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BRE2031 – Environmental Science:

Lecture 5 – Psychrometrics

Dr. Meng Ni (倪萌)

Associate Professor

Department of Building and Real Estate, The Hong

Kong Polytechnic University

Office: ZN713

Tel: 2766 4152

Email: bsmengni@polyu.edu.hk

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Content

1. Moist air and their properties

2. Psychrometric chart

3. Psychrometric process analysis

Objectives

After studying this lecture, you will be able to:

1. Understand the physical meaning of moist air properties

2. Read moist air properties from the psychrometric chart.

3. Determine heat addition for heating/cooling of moist air.

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Basic concepts

• The surrounding air is a mixture of dry air and water

vapor – moist air, plus some pollutants;

• Psychrometrics is a subject of studying how moist air

behaves when it is cooled or heated;

• It is a tool to analyze typical air-conditioning processes.

Atmospheric

air

Dry air

Water vapor

Various pollutants

Moist air

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Illustration of Dalton’s law – Concept of partial pressure

• For the mixture of gases A and B, pmix = pA + pB

• pA is the pressure gas

A would have if gas A

is alone in the tank.

For a mixture of gases occupying a given volume at a certain temperature, the total pressure

of the mixture is equal to the sum of the partial pressures of the constituents of the mixture.

From ideal gas law, the

pressure of gas depends on

its number of moles (at a

given temperature and

volume)RT

P nV

=

mixture A Bn n n= +

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Atmospheric pressure

Standard atmospheric pressure at the sea level:

P=101.325 (kPa) = 14.696 (psi) = 30 (In. Hg)

For the elevation above the sea level H<1220 m (or <4000 ft):

P=101.325-0.01153H (kPa)

P=29.92-0.001025H (In. Hg)

For H>1220 m (or >4000 ft):

P=99.436-0.010H (kPa)

P=29.42-0.0009H (In. Hg)

psi: pound force per square inch

Pressure can be measured by height of water or other liquid.

An elevation increase by

1m, how many Pascal

decrease?

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Composition of Dry and Moist Air

• Atmospheric air contains many gaseous components as well as

water vapor and miscellaneous contaminants (e.g., smoke, pollen, and

gaseous pollutants not normally present in free air far from pollution

sources).

• Dry air is atmospheric air with all water vapor and contaminants

removed. Its composition is relatively constant, but small variations in the

amounts of individual components occur with time, geographic location, and

altitude.

• Moist air is a binary (two-component) mixture of dry air and water

vapor.

• Saturation is a state of neutral equilibrium between moist air and the

condensed water phase (liquid or solid); Air sample in a saturation state

– contains the maximum amount of vapor possible at a given

Temperature.

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A Microscopic Comparison of Gases, Liquids and Solids

In a liquid

• Molecules are in random

motion;

• There are appreciable

intermolecular forces holding

molecules close together

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LIQUID VAPOR

break IM bonds

make IM bonds

Add energy

Remove energy

<---condensation

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On the physics of saturation – in a closed system

Evaporation process

• Molecules in liquid water

attract each other

• In motion

(courtesy T.S. Zhao)

• Collisions

• Molecules near surface

gain velocity by collisions

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Evaporation

• Fast moving molecules leave

the surface (Only those with

enough KE escape)

• Evaporation

Twater

• Rate of evaporation – Function of water temperature

• Evaporation is a cooling

process.

• It requires heat.

• Endothermic.

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Evaporation

• Soon, there are many water

molecules in the air

• Slower molecules return

to water surface

• Condensation

Change from gas to liquid

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Evaporation

• Net Evaporation

– Number leaving water surface is greater than the number returning

– Evaporation greater than condensation

– Evaporation continues to pump moisture into air

– Water vapor increases with time

Net Evaporation

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Equilibrium

• Eventually, equal rates of

condensation and evaporation

• “Air is saturated”

• Equilibrium

At Equilibrium – Rate of evaporation is a

function of temperature;

Rate of condensation depends on water

vapor mass; Also a function of temperature

Rate of

evaporationRate of

condensation=

Evaporation = f(T)

Condensation = f(T)

Tair = Twater

Achieves a dynamic equilibrium

with vaporization in a closed

system.

A closed system means matter can’t

go in or out.

Any difference between evaporation and boiling?

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• Vaporization is an endothermic process - it requires heat.

• Energy is required to overcome intermolecular forces

• Responsible for cool earth.

• Why we sweat ?

Vaporization (evaporation)

• A liquid boils when its temperature reaches its boiling point

• Normal Boiling point is the temperature a substance boils at

1 atm pressure.

• The temperature of a liquid can never rise above its boiling

point.

Boiling

Changing the Boiling Point

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• Lower the pressure (going up into the mountains).

• Lower vapor pressure means lower boiling point.

• Food cooks slower.

• Raise the external pressure (Use a pressure cooker).

• Raises the boiling point.

• Food cooks faster.

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Moist Air

Dry air + Water vapor = Moist air

V

Moist Air: Mda+Mw=m

T Pda

da w

nRTp p p

V= = +

w daM M<<

w dap p<<

Typically

thus

Pw is called partial pressure

of water vapor.

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Humidity Parameters

( )

( )

mass of water vapor

mass of dry air

18.015268 /( )

( ) 28.966 /

18.0152680.622

28.966

w

da

total w

total da

w w

da da

MW

M

g moleN X mole

N X mole g mole

X X

X X

= =

= ×

= × =

0.622 0.622 0.622( )w w w

da da w

X p pW

X p p p= = =

−

Humidity ratio W (alternatively, the moisture content

or mixing ratio) of a given moist air sample is defined

as the ratio of the mass of water vapor to the mass of

dry air in the sample:

is the molar fraction ratio of water vapor and dry air.w

da

X

X

ww

total

NX

N=

Molar fraction of

water vapor.

Number of moles of water vapor

over the total number of mole of

the gas mixture.

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Humidity ParametersSpecific humidity γ is the ratio of the mass of water vapor to total mass

of the moist air sample:

1

w

w da

w daw da

da da

M

M M W

M MM M W

M M

γ = = =+ +

+

Absolute humidity (alternatively, water vapor density) dv is the ratio of

the mass of water vapor to total volume (V) of the sample:

/v wd M V=

Density ρ of a moist air mixture is the ratio of total mass to total volume:

( ) /da wM M Vρ = +

Saturation humidity ratio Ws(t, p) is the humidity ratio of moist air saturated

with respect to water (or ice) at the same temperature t and pressure p.

Degree of saturation µ is the ratio of air humidity ratio W to humidity

ratio Ws of saturated moist air at the same temperature and pressure:

,s t p

W

Wµ = 0.622 ws

s

ws

pW

p p=

−

Very close to W

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Relative HumidityRelative humidity (RH) ϕ is the ratio of the mole fraction of water

vapor xw in a given moist air sample to the mole fraction xws in an air

sample saturated at the same temperature and pressure:

, ,

w w

ws wst p t p

x p

x pφ = =

Change in RH can come from:

–Change in air temperature or pressure

–For a constant air temperature, RH increases with addition of water vapor

–For a constant water vapor pressure, increase in air temperature lowers the RH

–Highest RH occurs in the early morning (coolest time corresponds to the highest

RH); lowest RH occurs during the warmest part of the afternoon.

xws increases with T;

warmer air can hold more

moisture than cold air.

The ratio of actual water vapor content to the maximum possible moisture content at a

given temperature and pressure.

pws is the saturated vapor pressure (SVP) – the vapor pressure of the water vapor in an

air sample that contains the maximum amount of vapor possible at that temperature.

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RH highest in the cool morning; lowest in warmest time

Why droplets are usually

observed in the morning time?

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Saturated Vapor Pressure (SVP) of water vapor

Temperature (oC) SVP (Pa) Temperature (oC) SVP (Pa)

0 610 13 1497

1 657 14 1598

2 705 15 1704

3 758 16 1818

4 813 17 1937

5 872 18 2063

6 935 19 2197

7 1001 20 2337

8 1072 25 3166

9 1148 30 4242

10 1227 40 7375

11 1312 50 12351

12 1402 100 101325

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An exampleA sample of air has an RH of 40% at a temperature of 20oC. Calculate

the vapor pressure of the air.

What we know?

RH = 40% T = 20oC → SVP = 2337 Pa

, ,

40%2337

w w w

ws wst p t p

x p p

x pφ = = = =

2337 40% 934.8wp Pa= × =

1 millibars (mb) = 100 Pa

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Humidity Parameters

The mixture also obeys the perfect gas equation:

pV nRT=

( ) ( )da w da wp p V n n RT+ = +

where p = pda + pw is the total mixture pressure and n = nda + nw is the total

number of moles in the mixture.

Dew-point temperature td is the temperature of moist air saturated at

pressure p, with the same humidity ratio W as that of the given sample of

moist air. It is defined as the solution td( p, W) of the following equation:

( ),s dW p t W=

When moist air is considered a mixture of independent ideal gases (i.e., dry

air and water vapor), each is assumed to obey the ideal gas equation of state

as follows:Dry air:

da dap V n RT=

Water vapor:w wp V n RT=

V: total volume of the mixture; pda and pw are partial pressures of dry air and water vapor;

nda and nw are the number of moles of dry air and water vapor.

Dew or condensationAt which a fixed sample of air becomes

saturated (condensation occurs).

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How to measure the energy of moist air ? –

Enthalpy

The unit of enthalpy (H) is J or kJ.

Specific enthalpy (enthalpy per unit mass): h = H/m (J/kg, or kJ/kg)

Sensible energy (sensible heat)

Latent energy (latent heat)

The specific volume v of a moist air mixture is expressed in

terms of a unit mass v = V/Mda (m3/kg)

Mass of dry air1

da

vρ

= Density of dry

air (kg/m3)

How to measure the amount of moist air?

Why is v based on mass of dry air?

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Hygrometry (Psychrometry) – measurement of humidity

Hygrometers (Psychrometers) – instruments that measure the humidity of air

Hair/paper hygrometers: make use of the fact that hair or paper

change their dimensions with changing moisture content.

Dew-point hygrometer: measuring the dew-point temperature by

cooling a surface until water vapor condenses on it (then use the dew-point

T and room T to obtain an RH from tables/charts).

Wet-and-dry-bulb hygrometer: measuring the difference between the

wet and dry thermometers, as it indicates the relative humidity (from tables).

Electronic hygrometer: electronic sensors whose resistance changes with

changing humidity (this affect the amount of water absorbed by the sensors

from the air).

Psychrometric Chart

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Content

1. Moist air and their properties

2. Psychrometrics chart

3. Psychrometric process analysis

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Psychrometric Chart• Dry-bulb temperature t: ordinary air temperature

• Wet-bulb temperature twb: temperature read by a thermometer

wrapped in a wet fabric.

• Humidity ratio W

• Relative humidity φ

• Partial pressure of water vapor pw

• enthalpy of moist air: h = hda + Whw

Once two of the above properties are

known, the others can be readily

found out from a two-dimensional

chart – Psychrometric Chart.

hda is the specific enthalpy of dry air (kJ/kgda)

hw is the specific enthalpy of water vapor (kJ/kgda)

Approximately, 1.006dah t≈

t is in oC 2501 1.86wh t≈ +

At 0oC, the specific enthalpy of dry air is set as 0.

The specific enthalpy of water vapor at

0oC is 2501 kJ/kg.

Are they the same?

Which one is larger?

Why is the specific enthalpy based

on kg dry air?

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Psychrometric Chart

Dry-bulb temperature

Wpw

twb

twb

twb

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Psychrometric Chart

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Psychrometric chart

Linear

interpolation

should be done

when

necessary.

There are 2 horizontal lines:1. Solid lines are moisture content

(humidity ratio)

2. Dashed lines are vapor pressure

Search the interception of two

curves to find the state of

moisture air.

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Examples

Moist air sample has a dry

bulb temperature of 10oC,

humidity ratio (moisture

content) of 0.003 kg/kg

dry air, find (1) the wet-

bulb temperature; (2)

vapor pressure (millibars);

And (3) relative humidity

Answer:

(1) Wet-bulb T: 4.5oC

(2) Vapor pressure:

About 4.85millibars

(3) Relative humidity:

About 40%.

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Examples

Moist air sample has a dry

bulb temperature of 15oC,

wet-bulb temperature of

11oC, find (1) humidity

ratio (moisture content);

(2) vapor pressure

(millibars);

And (3) relative humidity

Answer:

(1) Humidity ratio:

0.0066 kg/kg dry air

(2) Vapor pressure:

About 10.5millibars

(3) Relative humidity:

About 60%.

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Another example - Textbook

(1) The RH of the heated air;

(2) The RH of the heated air

if 0.005kg/kg of moisture

is added;

(3) The temperature at which

this moistened air would

first condense.

External air at 0oC and 80% RH is

heated to 18oC. Use the

Psychrometric chart to determine

the following information:

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Another example - Textbook

(1) When the moist air is heated, the

mass of dry air and vapor does not

change. So the moisture content

remain unchanged!

RH: 23%

(2) When 0.005kg/kg of moisture is

added, the moisture content become:

0.003 + 0.005 = 0.008 kg/kg

RH: 62%

(2) For condensation

(dew), RH = 100%,

moisture content of

0.008kg/kg

Dew point: 10.8 oC

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An exercise

Moist air sample has a dry

bulb temperature of 15oC,

relative humidity of 90%,

find (1) wet-bulb

temperature;(2) humidity

ratio (moisture content);

(3) vapor pressure

(millibars); and (4) dew

point.

Answer:

(1) wet-bulb T:

14.2oC;

(2) Humidity ratio:

0.0097 kg/kg dry air;

(3) Vapor pressure:

About 15.3millibars

(4) Dew point:

14.3oC.

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Psychrometrics

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An example - ASHRAEMoist air exists at 40°C dry-bulb temperature, 20°C

thermodynamic wet-bulb temperature, and 101.325 kPa

pressure. Determine the humidity ratio, enthalpy, dew-point

temperature, relative humidity, and specific volume.

Why is the specific volume based on dry air?

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The example

Humidity ratio: 0.0065 kg/kg dry air

Enthalpy: about 56.7 kJ/kgda

Dew point temperature: 7oC

Relative humidity: 14%

Specific volume: 0.896 m3/kgda

Moist air exists at 40°C dry-bulb temperature, 20°C

thermodynamic wet-bulb temperature, and 101.325 kPa

pressure. Determine the humidity ratio, enthalpy, dew-point

temperature, relative humidity, and specific volume.

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Content

1. Moist air and their properties

2. Psychrometrics chart

3. Psychrometric process analysis

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Psychrometric processes

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Psychrometric

Processes

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Analysis of Psychrometric Systems

1 da1 w1m m m= +ɺ ɺ ɺ

Moist air

wmɺLiquid or vapor

QɺHeat

Moist air

2 da2 w2m m m= +ɺ ɺ ɺ

The property with a dot above it means the rate form.

m :i

:m mass (kg)

rate of mass flow (kg/s) E :i

:E Energy (kJ)

Rate of energy flow (kJ/s, or kW)

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Mass & Energy Conservation Equations

Mass

C.V. INTO

RATE

FLOWMASS RATE OF

MASS Generation/Consumption

in THE C.V.

=

−

C.V. OUT

RATE

FLOWMASS

Energy

C.V.

INTO ENERGY

OFRATE RATE OF ENERGY

Gen/Consumption

in the C.V.

=

−

C.V.OF

OUT ENERGY

OFRATE

Typical Assumptions• Steady state, steady flow

• No work

• Kinetic energy change is zero - usually

• Potential energy change is zero - usually

C.V. – Control volume, in

thermodynamics, it is also

called “system”

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Mass Equation

w da 2 1m m ( )W W= −ɺ ɺ

w1 w w2m m m+ =ɺ ɺ ɺWater(vapor):

dam= ɺda1 da2m m=ɺ ɺDry air:

w1 da 1m m W=i iw

da

MW

M=

w2 da 2m m W=i i

and

Then we

can get:

No mass gen/con in the C.V.

Mass flowing into the CV = Mass flowing out of the CV

da1 w1 w da2 w2m m m m m+ + = +i i i i i

W2>W1: humidifying

W2<W1: dehumidifying

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Energy Equation

=

OUT RATE

ENERGY

IN RATE

ENERGY

QIN RATE

ENERGYɺ=

w wm h+ ɺ da da1 da 1 w1m m W hh+ +ɺ ɺ

=

OUT RATE

ENERGY

da da2 da 2 w2m m W hh +ɺ ɺ

Thusw wmQ h+

i

ɺ da da1 da 1 w1m m W hh+ +ɺ ɺ

da da2 da 2 w2= m m W hh +ɺ ɺ

Energy (specific heat) added

when adding water vapor

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Energy Equation

w da 2 1m m ( )W W= −ɺ ɺ

w wmQ h+i

ɺ da da1 da 1 w1m m W hh+ +ɺ ɺ

da da2 da 2 w2= m m W hh +ɺ ɺ

( ) ( ) ( )da 2 1 w da da1 1 w1 da da2 2 w2 m W W m W h m W h 0Q h h h+ − + + − + =i

ɺ ɺ ɺ

Read from ASHRAE Psychrometric chart

Why is the specific enthalpy based on the mass of dry air?

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Classic Moist Air Processes

� Heating of moist air;

� Cooling of moist air;

� Cooling and dehumidification of moist air;

� Heating and humidification of moist air;

� Mixing process of moist air streams.

The following typical processes of moist air are

frequently encountered in air conditioning applications:

In each of the following examples, the process takes

place at a constant total pressure of 101.325 kPa.

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Moist Air Heating or CoolingAdding heat alone to or removing heat alone from moist air is represented by

a horizontal line on the Psychrometric chart, because the humidity ratio

remains unchanged.

Below figure shows a device that adds heat to a stream of moist air. For

steady-flow conditions, the required rate of heat addition is

( )2 1daQ m h h= −i i

( ) ( ) ( )da 2 1 w da da1 1 w1 da da2 2 w2 m W W m W h m W h 0Q h h h+ − + + − + =i

ɺ ɺ ɺ

1 da1 1 w1W hh h= + 2 da2 2 w2W hh h= +

1 2

Heating coil

Qɺ

damɺ

h1

W1

damɺ

h2

W2

w da 2 1m m ( )W W= −ɺ ɺ

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Heating of moist air

tdb

W

t1

W1

t2

h1

h2

2 1( )daQ m h h= −ɺ ɺ

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An example – heating/coolingMoist air, saturated at 2°C, enters a heating coil at a rate of 10

m3/s. Air leaves the coil at 40°C. Find the required rate of heat

addition.

State 1 is located on the saturation curve at 2°C. Thus, h1 =

13.0 kJ/kg, W1 = 4.38 gw/kg, and v1 = 0.785 m3/kgda. State 2 is

located at the intersection of t = 40°C and W2 = W1 = 4.38

gw/kg. Thus, h2 = 51.5 kJ/kg. The mass flow is:

3

3

10 // 12.74 /

0.785 /da

da

m sm Q v kg s

m kg= = =

i

( ) ( )2 1 12.74 / 51.5 13.0 / 490kWda da daq m h h kg s kJ kg= − = × − =i

Why is the specific volume based on mass of dry air?

52

Cooling of Moist Air

1 2

Cooling

coilQi

damɺ

h1

W1

damɺ

h2

W2

da1 da2 dam m m= =ɺ ɺ ɺMass balance: w1 w2m m=ɺ ɺ

1 2( )daQ m h h= −ɺ ɺEnergy balance:

53

Cooling of Moist Air

tdb

W

t2

W1

t1

h2

h1

1 2( )daQ m h h= −ɺ ɺ

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An exerciseOutside air at dry bulb temperature of 5oC, 80% RH enters a preheater

coil and leaves at 24oC (dry bulb). The air volume flow rate is 6.5

m3/s. Find (a) the outdoor air wet-bulb temperature and specific

volume; (b) the heated air moisture content and RH; and (c) the

heating coil power (rate of heating addition to the coil)

Find the enthalpy of the air sample before the coil and after the

coil, the change in enthalpy is caused by the preheater.

55

The exercise

56

Cooling and Dehumidification

1 2

3

damɺ

h1

W1

damɺ

h2

W2

Cooling

coilQi

57

Cooling and Dehumidification

What if the supply air temperature is higher than 14oC?

58

Heating and Humidifying

1 2Qi

Heating coil

damɺ

h1

W1

damɺ

h2

W2

wmɺ hwa

59

Heating and Humidifying

tdb

W

t1

W1

t2

h1

h2

W2

2

1a

ha

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Adiabatic mixing of two moist air streams

,1 1 ,2 2 ,3 3da da dam h m h m h+ =ɺ ɺ ɺ

,1 ,2 ,3da da dam m m+ =ɺ ɺ ɺ

,1 1 ,2 2 ,3 3da da dam W m W m W+ =ɺ ɺ ɺ

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( )

( ) ( )

( )

( )

,1 1 ,2 2 ,1 ,2 3

,1 1 3 ,2 3 2

,13 2

1 3 ,2

da da da da

da da

da

da

m h m h m m h

m h h m h h

mh h

h h m

+ = +

⇒ − = −

−⇒ =

−

ɺ ɺ ɺ ɺ

ɺ ɺ

ɺ

ɺ

( )

( )

( )

( )

( )

,1 1 ,2 2 ,1 ,2 3

,13 2 3 2

1 3 ,2 1 3

da da da da

da

da

m W m W m m W

mW W h h

W W m h h

+ = +

− −⇒ = =

− −

ɺ ɺ ɺ ɺ

ɺ

ɺ

62

Thank you very much for your attention!

Please feel free to contact me if you have any

questions.

Meng Ni

ZN713

Tel: 2766 4152 (office)

Email: bsmengni@polyu.edu.hk