Thermal System Design and Simulation

Thermal System Design and Simulation(열시스템 설계 및 시뮬레이션)

Prof. Tae-Kuk Kim (김태국)

Spring 2014

Dep’t of ME, ChungAng Univ.

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Energy & Spectral Radiation Research Lab / 396 에너지및복사특성연구실

=========== SYLLABUS ===========

◈ Course Title : Thermal systems Design (열시스템설계)◈ Class hours : Mon 13:00 - 14:50, Weds 14:00 - 14:50◈ Open for : Senior ME students(3,4학년)◈ Class room : Rooms 606 (Mon & Weds)◈ Lecturer : Prof. Tae-Kuk Kim(김태국), Office : 02)820-5282

e-mail : [email protected] , HomePage : http://web.cau.ac.kr/energy◈ Materials : Reference 1 “열시스템설계및시뮬레이션(Thermal System Design

and Simulation” 김태국저, 도서출판 Intervision, 2000.Reference 2 “Principles of Energy Conversion”, A.W.Culp Jr., 1979.

1. Course descriptions and objectivesThermal systems are appeared in many fields of human life and in various types. These include the

plants for power generation and for industrial goods production. Also thermal systems are found in various household goods such as air conditioner, refrigerator, cooking appliances, heat supply systems, automobile systems and so forth. In this course some of the fundamental theories covered in other classes such as thermodynamics, fluid

mechanics and heat transfer are briefly introduced for summarizing and reviewing basic theories required for this class. Various types of energy systems utilizing the thermoelectric effect, the photovoltaic effect, the fuel cell, the MHD power generation etc are introduced by explaining the operating principle and applicability in daily life.

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Week Contents Ref.

1Course description(강의 소개)Thermal systems previews(열시스템 개요)Thermodynamic laws(열역학 기본 법칙)

Chap. 1Chap. 2

2 Cycles - Power cycles- Refrigeration cycles(Heat Pump) Chap. 2

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- Absorption refrigeration(흡수식 냉동)Other effects- Thermoelectric effect(열전기 효과)- Photoelectric effect(광전기 효과)- Fuel cell(연료전지)- Magnetohydrodynamic power(MHD 발전)

Chap. 2

4Heat transfer fundamentals(열전달 기초)- Basic heat transfer mechanism(기본 열전달 방정식: Conduction(전도), Convection(대류), Radiation(복사)

Chap. 3

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- Thermal resistance(열저항)- Overall heat transfer coeff.(총괄열전달계수)- Empirical correlations for convection heat transfer coeff.(대류열전달계수)Boiling heat transfer fundamentals(비등열전달 기초)- Basic concepts and empirical relations(기본개념 및 경험식)

Chap. 3Chap. 4

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Flow through piping systems(관로내부의 유동)- Pressure losses(관 내부 압력강하)- Piping systems design- Hardy-Cross method

Chap. 5

7 Properties of various media(여러 가지 매질의 물성)Data Curve Fitting

Chap. 6Chap. 7

8 Midterm exam(중간고사)

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Heat exchanger(열교환기)- Types of heat exchanger(열교환기의 형식)- Heat transfer area(전열면적)- Fouling of heat transfer area(전열면의 오염)

Chap. 8

10,11 - Overall heat transfer coeff. : heat exchanger(열교환기의 총괄열전달계수)- LMTD method. Chap. 8

12,13- NTU of heat exchanger- Effectiveness of heat exchanger- Effectiveness-NTU method.

Chap. 8

14 Shell&Tube heat exchanger design Chap. 9 <참고>15 Shell&Tube heat exchanger design Chap. 9 <참고>16 Final exam(기말고사)

3. Evaluation : subject to be changedAttendance(출석) 10점, Homeworks(숙제) 10점, Midterm(중간고사) 30점, Final(기말고사) 40점, Design(설계) 20점 : Total score 110점

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Chapter 1 Introduction

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1.1 Introducing remarks

• Modern human life style : in pursuit of convenience and comfort.→ Heat supply during winter→ Cooling by air-conditioner during summer

• Various types of energy systems used for office buildings, hospitals, residential buildings etc.

→ consumes large amount of electric power→ requires additional power plants

• Shortage in energy resources and environmental problems: needs efforts to minimize energy consumption and to use energy efficiently.→ requires advanced energy system design technology.

• Energy system : consisted of a variety of heat exchangers.


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1.1.1 Examples of energy systems1 ) Steam boiler

- utilizes the heat of fuel combustion in evaporating water- produces electric power through the Rankine cycle

Fig 1.1 An industrial package boiler with 2 drums


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Fig 1.2a Gas fired power boiler – small furnace


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Fig 1.2b Coal fired power boiler – large furnace


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2) Reciprocating Internal combustion engines- used for automobiles and ships- Gasoline engine : low compression ratio- Diesel engine : high compression ratio

(a) Gasoline engine (b) Diesel engine(Suzuki motor Co. Ltd.) (Wolkswagen motor Co. Ltd.)

Fig 1.3 Internal combustion engines


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3) Gas turbines- used for power generation plants, airplanes, ships, and automobiles

(a) Schematics of Turbo Jet (b) Turbo Jet for airplane (GE)

Fig 1.4 Gas turbines


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4) Refrigerators and air conditioners- for home and industrial purposes

Fig 1.5 Principles of vapor compression type refrigeration systems


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5) Nuclear reactors- very high energy density from nuclear fuel- applied for nuclear power plant, spacecraft, satellite, nuclear submarine

Fig 1.6 Schematics of a nuclear steam generator


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6) Solar energy utilization system- Energy from the sun : infinite amount but low energy density.Ex) Solar heat power plant (Fig 1.9a), solar cell power (Fig 1.9b), solar house(Fig 1.9c),

solar fish farm, green house etc.

(a) Solar heat power plant (b) ) Solar cell


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(c) Solar house

Fig 1.9 Solar utilization systems


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1.1.2 Types of energy and energy resources

1) Types of energy- Thermal energy(열에너지) : energy due to molecular or atomic vibration- Mechanical energy (기계에너지) : energy required for lifting an object- Electrical energy (전기에너지) : energy due to electric potential difference- Chemical energy (화학에너지) : energy from chemical material such as fuel- Electromagnetic energy (전자기에너지) : energy such as light or electromagnetic waves- Nuclear energy (원자력에너지 ) : energy from nuclear fusion or fission

2) Energy conversion- converting to a usable type of energy

Ex) Thermal energy → Mechanical energy : by enginesMechanical energy → Electric energy : by electric generatorsChemical energy → Electric energy : by electric cellsElectric energy → Thermal energy : by electric heater


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3) Categories of energy resources

- Energy resources stored in the earth : • Fossil fuels such as petroleum, coal, natural gas etc.• Geothermal energy stored as the magma in the mentle• Nuclear fission energy from uranium• Nuclear fusion energy from deuterium (heavy hydrogen)

- Energy resources from the outer space : • Intermittent(periodic) energy amount• Infinite amount and non-extinctive energy resources lasting forever• Mostly contains low energy density

예) Solar energy → solar heat and solar cell power generationLunar attraction → power generation by using the tidal flow

- Other types : • Intermittent(non-periodic) and variable energies such as wind power and hydraulic power


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1.2 Units

- Energy system design : usually the SI(System International) unit system is used.

- Newton’s 2nd law of motion : force on an object is proportional to the time derivative of the momentum (mass multiplied by velocity).

(1.1)

For constant object mass,

(1.2)

where, : acceleration

- Proportional constant in the SI(System International) unit system :

1.2 Units

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- SI unit systemForce : N (Newton)Mass : kgWork (Joule ) : work done in moving an object over a distance of 1[m] with the force 1[N].

(1.3)

- Heat content in calorie is converted as(1.4)

- Work done per unit time or heat transfer rate(1.5)

- Absolute temperature (Kelvin , ) and relative temperature (Celsius, ℃) scales are used in SI.(1.6)

- Pressure (SI unit : Pascal)

- Absolute pressure of the standard atmosphere

(1.7)

1.2 Units

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- SI units used for various physical quantities used in the thermal system design

Table 1.1 SI units used for various physical quantities

Force (Newton) Heat or work (joule)

Mass (kilogram mass) Power (watt)

Time (second) Thermal conductivity

Length (meter)Heat transfer coefficient

Temperature 또는 Heat capacity

Heat flux

1.2 Units

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Chapter 2 Thermodynamics review

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2.1 Introduction

2.1 Introduction

◆ Thermodynamics : study on the change of the system state due to heat or work transfer through system boundaries→ introduces fundamental laws required for the thermal system design

- law of mass conservation- law of energy conservation

→ concepts are applicable for the fluid mechanics and heat transfer

◆ Topics covered in this chapter- Introduction- Laws of thermodynamics- Thermodynamic cycles- Principles of thermo-electrics- MHD(Magneto-Hydro-Dynamics)- Fuel cells

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2.2 Laws of thermodynamics


2.2.1 Zeroth law of thermodynamics

: When the two systems A and B are at the same temperature with the third system C, the temperature of system A is the same as that of system B.→ This is the basic law for temperature measurement.

2.2.2 First law of thermodynamics

- A conservation law for mass or energy related to a system

1) For cyclic change of a closed system,

(sum of all work transfer (W) ) = (sum of all heat transfer (Q))→ indicates the energy conservation relation of a closed system.

(2.1)

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2) Definition for Energy (E)(2.2)

→ Eq. (2.2) is correct for an integrand which is a point function, and we call this new function as the energy (E) expressed as

(2.3)

where, (2.4)

내부에너지(internal energy, U)운동에너지 (Kinetic energy, K)위치에너지 (Potential energy, P)전자기에너지 (M)

3) The perpetual machine of the first kind which does not require energy input is impossible.


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2.2.3 Second law of thermodynamics

- First law of thermodynamics→ interested in the overall energy balance of the system.→ direction of the heat or work transfer is not addressed.

- Second law of thermodynamics→ direction of the heat or work flow is determined by this law.

1) For a system with no work input→ heat flows from the hot region to the cold region – Clausius statement

2) The perpetual machine of the second kind(a machine operating at thermal efficiency of 100%) which receives heat from a single reservoir is impossible - Kelvin-Planck statement.


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3) Carnot cycle : defines the highest limit of the thermal cycles operated between to thermal reservoirs.

→ Efficiency of the Carnot cycle is expressed by the temperatures of the two thermal reservoirs - Carnot.

(2.5)

4) Entropy of an isolated system is increased or unchanged→ principle of entropy increase.

2.2.4 Third law of thermodynamics

: A thermodynamic law for quantitative definition of the entropy→ sets the entropy of the perfect crystal at absolute temperature of zero K equal to zero.


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2.3 Energy equation by the 1st law

2.3 Energy equation by the 1st law: Energy equation for an open system

Fig 2.1 Energy balance for an open system

1

2

m.

hVz

q

W

E

m.zVh

1

1

1

2

2

2

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◆ For an unsteady system with an inlet with inflow and an outlet with outflow → inlet : 1 → outlet : 2

(2.6)

◆ For steady state,

(2.7)

2.3 Energy equation by the 1st law

/ 0dE dt

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2.4 Thermodynamic cycles


2.4.1 Performance of thermodynamic cycles

◆ Important application of the thermodynamics : continuous conversion from heat into work→ for this purpose the thermodynamic power cycle is considered

◆ Thermodynamic cycles are consisted by a series of processes as ;

- 등온과정 (Isothermal process) : T = constant- 등압과정 (Isobaric process) : p = constant- 등적과정 (Isometric process) : v = constant- 등엔트로피과정 (Isentropic process) : s = constant- 단열과정 (Adiabatic process) : Q = 0- 교축과정 (Throttling process) : h = constant

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선도 선도Fig 2.2 Examples of constant properties

p

v

T

s

s

T vp

= constant= constant

= constant

= constant

= constantp

= constantT

= constantv

s= constant


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◆ Thermal efficiency of a power cycle, → performance criterion for heat engines (power cycles)

(2.8)

◆ Coefficient of performance, COP, of a refrigeration cycle→ performance criterion for refrigerators or heat pumps (refrigeration cycle)

(2.9)


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2.4.2 Ideal gas power cycles

◆ Types of ideal gas power cycles

→ Standard cycles consisted of reversible processes and these are ;- Carnot cycle- Ericsson cycle- Stirling cycle

◆ Common features of ideal cycles→ result in a maximum thermal efficiency for those given high temperature(TH) and low

temperature(TL) reservoirs.

(2.10)

: high absolute temperature.: low absolute temperature.

→ all the processes of the cycles are consisted of the reversible processes


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◆ Reasons why the real thermodynamic processes become irreversiblei) Frictional lossii) Restricted expansioniii) Mixing of the operating media with othersiv) Heat loss


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◆ Types and compositions of ideal reversible gas power cycles

가) Carnot cycle

- 2 reversible constant temperature processes with heat input and out put- 2 reversible adiabatic processes (constant entropy process) with work input and output

〈T-s 〉〈p-v 〉Fig 2.3 T-s and p-v diagrams of the Carnot cycle.

H

max

L

min v

p

Isothermal

Isentropic

Isothermal1

2 3

4

s

T

ss

Isentropic

T

T

Isothermal

Isothermal

IsentropicIsentropic

1

23

4

smin maxs

TH

LT


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나) Ericsson cycle- 2 reversible constant temperature processes with heat input and out put- 2 reversible constant pressure processes with perfect regeneration

〈T-s 〉〈p-v 〉Fig 2.5 T-s and p-v diagrams of the Ericsson cycle .

H

L

LH

min

max

max

min

in

out

inout

REGENERATION

p

Isothermal

Isothermal

IsothermalIsothermal

IsobaricIsobaric

Isobaric

Isobaric

T

T

T

s

QQ

Q

Q

p

p

p

p

TT

1 12

23 34

4

v


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◆A perfectly regenerated Ericsson gas turbine cycle→ Heat rejected from the gas turbine is perfectly regenerated to heat up the turbine

inlet gas – nearly impossible to be realized.

Fig 2.6 A perfectly regenerated gas turbine system

out in

REGENERATOR

TURBINECOMPRESSOR

1

2 3

4

QQ

W


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다) Stirling cycle- 2 reversible constant temperature processes with heat input and out put - 2 reversible constant volume processes with perfect regeneration→ This cycle can be used as a basis of a real operable system

Fig 2.7 T-s and p-v diagrams of the Stirling cycle.

H

L

L

H

minmax

min

max

in

out

in

out

REGENERATION

v

Isothermal

Isothermal Isothermal

Isothermal

IsometricIsometric

Isometric

Isometric

T

T

T

s

Q

Q

Q

Q

p

v

v

v

T

T11

22

3

3

4

4

v


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2.4.3 Steam power cycles

◆ Steam power cycle→ Generates power while the operating medium changes its phase between vapor and liquid→ This cycle is called as the Rankine cycle.→ Typical operating medium : water, Freon, mercury, etc.→ Rankine cycle

: 2 constant pressure processes with heat input and output: 2 adiabatic processes(one for liquid pump and the other for steam turbine).

◆ Energy balance equation for the constant pressure process(boiler, superheater, reheater, condenser, etc) becomes ;

(2.11)

During this constant pressure process no work is involved ( ) and Eq. (2.11) is(2.12)

◆Adiabatic process (pump, turbine) : )(2.13a)

→ (2.13b)


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1) Basic Rankine cycle

Fig 2.8 Schematic drawing and T-s diagram of a basic Rankine cycle

T

s

1

2

3 4

Isothermal

Isothermal

Isentropic

Isentropic

SATURATION LINE

LOAD4

3

2 1

CONDENSER

TURBINE

BOILER

PUMP

Qout

WoutQ

in

Win

5

inQ

outQ

outWWin

5"condenced water"

"superheated steam"

"saturated water+saturated steam"


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: pump work per unit mass of operating medium.: boiler heat input per unit mass.: condenser heat rejection per unit mass.: turbine work generated per unit mass.

→ adiabatic expansion in turbine 4→5 : Since the operating medium in this process is in the state of saturated water and steam mixture, corrosion of the turbine parts can be worried.

→ Thermal efficiency of the Rankine cycle with no superheater,

(2.14)


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2) Rankine cycle with superheater: For the adiabatic process 4→5, low moisture content is required to prevent corrosion

Fig 2.9 Schematic drawing and T-s diagram of a Rankine cycle with superheater

◆ Thermal efficiency for the Rankine cycle with superheater ;

(2.15)

T

s

1

2

3

4

Isothermal

Isothermal

Isentropic

Isentropic

SATURATION LINE

LOAD4

3

2 1

CONDENSER

TURBINE

BOILER

PUMP

Qout

WoutQ

in

Win

5

inQ

outQ

outW

Win

5

SUPERHEATER

3'3'


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3) Rankine cycle with superheater and reheater: To improve the efficiency of the Rankine cycle and to reduce the moisture content in the operating medium → reheater can be applied for the inter-stage of the turbine.

Fig 2.10 Schematic drawing and T-s diagram of a Rankine cycle with superheater and reheater

◆ Thermal efficiency of the Rankine with superheater and reheater ;

(2.16)

T

s

1

2

3

4

Isothermal

Isothermal

Isentropic

Isentropic

SATURATION LINE

LOAD

4

3

2 1

CONDENSER

TURBINE

BOILER

PUMP

Qout

Wout

Qin

Win

5

inQ

outQ

outW

Win

6

SUPERHEATER

6

5

REHEATER


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<Ex 2.1> Efficiency of a Rankine operated by steam

- Boiler operating pressure = 300 psia- Turbine exit pressure = ①Atmospheric pressure (=14.7 psia),

② Condenser at vacuum(=2 psia)

①When turbine exit is maintained at atmospheric pressure (= 14.7 psia) ;

Fig 2.11 T-s diagram of a Rankine cycle with a condenser operating at atmospheric pressure

T

s

1

2

3 4

SATURATION LINE

5

inQ

outQ

outWWin

2 MPa

101.3 kPa

s fg


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From the steam table :

Saturated water at : : 엔탈피.: 엔트로피.

Condensed water at : : 엔탈피.

Saturated steam at : : 엔탈피.: 엔트로피.


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◆At point 5 the steam properties are estimated by using the heat of evaporation ;→ from the steam table at :

- Entropy change for different steam quality

→

- Enthalpy change for different steam quality

◆ Therefore, the efficiency is


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②터빈출구압력이진공상태 (= 2 psia) 로유지되는경우

Fig 2.12 T-s diagram of a Rankine cycle with a condenser operating at vacuum pressure

◆증기표로부터 :인포화수 : , 인응축수 :

인포화증기 : ,

<참고> 등엔트로피 펌핑 과정 (1→2)에서는 이므로여기서,

T

s

1

2

3 4

SATURATION LINE

5

inQ

outQ

outW

Win

2 MPa

101.3 kPa15 kPa

1

2


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◆위치 5에서의물성값은증기표로부터직접구할수없음→ 증발잠열을이용하여산출

증기표로부터 에대하여,

혼합유체의엔트로피변화 :

→

혼합유체의엔탈피변화 :

◆따라서,

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ <끝>


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4) Rankine cycle with regenerator- a portion of the steam flowing in the turbine is extracted for heating the feed water

(regeneration process)

Fig 2.13 Schematic drawing and T-s diagram of a Rankine cycle with regenerators

T

s

1

2

SATURATION LINE

LOAD8

7

1

CONDENSER

TURBINE

BOILER

PUMP

Qout

WoutQ

in11

HEATER

1 kg

x kg y kg 1-x-y kg

23456

9 10

1 kg

x kg

y kg

1-x-y kg

34

5

67 8

9

10

11

1-x kg 1-x kg


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◆ The extraction steam flow x is determined for the state at point 5 to be the saturated water.And y is determined for the state at point 3 to be the saturated water at that pressure.

→ x and y are determined from the energy balance equations at the corresponding pointsi.e. (heat provided by the extracted steam) = (heat received by the feed water)

At point 5 : (2.17a)

→ (2.17b)

At point 3 : (2.18a)

→ (2.18b)


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◆ Pump work(2.19)

◆ Turbine work(2.20)

◆ Heat supplied by the boiler(2.21)

◆ Heat rejected by the condenser(2.22)

◆ Thermal efficiency ( )

(2.23)


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<Ex 2.2> Rankine cycle with a 1st stage steam extraction

Boiler pressure : 300 psiaTurbine exit pressure : 2 psiaSteam extraction pressure : 20 psia

Fig 2.14 Rankine cycle with a 1st stage steam extraction

T

s

1

2

SATURATION LINE

1 kg

x kg

1-x kg

3

45 6

7

8

2 MPa

150 kPa

15 kPa


50


> From the steam table :인포화수 :

인응축수 :

인포화수 :

인응축수 :

인포화증기 :


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◆ Properties at point 7 are determined by considering the heat of eqaporation

At :

→

◆ Extraction flow

→


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◆ Properties at 8 can be estimated by considering the mixture fraction of steam and water→ heat of evaporation is considered for the saturated water and saturated steam

From the steam table at ; ,

And, →

◆ Cycle efficiency :

━━━━━━━━━━━━━━━━━━━━━━━━━━━━ <끝>


53


5) Actual Rankine cycle- 터빈의비등엔트로피팽창→이론적인싸이클일보다작은일을얻게하는요인

- 비등엔트로피팽창의원인→ 마찰손실, 난류손실등과같은현상

Fig 2.15 Entropy changes within water pump and steam turbine

T

s

3

4

4'

1

22'


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이상적인터빈일 = 실제의터빈일 =

여기서,

터빈의효율 :(2.24)

- 또한, 펌프의비등엔트로피압축(그림2.15 참조) 현상→ 마찰손실, 난류손실등으로인하여발생

펌프의효율 :(2.25)


55


2.4.4 Gas power cycles

- 가스동력사이클은내연기관의기본사이클

1. 오토사이클(가솔린기관) → spark 점화, 왕복엔진

2. 디젤사이클(디젤기관) → 압축점화, 왕복엔진

3. 브레이튼사이클(가스터빈) → 가스터빈엔진

→ 연소에의한고압고온의연소가스를팽창일로써동력을발생시키는사이클.


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1) Otto cycle (Fig 2.16)2개의과정 : 등엔트로피과정( =상수)2개의과정 : 등적과정( =상수)→열을출입시킨다.

0→1 ; 연료가스와공기가실린더내에흡입1→2 ; 혼합기체(연료+공기)가피스톤에의하여압축2→3 ; 혼합기체는점화기에의하여점화되고, 연소가진행3→4 ; 고온연소가스가팽창하여피스톤을밀어냄4→1 ; 배기밸브가열리고, 연소가스가대기로방출됨1→0 ; 피스톤이배기가스를밀어냄.

Fig 2.16 T-s and p-v diagrams of Otto cycle.

in

in

ou t

T

s

Q

Q

Q

p

11

22

33

4

4

v

0

s

Q out

v팽 창

점 화

압 축

배 기

= constan t

= constan t

= constan ts= constan tv


57


- 이러한과정들은 4개의행정으로구성→ 4 행정엔진(Four-stroke engine).→ 4개의행정동안 2번의피스톤왕복또는 2번의크랭크축회전이발생.

- Otto 사이클 : Spark 점화엔진(Otto cycle)

- 효율•등적과정( )에열역학제 1 법칙을고려하면;

(2.26)•따라서, Spark 점화엔진(Otto cycle)의효율은;

(2.27)


58


- 따라서, Otto 사이클의효율은압축비( )만으로표시가능;

(2.28)

여기서, 압축비 의정의는

그리고비열비 는


59


그림 2.17 비열비 및압축비 에따른 Otto 사이클의효율( )

그림 2.17에서 ; - 비열비( )가증가하면→ 가증가- 압축비( )가증가하면→ 가증가

·가솔린기관의연소가스에대한비열비→ 공기의비열비인≒1.4로고려함

· 압축비→ detonation을유발하기때문에너무높게할수없음. * detonation : 피스톤이상사점에도달하기전에급속한연소가발생하는이상연소현상

엔진은 knocking을하기시작함.

0.2

0.3

0.4

0.5

0.6

4 5 6 7 8 9 10

= 1.4

= 1.3

= 1.2

rv


60


<Ex 2.3>━━━━━━━━━━━━━━━━━━━━━━━━━━

이상적인 Otto 사이클 (그림2.18 참조)ㅇ. 작동유체 : , ㅇ. 압축비 8 , 1200ㅇ. ㅇ. →

그림 2.18 Otto 사이클.→ 열효율?

1

2

3

4

v = v2 3 v = v1 4

v

등엔트로피

등엔트로피 등적

등적

p


61


Sol) 열효율

: 압축비를이용하여

━━━━━━━━━━━━━━━━━━━━━━━━━━━━<끝>


62


2) Diesel cycle (compression ignition) (Fig2.19)

1개의과정 : 등압과정 (흡열과정)1개의과정 : 등적과정 (방열과정)2개의과정 : 등엔트로피과정 (가역단열과정)

Fig 2.19 T-s and p-v diagrams of Diesel cycle

in

out

T

s

Q

Q

p

11

2

2 33

4

4

v

0

(점화)

(압축)

(배기)

등압

(흡입)

(배기)

등엔트로피

연료분사완료연료분사시작

등엔트로피

등압

등적


63


- 등적과정 1-4 사이의방열량→ 열역학제 1 법칙에의하면;

(2.29)- 등압과정 2-3 사이의입열량→ 열역학제 1 법칙을적용하면;

(2.30)

- 따라서, Diesel 사이클의효율은;

(2.31)

여기서 , : 비열비


64


(2.32a)

여기서, 는 cut-off ratio→ 등압상태로연료가연소되기시작하는때의비체적에대한연소가완료되는때의비체적의비를나타냄

따라서,

(2.32b)

→ 이어야함 : 연료를분사하는시간이필요하기때문.


65


- 따라서 Diesel 싸이클의효율은 ; → 압축비 및 cut-off 비 의함수로표현가능.

(2.33)

그림 2.20 Cut-off ratio 및압축비 의변화에따른이상적인 Diesel사이클효율(비열비 에서)

- 압축비( )가증가 : 효율 증가- Cut-off 비( )가증가 : 효율 감소

0.4

0.5

0.6

10 12 14 16 18

r = 1.1cr = 2.0cr = 3.0cr = 4.0c

r > 1.0c

rv

= 1.35


66


- Otto 싸이클과 Diesel 싸이클의효율비교;

(2.34)

(2.35)

식 (2.52)에서;이므로 :

→ 주어진압축비에대하여 Diesel사이클은 Otto사이클보다효율이낮다.→ Diesel사이클에서는 Otto사이클과는달리공기만을압축하기때문에압축

도중에이상연소(detonation)의염려가없다. : Diesel 사이클은높은효율을얻도록높은압축비에서작동시킴.→ 일반적으로

: Otto 싸이클은: Diesel싸이클은


67


3) Brayton cycle(gas turbine cycle)2개의등압과정

: 하나는흡열과정→연소기: 하나는방열과정→대기방출

2개의등엔트로피과정: 하나는등엔트로피압축→compressor: 하나는등엔트로피팽창→터빈

Fig 2.21 Schematics of a gas turbine system1

2 3

4

Work

연소기

압축기 터빈


68


Fig 2.22 T-s and p-v diagrams of a Brayton cycle

2 3

41 12

3

4

P T

s

Qin

Qin

outQ outQ

v

등압(흡열)

등엔트로피(팽창)

등압 (방열)

등엔트로피(압축)

등압

압축기

터빈

등압


69


- Brayton 싸이클의효율은

(2.40)

- 또는, Brayton 싸이클의효율은 ; 압축기의압력비만의함수로표현가능

(2.41)

여기서, : 압축기입구및출구의압력비- 가증가하면→ 가증가

→ 압축기일의증가→ 압축공기의과열에의한압축효과의감소를초래

: 중간냉각기(inter cooler) 설치가요구.


70


- Brayton 사이클에서주어진최고온도 와최저온도사이에서최대의출력을얻을수있는압력비( )→ 최고온도 를고온에노출되는가스터빈부품이견딜

수있는온도로유지하고,→ 최저온도 을대기온도로→ 그리고최저압력 을대기압으로가정.

그림 2.23 Brayton사이클의최대출력

T 3

T 1

T

s

2

3

412

3

4

2

3

4

P 1

최대출력(면적 1-2''-3''-4''가 최대)

(압축기 입구압력)


71


- Brayton 사이클의압력비는다음과같이정의;

(2.42)

- 여기서, → 이면 는최대가됨.

(2.43)

- 그러나, → 이면 T-s 선도에서면적(1-2-3-4)이 0으로되므로사이클의유효일은 0이됨.

- 또한, 또한인경우(압축기에서공기를압축하지않는경우)에도 T-s 선도상의면적(1-2'-3'-4')는 0으로되고, 유효일을하지않음.

- Brayton 사이클의유효일량( )은→ 터빈일에서압축기소모일을제외한값

(2.44)


72


- 여기서, 이므로;

(2.45)

- 최대출력을내는 값(T-s선도에서면적 가최대인경우의 값)은다음조건을만족하는값이됨;

(2.46)

- 따라서, Brayton 싸이클의최대출력을나타내는 값은;

(2.47)


73


2.4.5 Refrigeration cycles

- 냉장또는냉동 : 어떤공간을주위의 (대기)온도보다낮은온도가되도록하는것.

→ 얼음등을이용하거나, → 또는작동유체(냉매)에일을가하여저온영역에서고온영역으로열을이동시기는열역학적냉동사이클을활용함으로써가능.

- 냉동사이클의성능계수(COP , Coefficient of performance):→ 냉동기의성능을나타내는지표로서다음과같이정의


74


1) Carnot refrigeration cycle→ 최고의 COP 를갖는이상적인가역냉동사이클.

Fig 2.24 Carnot refrigeration cycle

- 저온열원으로부터제거되는열량은;

(2.48)- 단열압축과정 1→4 (등엔트로피과정)에서동안에싸이클에공급되는일은 ;

(2.49)

T high

low

out

abstracted

T

Q

Q

Win

T

s

3 4

2 1

A B

등온 (방열)

등엔트로피등엔트로피

등온 (흡열)

팽창압축


75


- 따라서, Carnot냉동싸이클의성능계수는 ;

(2.50)


76


2) Vapor compression refrigeration (or vapor compression heat pump) cycle→ 사이클에일을공급하여냉동(또는 heating)효과를얻으며실용적인사이클.

- 냉동효과는증발기에서냉매가증발될때필요한증발잠열을주위에서흡열하는현상을통하여얻음.

Fig 2.25 Vapor compression refrigeration system

valve Work

inQ

outQ3

4

2

1

교축압축기

증발기

응축기


77


Fig 2.26 T-s and P-h diagrams of the vapor compression refrigeration cycle

- 그림에서과정 1→2 : 등엔트로피압축(압축기)과정 2→3 : 등압방열 (응축기)과정 3→4 : 교축과정(등엔탈피과정) : 교축밸브(throttling valve)과정 4→1 : 등압흡열(증발기) : 냉동효과를얻음

Qin

1

2

3

4 4 1

23

T

s

P

h

Qout

Win


78


- 증기압축냉동기의성능계수는 ;

(2.51)

- 한편, 동일한싸이클에서응축기의방출열을이용하는 Heat pump의성능계수는 ;

(2.52)


79


<예 2.4> ━━━━━━━━━━━━━━━━━━━━━━━━━

증기압축냉동시스템 (그림2.27 참조).용량 : 8tons , Freon-12 작동매체응축기온도 : 95℉ ← ; 증발기온도 : 32℉ ←

From table(Freon-12):, ,

:

,

Fig 2.27 Example of a vapor compression refrigeration system━━━━━━━━━━━━━━━━━━━━━━━━━━━ <끝>

P=const

3 4

T

s

1

2

3

4

1 및 3 점 : 포 화 상 태1 2 : 등 엔 트 로 피 압 축1 4 : 등 에 트 로 피 교 축

( h = h )


80


3) Absorption refrigeration system→ 사이클에일대신에열을공급함으로써냉동또는 heating효과를얻는장치

Fig.28 Schematics of an absorption refrigeration system

Qcon

Qeva

Qgen

Qabs

10

9

8

6 7

4 5

321

열교환기

용액펌프

감압밸브

냉각수

흡수기

냉매증기

Steam 으로

가열

고압

저압

응축기

교축

증발기

발생기


81


- 흡수식냉동시스템의작동원리 ;

1→3 과정 : 증발기로부터흡수기에공급된냉매2→3 과정 : 발생기에서농도가떨어진용액이열교환기와

감압밸브를통과하여흡수기에공급된용액(6→4→2 과정 : 냉매농도가낮은용액)

→ 흡수기에서냉매가용액을흡수하면서흡수열발생→ 냉각수로서흡수열을제거

3→5→7→8 과정 : 냉매의농도가높은용액이펌프와열교환기를통과하여발생기로보내진다. 발생기에서는이러한용액이가열되어비등점이낮은냉매는증발되어응축기로보내지고 ,비등점이높은용액은다시열교환기ㆍ감압밸브를통하여흡수기로보내진다.

9→10 과정 : 교축밸브를통과하면서등엔탈피로감압. 감압된냉매는증발기에서증발되면서주위로부터증발잠열을흡열한다.


82


- 흡수식냉동기의성능계수는 ;

(2.53a)

- 또한, 흡수식 Heat Pump의성능계수는 ;

(2.53b)

- 흡수식시스템의작동매체 ;→ 2 성분유체(또는다성분매체)로고려

ex) 1. 2. 3. 4. 등등ㆍㆍㆍ


83


2.5 Power generation other than cycles2.5 Power generation other than thermodynamic cycles→ 열사이클을따르지않는전기발생장치→ Carnot사이클의효율에제한을받지않음.

2.5.1 Thermoelectricity- 열전기효과는 ; → 열전대온도계

→ 열전기발전장치→ 열전기냉동장치등에응용

i) Seebeck effect : 두개의다른물질이연결되어환을이룰때이들의두접점들이각각다른온도로유지되면전위차가발생(그림 2.29 참조).

→ 전위차와온도차사이에비례관계성립 : 비례상수를 Seebeck상수.

(2.54)

Fig 2.29 Ring type connection of two different wires

V

T hot coldT

물 질 1

물 질 2

84


Table 2.1 Seebeck constants for various materials

ex) Combined Seebeck constantㅇ. Iron-Constantan→ (13.6+47.0)×E-6 =60.6

ㅇ. Germanium-silicon → (375.0+455.0)×E-6 =830

물질

금속

Aluminum -0.2 E-6

Constantan -47.0 E-6

Copper +3.5 E-6

Iron +13.6 E-6

Platinum -5.2 E-6

반도체Germanium +375.0 E-6

Silicon -455.0 E-6

2.5 Power generation other than cycles

85


ⅱ) Pelier effect : 두개의다른물질이연결되어있는환에전류를통과시키면연결점에서는열을흡수또는열을발생한다. (그림 2.30 참조)

그림 2.30 두가지물질로연결된환에가한전류

→ Peltier 효과는 Seebeck 효과에반대되는효과→ 통과시킨전류의량과열량은서로비례: 비례상수를 Pelier 상수

(2.55)

ⅲ) Thomson effect : 한가지전도체가환을이루고있을때그환의일부에온도차가주어지고동시에전류 를흘리면열을흡수또는방출.

→ 는흡수또는방출되는열량 에비례 : 비례상수 를 Thomson계수. (2.56)

Q

물질 1

물질 2


86


2.5.2 MHD (Magneto-Hydro-Dynamics) power generation→ Faraday 효과를이용한전기발생장치→ 유체의 kinetic(운동)에너지를전기에너지로직접전환→ 열역학싸이클에의존하지않는발전방식(그림 2.31참조).

Fig 2.31 Principle of MHD power generation◆ Principle of MHD power generation;

- 전기전도도가큰유체의유동속도벡터 : - 유체의흐름과직각방향으로자장(자속밀도 )을걸면→ 유체의흐름방향및자속의방향에각각직각이되는방향으로전류( )가발생(전위차 를형성).

BJ

VJ B

V전기전도도가 큰 유체의 흐름

전극 (+)

전극 (-)

부하 (+)

(-)


87


◆ Farady 법칙;

(2.57)◆ Ohm의법칙

(2.58)

여기서, : 유체의전기전도도(electric conductivity).

◆ MHD발전장치에서많이사용되는작동유체의종류및특징1. 액체금속(수은,용융나트륨):→높은전기전도도를갖음.→유동저항때문에를크게하기어렵다.

2. 이온화가스(고온의연소가스):(gas plasma)→전기전도도가액체금속에비하여낮다.

(보통전기전도도를높이기위하여첨가제를섞는다.)→유동저항이적으므로를크게할수있다.

◆고온의이온화가스를이용한 MHD발전과 Rankine사이클의복합시스템은이론적으로총괄효율이 50～60%에이른다.


88


2.5.3 Fuel Cell→ 연료전지는건전지와같은원리→ 화학적에너지로부터전기에너지를발생(60～80%의효율)→ 건전지는아연,납,수은등의비싼연료를이용하며일정량의화학에너지를저장→ 연료전지는연속적으로값싼연료를공급하여화학에너지를전기에너지로

전환하는시스템

◆연료전지에서많이사용되는연료 : 수소 , 일산화탄소등: 음극을띠는다공질전극(porous electrode)에공급

◆산화제 : 산소또는공기: 양극을띠는다공질전극에투입.

◆연료전지의전극은다음과같은특징을갖는것이요구됨.①연료및전해질이통과될수있는다공질이어야한다.②다공질전극의미세한구멍들은적당한크기이어야하며,이구멍들의크기가연료전지의성능을좌우한다.

③다공질전극의재료는공급되는연료가미세구멍들을통과할때반응이쉬운전자상태로되도록촉매역할을하는재질이어야한다.ex)백금, 소결된니켈

④전극은전기전도도가높아야한다.


89


◆작동원리→ 연료또는산화제는각전극의다공부를통과하면서이온화하게되고

각각의이온들은상대방전극으로전해질을통하여이동: 기전력의차이가발생(그림 2.32 참조).

→ 두전극사이의전해질은 이온과 이온을쉽게상대되는전극으로통과시킬수있는용액으로서 수용액이많이사용.

(1) 이온의반응(음극에서양극으로이온이흐름)

음극에서의반응 : (2.59)

양극에서의반응 : (2.50)

(2) 이온의반응(양극에서음극으로이온이흐름)

음극에서의반응 : (2.51)

양극에서의반응 : (2.52)


90


Fig 2.32 Schematics of a fuel cell system

2H O2

H2 O2e4 - e4 -

H4 + H4 +2H2

공급

부하

다공질전극

음극 양극공급

제거

"전해질"

(-) (+)


91


<Chapter 2 Problem set>2.1 In Rankine cycle what effect can be obtained from the following processes?

① Superheat process ② Reheat process ③ Regeneration process

2.2 For Brayton cycle, explain those methods for improving efficiency.

2.3 Answer the questions for the following Rankine cycle with superheater and regenerator.

Fig 2.33 Steam turbine system with superheater and regenerator

When the minimum and maximum system pressures are maintained at and compare the system efficiencies for three different steam extraction pressures of .

T

s

12

3

4

SATURATION LINE

LOAD

95

1

CONDENSER

TURBINE

BOILER

PUMP

Qout

Wout5

1 kg

PUMP

HEATER

234

7

8

1 kg

x kg

1-x kg

7

8

9

1-x kgx kg

6

(500 )o F260 o C

2 MPa (21.09 at)

150 kPa (1.406 at)

15 kPa (0.1406 at)

s1 s3 s7

2 MPa

150 kPa15 kPa

Chapter 2 Problem set

92


2.4 Answer for the following Brayton cycle.

(1) Compressor inlet air : 65℉, 14.7Compressor exit air : 70Maximum allowable cycle temperature : 1600℉ (to prevent material failure)

1) Draw T-s and P-v diagrams.2) Find unknown temperatures and pressures at each points.3) Find available heat rate.4) Find cycle efficiency.5) Find the ratio of the compressor deriving work from the total turbine work.6) Find the pressure ratio ( ) of the compressor and the cycle efficiency at the maximum power of the

Brayton cycle operated between 65℉ and 1600℉.

(2) Draw an efficiency chart of the Brayton cycle for different pressure ratio ( ) and heat capacity ratio( ) and explain the results for the ranges of and .

2.5 For fuel cells, search related materials through various sources including library, internet sites, etc. And explain the principles of the fuel cells, categorize the types, list some application examples, and comment on the future developments.

2.6 For MHD power generation, search related materials through various sources including library, internet sites, etc. And explain the principles of the fuel cells, categorize the types, list some application examples, and comment on the future developments


93


Chapter 3 Fundamentals of Heat Transfer

94


3.1 Introduction

3.1 Introduction◆ Thermodynamics and heat transfer

- Heat transfer : studies the physical phenomena of energy transfer due to the temperature difference→ interested in both the heat transfer rates and the time required for the heat transfer→ interested in the intermediate processes in reaching at the equilibrium states.

- Thermodynamics : studies the change of a system at an equilibrium state to another equilibrium state due to the heat and/or work transfer.→ the time required for the energy transfer is not asked.→ interested in the final equilibrium state after an infinitely long time.

- In heat transfer analysis we use the basic laws of thermodynamics→ 1st law of thermodynamics (heat and mass balance laws)→ In the heat transfer analysis, the thermodynamic laws are applied for a system at a quasi-

equilibrium state.

◆ Contents covered- Introduction of the heat transfer laws- Determination of the heat transfer coefficients- Estimation of the heat transfer rates.

95


3.2 Fundamentals of heat transfer


3.2.1 Fourier’s law of conduction

◆ Heat transfer mechanism by conduction- Heat transfer phenomena by molecular vibration.- Heat transfer due to a temperature gradient within the matter without mass transfer.- Needs heat transfer medium.

◆ Fourier’s law of conduction

- 전도열전달의관찰결과 : 열유속(heat flux)과온도구배가비례

(3.1)- 비례상수를도입하면;

: Fourier 의열전도법칙 (3.2)

- 식 (3.2)에서 (-)부호는온도구배가항상 (-)값을갖기때문→ x 증가 : T는감소, 따라서 는 (-)값이됨.

96


(a) 1 차원평판 (b) 평판내의온도분포그림 3.1 열전도의개념도.

T

q

dx x

Tx

dx

k

q


97


- 비례상수 : 열전도계수또는열전도율(thermal conductivity)→ SI 단위 : 또는→ 두께 1m인평판의양면이 1 의온도차로유지될때단위시간당전달되는열량으로정의.

→ 단위면적당의열전달율( )과 만큼떨어진두점에서의온도차 를각각측정함으로써산출

(3.3)

→ (3.4)

그림 3.2 평판에서의 1차원열전도 .

T1 T2q

x


98


표 3.1 Conductivities of typical matters,

Metals

Copper(구리) 385

Aluminium 202

탄소강(1%C) 43

Non-metals

Quartz 41.6

유리 0.78

Glass wool(보온재) 0.038

Liquids

수은 8.21

물 0.556

Freon12 0.073

Gases

수소 0.175

공기 0.024

수증기(포화) 0.0206

0.0146


99


Fig 3.3 Conductivities of various solid matters


100


a) Gases b) Saturated liquids

Fig 3.4 Temperature dependency of conductivities for various gases and saturated liquids


101


Table 3.2 Conductivities of insulators

Types of insulators Density Conductivitry Applicable temperature range

Filled gases Remarks

Glass wool (board) 40 573 K 이하 공기

Glass wool (roll) 32 " 공기

Glass wool (단섬유, 4.7μ) 166 그림 3.5 " 공기

Foam glass 200 73～373 K

Foam glass 154 그림 3.5 " 공기

염기성 탄산마그네슘 300 그림 3.5 ～540 K

Castable 내화물 (90)

(1)～1300 K

Castable 내화물 (27)

(51)"

고알루미나 내화벽돌 1120 ～1900 K

Foam polystylene 30 120～300 K

" 15.9 "

경질 우레탄 foam 30～35 그림 3.5 150～350 K 프레온11

" 30～35 273～350 K 프레온11

염기 foam 70 ～343 K

염기 foam 30～70 "

Phenol foam 30～40 223～393 K


102


Fig 3.5 Conductivities of thermal insulators

100 200 300 4001

2

3

4

5

6

7

8

K

W/(m

K)

발포

glass

염기성 탄산마그네슘

Glass wool

경질 우레탄 foam


103


◆정상상태의 1차원열전도→ 넓이에비하여두께가상대적으로매우얇은매질

1) 무한히큰평판의두께방향으로만온도변화가있는경우두께방향으로의열전달, .

2) 온도구배가방향으로만있는무한히긴원통의방향으로의열전달,

3) 온도구배가방향으로만있는구의방향으로의열전달, .


104


1) Heat transfer through a plane wall

Fig 3.6 Heat transfer through an infinitely large plain plate with thickness L.

- For constant thermal conductivity ;

(3.6)

k

s,1T

s,2T

Lx

(x)


105


- Introducing the thermal resistance ;

(3.7)

[ ] [ ] : conduction thermal resistance

- Analogy between the heat transfer and the electric resistor circuit ;

Table 3.3 Comparison between the heat transfer and the electric current flow.

Fig 3.7 Conduction thermal resistance

Heat transfer Electric circuit

Heat flux ( ) Curreent ( )

Temperature difference( )

Voltage difference( )

Thermal restance( )

Electric resistance( )

Ts,1

Ts,2q

q

RtTs,1 Ts,2


106


2) One-dimensional cylindrical system→ 실린더에서의 1차원열전도 : 반경방향으로의열전달

Fig 3.8 One-dimensional cylindrical system.

T=T(r)

-

r

x

+q

q

Ts,2

Ts,1

r2

r1 r

dr

q q

r+dr

r+dr

r

r


107


- 반경방향으로의열유속 은 ;

(3.8)

여기서, 실린더의길이 : 반지름 인위치에서의단면적 :

- 전도열저항을이용하여나타내면 ;

(3.9)

여기서, : 원통시스템에대한전도열저항

Fig 3.9 A cylinder with single layer


108

Ts,2

Ts,1

r2

r1

q

qTs,2Ts,1

Rt

r

r


3.2.2 Convection heat transfer→ 대류열전달은물질의이동에의하여수반되는열전달현상→ 매질내에온도차가존재할때유동은열전달을야기→ 대류열전달은열을이동시키는전달매개체가반드시필요

: 진공속에서는대류열전달이발생될수없음.

◆Newton's law of cooling→ 대류열전달을나타내는기본법칙

- 물리적현상 ;(온도 인물체주위에열경계층(thermal boundary layer)을형성하면서흐르는온도 인유체에의한열전달량) ∝ (유체와벽면의온도차이에비례)

Fig 3.10 Thermal boundary layer near an object surface.qTw

Tflow

boundarylayer


109


- 수학적표현 ;

1) 인경우 :

Fig 3.11 Heat transfer within the thermal boundary layer ( )

(3.10)

where, : average convective heat transfer coefficient → determined according to the fluid properties and the flow conditions

Tw q

T


110


2) 인경우 :

Fig 3.12 Heat transfer within the thermal boundary layer ( )

(3.11)

Tw q

T


111


- Local heat transfer rate over an infinitesimal surface area is ;

(3.12)where, : local convection heat transfer coefficient

Fig 3.13 Local heat transfer surface

- Total heat transfer rate over the whole surface is ;

(3.13)

h

T

T dA

s


112


- If is assumed to be constant ;

(3.14)

where , : average heat transfer coefficient over the whole surface determined as

(3.15)

* Generally almost all of the convective heat transfer coefficient data appeared in the literature are the averaged values.


113


- Classification of the convective heat transfer

· Forced convection : convective flow induced by external force.

· Natural convection or free convection : convective flow induced by the buoyant force.

· Boiling or condensation heat transfer : latent heat involved in the phase change

Table 3.3 Typical convection heat transfer coefficient .

Natural convectiongases 2～25

liquids 50～1000

Forced convectiongases 25～250

liquids 50～20000

Boiling or Condensation 2500～100000


114


- Introducing the convective thermal resistance ;

(3.16)

where서, [ [℃/W] : convective thermal resistance (3.17)

Fig 3.14 Convective thermal resistance.

Ts,1

q

Rt

hTs,1 T

T


115


- Empirical correlations for the convective heat transfer

→ Mostly expressed by using the Reynolds No. ( ) and the Prandtl No.( ) as

(3.18)

where, , , are determined by using the experimental data.


116


1) Empirical correlations for convection by internal tube flows

-Velocity profile for the internal tube flow

Fig 3.15 Velocity profile for the internal tube flow

- Average fluid velocity ( ) and average fluid temperature ( )

(3.19)

(3.20)

Laminar sublayer

Turbulent core

u


117


- Conditions for fully developed flows1) Conditions for fully develop velocity profile :

: laminar flow (3.21a)

: turbulent flow (3.21b)

2) Condition for fully developed temperature profile :(3.22)

Fig 3.16 Entrance region of a tube

dL i


118


Table 3.4 Empirical correlations of convection heat transfer coefficient and friction factor for internal tube flows

(주) 표 3.4에주어진수식을사용할때물성값은유체의평균온도 으로결정된값을사용하며,Reynolds수( )는관의단면이원형이아닌경우수력지름( )을이용하여산출하여야한다.

관계식 연구자 적용 조건층류유동 이론 완전히 발달된 층류유동, constant ,

이론 완전히 발달된 층류유동, constant ,

또는,

Hausen

Sieder &Tate

층류의 입구영역( 또는 비가열 입구부)일정한

층류상태의 입구 영역 ( 일정)

,난류유동 (±25% error)

또는,

(±25% error)

(±10% error)

Dittus &Boelter

Sieder &Tate

Petukhov

완전 발달 난류유동, ,

: ; :

완전 발달 난류유동, ,

완전 발달 난류유동관마찰계수 :n = 0.11 : , n = 0.25 ;n = 0.0 : =constant or gas flow

액체금속 Skupinski등 액체금속의 완전히 발달된 난류유동( 일정),

Shimazaki등완전 발달된 난류 액체금속 유동( 일정),

현재 이 이미지를 표시할 수 없습니다 .


119


- Heat transfer coefficient for noncircular tubes → Hydraulic diameter ( ) is used instead of the tube inside diameter ( ) while the correlation for circular tubes are used.

(3.23)

- Examples of calculating the hydraulic diameter ;i) For a rectangular tube :

Fig 3.17 Rectangular tube.

Area ( ) , wetted perimeter ( )

- Hydraulic diameter ; (3.24)

- For a square tube ( ) ; (3.25)

a


120


ii) For a concentric double tubes :

Fig 3.18 Concentric double tubes

- For estimation of the pressure loss between two tubes→ fluid friction occurs at surfaces with diameter and

(3.26)

d

D


121


- For estimation of the heat transfer between the inner tube fluid and the fluid between two tubes → for determining the heat transfer coefficient on the outer surface (diameter d) of the inner tube which is the reference area for the heat exchange

(3.27)

* For tubes with non-circular cross sections, the hydraulic diameter ( ) determined as above can be used for those empirical correlations developed for circular tubes shown in Table 3.4 .


122


<Ex 3.1>━━━━━━━━━━━━━━━━━━━━━━━━━

Determine the convective heat transfer coefficient for the following internal tube flow.

Fig 3.19 Het transfer within a circular tube

ㆍDetermination of fluid (water) properties

T

T

W C

Water(1 atm)

= 150

= 20 cm

= 50

= ?= 1 m/s

h

C

vdi

i


123


ㆍReynolds Number

→ flow is turbulent.

ㆍCorrelation for fully developed turbulent flow (since water is cooled ; n=0.3)

⇒

━━━━━━━━━━━━━━━━━━━━━━━━━━━━<end>


124


2) Empirical correlations for external tube flows.→ for different tube arrangements and flow conditions

i) Cross flow across a tube bank : Correlation suggested by Zukauskas

◆ Heat transfer coefficient

(3.28)

-.

-. All the fluid properties are determined at (but is determined at )

-. (3.29)

-. : the maximum fluid velocity through the tube bank→ determined according to the different tube arrangements


125


ㆍAligned (A1 in Fig 3.20) ;

(3.30)

where, : fluid velocity before entering the tube bank

ㆍ Staggered : occurs at the smaller flow area among A1 and A2

(3.31a)

(3.31b)


126


Fig 3.20 Types of tube arrangement

(in-line)

S

D

S A

V,T

Aligned tubes

T

L

1

o

o

Staggered tubes(zig-zag)

S

V,T oo

S T

L

A1

D

2A

S D


127


-. C and m values : according to the tube arrangement and Reynolds number ;ㆍAligned ;

ㆍStaggered ;

-. For tube rows less than 20 (N<20) ; Correction factor( ) needs to be multiplied as

(3.32)

Table 3.5 Correction factor ( ) for tube rows less than 20.Rows # 2 4 6 8 10 12 14 16 18 20

Aligned 0.80 0.90 0.95 0.97 0.97 0.98 0.98 0.99 1.00 1.00

Staggered 0.77 0.89 0.94 0.97 0.97 0.98 0.98 0.99 1.00 1.00


128


◆ Frictional loss-. Pressure drop through tube banks [Zukauskas].

(3.33)where, : number of tube rows

and : determined from Figs. 3.21 and 3.22 .

Fig 3.21 Friction factors and correction factors for aligned tube arrangements


129


Fig 3.22 Friction factors and correction factors for staggered tube arrangements


130


ii) Heat transfer coefficients for natural flow around a horizontal circular tube(Fig 3.23)

(3.34a)

(3.34b)

where, : convection heat transfer coefficient, [ ], [℃]

: outside tube diameter , [ ]

Fig 3.23 Natural flow around a horizontal circular tube.

TW

ho

T


131


iii) Heat transfer coefficients for natural flow around a vertical circular tube(Fig 3.24)

(3.35a)

(3.35b)

Fig 3.24 Natural flow around a vertical circular tube.

LT w

d

T


132


iv) Heat transfer coefficient for a horizontal tube outer surface with film condensation

(3.36)

where, : density of vapor , [ ]: heat of evaporation , [ ]: conductivity of the condensing liquid, [ ] , : saturated vapor temperature and wall temperature, [ ]

v) Heat transfer coefficient for a vertical tube outer surface with film condensation

(3.37)

where, : density of vapor, [ ]: heat of evaporation , [ ]


133


3.2.3 Stefan-Boltzmann’s law of radiation→ heat transfer phenomena in the form of electromagnetic waves → wavelength region important for heat transfer

: wavelength region between 0.1～100 including visible light, infrared & ultraviolet

◆ Features of radiative transfer- Electromagnetic wave transmission through the medium is required.- Electromagnetic energy transmission through the vacuum is possible without any loss.

◆ Governing law for the radiative transfer : Stefan-Boltzmann’s law of radiation→ the radiative heat flux ( ) emitted from a blackbody at an absolute temperature is proportional to the 4th power of the absolute temperature as

(3.38)

where, : Stefan-Boltzmann constant


134


◆ Effective radiative transfer equations for various geometric shapes

1) Effective sradiative transfer between two plain parallel black plates with infinitely large size (Fig 3.25)

Fig 3.25 Two plain parallel black plates with infinitely large size.

(3.39)

qT1T 2

(vacuum)

black body


135


2) Effective radiative transfer between two plain parallel black plates with finite size (Fig 3.26)

Fig 3.26 Two plain parallel black plates with finite size .

(3.40)

where, view factor ( )→벽1이벽2를보는때의차단율을나타냄.

qT1T2

(vacuum)


136


3) For non-black surfaces Eq. (3.50) becomes

(3.41)

where, emissivity factor ( )

◆ Rewrite Eq. (3.51) by introducing the radiation thermal resistance (factorization) ;

(3.42)

where, : radiation thermal resistance [ ]

(3.43)

: radiation transfer coefficient [ ]→ proportional to the 3rd power of the absolute temperature as

(3.44)


137


3.3 Combined mode heat transfer

→ In most cases heat transfer is occurred in a combined mode including conduction, convection and radiation.→ The combined mode heat transfer can be expressed by using the overall heat transfer coefficient U.

1) Overall heat transfer coefficient ( )→introduced by expressing the heat transfer through the heat transfer area in the

following form ;

(3.45)

where, : overall heat transfer coefficient ,

3.3 Combined heat transfer

138


i) Convection-conduction-convection system (Fig 3.28)

Fig 3.28 Convection-conduction-convection system .

ㆍ Heat transfer equation expressed by the overall temperature difference and the total thermal resistance

(3.46)

where,

q

T ,2

T ,1s,1T

s,2T

2

1h

h

L


139


ㆍ Or Eq (3.58) becomes ;

(3.47)

where, : 총열저항 (total thermal resistance)

(3.48)

ㆍ Heat transfer equation expressed by the partial temperature difference and the corresponding partial thermal resistance

(3.49)


140


ㆍ Comparing Eq (3.59) to Eq (3.57), we get the overall heat transfer coefficient as following

(3.50)

ㆍ Comparing Eq (3.60) to Eq (3.62), we get the following relation between and

(3.51)


141


ii) Plain wall with multi-layer (Fig 3.29 )

Fig 3.29 Thermal resistance of a plain wall with multi-layer.

q T

T

3T

2

1h

h

T4

T2Ts,1

k k kA

LA L L

B C

B C

xA B C

T Ts,1 T2 T3 T4 Tq q q q q

R1 R2 R3 R4 R5

4

1

41


142


(3.52)

(3.53)

Therefore, the overall heat transfer coefficient is

(3.54)


143


iii) Cylindrical system with conduction-convection+radiation (Fig 3.30 )

Fig 3.30 Conduction-convection+radiation system.

ㆍThermal resistance ;

where, : summed heat transfer coefficient including convection and radiation effects

Ts,2Ts,1

ho

hi

ri T

ro

Ti


144


ㆍ Radial heat transfer rate ;(3.55)

ㆍ Introducing the overall heat transfer coefficients ;(3.56a)(3.56b)

ㆍ Comparing Eq (3.55) to Eq (3.56) , the overall heat transfer coefficient is(3.57a)

: Ai as the reference surface → used for Eq (3.56a).(3.57b)

: Ao as the reference surface → used for Eq (3.56b).


145


iv) Composite cylinder (Fig 3.31 )

Fig 3.31 Composite cylinder.

Ts,4Ts,3

Ts,2Ts,1

T

T

AB

C

r4

r3r2

r1

TTh4

h1

T Ts,1 s,2T Ts,3 Ts,4 Tqr

ln( r2 /r)12 k LA2 r

1Lh1

ln(2

/r)k

r3LB

2 ln(2

/r)k

r4LC

3

1 Lh1

2 r 4 4


146


ㆍ Radial heat transfer rate ;

ㆍ Introducing the overall heat transfer coefficient

(3.58)

where,

, , : can be expressed in the similar form as .


147


2) Heat transfer from the finned surfaces

ㆍDefinition of the fin efficiency :

where, : actual heat transfer from the finned surface.: heat transfer from the finned surface when the fin surface temperature

is equal the base temperature. : fin surface area.: fin base temperature.: surrounding fluid temperature.: convective heat transfer coefficient for the fin surface.


148


ㆍ Fin efficiency : rectangular fin with constant cross sectional area (Fig 3.32 )

(3.59)

where, (3.60)

: perimeter of fin cross section : cross section al area of fin: corrected length

i) For insulated fin end : ii) For fin end with convection⒤ Rectangular type fin : (ii) Circular bar type fin :


149


Fig 3.32 Rectangular type fins mounted on a flat surface

S

W

T X

W

T

L b

t

t

oo


150


Fig 3.33 Efficiency of the fins mounted on a flat surface


151


Fig 3.34 Efficiency of the fins mounted on a circular surface


152


ㆍ Total heat transfer rate from the finned surface,→ (Heat transfer rate from the fin surface)

+ (Heat transfer rate from the bare surface)(3.61)

where,(3.62a)(3.62b)(3.62c)

and,A : total heat transfer area

: total surface effectiveness: fin surface area

ㆍ Total heat transfer area : for rectangular fins on a flat surface(Fig 3.32)(3.63)


153


ㆍ Substituting Eq (3.62) into Eq (3.61) ;

(3.64)ㆍ Rearranging Eq (3.64) ;→ Total surface efficiency can be derived as

(3.65)

ㆍAnd the heat transfer from the finned surface can be expressed by using the overallsurface efficiency as ;

(3.66)

where, : thermal resistance for the finned surface(3.67)


154


1) Overall heat transfer coefficient for plane composite finned wall (Fig 3.35)

(3.68)

where, : total fouling factor (includes the fouling effects of inside and outside surfaces)

Fig 3.35 Composite finned wall

2l

l 3l 1


155


2) Overall heat transfer coefficient for cylindrical composite finned wall (Fig 3.36)

(3.69)

where, : the reference surface area used to define

Fig 3.36 Finned composite cylinder

l

rr

rr

12

34


156


3.4 Energy balance equation

◇ Energy balance within a control volume

→ A control volume is introduced.→ Apply the 1st law of thermodynamics(energy balance law) for the control volume.

Fig 3.37 Control volume for a closed system

Ed t.

dt

.Ein

.Eout

.Eg


157


◇ Energy balance equation for the control volume shown in Fig. 3.37 ;

(3.70)

Ein : heat or work input through the boundary of the control volumeEout : heat or work output through the boundary of the control volume Eg : energy generation within the control volumedEst/dt : change of the system energy with time

◇ Procedure of applying the energy balance law in the system design ;i) Set a proper control volume. Or draw the system boundary.ii) Identify every energy flow contents per unit time through the system boundary.iii) Directions of all the energy flows are indicated using arrow mark.iv) Energy balance equation is derived by considering inflows and outflows of energy.


158


◇ Plane wall with conduction-convection-radiation(Fig 3.38) : steady state heat transfer

Fig 3.38 Energy balance on a wall surface.

qTw

T

flow

conduction

rad ia tionqconvectionq

convection

rad ia tion

conduction


159


ㆍ Set the control volume (dotted line)ㆍ Heat flow contents expressed in equations using the basic law of each heat flow;

(3.71a)

(3.71b)

(3.71c)ㆍ Heat inputs and heat outputs are balanced as

(Heat conducted from the wall) = (Heat lost by convection and radiation from the surface)

ㆍ Energy balance equation derived

(3.72)

where, : absolute free stream temperature: absolute wall temperature

: absolute temperature of a virtual surface exchanging radiative heat with the wall .


160


3.5 Boundary and initial conditions

Energy system simulation requires ;1) Governing equation (mostly differential equations)2) Boundary conditions3) Initial conditions for unsteady problems.

1) Types of boundary conditionsi) Dirichlet condition : boundary with given temperature(or solution).

ii) Neumann condition : boundary with given temperature gradient(or heat flux).For example, in case of insulated boundary :

→ (3.73)

iii) Convection condition : boundary is cooled by a fluid flow


161


Fig 3.39 Convection boundary condition.

The heat transfer at the boundary shown in Fig 3.25 can be expressed by the following balance equation ;

.

(3.74)

2) Initial condition : indicates the system condition at time (Ex, initial temperature distribution)

hTs,1

T

x = L


162


<Chapter 3 Problem set>

3.1 For tube banks with , , , N=10 and , determine the pressure drops and the heat transfer coefficients for ①Aligned tubes and ② Staggered tubes, respectively. The properties of the fluid considered are as follows.

3.2 Estimate the effective radiative heat transfer rate between the two parallel flat plates with the same emissivity of ε=0.3 at and . The surface area of each plate is 1m2 and assume .

3.3 In problem 3.2 , estimate the heat transfer rate when a radiation shield is applied between the two plates. Emissivity of the shield is given as εs=0.5.

3.4 Estimate the overall heat transfer coefficient U4 and the heat transfer rate qr for the composite cylinder shown in Fig 3.31. Parameters applied are as follows;

3.5 Estimate the overall heat transfer coefficient U4 and the heat transfer rate qr when the composite cylinder shown in Fig 3.31 is installed with the fins of the dimensions as ; , , .The convection heat transfer coefficient on the fin surface is given as . All parameters given in Problem 3.4 are applied.


163


3.6 For the following finned flat surface, answer for the questions., , , ,

, , ,, ,

① Estimate the fin effectiveness and the fin efficiency and the overall surface efficiency .③ Estimate the overall heat transfer coefficient U.④ Using U above, estimate the heat transfer rate q [W].

Fig 3.40 Finned flat surface3.7 By evaluating the effectiveness of the insulation applied for the outer surface of a spherical container with a radius of

ro , the minimum effective insulation thickness can be determined. Derive the equation for evaluating the minimum effective insulation thickness by using the given convective heat transfer coefficient , conductivity of the insulator , , surface temperature and temperature of the surrounding fluid . --- Explain the results by using the related drawings.

t

L 2L

b

Tq

h

Th

1

21

2

p


164


3.8 The furnace wall designed as following figure is receiving a constant heat flux of from the furnace, answer for the following questions to determine the wall surface temperature .

① Draw an equivalent electric circuit and express all the thermal resistances.

②Write the heat transfer equation.

③Write the equation to determine the wall temperature and estimate it by using ;

, , , , ,

3.9 Consider the following pin shaped fin with ;( , , ,

, , )① Estimate heat transfer rate through the fin .② Estimate the fin effectiveness .③ Estimate the fin efficiency .④ Estimate the overall surface efficiency .⑤ Explain if the installation of this fin is recommended or not.

qTs,1

kAxA xB

Bk

s,2T s,3T Too

h

D insulatedtipL

Too

bT

LxL square plate


165


3.10 Answer for the following finned flat plate., ,

,, ,

, ① Estimate the fin effectiveness and

the fin efficiency .② Estimate the overall surface efficiency .③ Draw the equivalent electric circuit with

the thermal resistances.④ Estimate the overall heat transfer coefficient .⑤ Estimate the heat transfer rate .

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

t

L 2L

b

Tq

h

Th

1

21

2

p


166


Chapter 4 Boiling heat transfer

167


4.1 Introduction

4.1 Introduction

◇에너지시스템에서의상변화예ㆍ Rankine 사이클에서의작동유체 : 액체→기체→액체ㆍ보일러증발관ㆍ냉동기의증발기등

◇응축및비등과정ㆍ열전달이증대됨ㆍ과열등에의한시스템의손상이우려됨

4.2 Boiling Heat Transfer: 액체가열을받아서기체로변하는현상.

ㆍ비등곡선(boiling curve) : 벽면열유속과벽면과열도의관계를나타내는곡선

ㆍ벽면의과열도(wall superheat) : : 벽면의온도, : 액체의평균온도

ㆍ열유속(heat flux, ) : 비등이발생되는벽면에단위면적당단위시간당공급되는에너지

168


◇비등곡선(boiling curve)

그림 4.1 비등곡선 - 열유속과과열도

4.2 Boiling heat transfer

169


ㆍAB영역 : 단상류의대류열전달영역( )ㆍ BS영역 : 액체의평균온도가포화온도이하( )로국부비등이발생하는영역

→ 과냉핵비등(subcooled nucleate boiling)이발생ㆍ SC영역 : 액체의평균온도가포화온도이상( ㅍ )으로액체내부에서전반적으로

비등이발생하는영역→ 핵비등(nucleate boiling)이발생

ㆍ C점 : 핵비등이끝나고막비등(film boiling)으로천이되는지점→ 이때가해지는열유속을임계열유속(Critical Heat

Flux, CHF)이라함→ 이지점에서부터는관벽의온도가급격히상승

: 설계에있어서세심한주의를필요로함ㆍ CD영역 : 막비등이개시되는영역

→ 열유속의미세한변화에대하여관벽의온도가급격히변하는불안정한영역

→ 즉, D상태로가해지는열유속을유지하는경우관벽의온도는 D'로급격히증가할수가있음

→ 열유속을조절할경우 E상태가되도록할수있음→ E상태가되도록열유속을조절할경우에는최소의열유속으로막비등이가능하게됨

ㆍ EF영역 : 안정된막비등이발생→ 열유속의증가에따라관벽온도는서서히증가


170


◇수직관내부를흐르는액체의흐름비등(flow boiling) 특성 : 그림 4.2→ 관내부유체의온도변화, 건도변화및관벽의온도변화등을나타내고있음→ 유체의상태, 비등양상및 CHF의위치등이표시

그림 4.2 수직원관내부에서의비등현상


171


◇비등에따른관벽의온도변화: 유체의건도변화와벽면에가해지는열유속값에따른영향

ㆍ벽면에가해지는열유속값이낮은경우→ 유체의건도가 0%에서 100%로증가하는동안

: 유체의평균온도는일정하게유지: 동시에관벽의온도도거의일정하게유지

ㆍ벽면에가해지는열유속값이높은경우→ 유체의건도가 0%에서 100%로증가하는동안

: 유체의평균온도는일정하게유지: 해당건도값에대한임계열유속(CHF)을초과하는상태가될수있음→ 관벽온도는급격히상승


172


그림 4.3 건도변화및열유속값에따른관벽의온도변화


173


4.2.1 흐름비등(Flow boiling) 열전달계수: Chen방법(Chen Correlation,1963)이가장정확한것으로알려져있음

◇흐름비등열전달계수 : 핵비등과강제대류두가지의영향을고려하여결정

(4.1)

여기서, : 핵비등(nucleate boiling) 열전달계수: 강제대류(forced convection) 열전달계수

◇핵비등열전달계수,

(4.2)

여기서, : 억제지수 (suppression factor)→ 증기건도의증가에따라 1에서 0으로감소: 핵비등열전달계수


174


◇핵비등열전달계수, [ ] : Forster-Zuber 식

(4.3)

여기서, : [bar] : 온도인포화압력과온도인포화압력의차이: 포화액의비열[ ]: 포화액의밀도[ ]: 포화액의열전도율[ ]: 포화액의점성계수[ ]: 포화액의표면장력[ ]: 증발잠열[ ]


175


◇강제대류 (forced convection) 열전달계수,

(4.4)

여기서, : 이상유동열전달계수승수(two-phase heat transfer coefficient multiplier)

: 단상유동대류열전달계수

◇단상유동대류열전달계수, : Dittus-Boelter 식

(4.5)

여기서, ,


176


◇ 인자 : Martinelli 인자 의함수

(4.6)

여기서, : 액체유동에의한마찰손실압력구배: 기체유동에의한마찰손실압력구배

ㆍ인자 의계산 : 액체및기체의마찰계수가각각 에비례한다고가정하면

(4.7)


177


ㆍ 에대한인자 의변화는그림 4.4로부터구함.

그림 4.4 Chen 의 계수


178


◇억제지수ㆍ이상유동 Reynolds 수 ( )

(4.8)

ㆍ 에대한억제지수의변화는그림 4.5로부터구함

그림 4.5 Chen 의 계수


179


4.2.2 흐름비등(Flow boiling)에대한임계열유속

◇임계열유속(critical heat flux, ) : 열전달계수가급격히감소하고, 벽면의온도는반대로급격히증가하게되어전열면재료에심한열적손상을초래할수있는열유속

→ 전열면의조건(열유속)은임계열유속이하가되도록하는것이바람직

◇흐름비등에서임계열유속의두가지구별되는특징

i) 증기건도가낮은경우(subcooled boiling 또는 saturated nucleate boiling)→ 풀(pool) 비등임계열유속과유사한특성을보임→ 비등곡선상에 hysterisis영향이나타남(그림4.6a참조)

ii) 증기건도가높은경우 (anular flow)→ 비등곡선상에 hysterisis영향이나타나지않음(그림4.6b참조)


180


(a) 증기건도가낮은경우 (b) 증기건도가높은경우그림 4.6 흐름비등곡선

q

Tsat satT

q

CHFq

qCHF


181


◇임계열유속 : 유체의건도, 질량유속, 압력등에관계됨

(i) 유체의건도가클수록작은값을갖음(ii) 유체의건도및질량유속이일정한경우압력이높을수록작은값을갖음(iii) 유체의건도및압력이일정한경우질량유속이증가할수록큰값을갖음

(1) 관길이에대한평균건도를이용한임계열유속(Katto 식)[Katto and Ohne, 1984]

(4.9)

여기서, : 총질량유속 [ ]: 증발잠열 [ ]: 입구부의과냉 (inlet subcooling, [ ])및 : 다음과같이정의되는무차원양들의함수

, 및 .여기서, : 관의길이 [ ]

: 관의지름 [ ]


182


에대한다섯가지의표현 :

값에대한세가지의표현 :


183


값 : 값에따라결정

및 값의결정 : 값의크기에따라결정인경우 :

인경우 :


184


(2) 국부임계열유속 (Groeneveld 식) : 국부적인건도값을이용→ 임계열유속값 : 유체의온도, 압력, 유량및건도등에따라다른값을갖음

여기서, : Data Curve Fit 으로구함 [Groeneveld, 1986]


185


또는

여기서,

BLA : boiling length average


186



4.1 A mixture of water and steam flows through a circular tube with the inside diameter , length at a total flow rate and at a steam .

Fluid status : , , ,,

Properties : , ,, ,

,

1) By using the Chen correlation find the flow boiling heat transfer coefficient .2) By using the Katto correlation determine the critical heat flux .

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


187


Chapter 5 Flow through pipe network

188


5.1 Introduction

ㆍ작동유체가통과하는유로→ 원형관, 또는임의단면형상을한유로로설계

ㆍ유로를통과하는동안의압력강하→ 마찰에의한손실, 축소, 확대, 오리피스및벤드등에의한압력의손실

ㆍ보일러와같은복잡한에너지시스템또는플랜트등의경우→ 관로의연결망이매우복잡하여압력강하및유량배분에대한예측은매우중요한설계자료임

→ 특히, 보일러및냉동기등에서는작동유체가기체및액체혼합물인상태로흐르는영역도있음

ㆍ본장에서는1) 관로내부를흐르는단상유체및 2상유체의압력강하를예측하기위한다양한경험식들을소개

2) 복잡한관로망의해석에적용되는 Hardy-Cross법을소개.

5.1 Introduction

189


5.2 Single phase pressure loss

5.2 Single phase flow pressure losses

ㆍ관로에서의손실1) 관표면과유체사이의마찰에의한손실 (major loss)2) 밸브, 곡관, 열교환기등의설치로인한관로단면적의축소-확대에의한부차적인손실 (minor loss)

1) Frictional loss for single phase flow through pipes : Darcy equationㆍ지름 : D ㆍ길이 : L ㆍ평균속도 : V

◇마찰압력손실

(5.1)

여기서, 만약관로의단면이원형이아닌경우→ 등가지름또는수력지름(hydraulic diameter) 를 대신사용하여압력손실을계산

(5.2)

여기서, : 관로의단면적: 접수길이

190


◇마찰계수 : Reynolds수( )와상대조도( )의함수→ Moody 선도 [Chisholm, 1983]의사용→ 컴퓨터시뮬레이션을위한보간식의사용

◇마찰계수에대한보간식

i) Blasius식(매끈한관내부의난류흐름)[Chisholm, 1983]

(5.3)여기서, :

: 난류유동범위 :

ii) Churchill식 [Churchill,1977] → 모든 Reynolds수범위에서상당히정확

(5.4)

여기서,


191


iii) Haaland식 (Haaland, 1983)

여기서, n ≈ 3.

→ 위수식들중에서 Churchill식이정확도면에서가장우수


192


2) Minor losses for single phase flow: Bend, 축소, 확대, orifice, 밸브등을통한단상류압력손실 (그림 5.1 참조).

(5.7)

여기서, 손실계수 : Chilton handbook[1963], Hodge[1990] 등에정리

Fig 5.1 Piping parts of sudden expansion, sudden contraction, orifice, bend .


193


i) 급격한확대에따른압력손실계수 (그림 5.1a).(5.8)

ii) 급격한축소에따른압력손실계수 (그림 5.1b).

iii) Orifice의압력손실계수[Whalley,1987] (그림 5.1c 및 d).

(5.9)

여기서,

: 상류측관의안지름: orifice 의안지름.

0.0 0.2 0.4 0.6 0.8 1.0

0.5 0.45 0.36 0.21 0.07 0.0


194


iv) bend 에서의손실 (그림 5.1e).ㆍ ㆍ : 관의중심선에서의굽힘반지름[ ]

ㆍ : 관의안지름[ ]

Fig 5.2 Minor loss factor for bends


195


5.3 Two phase flow pressure losses

5.3.1 물성및기본물리량의정의 : SI 단위계

◇물성: 밀도 [ ] : 점성계수 [ ]: 열전도율 [ ] : 비열 [ ]

◇첨자첨자 L : 액체상태 첨자 G : 기체상태첨자 LO : 액체상태만의유동 첨자 GO : 기체상태만의유동

◇유량: 총질량유속[ ]

- 액체및기체전체의총질량유량을유로의단면적으로나눈값.: 액체의질량유속[ ]

- 액체의질량유량을유로의단면적으로나눈값.: 기체의질량유속[ ]

- 기체의질량유량을유로의단면적으로나눈값.

5.3 Two phase pressure loss

196

G

LG

GG


◇증기의건도

ㆍ건도의정의 : 유체의총질량에대한기체의질량비(5.10)

ㆍ건도의변화 : 열밸런스이용(5.11)

여기서, : 관의단위열전달면적당열전달량: 관입구에서의건도: 관출구에서의건도: 물의증발잠열 [ ]: 유로의 perimeter [ ]: 유로의단면적[ ]: 유로의길이 [ ].

ㆍ건도의변화율 : 가 0 에근접할경우(5.12)

x

GGxG


197


◇겉보기속도 (superficial velocity, [ ]) → 기체및액체의겉보기속도는유로에기체또는액체만흐르는것으로가정하였을경우의속도

(5.13a)

(5.13b)

◇기공율(Void Fraction), → 유로의총단면적(또는체적) 중에서기체가차지하는면적 (또는체적) 의비→ 기체또는액체의실제속도( 또는 ) 및겉보기속도( 또는 )를이용하여표시가능

(5.14a)

또는, (5.14b)◇혼합유체의물성→ 균일하게혼합된유체의평균밀도및평균점성계수

(5.15)

(5.16)


198


5.3.2 이상유동압력손실성분

◇이상유동압력강하

ㆍ마찰에의한손실,ㆍ기체-액체혼합물의유동에따른가속손실,ㆍ수두차이에따른손실, ㆍ관로의부차적손실(축소, 확대, bend 등), ㆍ관로에서의총압력강하,

(5.17)

( / )Fdp dz

( / )Adp dz

( / )Zdp dz

( / )Mdp dz

( / )TPdp dz

TP F A Z M

dp dp dp dp dpdz dz dz dz dz


199


Fig 5.3 Pressure drop for two phase flow


200


1) 이상유동마찰압력손실 (two-phase flow frictional loss).ㆍ단상유동압력강하계산식에승수(multiplier)를곱하여산출

(5.18)여기서, : 이상유동마찰압력강하승수(Two-phase frictional multiplier)

ㆍ정확도가높은것으로알려져있는 Friedel[1979] 관계식

(5.19)여기서, ,

, : Froude number

: Weber number

: Homogeneous mixture density및 : 총유량(물+증기, )에해당하는액체또는기체가단상으로

흐른다고가정할경우의마찰계수: 표면장력 [ ]: 중력가속도, [ ]

2LO

2

,LO

TP F LO

dp dpdz dz

20.045 0.035

3.24LO

F HEFr We


201


ㆍ 에대한여러가지다른표현표 5.1 물-증기혼합유체에대한이상유동압력강하관계식.

2LO

관 계 식

Homogeneous Collier

Baroczy where

Chishoim where

CISE

Lombardiwhere

,

Martinelli

-Neison

Smith-

Macbeth where


202


ㆍ 는총유량 (물+증기, , ) 에해당하는물이단상으로흐르는경우의압력강하로서다음과같이계산

(5.20)여기서, : 액체만의유동에대한마찰계수

ㆍ위치 1 과위치 2 사이의이상유동마찰에의한압력차→ 사각형법칙 (trapezoidal rule) 을적용하여계산

(5.21)


203


2) 가속압력손실 (acceleration loss)·□혼합유체내의액체속도와가스속도가각각다른경우가스는압축될수있으며, 이로인한손실을가속손실이라함

(5.22)여기서, : 총질량유량 (total mass velocity)

: 증기의건도 (steam quality): 기공율 (void fration) (표 5.2 참조)

, : 증기및물의비체적

□기공율 에대한관계식▪미끄럼비 (slip ratio) : 기체와액체의속도비

(5.23)

▪기공율 는 : 윗식으로부터(5.24)

Gxa

G L

a

a


204


▪미끄럼비 에대한관계식→ CISE관계식[Premoli et al., 1970]이정확도면에서우수 [Whalley, 1987]

(5.25)

여기서, , ,

,

S


205


▪기공율 또는미끄럼비 에대한다른관계식

Table 5.2 Other correlations for void fraction of steam-water mixture

Sa

관 계 식

Lockhart &

Martinelliwhere

Hughmark where

Smith where

CISE Premoli

Chisholm

Thom

Bankoff-Jones where

Bryce where


206


□균일단면을갖는관의위치 1 과위치 2 에서혼합유체의건도가 에서 로변하고이와함께기공율이 에서 로변하는경우의가속에의한압력차→ 식 (5.22)를두위치사이에서적분하여

(5.26)

1x 2x

1a 2a


207


3) 높이차이에의한수두손실 (hydrostatic head loss).□수평으로부터 인각도로기울어져있는관로

(5.27)

□위치1 과위치2 사이에서유체의수두차이에의한압력차→ by using the trapezoidal rule

(5.28)


208


4) 이상유동에의한관로의부차적손실 (bend, 축소, 확대, orifice 등에서의손실)

(5.29)

i) 급격한확대에따른압력손실승수 (그림 5.1a)

(5.30a)

(5.30b)

여기서, = (작은단면적/큰단면적)


209


ii) 급격한축소에따른압력손실승수 (그림 5.1b).

(5.31a)

(5.31b)

여기서, : 상류측관의단면적: 하류측관의단면적

iii) Orifice에의한압력손실승수[Whalley,1987](그림 5.1c 및 d)

(5.32a)

(5.32b)

2A1A


210


iv) bend 에서의손실 (그림 5.1e)

(5.33a): 그림 5.2 참조 (5.33b)

여기서,

: 관의중심선에서의굽힘반지름 [ ]: 관의안지름 [ ]

K

mmid

R


211


5.4 Piping network analysis

□관로망을분석하는목적1) 관로를구성하는파이프의지름결정,2) 적절한유량분배가되도록하여시스템을고성능화

□관로망의설계→ 관로연결점에서의질량밸런스및압력의일치등을고려

□ Hardy-Cross 법→ 관로망을해석하는체계적인방법 : 다음 2가지방법이있다.

▪마찰계수의속도(유량)에대한함수관계를무시하는방법(friction factor based Hardy-Cross method)▪마찰계수의속도(유량)에대한함수관계를고려하는방법


212


□ Hardy-Cross 법: 마찰계수의속도(유량)에대한함수관계를무시하는방법

▪연속방정식(5.34)

여기서, 유체의유량 : 유체의속도 : 유로의단면적 :

▪관로를통한마찰압력강하 : Darcy 방정식

(5.35)

▪밸브, orifice, 축소, 확대, 곡관등에의한부차적인손실

(5.36)

3[ / ]Q m s[ / ]V m s

2m


213


▪압력손실관계식들의일반적인형태

(5.37)여기서, : 마찰손실

: 부차적인손실▪식 (5.37)을유량 에대하여 Taylor 전개하면→ 유량 의미소변화량 에대하여

(5.38)▪각관로에대한유량들이수렴할경우→ 는매우작은값→ 식 (5.38)의고차항들을무시하여다음과같이근사화

(5.39)▪여기서, 는식(5.37)의일반식으로부터

(5.40)▪식 (5.37) 및 (5.40)을 (5.39)에대입하면

(5.41): 임의관로에서의유량를구하는데사용할수있음

QQ Q

Q


214


▪기호설명- 번째관로(loop)의 번째관요소를흐르는유량값에대한 n번째가정값 : - 번째관로(loop)의 n번째수정유량 :

그림 5.4 일반적인관로망의일부분

i j ( )nijQ

( )nijQi


215


▪수정유량 를구하는조건→ 관로의연결점압력은유일한값

▪식 (5.41)을그림 5.4의 번째관로(loop)에적용하면→ 연결점압력이유일한값으로되기위하여

(5.42)

▪식 (5.41)을식 (5.42)에대입하면 ;(5.43)

( )nijQ

i


216


▪식(5.43)을 번째관로(loop)의수정유량 으로나타내면

(5.44)

여기서,1) 관로 의방향과유량 의방향이일치하면 :

2) 반면에방향이일치하지않으면 :

▪ 유량의보정 : 는 관로 의모든가정된유량 에더함

for j=1,2,.....,M (5.45)

( )nijQi

( ) 0nijQ

( ) 0nijQ

( 1) ( ) ( )n n nij ij iQ Q Q

( )nijQ

( )nijQ ( )n

ijQi


i

217


<Chapter 5 Problem set>5.1 Water is flowing through a vertical tube from the bottom to the top . A uniform heat flux is applied over the whole

outside surface of the tube. When the water entering from the bottom end of the tube is at the saturated liquid, determine the quality of the exit water at the top, .

, , , ,

Fig 5.5 Boiling of water flowing through a vertical tube

0x

Hq

G

d

xo

x i

i

d o


218


5.2 The following figure indicates a simple circulation loop to be used to produce steam. Determine the natural circulation flow rate, the steam generation rate, and the average riser tube steam quality at the drum inlet.

Fig 5.6 Natural circulation loop for steam generation

L1 L2 L3

L4

T

d

G

G

q

L5

od

fw

s

o

D

R


219


5.3 For the following piping system apply the Hardy-Cross method to determine the flow rates.(consider the losses for T, bend, straight parts of the piping system)

Fig 5.7 Simple piping system

5.4 In analyzing the piping systems the Hardy-Cross method is used frequently. In this method if the pressure

drop and the friction factor are expressed as and . Derive the flow rate

correction equation for for the given friction factor relation.

Q 1

2Q

= 1000 l/hr

(20 C)

15 mm ID

19 mm ID

Q물


220


5.5 Sub-cooled water below the saturated condition is entering into the vertical tube shown in the figure from the bottom end. When a uniform heat flux is applied on the outer surface of the tube derive an equation to determine the height of the location measured from the bottom where the water reaches at the saturated water state.

기호 : : 열유속[ ]: 관의바깥지름[ ]: 관의안지름[ ]: 질량유속[ ]: 비열[ ]: 증발잠열[ ]

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

q0did

pCm

LG

m

2kW m

2kg m s.kJ kg K

kJ kg

mq

d

rsat

Tin

i

do

m


221


Chapter 6 Properties of fluids

222


6.1 Properties of gases and liquids

ㆍ단일성분으로구성된기체및액체의온도변화에따른열전도율및점성계수

→ 부록 C 의표 C1, C2 및 C3에각각참고로주어짐.

6.1 Properties of gases and liquids

223


6.2 Compositions and properties of combustion gases[Ferguson, 1986].

ㆍ온도및압력이주어지면→ 각화학종들의성분율을계산→ 혼합유체의비열, 점성계수및열전도율등을구함

6.2.1 연소가스의조성

가. 이론화학반응에대한가스조성ㆍ탄화수소계연료의일반적인화학구조 : ㆍ이론화학반응 (stoichiometric reaction)→ 생성물질이이산화탄소와물만으로된다고가정하면

(6.1)

여기서, : 이론연료-공기 mole 비: 몰수

6.2 Properties of combustion gases

224


ㆍ원자수(atoms)는보존되기때문에→ 다음과같은원자량보존식이가능

여기서, 등 : 연료의조성을나타내는상수

ㆍ위의방정식으로부터각변수 , 를구하면 ;

(6.2)

ㆍ이론연료-공기비(stoichiometric fuel-air ratio)→ 이론화학반응인경우→ 공기질량에대한연료질량의비율로정의

(6.3)


225


Table 6.1 Typical hydro-carbon fuels

Chemical Form Name Stoichiometric constants

Hydrogen ,

Methane ,

, Methanol ,

Acetylene ,

Ethanol ,

Propane ,

Benzene ,

Octane ,

Gasoline ,

Diesel fuel ,

Nitromethane ,


226


ㆍ연료-공기당량비 (fuel-air equivalence ratio)→ 이론연료-공기비( )에대한실제연료-공기비( )의비율로정의

(6.4)여기서, 만약 이면 : 연료희박반응

이면 : 연료과잉반응이면 : 이론연료-공기반응

ㆍ이론산소-연료비 (stoichiometric oxidant-fuel ratio) 는 ;(6.5)


227


나. 연료과잉( ) 또는연료희박( ) 연소.ㆍ낮은온도및탄소/산소비 (C/O ratio) 가 1이아니면

→ 연소반응식은다음과같다.

(6.6)

ㆍ식 (6.6)은연료과잉및연료희박연소에대하여→ 연료희박연소인경우 : → 연료과잉연소인경우 :

ㆍ연료희박( ) 또는이론연료-공기비반응( )인경우→ 원자수의평형만을고려하는것으로충분

ㆍ연료과잉반응 ( ) 인경우→ 반응평형상수(Equilibrium constant)의도입이필요

(6.7)


228


ㆍ연료과잉, 연료희박및이론연료-공기비연소에대한결과

Table 6.2 Composition of combustion , [moles/mole of air]

→ 연료과잉연소의경우, 계수 는 2차방정식의해로부터

(6.8)

여기서,(6.9)

1

2

3

4 0

5 0

6 0


229


6.2.2 다성분이상유체(ideal gas)의비열및엔탈피→ 종류의이상기체들로구성된다성분혼합유체

ㆍ질량분율 와몰분율 사이의관계

(6.10)

또는, (6.11)

여기서, : 각성분들의분자량

ㆍ각성분의몰수 와각성분의몰분율관계식

(6.12)

ㆍ연소생성물의각성분들에대한몰분율의총합(6.13)

iM

im iy

J

iv


230


ㆍ여러가지종류의가스성분에대한열역학적물성자료

→JANAF Tables[1985] : 미국표준국 (National Bureau of Standards)발행

→전산처리를위하여도표형태의자료를수식화: 가스성분( )에대하여다항식으로수식화된결과: 온도범위 : 표 6.3: 온도범위 : 표 6.4

(6.14)

(6.15)

ㆍ따라서, 개의성분을갖는혼합가스인경우

(6.16)

(6.17)

i300 1000T K 1000 4000T K

J


231


표 6.3 열역학자료 (300 ≤ T ≤ 1000K) [Gordon 등, 1973]

products

1 .24007797E+01 .87350957E-02 -.6607878E-05 .20021861E-08 .63274039E-15 -.48377527E+05 .96951457E+01

2 .40701275E+01 -.11084499E-02 .41521180E-05 -.29637404E-08 .80702103E-12 -.30279722E+05 -.32270046E+00

3 .36748261E+01 -.12081500E-02 .23240102E-05 -.63217559E-09 -.22577253E-12 -.10611588E+04 .23580424E+01

4 .36255985E+01 -.18782184E-02 .70554544E-05 -.67635137E-08 .21555993E-11 -.10475226E+04 .43052778E+01

5 .37100928E+01 -.16190964E-02 .36923594E-05 -.20319674E-08 .23953344E-12 -.14356310E+05 .2955535E+01

6 .30574451E+01 .26765200E-02 -.58099162E-05 .55210391E-08 -.18122739E-11 -.98890474E+03 -.22997056E+01

fuels

7 .38261932E+01 -.39794581E-02 .24558340E-04 -.22732926E-07 .69626957E-11 -.10144950E+05 .86690073E+00

8 .14102768E+01 .19057275E-01 -.24501390E-04 .16390872E-07 -.41345447E-11 .26188208E+05 .11393827E+02

9 .14256821E+01 .11383140E-01 .79890006E-05 -.16253679E-07 .67491256E-11 .53370755E+04 .14622819E+02


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표 6.4 열역학자료 (1000 ≤ T ≤ 4000K) [Gordon 등, 1973]

product

1 .44608041E+01 .30981719E-02 -.12392571E-05 .22741325E-09 -.15525954E-13 -.48961442E+05 -.98635982E+00

2 .27167633E+01 .29451374E-02 -.80224374E-06 .10226682E-09 -.48472145E-14 -.29905826E+05 .66305671E+01

3 .28963194E+01 .15154866E-02 -.57235277E-06 .99807393E-10 -.65223555E-14 -.90586184E+03 .61615148E+01

4 .36219535E+01 .73618264E-03 -.19652228E-06 .36201558E-10 -.28945627E-14 -.12019825E+04 .36150960E+01

5 .29840696E+01 .14891390E-02 -.57899684E-06 .10364577E-09 -.69353550E-14 -.14245228E+05 .63479156E+01

6 .31001901E+01 .51119464E-03 .52644210E-07 -.34909973E-10 .36945345E-14 -.87738042E+03 -.19629421E+01

fuel

7 .15027072E+01 .10416798E-01 -.39181522E-05 .67777899E-09 -.44283706E-13 -.99787078E+04 .10707143E+02

8 .45751083E+01 .51238358E-02 -.17452354E-05 .28673065E-09 -.17951426E-13 .25607428E+05 -.35737940E+01

9 .34552152E+01 .11491803E-01 -.43651750E-05 .76155095E-09 -.50123200E-13 .44773119E+04 .26987959E+01


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6.2.3 혼합기체의점성계수및열전도율→ 기체의열전도율및점성계수등을정확하게계산할수있는가스의운동역학이론(kinetic theory) 은상당히발달[Hirschfelder 등, 1954]

ㆍ분자량 , 절대온도인순수성분가스의점성계수 : 단위 [g/(cm‧s)] 또는 [poise]

(6.18)여기서, : 분자의지름 [angstrom]

: 분자간충돌에대한영향을나타내는무차원인수

(6.19)여기서, 무차원온도는

(6.20)여기서, : Boltzmann 상수 [J/K]

: 분자간충돌시최대인력(maximum attraction energy, [J])


234


ㆍ여러가지가스성분의 값 및 값

표 6.5 가스성분별충돌상수 [Svehla, 1962].

성분명 [angstrom] [K]

1 3.941 195.2

2 2.641 809.1

3 3.798 71.4

4 3.467 106.7

5 3.690 91.7

6 2.827 59.7

7 2.708 37.0

8 3.050 106.7

9 3.147 79.8

10 3.492 116.7

11 3.758 148.6


235


ㆍ가스의열전도율 ( )→ 병진운동에너지 (translational energy) 를고려한→ 내부에너지 (internal energy) 를고려한 로구분

(6.21)

ㆍ병진운동에너지 (translational energy) 를고려한은

(6.22)

여기서, 는일반기체상수.ㆍ내부에너지 (internal energy) 를고려한 는

(6.23)

ㆍ혼합가스에대한점성계수 ( ) 는(6.24)

ik'ik

''ik

''ik


236


여기서, 계수 는

(6.25)

ㆍ또한, 혼합가스에대한이동성분열전도율 ( ) 은

(6.26)

여기서, 계수 는(6.27)

ㆍ또한, 혼합가스에대한내부성분열전도율 ( ) 은

(6.28)


237


ㆍ반응하고있는혼합물의열전도율은분해및재결합등으로인한고려

(6.29)

여기서, : 평형상태의비열: : 성분의변동이없는상태의비열 (frozen specific heat)


238


그림 6.1 혼합가스의물성계산프로그램.


239



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243


6.3 Properties of water and steam

- 에너지시스템을설계하기위하여는열전달, 유체의순환및전체시스템의성능등을해석하여야함.

- 물(water) : 쉽게구할수있으므로여러가지에너지시스템의작동유체로사용됨.

- 이러한해석을위하여: 물및증기의

엔탈피(enthalpy), 엔트로피(entropy), 비체적(specific volume), 밀도(density), 비열(specific heat), 점성계수(viscosity)열전도계수(thermal conductivity)등과같은물성값을필요로함.


244


- 물성값의계산 : 열역학적상태량들의상관관계를이용.

예를들어 : > 응축수또는과열증기영역에서는온도 ( ) 와압력 ( ) 이주어지면

위에서언급된물성값들이계산될수있음.> 포화수와포화증기가혼합된상태영역에서는온도( ) 와압력( ) 그리고

건도 ( ) 가주어지는경우이러한물성값들이계산될수있음.> 포화수 ( ) 또는포화증기 ( ) 의경우에는온도 ( ) 또는압력 ( )

중하나만주어지면위에서언급된물성값들이계산될수있음.

- 물및증기의물성값은: 부록 D에소개된바와같은 IFC (International Formulation Committee)에서제시하는수식들을이용하여계산하고있다 [일본기계학회, 1968].

: 그림 6.2에는물및증기의물성계산 FORTRAN프로그램이주어져있음.


245


Fig. 6.2 Computer program for water and steam


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269



6.1Considering the following chemical reaction, determine the mole fractions of combustion gases , , .

6.2 For the chemical reaction in Eq. (6.6) , determine the mole fractions of the products for , ,, and .

6.3 For given mole fractions , , and 인, determine the corresponding mass fractions , and .

6.4 For given mole fractions , , and of the combustion products at 400 [K] 700 [K] and 1300 [K] , determine the mixture properties of , , and at those three different temperatures.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


270


Chapter 7 Curve fitting

271


7.1 Introductionㆍ에너지시스템의열및수력학적분석및최적화→ 컴퓨터를이용한시뮬레이션이많이고려되고있음

ㆍ컴퓨터를이용한시뮬레이션→시스템의열밸런스및최적구성분석→ 시스템을구성하는요소(열교환기류)들의설계→ 연결하는관로망의설계등

ㆍ기존의여러문헌에서그림이나표로주어진설계자료→ 유체또는고체의물성→ 열.유동특성자료→ 요소기기의성능특성자료→ 요소기기의설계자료등

: 일반적으로불연속적인임의의간격으로값이주어짐: 컴퓨터시뮬레이션에위하여수식화(Curve fitting)

ㆍ본장에서는 Curve fitting에의한수식화기법을설명

7.1 Introduction

272


7.2 Polynomial curve fitting: 다항식을이용한 curve fitting 은가장널리사용: 주어진자료값을정확하게만족하는다항식을구하는것: 자료의수에의하여다항식의차수가결정

7.2.1 단일독립변수를갖는자료 (Single data set)ㆍ 1개의독립변수 : ㆍ data set : 인형태의다항식으로표현가능

Fig 7.1 Single data set.

y=f(x)y

x

7.2 Polynomial curve fitting

273


□주어진 (n+1) 조의 data set: ( ), ( ), ( ), , ( )

□다항식(7.1)

여기서, , , , ㆍㆍㆍㆍㆍ, : 보간상수□보간상수의결정ㆍ주어진 (n+1) 조의 data set에식 (7.1)을적용하여구함

ㆍ (7.2a)ㆍㆍ


274


ㆍ식 (7.2a)를 matrix의형태로나타내면

(7.2b)

ㆍ식 (7.2b) 로부터계수 을구하면: 주어진 data point 를정확하게지나는다항식이구해짐

여기서, n = 1 : linear curve fitting.n = 2 : qudratic curve fittingn = 3 : cubic curve fitting


275


□여러가지다른함수들을이용하여다항식을나타낼수도있음: 삼각함수및지수함수등

1) 삼각함수

(7.3a)

여기서, 계수 은다음과같이결정된다.

(7.3b)

2) 지수함수(7.4a)

여기서, 계수 은다음과같이결정된다.

(7.4b)


276


: 많은종류의자료들은선형보간으로표현하는것으로서충분함→ 표 7.1 참조

Table 7.1 Frequently used linear interpolation equations

명 칭 수식의 기본 형태 선형 수식의 형태 기울기 교차점

선형 방정식

Hyperbolic여기서,

Exponential

Inverseexponential 여기서,

Power law

Saturationgrowth rate 여기서, ,


277


7.2.2 2개의독립변수를갖는 data (multiple data set)

ㆍ 2개의독립변수 및 를갖는 data set : → 이중다항식으로표현

그림 7.2 Multiple data set.ㆍ 에대한 차다항식

(7.5)

=f(x,y)

y=y1

y=y

y=y

2

3

x


278


ㆍ계수 : 변수 의 차다항식으로표현

⋅ (7.6)⋅

ㆍ식 (7.6)을식 (7.5)에대입하면(7.7)

→ : 개의계수→ 방향으로 개 data set→ 방향으로 개의 data set 이필요→ 에대한연립방정식은 matrix


279


<예 7.1> Pump head : → 유량 및 회전수 의함수

Fig 7.3 Pump head.

ㆍ Pump head를다음과같이유량 의 2차함수로서표현

(7.8)

H=H(Q,N)H

N=N

N=NN=NQ

1

2

3


280


ㆍ계수 는회전수 의 2차함수로표현

(7.9)

ㆍ Pump head는유량 및회전수 의이차함수로서표현

(7.10)

→ 식(7.10)으로부터 9개의 data point 가주어지면→ 미지계수 가구해짐

━━━━━━━━━━━━━━━━━━━

<끝>


281


7.3 Least squares curve fitting□불규칙적인 data : 실험의결과로얻어진자료등

→ 주어진 data 값을정확하게만족하는다항식보다→ 주어진 data의변화경향을나타내는것이필요

° : data point.── : data point에가장근사한 simple function.· · · · : 고차의다항식으로 data point 를정확하게지나도록구한방정식.

그림 7.4 실험 data point 의수가많은경우의 curve fitting

f(x)

x

7.3 Least squares curve fitting

282


□주어진 data의변화경향을나타내는방법→ 최소자승법또는최소제곱회귀분석(Least squares curve fit)이고려됨

ㆍ총 개의 data set에대하여 → 다음과같은차방정식을고려

(7.11)

ㆍ여기서, 만약기초함수가 인경우

(7.12)

ㆍ 차최소자승법( order least squares curve fitting)→ 주어진 data point 값과 order polynomial함수의값사이의차이가모든

data point 들에서최소가되도록 polynomial 의계수 을구함→ 이렇게구한 polynomial 은주어진 data point 들을정확하게지나지않는경우가대부분

→ 필요한 data set의수 은 을만족하여야함


283


ㆍ 번째 data point에대하여 → 실제값과 curve 사이의차이의제곱은;(7.13)

여기서, : data point에대한잔류오차(residual)

ㆍ모든 data point에대한 residual의총합을나타내면 ;

(7.14)

ㆍ여기서, 미지계수 → 총잔류오차 값이최소가되도록결정ㆍ이러한최소 residual은 의각계수 들에대한미분이영(zero)이되도록함으로써구할수있음

(7.15)


284


ㆍ여기서, 이므로 → 미지계수에대한다항식으로표시

(7.16)

ㆍ식(7.16)를 matrix 의형태로나타내면 ;

= (7.17)

여기서, n＞m+1 n : 주어진 data point 의수m : polynomial의차수


285


ㆍ식(7.11)의다항식→ 기초함수가임의의함수형태→ 다음과같은일반형

= (7.18)

ㆍ여기서, 기초함수(7.19a)(7.19b)


286



7.1 Using the following computer program, exercise the polynomial curve fit in the form .

Fig 7.5 A computer program for curve fit of the function with a single variable.


287



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291



292


7.2 Using the following computer program, exercise the polynomial curve fit in the form (Fig 7.6).

Fig 7.6 A computer program for curve fit of the function with two variables.


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300


7.3 For a data set with , derive the least squares equations by using the polynomial with .

7.4 Using the following computer program (Fig 7.7), exercise the least squares curve fit in the form with mth order polynomial for a given data set.

Fig 7.7 A computer program for the least squares method.


301



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306



307



308


Chapter 8 Heat Exchangers

309


8.1 Classification of heat exchangers

□열교환기 : 고온유체와저온유체사이에열교환을하게하는장치

□유체와유체사이에전열벽이있는지없는지에따라

(1) 격벽식간접열교환기 : 열을받는유체와열을주는유체사이에격벽(일반적으로전열면이라고함) 이설치되어두유체는서로직접적으로접촉을하거나섞일수없음.

(2) 직접접촉식열교환기 : 직접접촉식열교환기는두가지다른유체가직접적인접촉에의하여한유체가갖는열을다른유체로전달되게하며두유체는서로혼합될수가있음.

→ 일반적으로열교환기는격벽식열교환기가대부분→ 격벽식간접접촉열교환기를대상으로공부


310


8.1.1 격벽식간접열교환기

․관내부 (또는좁은유로) 를흐르는유체 : 유체A․관바깥의넓은유로를흐르는유체 : 유체 B

□열교환기의호칭법또는분류방법i) 한쪽의유체가공기인경우 : 공기열교환기.

ii) 유체의상변화또는열교환기의기능을고려하여: 증발기, 응축기, 단상열교환기, 재생기, Cooler등의호칭.

iii) 전열면의밀도가 700 [ ]초월 : Compact 열교환기로분류.

iv) 회전식축열열교환기 : 융그스트롬열교환기로도칭함.

→ 표 8.1 및그림 8.1 참조


311


표 8.1 열교환기의분류.유체B의 유로 형 식 명 개략도

원관

원관이중관식 열교환기

(Double-pipe H.E.)그림 8.1(a)

쉘(shell)쉘 앤드 튜브 열교환기

(Shell and tube H.E)그림 8.1(b)

관열 사이

나관

(Bare tube)

관형 열교환기

(Tubular H.E.)그림 8.1(c)

원형 fin

부착관

Fin부착 관형 열교환기

(Finned tubular H.E.)그림 8.1(d)

Plate fin

및 관

Fin and tube 열교환기

(Tube-in-fin H.E.)그림 8.1(e)

장방형 관 fin 열 Corrugated fin 열교환기 그림 8.1(f)

평행 평판 평행 평판Plate 형 열교환기

(Plate H.E.)그림 8.1(g)

평판+fin 열 평판+fin 열Compact 열교환기

(Compact H.E.)그림 8.1(f)

벌집모양 판벌집모양 판

(Matrix)

축열식 열교환기

(Regenerator)

회전 matrix

(Rotary H.E.)

고정 matrix

+유로 교환

그림 8.1(h)전 열교환기

(Total H.E.)


312


(a) 이중관식열교환기 (b) Shell & tube형열교환기

(c) 관형열교환기 (d) 환형 Fin 부착열교환기

(e) 판형 Fin and tube 열교환기 (f) Corrugated fin 열교환기

(g) Plate 형열교환기 (h) 축열식재생열교환기그림 8.1 여러형태의열교환기


313


□열교환기단위체적당설계될수있는전열면적의크기 : 전열면의밀도→ 전열면의밀도가 700[ ]을초과하는열교환기

: Compact 열교환기라부름

그림 8.2 각종열교환기의전열표면밀도.

Area/Volume [m / m ]

50 100 500 1000 5000 100002 3

Classified as Compact H.E.

Corrugated fin H.E.

Fin & Tube H.E.

Compact H.E.

Plat H.E.

Shell & Tube


314


□표준형열교환기

1) 이중관식열교환기( double pipe heat exchanger)

․두유체모두의흐름방향이관의길이방향으로흐름→ 열전달율이낮다.

․열교환기의설계에서표준형으로고려됨→ 이론적수식의유도를위하여고려

․이중관식열교환기의구분 : 두유체의흐름방향에따라병류형(parallel flow type)향류형(counter flow type)

→병류형의경우열교환기내부의각위치에서두유체의온도차이가크기때문에향류형에비하여열응력이발생한다든지열전달효과가떨어지는문제가있다.


315


a) 병류형 b) 향류형

그림 8.3 2중관식열교환기의유동형식및온도분포.

L

"parallel flow "

Fluid A

Fluid B

Hot Fluid

Cold Fluid

T

L

"C oun ter flow "

F lu id A

H ot F lu id

C old F lu id

T

F lu id B


316


2) Shell and tube 형열교환기․다목적다용도의열교환기→ 산업용으로가장널리사용․Shell and tube 형열교환기내부의대표적인유체유동→ shell 측유체는차단막(baffle)에의하여관과거의직교류가형성되도록하고있다.

그림 8.4 Shell and tube형열교환기의개략도

․Shell and tube 열교환기내의유체흐름병류형(parallel flow type)향류형(counter flow type)혼합류형(mixed flow type)

F l u i d A

B a f f l e

F l u i d B


317


․Shell측유체및 tube측유체의특징- Shell측유체(유체A) : → 차단막에의하여유체의흐름방향을바꾸어가능한한유체의흐름을

tube에직각이되도록설계하여열전달을높이고있다.

- Tube측유체(tube 내부를흐르는유체 B) : → 이중관식열교환기에서와같이관의길이방향으로유체가흐르고있다.

․Shell and tube 형열교환기의구분: tube측유체의흐름 pass수에따라- Single pass : tube 내부의유체 B가모두같은방향으로흐름.- Double pass : 총 tube 수를반으로나누어유체 B를흐르게함.

→ 반은유체A에대해 count flow가되고하고,→ 나머지반은 parallel flow가되게함. → 이때일차pass를통과하면서가열(또는냉각)된유체는열교환기의 header부분에설치된격막에의하여 2차 pass로유입되어재차가열(또는냉각).


318


3) 직교류형열교환기 (Cross flow heat exchanger)

→ 관내부의유체는관의길이방향으로관과나란히흐름→ 관외부의유체는관과직각이되도록흐름방향이구성

그림 8.5 직교류형열교환기에서의유체흐름방향

유체 A

유체 B


319


□그림 8.5에서 ;

․유체A : 관에직각되는방향으로흐르며정해진통로가불분명하여관의전체길이에걸쳐서혼합됨(Mixed flow).

→ 관의길이방향에대하여혼합되므로관길이방향으로의온도차이(온도구배)는상당히작다.

․유체 B : 관내부에서관에평행하게흐르며, 유체의통로는주어진관에국한되기때문에열교환기를완전히통과하기이전에는서로혼합되지않음(Unmixed flow).

□그림 8.6에서와같이․외부의유체A의흐름에대하여도판형 fin 등의설치에의하여 : 관의길이방향으로서로혼합이

일어나지않도록개별적인유로를형성할수도있음 (관외부의유체A도 unmixed flow)

그림 8.6 차단막이있는직교류형열교환기.

유체 A

유체 B


320


□이와같이관외부의유체흐름이mixed 인지 / unmixed 인지 : 열전달에상당한영향을미침.

→ 유체가혼합될때(mixed flow) 혼합되는유체의흐름에직각인방향으로의온도구배는상당히작으므로유체흐름방향으로의온도구배만중요한인자로서고려.

→ 반면에, 유체가혼합되지않을때(unmixed flow)는유체의흐름방향및흐름에직각되는방향으로각각온도구배가크게되는것을열설계에서고려하여야한다.


321


4) 각종열교환기의전열효과비교.□열교환기의총괄열전달계수 : 다음과같은인자들에관련(1) 유체의유속.(2) 전열면의재질.(3) 전열면의오염정도.(4) 전열면의확장(fin 부착등), 난류촉진체등의채택유무.

표 8.2 각종열교환기의대략적인총괄열전달계수및용도.

열교환기 형식유체 A

(관내부의 유체)유체 B

총괄열전달계수(관 내면 기준)

용도 (예)

Shell & tube형

공기 연소가스 40-100 재생기, 배열회수

물 비등하는 냉매 2000-5000 냉동기,Binary

cycle물 응축하는 냉매 2000-6000

상변화유기매체 상변화유기매체 200-700 화학 plant

해수 증발하는 LNG 1500-3500 LNG 기화기

증발하는 물 고온 헬륨(helium) 700-1000 高溫가스爐 증기발생기

Plate 형공기 Oil 30-100 Oil cooler

물 물 2300-5800 Water cooler

Compact 형 공기 공기 50-200 Air cooler

Fin 부착관 유기액체 공기 150-230 화학 plant cooler

Fin & tube 형 상변화냉매 공기 400-1000 공조기

Corrugated 형 물 공기 200-1000 자동차용 라디에타


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8.1.3 열교환기의선정요령

□열교환기의선정에있어서고려하여야하는주요사항들(1) 주유체의종류와상(phase)의변화 (응축및비등유체). (2) 고온.고압유체.(3) 열부하에대응하는유연성.(4) 전열면의청소의용이함.(5) 사용유체의점도및오염도.(6) 사용유체의부식성.

표 8.3 각종열교환기의특징및용도열교환기 형식 특 징 용도 (예)

Shell & tube형- 보편적이며 가장 많이 사용되는 열교환기.- 응축 및 비등 유체에의 적용이 용이.- 고온.고압 유체에 유리.

재생기, 배열회수, 냉동기, Binary cycle, 화학 plant, LNG 기화기, 高溫가스爐, 증기발생기

Plate 형- 열부하에 대응하는 유연성이 큼.- 주로 저온.저압 유체에 적용.- 오염도 또는 점도가 큰 유체에 유리

Oil cooler, Water cooler, 식품처리

Spiral 형- 오염도가 큰 유체에 적합.- 고점도, 섬유질 또는 slurry등의 유체

Slurry, pulp, 섬유질

Fin 부착관 - 공기, 연소가스 등과 같은 기체에 응용 화학 plant cooler, 폐 연소가스 열회수, 공조기

Corrugated 형 - 전열면적의 집적도가 높음 자동차용 라디에타, 공조기


323


기타고려하여야할사항들로는

(1) 열교환기의최대온도차(2) 가격(3) 내구성(요구내용년수)

→ 제작의용이성, 사용재료및구조등도참고하여결정.


324


8.2 Fouling of heat transfer area

→ 열교환기의전열표면의오염: 깨끗하지않은유체의

- 직접적인접촉- 화학적물리적인요인

→ 전열면의오염 : 열저항의증가를초래

→ 오염의방지방법- 유체에포함된오염원을제거- 물리/화학적방법으로오염물질의부착또는퇴적을방지

: 필터또는침전: 열교환기의구조를고려하여오염물질이전열면에쉽게부착되지않도록유동을조절하는방법

: 유체내부에화학물질등을첨가


325


그림 8.7 열교환기전열면의오염.


326


□오염계수 (fouling factor, ): 깨끗한열교환기가장기간사용되었을때열전달표면이손상되거나표면에열전달을방해하는이물질이끼었을때열교환기의성능은크게저하→ 이러한영향을수치화한계수․깨끗한열교환기의총괄열전달계수 ( )

: 제 3 장에서와같은계산에의하여구할수있음․장기간사용된열교환기의총괄열전달계수 ( )

: 실제열교환기의설계에사용되는수정된열전달계수

(8.1a)

(8.1b)

□오염계수의고려- 유체의성질, 온도및유속등에관계- 또한전열면의온도, 재질, 표면상태등의운전조건에관계→ 오염계수의결정 - 전열성능등의악화속도

- 청소의주기, 청소의경비및설비비등경제적요인등을고려하여가장유리한상황이되도록결정하여야함


327


표 8.4 물의오염계수, .가열유체온도 115℃이하 115～205℃

물의 온도 50℃이하 50℃이하

물의 속도 0.09m/s이하 0.09m/s이상 0.09m/s이하 0.09m/s이상

증류수 0.00009 0.00009 0.00009 0.00009

염수 0.00035 0.00018 0.00035 0.00035

해수 0.00009 0.00009 0.00018 0.00018

냉각탑 또는 분수대물

처리된 물

처리되지 않은 물

0.00018

0.00053

0.00018

0.00053

0.00035

0.0009

0.00035

0.0007

수도물,우물물,큰 호수의 물 0.00018 0.00018 0.00035 0.00035

하수

최소치

평균치

0.00035

0.00053

0.00018

0.00035

0.00035

0.0007

0.00035

0.00053

진흙을 포함한 물 0.00053 0.00035 0.0007 0.00053

경수 0.00053 0.00053 0.0009 0.0009

Engine jacket 0.00018 0.00018 0.00018 0.00018

연화된 보일러 급수 0.00018 0.00009 0.00018 0.00018

boiler flow tank 수 0.00035 0.00035 0.00035 0.00035


328


표 8.5 각종유체의오염계수, .

유체명 오염계수 유체명 오염계수

가스 또는 증기

기관배기

증기(유류를 포함하지 않은 것)

폐증기(유류를 포함한 것)

냉매증기(유류를 포함한 것)

압축공기

공업용유기열매체

0.00180

0.00009

0.00018

0.00035

0.00035

0.00018

유(Oil)

연료유

변압기유

기관윤활유

소입유

유압용압력유

가솔린

석유

식물유

0.00090

0.00018

0.00018

0.00070

0.00018

0.00018

0.00018

0.00053액체

액체냉매

공업용유기열매체

전열용 용융염

0.00018

0.00018

0.00009가스(Gas)

천연가스 0.00018


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8.3 Theory of heat exchanger design

8.3.1 기본개념의이해․열교환기내에서발생되는열전달을해석하기위하여는

- 열역학적기초가되는에너지보존법칙- 열전달의기초가되는기본열전달방정식- 압력강하를예측하기위하여 : 수력학적기초이론

(1) 총괄열전달계수, : 전열면적의총괄적인전열성능을나타내는값

- 유체와전열면사이의대류열전달- 전열면을구성하는고체내부에서의열전도- 전열표면에부착된오염층에서의열전도등을망라

(8.2)

→ 그림 8.8 : 전열면의여러가지형태별수식→ , 및 중가장작은값이가장중요한역할을함.


330


표 8.6 여러가지전열면에대한총괄열전달계수, .

여기서, = 관내부또는평판의한쪽오염계수.= 관내부또는평판의한쪽열전달계수.= 관의안지름.= 관의 fin 기준면바깥지름.= 관또는평판의열전도계수.= 관내부또는평판의다른한쪽오염계수.= 관외부또는평판의다른한쪽열전달계수.

전열면의 형태 총괄열전달계수 식 열전달 식

평판보통 평판

Fin 부착

원관

나관 (bare)

안지름기준

바깥지름기준

Fin 부착

안지름기준

바깥지름기준


331


a) 평판 b) 원관

c) Fin 부착관그림 8.8 여러가지형태의전열면.

T

T

hh

kL

kri

ro

Th

Th

h

h

c

ci

i

o

o

tt

f

f0S

dH

fo Df

di


332


․fin 부착전열면의총괄면적효율,

(8.3)

여기서, Fin 부착관의전체열전달표면적,(8.4)

: 나관부의표면적: 단위 pitch당나관부의표면적

: fin 의총갯수.: fin 의표면적.

:단위 pitch당나관부표면적


333


․유체의물성을구하기위한기준온도i) 경막온도 (film temperature),

: 열전달방향으로의평균온도로정의: 주로국부대류열전달계수를구할때사용.

(8.5)

그림 8.9 열전달표면근처의온도

h

T

T dA

s


334


ii) 유체의평균온도, ( ).1. 향류형열교환기 ( )

(8.6)

- 저온유체및고온유체에대하여각각나타내면,

(8.7a)

(8.7b)

그림 8.10 향류형열교환기

Tc1

hT 1 hT 2

cT 2


335


2. : 한쪽유체의 가다른쪽에비하여아주클때.

→ 작은값의 을갖는유체의온도는지수함수에따라변한다.

(8.8)

3. 가아주큰유체의온도(상변화가있는유체)


336


<예제8.1> Counter flow

그림 8.11 향류형열교환기의입출구온도

Oil (0.1 m/s)

Water (1 m/s)

= 75

= 35

= 110= 90

Oil

Water

= 2 m

50 cm 30 cm

ho

h i

0.005 m

o

o

o

o

C

C

CCh2T

c2T

h1T

c1T

L


337


,

- 물성치

물 :

수력지름

Oil :


338


- 열전달계수의계산

Oil :

여기서, 로고려

→


339


물 :

→


340


그러므로

<끝> ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


341


8.3.2 열교환기의설계방법

□열교환기를설계하는방법

(1) 대수평균온도차법 (Log Mean Temperature Difference Method, LMTD법)

(2) 유효도-전열단위수법(Effectiveness-NTU Method)

(3) 법

(4) 법

등


342


(1) 대수평균온도차법 (Log Mean Temperature Difference Method, LMTD법): 열교환기내의고온유체와저온유체사이의온도차이는열교환기의위치에따라매우다를수있으며평균온도차는대수함수(log function)로나타남- 대수평균온도차 (LMTD)

: 이방법은열교환기의모든입구및출구에서의온도값들을알고서열전달량및전열면적을구할때많이사용.

i) 대수평균온도차(LMTD)의유도: 열교환기의입․출구에서각유체의온도를알고있는이중관식열교환기를고려(그림 8.12)

: 국부적인ΔT는전체길이걸쳐서위치마다크게다르다.- 열교환기의평균온도차 :- 열교환기를통한총열교환량 :

(8.9)

여기서, U : 총괄열전달계수A : U를계산할때기준이된전열면의면적

: 전체길이에대한평균온도차


343


그림 8.12 병류형열교환기및온도분포

Cold Fluid A

Hot Fluid B

q dq

dA

T

Th

T

T

h1

c1

TT

h2

c2

Cold Fluid

Hot Fluid

T T c

dq


344


․열교환기의미소면적 dA를통하여전달되는열량을고려→ (고온유체가빼앗긴열량) = (저온유체가받은열량)

(8.10)

․따라서, 고온유체및저온유체의온도변화는각각다음과같다.

(8.11a)

(8.11b)

여기서, : 질량유량, : 정압비열첨자 : 고온유체에대한값, 첨자 : 저온유체에대한값


345


․식 (8.11)로부터고온유체와저온유체의온도변화의차이는

(8.12)

․여기서, 미소전열면을통한열전달량은→ 총괄열전달계수 U를이용하여(8.13)

․식 (8.13)을식(8.12)에대입하면

(8.14)

․식 (8.14)를변수분리하여정리하면

(8.15)

․식 (8.15)의양변을적분하면

(8.16)


346


여기서, 열교환기전체길이에대하여및 : 일정

, 및 등 : 위치별로약간의차이가있으나일정하다고가정하면

(8.17)

․한편, 열교환기전체에걸쳐서고온유체가잃는열량은

(8.18a)

․또는열용량율에대하여나타내면

(8.18b)

․또한, 열교환기전체에걸쳐서저온유체가받는열량은

(8.19a)


347


․따라서, 열용량율은다음과같다.

(8.19b)

․식 (8.18b) 및 (8.19b)를각각식 (8.17)에대입하면

(8.20a)

․식 (8.20a)를다시정리하면

(8.20b)

․따라서, 고온유체와저온유체의평균온도차는

(8.21)

: 병류형이중관식열교환기의대수평균온도차(LMTD)


348


․대수평균온도차를유도하는데있어서가정된사항은①유체의비열이온도에따라변하지않는다고가정되었음.②총괄열전달계수(U)가열교환기의전체에걸쳐서상수로가정되었음.③이중관식열교환기로가정

․이중관식열교환기의병류형및향류형에대한 LMTD수식

(8.22)

여기서, 및 : 열교환기입구또는출구의최고및최저온도차를각각나타냄

- 향류형열교환기 (Counter flow H.E.) :(8.23a)(8.23b)

- 병류형열교환기 (Parallel flow H.E.) :(8.24a)(8.24b)


349


그림 8.13 이중관식열교환기내의온도변화.

A

1

1 2

T T TT

T

h out

c out

Th

Tc

d

d

d

2

Th in

Tc in

1 2

T

T1

T2d cT

d hT

(m C )p h.

p(m C ).

c

hT in

cT in

h outT

c outT (m C )p

p(m C )

cT in

.hT in

.

ccT out

h hT out

hT out

Tc in

Tc out

hT in

1

Tc in

hT in

h out

out

2

Tc

T

1incT

inhT

2

T

cT ou

h ouor

or

Parallel F low H eat E xchanger C ounter F low H eat E xchanger

a) N o flu ids under phase change

b) O ne flu id under phase change


350


ⅱ) LMTD 보정계수, F

․열교환기의형식이표준이중관식열교환기와다른경우나타나는 값의오차를보정하기위한수정계수 ( )를나타낸다.즉, 표준이중관식열교환기인경우 : (기준이되는형식)

․또한유체의상변화(기체→액체또는액체→기체)가수반될때에도해당유체의온도가일정하게유지되므로 LMTD의계산은간편, 보정계수 F는1이된다.

․열교환기에의한열전달은형식별로각각다음과같이구할수있다.: 표준이중관식열교환기 (8.25a): 이중관식이외의열교환기 (8.25b)

․그림 8.14 : shell & tube형열교환기에대한 F 값을나타냄그림 8.15 : 직교류열교환기에대한 F 값을나타냄


351


그림 8.14a Shell & Tube형열교환기의 LMTD 보정계수(1 shell - 2 tube passes).


352


그림 8.14b Shell & Tube 열교환기의 LMTD 보정계수(2 shell - 4 tube passes).


353


그림 8.14c Shell & Tube 열교환기의 LMTD 보정계수(3 shell-6 tube passes).


354


그림 8.14d Shell & Tube 열교환기의 LMTD 보정계수(1 shell - 2 tube passes) - Divided flow


355


그림 8.14e Shell & Tube 열교환기의 LMTD 보정계수(1 shell - 2 tube passes) - Split flow


356


그림 8.15a 직교류형열교환기의 LMTD 보정계수(both fluid unmixed).


357


그림 8.15b 직교류형열교환기의 LMTD 보정계수(1 mixed & 1 unmixed).


358


<예제. 8.2> Double Pipe Heat Exchanger(Counter flow) - 가열되는유체 : water - 가열하는유체 : oil

- U = 320 [ W / ℃] 일때, A = ?


359


그림 8.16 2중관식열교환기및입출구온도.

2

WaterOil

C

Th1

Tc1

Th2

Tc2

=110

= 75 = 75C C

= 35 C1 2

= 320 [W /m ]CU


360


◦.총열전달량

◦.LMTD

◦.

→

<끝> ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


361


<예제 8.3> Shell and tube type heat exchanger

그림 8.17 Shell and tube 열교환기의입출구온도

<예제 8.2>에서

Water Water

Oil

Oil

(35

(75 C)

)C (75 C)

(110 C)


362


◦.Correction factor F를구하기위하여 (그림 8.14a 참조)

→ F = 0.81

◦.열전달량.

<끝> ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


363


(2) 유효도-전열단위수법 (Effectiveness-NTU Method).: 열교환기를설계하는다른하나의방법: 열교환기내부를흐르는두유체사이의평균온도는고려하지않고열교환기로유입되는두유체사이에서전달될수있는최대가능한열전달과실제열교환기의열저항을고려하여해석하는방법

: 이방법은일반적으로열교환기를흐르는유체의유량과전열면적등을알고서열교환기출구의온도및예상되는열전달량을구하는데많이사용되는방법.

ⅰ) NTU : 열교환기의전열단위수(Number of Heat Transfer Unit)라부르는무차원수로서다음과같이정의.

(8.26)

여기서, : 고온및저온유체의열용량율(Heat capacity rate)및 중에서최소값을나타냄

: 전열면적에서의평균총괄열전달계수ⅱ) 열용량율비(Heat capacity rate ratio),

: 열용량율비는최대열용량율과최저열용량율의비

(8.27)


364


ⅲ) 유효도또는전열유효도 (Effectiveness), ․열교환기의전열유효도는최대가능한열전달량에대한실제열전달량의비로서정의

(8.28)

(8.29)

․또한고온측및저온측의온도효율및는각각 ;

(8.30)

(8.31)

․따라서, 전열유효도와온도효율들사이에는각각다음과같은관계가성립1) 고온유체의열용량율이최소일때 :

: (8.32)

2) 저온유체의열용량율이최소일때 :: (8.33)


365


iv) 유효도방정식의유도.

a) 병류형 b) 향류형그림 8.18 병류형및향류형열교환기의온도분포

A

1

1 2

T T TT

T

h o u t

c o u t

Th

Tc

d

d

d

2

Th in

Tc in

1 2

T

T1

T2d cT

d hT

(m C )p h.

p(m C ).

c

hT in

cT in

h o u tT

c o u tT (m C )p

p(m C )

cT in.

hT in.

ccT o u t

h hT o ut

hT o ut

Tc in

Tc o ut

hT in

1

Tc in

hT in

h o u t

o u t

2

Tc

T

1incT

inhT

2

T

cT o u t

h o u to r

o r

P a ra lle l F lo w H eat E xch an ger C o u n te r F lo w H ea t E x ch ang er

a) N o flu id s u n d er p h ase ch an g e

b ) O n e flu id u n d er p h ase ch an g e


366


․병류형(Parallel flow)의이중관식열교환기에대하여: 식 (8.17)로부터다음과같은수식을유도한바있음 (LMTD식의유도과정참조).

(8.34)

또는,

(8.35)

․또한, 저온유체의 ( )가최소일때유효도관계식은

(8.36)

․그리고, 고온유체와저온유체사이의총열전달량밸런스는

(8.37)


367


․식(8.37)를이용하여식(8.35)의좌변을다시쓰면 ;

(8.38)

․식 (8.38)를식 (8.35)에대입하면 ;

(8.39)


368


․따라서, 저온유체의열용량율이최소일경우( ) → 열교환기의유효도는다음과같이표현

(8.40)

․같은방법으로고온유체의가최소일경우 ( )

(8.41)

․식 (8.40) 및 (8.41)을하나의수식으로표시하면

(8.42)

여기서, , , .


369


․유사한방법으로향류형(Counter flow) 열교환기인경우에도

(8.43)


370


표 8.7 열교환기의형식별 과 사이의관계식

여기서, : 열용량율비

열교환기 형식 과 NTU 사이의 관계식

이중관식 열교환기Parallel flow :single pass

,

이중관식 열교환기Counter flow :simgle pass

,

( ) ; ,

Shell and Tube(one shell pass : 2,4,6,etc. , tube passes)

Shell and tube(n shelll passes : 2n,4n,6n,etc.,tube passes)

Cross flow(both streams unmixed) ,

Cross flow(both streams mixed)

Cross flow:streams mixed,: unmixed

,

Cross flow:streams unmixed,: mixed

,

All exchangers ( ) ,


371


ㆍ 인경우, 즉 인경우 (비등/응축이있는경우이에해당): 열교환기의형식및유로의방향에관계없이다음식

(8.44)


372


<예제8.4> Counter flow double pipe heat exchanger 물 : , , ,Oil: , , ,

, .

그림 8.19 향류형열교환기의온도분포

T

T

T

Th2

h1

c1

c2

1 2


373


ㆍ Heat capacity rate (열용량율)

따라서, 물(저온유체)이최소열용량유체이다.

ㆍ

ㆍ NTU

ㆍ


374


ㆍ

물의출구온도 :

ㆍ<끝>

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


375


<예제 8.5> 직교류형열교환기(fin부착관)(unmixed flows)공기 : , , ,

물 :,

,

그림 8.20 판형 fin이부착된직교류형열교환기

air

water


376


ㆍ공기유량 (at )

ㆍ열전달량

ㆍ공기또는물중에서어떤유체가최소열용량을갖는지모른다.→ trial and error법에의한계산을요함

ㆍ공기가최소열용량을갖는다고가정.

→이들에상당하는열용량비( )는구할수없다 ( )


377


ㆍ공기가최대열용량을갖는다.(E8.5a)

(E8.5b)

(E8.5c)

(E8.5d)

: (E8.5e)

표 8.8 반복계산에의한 effectiveness의계산

(가정값) 식(E8.6a) 식(E8.6b) 식(E8.6d)식(E8.6e) 식(E8.6c)

0.5 1452 1.452 27.78 0.65 0.417

0.25 726 2.905 55.56 0.89 0.833

0.22 639 3.301 63.13 0.92 0.947

0.222 645 3.27 62.54 0.93 0.938


378


⇒

or ←

<끝>━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


379


<예제 8.6> Shell and tube heat exchange

, ㆍEnergy balance

; ⇒공기가최소열용량을갖는다.


380


ㆍEffectiveness

ㆍ 식 (표 8.7 참조)

ㆍ<끝>

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━


381


(3) 여타의열교환기설계방법.ⅰ) 법.

ㆍ이방법은 법의변형된방법ㆍ 관계식은 Shell and tube 열교환기에서볼수있듯이관내부를흐르는

유체가 인지 인지에따라다른형태로나타남ㆍ 법에서는어느유체가값을갖는지혼동되는것을피하기위하여,

고온측또는저온측에관계없이어떤한쪽유체에대하여열교환기의 온도효율 를정의

ㆍ이때 값은대상유체의열용량율을이용하여결정ㆍ또한 은다른쪽유체의열용량율에대해고려하고있는유체의열용량율의

비로정의ㆍ일반적으로기준유체는임의로저온측유체또는 tube 측유체를많이채택ㆍ , 및 등은

(8.45)

(8.46)


382


(8.47)

여기서,

ㆍ온도효율 와열교환기의유효도 의정의를비교하면

(8.48)

ㆍ또한 는 을기준으로한 값으로정의

(8.49)


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ⅱ) 법. ㆍ LMTD 법으로성능해석을하거나 법으로열교환기의크기를결정하고자할경우에는반복계산을필요

ㆍ이와같은불편한반복계산을피하기위하여제안된방법

ㆍ LMTD 법과 법을조합하여만들어진방법 → 법이라한다.ㆍ 는두유체의입구온도차에대한실제평균온도차의비를나타내고다음과같이정의된다.

(8.50)

여기서,: LMTD 수정계수.

ㆍ이것은또한 및 와다음과같은관계를갖는다.(8.51)


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ㆍ이방법에서열전달량은

(8.52)여기서, → LMTD값 은불필요.

ㆍ예를들어, counter flow 열교환기의경우함수 는

(8.53)


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8.3.3 열교환기설계방법의선택요령

(1) 열교환기의입구및출구조건을알고서전열면적을구할경우.: LMTD 법을이용

(2) 전열면적,유량및입구조건을알고출구조건과열교환량을구할경우.: 유효도-전열단위수법( -NTU법)을이용


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<Chapter 8 Problem set>8.1 Derive the LMTD equation for the counter flow double pipe heat exchanger.

8.2 Derive the relation for the counter flow double pipe heat exchanger.

8.3 Answer for the double pipe heat exchanger operated at the following conditions.

water : oil :

ⅰ) Calculate the total heat transfer rate ( ).ⅱ)

ⅲ) From the results in ⅱ), explain the differences between the counter flow type and the parallel flow type.

8.4 Answer for the problem 8.3 ii)③ and iii) by using the method.

Counter Flow Parallel Flow

①Draw temperature profiles. ①Draw temperature profile.

②Calculate the LMTD. ②Calculate the LMTD를.

③Calculate the required heattransfer area A.

③Calculate the required heattransfer area A.


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Thermal System Design and Simulation

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