbine (approximately 500- 550 °C) are led through

a flue gas boiler, where saturated steam is generated and used in the production

processes.

The turndown ratio of these plants is poor, because of the need for the turbine to rotate

at a speed synchronized to the electrical frequency. This means that, without special

provisions like “Inlet Guide Vanes”, these plants run at full-load, providing the base load

of steam to the production processes.

When the load of the turbine is turned down the fuel supply will also be turned down

however the speed stays constant. This means that the amount of combustion air also

stays constant which results in high air factors and bad efficiencies.

In these plants the ratio of the steam and electricity production is fixed. Electricity that

can not be used on the site is exported to the national grid. For a flexible steam supply

often co-firing is installed on the waste heat boiler; the advantage of cofiring is that


efficiencies between about 95 and 105 % are obtained.

B5.2 Combined cycle plant (CCP)

A combined cycle employs more than one thermodynamic cycle. Gas turbines are only

able to use a portion of the energy their fuel generates.

Combining two or more "cycles" such as the Brayton cycle and Rankine cycle results in

improved overall efficiency.

In a combined cycle power plant a gas turbine generator generates electricity and the

waste heat is used to make steam to generate additional electricity via a steam turbine.

As a rule, in order to achieve high efficiency, the temperature difference between the

input and output heat levels should be as high as possible (see Carnot efficiency).

By combining both gas and steam cycles, high input temperatures and low output

temperatures can be achieved. The efficiency of the cycles add, because they are

powered by the same fuel source. So, a combined cycle plant has a thermodynamic

STEAMBOILER

Combus.chamber

Combus-

tion air

Compres

-sed air

Air com-

pressorGas

turbine

Natural

gas

Hot

gasses

Exhaust

gasses

Gener-

ator

Burnerco-firing

Super heater

Evaporator

Economizer

Flue gasses

Conden

-sing

turbine

Conden

sate

Back pr.

turbine

Natural

gas

Condensor

Boiler feed watertreatment

plant

HP

steamLP

steam


1.

4.3.

3.

2.

1".

5".

4". 5.

5.

4".

Figure B13 Combined Cycle plant with steam extraction from BP-turbine

http://en.wikipedia.org/wiki/Thermodynamic_cycle

http://en.wikipedia.org/wiki/Brayton_cycle

http://en.wikipedia.org/wiki/Rankine_cycle



cycle that operates between the gas-turbine's high firing temperature and the waste

heat temperature from the condensers of the steam cycle. This large range means that

the Carnot efficiency of the cycle is high. The actual efficiency, while lower than this is

still higher than that of either plant on its own.

The CCP can, like other CHP plants, be designed with co-firing (also called:

supplementary firing) of fuel after the gas turbine in order to increase the quantity or

temperature of the steam generated. Co-firing lets the plant respond to fluctuations of

electrical load.

The relatively low temperatures of the flue gas from the gas-turbine may require

additional heating to bring the steam up to the specification required for the steam

turbines.

A CCP plant in paper and board mills is composed of a gas-turbine, a flue gas boiler and

a back pressure steam-turbine and sometimes, additional, a condensing steam-turbine.

The process steam is extracted from the back pressure steam-turbine. The principle is

shown in figure B13.

The superheated steam is directed to steam turbines which drive the generator or, after

being cooled to about 15 oC above the saturation temperature, to the production

process.

The combined-cycle system includes single-shaft and multi-shaft configurations.

The single-shaft system depicted in figure B13 consists of one gas turbine, one steam

turbine, one generator and one flue gas boiler with co-firing. The key advantage of this

single-shaft arrangement is its operating simplicity with higher reliability than multi-

shaft blocks. Further operational flexibility is provided with a steam turbine which can

be disconnected for start up or for simple cycle operation of the gas turbine.

http://en.wikipedia.org/wiki/Waste_heat

http://en.wikipedia.org/wiki/Waste_heat

http://en.wikipedia.org/wiki/Carnot_efficiency


ANNEX C: CENTRIFUGAL FANS AND PUMPS

For more information about “Centrifugal fans and pumps” see handout “Themadag

verbetering energie-efficiency pompen, compressoren en vacuumsystemen” published

by Royal VNP on 1st august 2006.

CENTRIFUGAL FANS AND COMPRESSORS

C1 Types of centrifugal fans and compressors

Compressors and fans are similar to pumps, both increase the pressure on a fluid and

both can transport the fluid through a pipe. As gases are compressible, the compressor

also reduces the volume of a gas. Liquids are incompressible, so the main action of a pump is to pressurize and transport liquids.

One of the most common types of compressors is the centrifugal compressor.

Centrifugal compressors use a rotating disk or impeller in a shaped housing to force the

gas to the rim of the impeller, increasing the velocity of the gas. A diffuser (divergent

duct) section converts the velocity energy to pressure energy. Compressors with

multiple staging can achieve high output pressures.

A centrifugal fan is depicted in figure C1.

Figure C1: Components of a centrifugal fan (from Wikipedia)

The fan drive determines the speed of the fan wheel. When the fan drive is provided

with an electronic variable-speed control better overall energy efficiency at reduced

speeds may be obtained. This is important when the weights (g/m2) in the product

range produced on a machine differ.

See also Box C2 “Pump- and system characteristic and control pump capacity”.

Fan dampers, to control gas flow into and out of the centrifugal fan, may be installed on

the inlet side or on the outlet side of the fan, or both. Inlet dampers reduce fan energy

usage.

C2 Power consumption centrifugal fans and compressors

In thermo dynamics two compression processes are distinguished: isothermal

compression and isentropic compression.

Although neither of them model the real world exactly, each can be useful for analysis.

Isothermal compression

This model assumes that the compressed gas remains at a constant temperature

throughout the compression process. In this cycle internal energy is removed from the

system as heat at the same rate that it is added by the mechanical work of

http://en.wikipedia.org/wiki/Fluid

http://en.wikipedia.org/wiki/Pipe_(material)

http://en.wikipedia.org/wiki/File:CentrifugalFan.png


compression. Compressors that utilize inter-stage cooling between compression stages

come closest to achieving isothermal compression.

Isentropic compression

This model assumes that no energy (heat) is transferred to or from the gas during the

compression, and all supplied work is added to the internal energy of the gas, resulting

in increases of temperature and pressure. Theoretical temperature rise is:

Toutlet = Tinlet x (poutlet/pinlet)(k-1)/k (C1)

Where:

Toutlet = outlet temperature in K

Tintlet = inlet temperature in K



k = cp/cv, (= ratio of specific heats, approximately 1,4 for air and 1,3 for water vapor)

Isentropic (also called adiabatic) compression is more closely real life.

In practice there will always be a certain amount of heat flow out of the compressed

gas.

Power consumption centrifugal fans and compressors

Isothermal compression takes less work than isentropic (adiabatic) compression:

Pisentropic=)ηx(η

}1)p

p{(x

1k

kx (kPa)px /s)V(m

compressormotor

k

1k

inlet

outlet

inlet

3

kW (C2)

Pisothermic= Wk)ηx(η

p

p lnx (kPa) px /s)(mV

compressormotor

inlet

outlet

inlet

3

(C3)

Where:



ηmotor = efficiency electric motor in decimal

ηcompr. = efficiency compressor or fan in decimal

V = flow rate in m3/s (at inlet pressure)

Pisentropic = required power for isentropic compression in kW

Pisothermic = required power for isothermic compression in kW

k = cp/cv, (= ratio of specific heats, appr. 1,4 for air and 1,3 for water vapor)

The efficiency of multi-stage compressors can be increased by cooling the compressed

gas between stages; this makes the compression less isentropic (adiabatic) and more

isothermal.

C3 Relation between pressure and required power

When optimizing the energy efficiency of drying processes very often the dew point of

the exhaust air in the optimal situation will be higher than in the reference situation.


When this happens the amount of air will decrease and the absolute vapour content of

this air (g vapour / kg dry air) will increase. This may result in substantial electricity

savings for supply and exhaust air fans of the dryer.

Usually the electricity consumption in the reference situation is known. Hereafter will be

explained how the electricity consumption in the optimal situation can be calculated.

The power consumption of the centrifugal fan in case of isentropic compression:

Pisentropic=)ηx(η

}1)p

p{(x

1k

kx (kPa)px /s)V(m

compressormotor

k

1k

inlet

outlet

inlet

3

kW (C2)

Conversion of pressure (p in kPa= kN/m2) in head (m); see box C1:

p = ρ (kg/m3) x g (m/s2) x H (m) kPa (C4)

Where:

p = pressure in kPa; (1 bar = 105 N/m2 = 105 Pa; 1 N = 1 kg.m/s2)

ρ = fluid density in kg/m3; (ρwater = 1000 kg/m3; ρair = about 1,3 kg/m3)

g = acceleration due to gravity; (g = 9,81 m/s2)

H = head in m

The density of water is 1000 kg/m3, for air this is 1,29 kg/m3 at standard conditions (0 oC en 101,325 kPa).

At higher than standard elevation (sea level) and higher than standard temperature, air

density is lower than standard density. At higher temperatures the centrifugal fan will

create less pressure and will require less power. As the density of air is, compared to

water, very low the elevation head for air will also be low.

Note that a centrifugal fan is a constant volume device that will move the same volume

of air at two different temperatures.

The density ρ of the wet exhaust air depends on the vapour content:

ρwet.air = mwet .air / V (C5)

mwet .air = mdry.air + mvapour (C6)

Where :

mwet.air = mass wet air in kg/s

mdry.air = mass dry air in kg/s

mvapour = mass vapour in air in kg/s

V = flow rate in m3/s

The mass of the dry air and the mass of the vapour can be derived from:

mdry.air = (28,84/8,314) x (pdry.air x V /T) (13c)

mvapour = (18/8,314) x (pvapour x V /T) (13d)

Substitute formula (13c) and (13d) in formula (C6) :

ρ = {(28,84/8,314) x (pdry.air x V /T) + (18/8,314) x (pvapour x V /T)} / V

ρwet.air= (28,84 x pdry air+ 18,00 x pvapor)/(8,314 x (273,15 + θdry bulb)) (C7)


C4 Electrticity consumption supply, exhaust and circulation fans

In table C1 the formula’s for centrifugal fans are applied on supply air, exhaust air and

circulation fans for multi cylinder and yankee dryers for as well the reference as the

optimal situation.

The known data of the reference situation are applied to try to predict the electric

consumption in the optimal situation.

Total head

In box C1 is derived that for pumps the required power: P(kW)= Q (m3/s) x pman (kPa).

In this formula the required power for liquid pumps depends on the difference between the inlet and outlet pressure pman is:

pman = poutlet – pinlet (C17)

pman = ρ.g.Hgeo + (pa – pe) + ½ρ. )v(v 2

e

2

a + Cpipe .(½. ρ. vpipe2) (C18)

pman =elevation head + pressure head + velocity head + resistance head

The required power for gas compressors and fans for air depends on the pressure ratio:

“pman” = pinlet x (k/(k-1)) x (((poutlet/pinlet)^((k-1)/k))-1) (C9)

When the suction and discharge pipes of the fans are at the same level (Hgeo=0) and

the air velocities are also the same (va = ve) then the elevation head (ρ.g.Hgeo) and the

velocity head (½ρ. )v(v 2

e

2

a ) can be neglected. In these cases the power consumption

depends on the pressure head and the resistance head.

In yankee hoods pressure head is a result of the pressure drop in the plenum where air

velocities of more than 100 m/s are created. Also blow boxes in multi-cylinder dryers

may need pressure head.

Resistance head

In table C1 is assumed that there are only “other” and resistance losses due to friction.

The pressure loss in pipes and tubes due to friction depends on the flow velocity, pipe


the pipe or duct, and whether the flow is turbulent or laminar – the Reynolds Number of

the flow-. The pressure loss in a tube or duct due to friction can be expressed as:

ploss.resistance = λ * (l / dh) * (½ * ρ * v2 ) (C10)

Where:

ploss.friction = pressure loss due to friction in ducts (Pa)


l = length of duct (m)


vpipe = flow velocity in the duct (m/s)

For an existing ventilation system λ, l, and dh are, within certain limits, constant (Cpipe):

Cpipe = λ* (l / dh) (C11)

Substituting formula (C11) in (C10):

ploss.resistance = Cduct * (½ * ρ * vduct2 ) (C12)


Table C1 Electricity consumption supply, exhaust and recirculation fans dryers (rough estimation)

a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac ad ae af ag ah

Requir

ed

power

Dryin

g a

ir (

*3)

Dew

poin

t (*1)

Tem

peratu

re (

*1)

Absolu

te h

um

idity (

*2)

Enth

alp

y (

*2)

Rel.hum

idity(*2)

Partial vapour p

ressure

(*2)

Abs.

hum

. supply

air

(*2)

Product

wate

r e

vap.

(*3)

Inle

t pressure

(*1) (

*14)

Partial pressure

dry a

ir (

*4)

Density h

um

id a

ir (

*5)

Volu

me f

low

(*6)

Tem

peratu

re (

*7)

Partial vapour p

ressure

(*8)

Outlet

pressure (

*1)

(*15)

Partial pressure d

ry a

ir

(*9)

Dryin

g a

ir (

*10)

Density h

um

id a

ir (

*5)

Volu

me f

low

(*6)

Tota

l pressure losses

(*18)

Ele

vation,

pressure &

velo

city loss

Ele

vation,

pressure &

velo

city loss (

*19:r)

Resis

tance loss (

fric

tion

in d

ucts

etc

) (

*20)

Cross s

ectional area

duct

(*1)

Air

velo

city in d

ucts

etc

(*21)

Eff

icie

ncy e

lectr

ic m

oto

r

(*1;16)

Eff

icie

ncy c

om

pressor

(*1;17)

Requir

ed p

ow

er (

*11)

- - - mdry.air θdew θ wvap.out h φ pvapor wvap.in PWE pinlet pdry.air ρwet.air V θ pvapor poutlet pdry.air mdry.air ρwet.air V ∆ptot ∆pother ∆pother ∆pl.res. A v ŋmotor ŋcompres Pisentr

- - - kg/s °C °C g/kg kJ/kg % kPa g/kg kg/s kPa kPa kg/m3 m3/s °C kPa kPa kPa kg/s kg/m3 m3/s kPa % kPa kPa m2 m/s - - kW

Reference situation1 exhaust 52,9 57,0 80 128,7 422 36,4 17,3 - 6,5 101,3 84,0 0,93 56,8 83,6 18,0 105,0 87,0 52,9 0,96 55,4 3,7 60,0 2,2 1,5 3,86 14,4 0,95 0,85 255

2 supply 52,9 7,0 10 6,2 26 81,6 1,0 6,2 0,0 100,3 99,3 1,22 43,2 12,6 1,0 103,6 102,6 52,9 1,25 42,3 3,3 60,0 2,0 1,3 3,86 10,9 0,95 0,85 175

3 exhaust 52,9 57,0 80 128,7 422 36,4 17,3 - 6,5 101,3 84,0 0,93 56,8 83,6 18,0 105,0 87,0 52,9 0,96 55,4 3,7 60,0 2,2 1,5 3,86 14,4 0,95 0,85 255

4 supply 52,9 7,0 10 6,2 26 81,6 1,0 6,2 0,0 100,3 99,3 1,22 43,2 12,6 1,0 103,6 102,6 52,9 1,25 42,3 3,3 60,0 2,0 1,3 3,86 10,9 0,95 0,85 175

5 exhaust 52,9 57,0 80 128,7 422 36,4 17,3 - 6,5 101,3 84,0 0,93 56,8 83,6 18,0 105,0 87,0 52,9 0,96 55,4 3,7 60,0 2,2 1,5 3,86 14,4 0,95 0,85 255

6 supply 52,9 7,0 10 6,2 26 81,6 1,0 6,2 0,0 100,3 99,3 1,22 43,2 12,6 1,0 103,6 102,6 52,9 1,25 42,3 3,3 60,0 2,0 1,3 3,86 10,9 0,95 0,85 175

7 recirculation20,0 57,0 80 128,7 422 36,4 17,3 128,7 0,0 100,3 83,0 0,92 21,7 84,5 18,1 104,8 86,7 20,0 0,95 21,0 4,5 60,0 2,7 1,8 0,20 105,2 0,95 0,85 119

Optimal situation1 exhaust 30,7 66,0 90 217,6 671 37,1 26,2 - 6,5 101,3 75,1 0,87 35,1 92,8 26,9 104,1 77,2 30,7 0,89 34,5 2,7 80,6 2,2 0,5 3,86 8,9 0,95 0,85 118

2 supply 30,7 7,0 10 6,2 26 81,6 1,0 6,2 0,0 100,3 99,3 1,22 25,1 11,9 1,0 102,7 101,7 30,7 1,25 24,6 2,4 81,6 2,0 0,4 3,86 6,4 0,95 0,85 75

3 exhaust 30,7 66,0 90 217,6 671 37,1 26,2 - 6,5 101,3 75,1 0,87 35,1 92,8 26,9 104,1 77,2 30,7 0,89 34,5 2,7 80,6 2,2 0,5 3,86 8,9 0,95 0,85 118

4 supply 30,7 7,0 10 6,2 26 81,6 1,0 6,2 0,0 100,3 99,3 1,22 25,1 11,9 1,0 102,7 101,7 30,7 1,25 24,6 2,4 81,6 2,0 0,4 3,86 6,4 0,95 0,85 75

5 exhaust 30,7 66,0 90 217,6 671 37,1 26,2 - 6,5 101,3 75,1 0,87 35,1 92,8 26,9 104,1 77,2 30,7 0,89 34,5 2,7 80,6 2,2 0,5 3,86 8,9 0,95 0,85 118

6 supply 30,7 7,0 10 6,2 26 81,6 1,0 6,2 0,0 100,3 99,3 1,22 25,1 11,9 1,0 102,7 101,7 30,7 1,25 24,6 2,4 81,6 2,0 0,4 3,86 6,4 0,95 0,85 75

7 recirculation18,8 66,0 90 217,6 671 37,1 26,2 - - 100,3 74,1 0,86 21,7 94,5 27,3 104,7 77,4 18,8 0,89 21,1 4,4 61,5 2,7 1,7 0,20 105,3 0,95 0,85 116

Explanation ((*…r) = reference situation; (*…o) = optimal situation; (*…r+o) = both)): (*18:r+o) ∆ptot = poutlet - pinlet

(*1:r+o) White cells: data measured or calculated elsewhere (MED) (*11:r+o) Pisentr. = V x pinlet x (k/(k-1)) x (((poutlet/pinlet)^((k-1)/k))-1) / (ŋmotorXŋcompressor) ∆ptot = ∆pelevation + ∆ppressure + ∆pvelocity+ ∆presistance

(*2:r+o) Calculated with formula's from "Formula library" (*5:r+o) ρwet.air= (28,84 x pdry air+ 18,00 x pvapor)/(8,314 x (273,15 + θdry bulb)) ∆pother = ∆pelevation + ∆ppressure + ∆pvelocity

(*3:r) PWE = mdry.air x 0,001 x (wvap.out - wvap.in) (*6:r+o) V = mdry.air / ρwet.air (*19:r) ∆pother = % x ∆ptot (= rough estimation)

(*3:o) mdry.air = PWE / ((0,001 x (wvap.out - wvap.in)) (*7:r+o) Toutlet = Tinlet x (poutlet/pinlet) ^ ((k-1)/k) (*19:o) ∆pother(o) = ∆pother(r)

(*3:o) cell l3ref= cel l3opt (PWEoptmal.sit = PWEreference.sit.) (*8:r+o) pvapour = poutlet - pdry.air (*20:r) ∆pl.res = ∆ptot -∆pother

(*3:o) cell l5ref= cel l5opt (PWEoptmal.sit = PWEreference.sit.) (*15:o) poulet = pinlet + ∆ploss.friction + ∆pother (*20:o) ∆p(o)l.res = ((ρ(r).v(r)2)/(ρ(o).v(o)2)) x ∆p(r)l.res

(*3:o) cell p7ref= cel p7opt (Voptmal.sit = Vreference.sit.);(vimpingement) (*9:r+o) pdry.air.outlet = (poutlet/oinlet) * pdry.air.inlet (*21:r+o) v = V/A

(*14:o) pinlet.optmal.sit = pinlet.reference.sit. (*10:r+o) mdry.air.outlet = mdry.air.inlet (*16:o) ŋmotor.optimal.sit. = ŋmotor.reference.sit.

(*4:r+o) pdry.air = pinlet - pvapor; (*20:r) C duct can be derived from ploss.friction = C duct * ρwet.air * v2 (*17:o) ŋcompressor.optimal.sit. = ŋcompressor.reference.sit.

multi

cyl.

yan-

kee

cyl.

other

multi

cyl.

yan-

kee

cyl.

Fan n

um

ber

Multi cylinder o

r y

ankee d

ryer

Exhaust,

supply

or c

ircula

tion

air

Inlet fan (suction) Outlet fan (discharge)

other


Exercise 1C One stage compression

Given:

Q = 1 m3 air, pinlet = 0,85 bar , poutlet = 8,5 bar, ηmotor x η compressor = 0,60

Question:

Calculate Pisentropic and Pisothermic in Wh.

Calculate the savings when isothermic in stead of isentropic compression is applied.

Answer:

Apply formula’s (C2) and (C3); 0,85 bar = 85 kPa, 8,5 bar = 850 kPa):

Pop(isentr.)= Wk461 0,60

11,4

11,4

)85

850(x

11,4

1,4x kPa85x /s

31m

Pop(isoth.) = kW k 326 0,60

85

850 lnx kPa85x /s1m

3

Savings: (461-326)/461= 29%

Exercise 2C Two stage compression with and without intermediate cooling

Given

Two stage compression with and without intermediate cooling:

Q = 1 m3 air, pinlet.stage1 = 0,85 bar =(85 kPa) , poutlet = 8,5 bar (= 850 kPa),

poutlet.stage2 = 8,5 bar (=850 kPa), θinlet.stage1= 10oC; ηmotor x η compressor = 0,60.

Stage 1 is rootsblower, compression ratio= 2, intermediate cooler: θinlet.stage2 = 10oC,

Stage 2 is screw compressor, compression ratio = 5.

Question:

Calculate Pisentropic in kW for roots-blower, screw-compressor and total with and without

intermediate cooling.

Calculate savings in relation to one stage compression.

What can you tell about the capacity of the compressor ?

Answer:

Two stage compression without intermediate cooling:

P Isentropic.stage1 = Wk108 0,60

1)85

170(x

11,4

1,4xkPa85x /sm 1 1,4

11,4

3

Apply Poisson I for calculatig Vafter.stage1 :

pinlet x Vkinlet = poutlet x Vk

outlet


Vafter.stage1 : 0,85 x 1 = 1,7 x Vkafter.stage1; Vafter.stage1 = 0,61 m3

P(isentopic.stage 2= kW 353 0,60

1170

850x

1-1,4

1,4x kPa 170x /s0,61m

1,4

11,4

3

Pisentr.tot = (108 + 353=) 461 kW; also the same as in one stage compression (see

exercise 1C.

Two stage compression with intermediate cooling:

Pisentropc.stage1 = 108 kW

For calculation Θafter.stage1 apply Poisson II:

;170

85

θ

10273 ;

p

p

T

T

1,4

11,4

e1after.stag

k

1k

outlet

inlet

outlet

inlet

Θafter.stage1 = 72oC

Vafter.stage1 = 0,61 m3

For calculation Vbefore.stage2: 10293

V

72273

0,61;

T

V

T

V ge2before.sta

2

2

1

1

Vbefore.stage2 = 0,50 m3

Pisentropicstage2 = Wk289 0,60

1170

850x

1-1,4

1,4x kPa 170x m 0,50

1,4

11,4

3

Pisentropic.tot = (108 + 289=) 397 kW

Savings in relation to one stage compression: ((461-397)/461=)14% (maximum

feasible 29%)

When normally two screw compressors are running, one can be replaced by a

rootsblower.

PUMPS

C5 Types of pumps

Pumps fall into two major groups: positive displacement pumps and rotodynamic

pumps. Their names describe the method for moving a fluid.

Positive displacement pumps

Positive displacement pumps are "constant flow machines". Positive displacement

pumps will produce the same flow at a given speed. A positive displacement pump must

not be operated against a closed valve on the discharge side of the pump because it

might burst the discharge line.

Roto- dynamic pumps

The most common roto-dynamic pump is the centrifugal pump.


In roto-dynamic pumps kinetic energy is added to the fluid by increasing the flow

velocity. This increase in kinetic energy is converted to a gain in potential energy

(pressure) when the velocity is reduced prior to or as the flow exits the pump into the

discharge pipe. These pumps can operate under closed valve conditions.

This pump uses a rotating impeller to increase the pressure and flow-rate of a fluid.

A centrifugal pump containing two or more impellers is called a multistage centrifugal

pump. The advantages to using a multistage centrifugal pump or a single stage pump

include production of a higher pressure head, and to discharge a larger quantity of liquid.

C6 Head pressure

Total head

Head is a concept that relates the energy in a fluid to the height of an equivalent static

column of that fluid. Head is expressed in m of height.

The static head of a pump is the maximum height (pressure) it can deliver. The

capability of the pump can be read from its Q-H curve (flow vs. height).

Head is equal to the fluid's energy per unit weight.

Head is useful in specifying centrifugal pumps because their pumping characteristics

tend to be independent of the fluid's density.

There are four types of head used to calculate the total head in and out of a pump:

elevation, pressure, velocity and resistance head. The equation is shown in formula (15)

in box 1.

With formula (16) height can be converted in kPa (=kN/m2) and, as the difference

between pressure gauges on the discharge and suction side of the pump is equal to the

total head, formula (18) can be derived from formula’s (15) and (17).

Resistance head

The resistance loss in pipes and tubes due to friction depends on the flow velocity, pipe


the pipe or duct, and whether the flow us turbulent or laminar – the Reynolds Number

of the flow. The resistance loss is divided in major loss due to friction and minor loss

due to change of velocity in bends, valves and similar. The latter are also expressed in

equivalent friction losses.

The resistance loss in a tube or duct due to friction can be expressed as:

ploss.resistance = λ* (l / dh) * (½ * ρ * vpipe2 ) (C12)

Where:

ploss.resistance = pressure loss (kPa)


l = length of duct or pipe (m)


vpipe = flow velocity in pipe or duct (m/s)

For an existing pump system λ, l, and dh are constant (Cpipe):

Cpipe = λ* (l / dh) (C13)

Substituting formula (C13) in (C14):

ploss.resistance = Cpipe * (½ * ρ * vpipe2 ) (C14)


Box 1C. Head and required power pumps

Explanation symbols and units

p = pressure in N/m2 (=Pa)

(1 bar = 105N/m2; 1 kN = 1 kg.m/s2)

ρ = fluid density in kg/m3

(ρwater = 1000 kg/m3)

g = acceleration due to gravity

(g = 9,81 m/s2)

Pinput = required input power in kW

H = energy head added to the flow in m

Q = flow rate in m3/s or m3/h

ŋpump = total efficiency pump (as decimal)

ŋmotor = efficiency electric motor (as decimal)

Total head (Htot in m):

Htot = (h1 – h2) + resistance

2

2

2

121 H2g

vv

gρ

p p

(C15)

Elevation head pressure head velocity head resistance head

Static head (independent of flow) Dynamic head (depending on flow)

Resistance head is friction loss in pipes, ducts, bends etc.

Conversion of pressure (p in N/m2) in head (m):

p (N/m2) = ρ (kg/m3) x g (m/s2) x h (m) (C16)

(Note: N = kg.m/s2)

Difference between outlet and inlet pressure pump (= pman):

pman = poutlet – pinlet (C17)

Total head (pman = ρ.g.Htot = poutlet – pinlet in kN/m2)


2

2

1 + Cpipe .(½. ρ. vpipe2) (C18)

Required power electric motor when Q in m3 water/s and p in kPa

Pinput =)ηx(η

}kPa)p(px /s{(Q)m

pompmotor

inletoutlet

3 kW C19)

h1

h2

v2

v1

p1

p2

poutlet

outle

pintlet

outle

Hg

eo =

h

1 -

h2

pressure gauges are on the same height level


C7 Power consumption pumps

The power required to drive a pump is defined by formula (C19) in box C1.

Units check: kg/m3 x m/s2 x m x m3/s = (kg.m/ s2) x m / s = kNm/s =kJ/s = kW

Pump efficiency (ηpump) is defined as the ratio of the power imparted on the fluid by the pump

in relation to the power supplied to drive the pump.

Efficiency is a function of the discharge and therefore also operating head; this means its

value is not fixed for a given pump. One important part of system design involves matching

the pipeline headloss-flow characteristic with the appropriate pump which will operate at or

close to the point of maximum efficiency.

Pump efficiencies tend to decline over time due to wear.

The energy usage is determined by multiplying the power requirement by the length of time the

pump is operating.

C8 Pump- and system characteristics

In box 1 is shown that the total pump head is the sum of the static head and the dynamic

head. The static head is independent of the liquid flow while the dynamic head depends on it.

The relation between the heads and the liquid flow is shown in the pump characteristic (H/Q) and

system characteristic (HA/Q) in figure 2.

Figure 2 Pump (H/Q) and system (HA/Q) characteristics

The crossing of the characteristics is the working point. In the example in figure 2 at this

point Q = 160 m3/h, H = 50 m, P = 28 kW and η = 73%.

Pump capacity can be controlled by throttling valve in the discharge op the pump or by

variable speed drive.


In box 2 the differences are shown. Throttling causes pressure drop across the valve and

will result in considerable losses.

Box 2 Pump- and system characteristic and control pump capacity

a) throttling b) variable speed control

Control pump capacity:

a) throttling

- system characteristic changes

- throttling loss = Δ h

b) variable speed control

- pump characteristic changes

- no loss

- relation between Q, H and P when n (speed in rpm) changes:

Qx = (nx/n) * Q (C19)

Hx = (nx/n)2 * Q (C20)

Px = (nx/n)3 * Q (C21)


Excercise C3 Head and required power head box pumps

Given:

Head box pumppumps suspension from cleaners to wire section

Velocity from head box to wire = 20 m/s (=velocity wire).

Recirculation from head box = 10%

Pressure on liquid surface on suction side pump = 1,2 bar

Pressure on liquid surface on discharge side pump = 1 bar (=barmetric prssure).

Elevation head liquid levels = 2 m.

Friction losses in pipes = 15 m.

Mass flow = 3.701.064 ton bone dry per year

Consistence = 0,93 %.

Question:

Calculate with formula (16) total, elevation, pressure, velocity and resistance head.

Convert m in bar

Calculate the yearly consumption in MWh and in euro’s (ŋmotor= 80%, ŋpump=95%and

1MWh costs € 65)

Answer:

H=2m+ (1,0–1,2)x105N/m2:(1000kg/m3x9,8m/s2) +(202-02)m2/s2:(2x9,8m/s2)+15 m

H = + 2 m - 2,04 m + 20,41 m + 15 m = 35,37 m

Pman = 1000 kg/m3 x 9,8 m/s2 x 35,37 m = 346.626 N/m2 (= 3,47 bar)

Pop(man)={(1,1x3.701.064):0,0093}m3x3,47barx(1:36.000)MWh/bar:(0,80X0,95)=55.50

2 MWh

K = 55.502 MWh x 65 €/MWh = € 3.608.802


ANNEX D: TABLES AND UNITS

D1 STEAM TABLE: PROPERTIES OF STEAM AT VARYING PRESSURES AND TEMPERATURES

Absolute pressure (kPa, kN/m2)

Temperature (oC)

Specific Volume (m3/kg)

Density - ρ - (kg/m3)

Specific Enthalpy of Specific Entropy of Steam - s - (kJ/kgK)

Liquid - hl - (kJ/kg)

Evaporation - he - (kJ/kg)

Steam - hs - (kJ/kg)

0.8 3.8 160 0.00626 15.8 2493 2509 9.058

2.0 17.5 67.0 0.0149 73.5 2460 2534 8.725

5.0 32.9 28.2 0.0354 137.8 2424 2562 8.396

10.0 45.8 14.7 0.0682 191.8 2393 2585 8.151

20.0 60.1 7.65 0.131 251.5 2358 2610 7.909

28 67.5 5.58 0.179 282.7 2340 2623 7.793

35 72.7 4.53 0.221 304.3 2327 2632 7.717

45 78.7 3.58 0.279 329.6 2312 2642 7.631

55 83.7 2.96 0.338 350.6 2299 2650 7.562

65 88.0 2.53 0.395 368.6 2288 2657 7.506

75 91.8 2.22 0.450 384.5 2279 2663 7.457

85 95.2 1.97 0.507 398.6 2270 2668 7.415

95 98.2 1.78 0.563 411.5 2262 2673 7.377

100 99.6 1.69 0.590 417.5 2258 2675 7.360

101.3 100 1.67 0.598 419.1 2257 2676 7.355

110 102.3 1.55 0.646 428.8 2251 2680 7.328

130 107.1 1.33 0.755 449.2 2238 2687 7.271

150 111.4 1.16 0.863 467.1 2226 2698 7.223

170 115.2 1.03 0.970 483.2 2216 2699 7.181

190 118.6 0.929 1.08 497.8 2206 2704 7.144

220 123.3 0.810 1.23 517.6 2193 2711 7.095

260 128.7 0.693 1.44 540.9 2177 2718 7.039

280 131.2 0.646 1.55 551.4 2170 2722 7.014

320 135.8 0.570 1.75 570.9 2157 2728 6.969

360 139.9 0.510 1.96 588.5 2144 2733 6.930

400 143.1 0.462 2.16 604.7 2133 2738 6.894

440 147.1 0.423 2.36 619.6 2122 2742 6.862

480 150.3 0.389 2.57 633.5 2112 2746 6.833

500 151.8 0.375 2.67 640.1 2107 2748 6.819

550 155.5 0.342 2.92 655.8 2096 2752 6.787

600 158.8 0.315 3.175 670.4 2085 2756 6.758

650 162.0 0.292 3.425 684.1 2075 2759 6.730

700 165.0 0.273 3.66 697.1 2065 2762 6.705

750 167.8 0.255 3.915 709.3 2056 2765 6.682

800 170.4 0.240 4.16 720.9 2047 2768 6.660

850 172.9 0.229 4.41 732.0 2038 2770 6.639

900 175.4 0.215 4.65 742.6 2030 2772 6.619

950 177.7 0.204 4.90 752.8 2021 2774 6.601

1000 179.9 0.194 5.15 762.6 2014 2776 6.583

1050 182.0 0.186 5.39 772 2006 2778 6.566

1150 186.0 0.170 5.89 790 1991 2781 6.534

1250 189.8 0.157 6.38 807 1977 2784 6.505

1300 191.6 0.151 6.62 815 1971 2785 6.491

1500 198.3 0.132 7.59 845 1945 2790 6.441

1600 201.4 0.124 8.03 859 1933 2792 6.418

http://www.engineeringtoolbox.com/temperature-d_291.html

http://www.engineeringtoolbox.com/temperature-d_291.html


D2 UNITS Unit

(SI) unit

Explanation

Name

Symbol

Name

Symbol *

temperature mass force pressure work power amount substance molar volume molaire massa density Gross Heating Value Net Heating Value Normal volume

θ

m F p p A P n

Vm

M ρ

Hs

Hi

Vo

kelvin gram newton (N) pascal (bar) joule watt mol - - -

GHV NHV -

K

g

kg.m/s2

Pa (bar)

J

W

mol

m3/kmol

kg/kmol

kg/m3

MJ/ 3om

MJ/ 3om

m3

o

273 K = 0oC Mass (m) of a body is the constant ratio between the force (F) and acceleration (a) caused by the force on the body: m (kg) = F (N) : a (m/s2); F = m x a (kg . m/s2) The force of gravity (G) that gives an body with a mass (m) of 1 kg an acceleration (g) of 1 m/s2 is 1 Newton A body with a mass of m kg has a weight of: G = m x g N ; g = 9,8 m/s2 (gravity) 1 Pa = 1 N/m2 1 bar = 105 Pa Convert pressure in m height: p (N/m2) = h (m) x ρ (kg/m3) x g (m/s2) 1 J = 1 N . m 1 W = 1 J/s For gas at 273 K and 101,325 kPa: Vm = 22,4 m3/kmol N = 14, O = 16, C = 12, H = 1, S = 32; bijv.: CO2: (1 x 12 + 2 x 16 =) 44 kg/kmol. ρ = M : Vm

G-gas Hs = 35,17 MJ/ 3om

G-gas Hi = 31,65 MJ/ 3om

V(…) = V (0oC; 101,325kPa)

Pre-fix: P = 1015, T = 1012. G = 109, M = 106, k = 103, - = 100, m = 10-3, μ = 10-6, n = 10-9


D3 EMPIRICAL RELATIONS BETWEEN θ, θdew, pvapor AND wvapor

Dew

poin

t

Absolu

te

hum

idity

Part

ial vapor

pre

ssure

Rela

tive

hum

idity

Tem

pera

ture

Wet

bulb

tem

pera

ture

Enth

alp

y

Dew

poin

t

Absolu

te

hum

idity

Part

ial vapor

pre

ssure

Rela

tive

hum

idity

Tem

pera

ture

Wet

bulb

tem

pera

ture

Enth

alp

y

θdew wvapor pvapor φ θ θwet.bulb h θdew wvapor pvapor φ θ θwet.bulb h

°C g/kg kPa % °C °C J/g °C g/kg kPa % °C °C J/g

1,0 4,1 0,7 100 1,0 1,0 11 51,0 92,6 13,0 100 51,0 51,0 2922,0 4,4 0,7 100 2,0 2,0 13 52,0 98,0 13,6 100 52,0 52,0 3073,0 4,8 0,8 100 3,0 3,0 15 53,0 103,7 14,3 100 53,0 53,0 3234,0 5,1 0,8 100 4,0 4,0 17 54,0 109,8 15,0 100 54,0 54,0 3405,0 5,5 0,9 100 5,0 5,0 19 55,0 116,1 15,7 100 55,0 55,0 3586,0 5,9 0,9 100 6,0 6,0 21 56,0 123,0 16,5 100 56,0 56,0 3777,0 6,3 1,0 100 7,0 7,0 23 57,0 130,2 17,3 100 57,0 57,0 3978,0 6,7 1,1 100 8,0 8,0 25 58,0 137,9 18,2 100 58,0 58,0 4189,0 7,2 1,1 100 9,0 9,0 27 59,0 146,0 19,0 100 59,0 59,0 441

10,0 7,7 1,2 100 10,0 10,0 30 60,0 154,7 19,9 100 60,0 60,0 46511,0 8,3 1,3 100 11,0 11,0 32 61,0 163,9 20,9 100 61,0 61,0 49012,0 8,8 1,4 100 12,0 12,0 34 62,0 173,8 21,8 100 62,0 62,0 51713,0 9,5 1,5 100 13,0 13,0 37 63,0 184,3 22,9 100 63,0 63,0 54614,0 10,1 1,6 100 14,0 14,0 40 64,0 195,4 23,9 100 64,0 64,0 57615,0 10,8 1,7 100 15,0 15,0 42 65,0 207,3 25,0 100 65,0 65,0 60916,0 11,5 1,8 100 16,0 16,0 45 66,0 220,2 26,2 100 66,0 66,0 64417,0 12,3 1,9 100 17,0 17,0 48 67,0 234,0 27,3 100 67,0 67,0 68218,0 13,1 2,1 100 18,0 18,0 51 68,0 248,7 28,6 100 68,0 68,0 72219,0 14,0 2,2 100 19,0 19,0 55 69,0 264,4 29,8 100 69,0 69,0 76520,0 14,9 2,3 100 20,0 20,0 58 70,0 281,5 31,2 100 70,0 70,0 81121,0 15,9 2,5 100 21,0 21,0 62 71,0 299,8 32,5 100 71,0 71,0 86122,0 16,9 2,6 100 22,0 22,0 65 72,0 319,7 34,0 100 72,0 72,0 91523,0 18,0 2,8 100 23,0 23,0 69 73,0 341,3 35,4 100 73,0 73,0 97324,0 19,1 3,0 100 24,0 24,0 73 74,0 364,7 37,0 100 74,0 74,0 1.03725,0 20,3 3,2 100 25,0 25,0 77 75,0 390,2 38,6 100 75,0 75,0 1.10626,0 21,6 3,4 100 26,0 26,0 81 76,0 417,9 40,2 100 76,0 76,0 1.18127,0 23,0 3,6 100 27,0 27,0 86 77,0 448,2 41,9 100 77,0 77,0 1.26228,0 24,4 3,8 100 28,0 28,0 91 78,0 480,5 43,7 100 78,0 78,0 1.35029,0 26,0 4,0 100 29,0 29,0 96 79,0 518,6 45,5 100 79,0 79,0 1.45230,0 27,6 4,2 100 30,0 30,0 101 80,0 559,3 47,4 100 80,0 80,0 1.56231,0 29,3 4,5 100 31,0 31,0 106 81,0 604,7 49,3 100 81,0 81,0 1.68532,0 31,1 4,8 100 32,0 32,0 112 82,0 655,7 51,4 100 82,0 82,0 1.82233,0 32,9 5,0 100 33,0 33,0 118 83,0 713,2 53,4 100 83,0 83,0 1.97734,0 35,0 5,3 100 34,0 34,0 124 84,0 777,8 55,6 100 84,0 84,0 2.15135,0 37,1 5,6 100 35,0 35,0 130 85,0 852,4 57,8 100 85,0 85,0 2.35236,0 39,3 5,9 100 36,0 36,0 137 86,0 937,7 60,1 100 86,0 86,0 2.58137,0 41,7 6,3 100 37,0 37,0 144 87,0 1.036,4 62,5 100 87,0 87,0 2.84638,0 44,1 6,6 100 38,0 38,0 152 88,0 1.152,4 65,0 100 88,0 88,0 3.15839,0 46,8 7,0 100 39,0 39,0 160 89,0 1.290,7 67,5 100 89,0 89,0 3.53040,0 49,5 7,4 100 40,0 40,0 168 90,0 1.458,7 70,1 100 90,0 90,0 3.98241,0 52,5 7,8 100 41,0 41,0 177 91,0 1.665,9 72,8 100 91,0 91,0 4.53842,0 55,6 8,2 100 42,0 42,0 186 92,0 1.927,2 75,6 100 92,0 92,0 5.24043,0 58,8 8,6 100 43,0 43,0 195 93,0 2.268,3 78,5 100 93,0 93,0 6.15744,0 62,3 9,5 100 44,0 44,0 205 94,0 2.731,7 81,5 100 94,0 94,0 7.40145,0 65,9 9,6 100 45,0 45,0 216 95,0 3.396,1 84,6 100 95,0 95,0 9.18546,0 69,8 10,1 100 46,0 46,0 227 96,0 4.426,7 87,7 100 96,0 96,0 11.95347,0 73,9 10,6 100 47,0 47,0 239 97,0 6.248,4 91,0 100 97,0 97,0 16.84448,0 78,1 11,2 100 48,0 48,0 251 98,0 10.303,1 94,3 100 98,0 98,0 27.73149,0 82,7 11,7 100 49,0 49,0 264 99,0 27.145,9 97,8 100 99,0 99,0 72.95350,0 87,5 12,3 100 50,0 50,0 277 100,0 - 101,4 100 100,0 100,0 -


D4 AVERAGE SPECIFIC HEAT

'Cp.av. in kJ/(kg.oC) of components in air or flue gas

temp

range °CCO2 Ar O2 N2 H2O Dry air

0 0,819 0,519 0,917 1,041 1,83 1,006

0 - 100 0,870 0,519 0,923 1,041 1,84 1,009

0 - 200 0,915 0,519 0,935 1,045 1,87 1,012

0 - 300 0,953 0,519 0,950 1,048 1,89 1,019

0 - 400 0,989 0,519 0,964 1,055 1,92 1,028

0 - 500 1,021 0,519 0,978 1,065 1,95 1,038

0 - 600 1,046 0,519 0,993 1,075 1,98 1,048

0 - 700 1,072 0,519 1,005 1,087 2,01 1,061

0 - 800 1,092 0,519 1,017 1,098 2,04 1,071

0 - 900 1,112 0,519 1,029 1,109 2,07 1,080

0 - 1000 1,129 0,519 1,037 1,118 2,11 1,090

0 - 1100 1,144 0,519 1,045 1,129 2,14 1,100

0 - 1200 1,158 0,519 1,052 1,139 2,17 1,109

0 - 1300 1,170 0,519 1,060 1,145 2,20 1,118

0 - 1400 1,181 0,519 1,066 1,152 2,23 1,125

0 - 1500 1,192 0,519 1,072 1,160 2,26 1,132

0 - 1600 1,203 0,519 1,078 1,168 2,29 1,139

0 - 1800 1,219 0,519 1,090 1,182 2,34 1,153

0 - 2000 1,234 0,519 1,101 1,192 2,39 1,162


D5 MOLLIER CHARTS FOR HUMID AIR

MED PART I OPTIMIZING ENERGY EFFICIENCY AND HEAT … · 10.1 Purpose of heating process water 10.2...

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Transcript of MED PART I OPTIMIZING ENERGY EFFICIENCY AND HEAT … · 10.1 Purpose of heating process water 10.2...