Lecture note psycrometry air-conditioning

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Psychrometry & Air-

Conditioning

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Introduction

Moist air (atmospheric air) is a mixture of twogases:

� dry air – mixture of a number of gases(nitrogen, oxygen and traces of argon, carbondioxide, neon etc.)

� water vapour (which may exist in a saturatedor superheated state – can be treated as anideal gas).

Although the amount of water vapour is small, itplays a major role in human comfort.

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2

� In psychrometry analysis, the water vapour is

treated as the variable component since the

amount of water vapour changes as a result of

condensation and evaporation from oceans,

lakes etc.

� The dry air is treated as the fixed component.

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AssumptionsAssumptions:

� Water vapour contains no dissolved gases.

� Both dry air and water vapour can be

considered as an ideal gas mixture since both

exist in the atmosphere at low pressures.

� No interaction between components and all

components are at the mixture temperature.

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Psychrometric Properties

(Properties of Moist Air)

� Dry air and water vapour form a binarymixture.

� A mixture of two substances requires 3properties to completely define itsthermodynamic state, unlike a pure substancewhich requires only 2.

� One of the three properties can be thecompositions.

� The properties of moist air (atmospheric air)are called psychrometric properties.

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Psychrometric Properties

(Properties of Moist Air)

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Volume V

Mass m

Pressure P (barometric pressure)

Temperature T

(T is the dry bulb temperature (DBT))

For defining and calculating the relevant

psychrometric properties, we may consider a

certain volume V of moist air at pressure P and

temperature T, containing ma kg of dry air and mv

kg of water vapour.

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Specific (or absolute) humidity of air

� The amount of water vapour can be specified

by the mass of water vapour present in a

unit mass of dry air.

� This is called humidity ratio, moisture

content, specific or absolute humidity and is

denoted by ω:

(kg water vapor/kg dry air)v

a

m

mω =

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Specific (or absolute) humidity of air

It can also be expressed as:

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air)dry our/kg water vap(kg 622.0

or

ly.respectiveour water vapandair dry toreferring and Subscript

622.0/

/

/

/

v

v

a

v

aa

vv

aa

vv

a

v

PP

P

va

P

P

RP

RP

TRVP

TRVP

m

m

−=

====

ω

ω

Since:

P = Pa + Pv (kPa) (Where P = atmospheric pressure)

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Relative humidity

� The amount of moisture in the air has a

definite effect on how we feel in an

environment.

� This comfort level depends more on the

amount of moisture the air holds (mv)

relative to the maximum amount of moisture

the air can hold at the same temperature

(mg).

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Relative humidity

The ratio of these two quantities is called the

relative humidity, φ.

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g

v

vg

vv

g

v

P

P

TRVP

TRVP

m

m===

/

/φ

Note:

Pg = Psat @T

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Relative humidity

� The relative humidity and specific humidity can also be

expressed as,

� The relative humidity ranges from 0 for dry air to 1 for

saturated air.

� The amount of moisture air can hold depends on its

temperature.

� Therefore, the relative humidity of air changes with

temperature even when its specific humidity remains constant.

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( )

g

g

g

PP

P

P

P

φ

φω

ω

ωφ

−=

+=

622.0

and

622.0

Specific enthalpy

� The enthalpy of moist air is expressed in terms of the

enthalpies of the dry air and the water vapour.

� The total enthalpy (an extensive property) of

atmospheric air (moist air) is,

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( )airdry kJ/kg

gives, by Dividing

va

v

a

va

a

a

vvaava

hhh

hm

mh

m

Hh

m

hmhmHHH

ω+=

+==

+=+=

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Specific enthalpy

The enthalpy of dry air can be determined from,

where, T is the dry bulb temperature.

The enthalpy of water vapour can be expressed by,

where hg is the enthalpy of saturated vapour at the

same temperature.

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( ) (kJ/kg) CkJ/kg.005.1 o TTch pa ==

( ) ( ) (kJ/kg) low, ThPTh gv ≅

Specific enthalpy

hg, can also be determine approximately from,

in the temperature range of -10oC to 50oC with negligible error.

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( ) (kJ/kg) 82.19.2500 TThg +≅

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Example 1Example 1

A tank contains 21 kg of dry air and 0.3 kg of

water vapour at 30oC and 100 kPa total

pressure. Determine,

a) the specific humidity;

b) the relative humidity;

c) the volume of the tank.

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Example 2Example 2

A room contains air at 20oC and 98 kPa at a relative

humidity of 85 percent. Determine,

a) the partial pressure of dry air;

b) the specific humidity of the air;

c) the enthalpy per unit mass of dry air.

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Dew-Point Temperature

� The dew-point temperature Tdp

is defined as the temperature at

which condensation begins

when the air is cooled at

constant pressure.

� Or, the saturation temperature of

water corresponding to the

vapour pressure.

vPsatdp TT @=

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Constant-pressure cooling of

moist air and the dew-point

temperature on the T-s

diagram of water.


� As the air cools at constant

pressure, the vapor

pressure Pv remains

constant.

� Therefore, the vapor in the

air (state 1) undergoes a

constant-pressure cooling

process until it strikes the

saturated vapor line (state

2).

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diagram of water.

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� The temperature at this

point is Tdp, and if the

temperature drops any

further, some vapor

condenses out.

� As a result, the amount of

vapor in the air decreases,

which results in a decrease

in Pv.

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diagram of water.


� The air remains saturated

during the condensation

process and thus follows a

path of 100 % relative

humidity (the saturated

vapor line).

� The ordinary temperature

and the dew-point

temperature of saturated air

are identical.

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diagram of water.

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Example 3Example 3

In cold weather, condensation frequently occurs on

the inner surfaces of the windows due to the lower air

temperatures near the window surface. Consider a

house that contains air at 20oC and 75% relative

humidity. At what window temperature will the

moisture in the air start condensing on the inner

surfaces of the windows.

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Adiabatic Saturation And Wet-Bulb

Temperatures

� One way to determine relative humidity and specific humidity is by determining the dew-point temperature of air, viz.,

1. knowing Tdp,

2. determine the Pv,

3. able to determine the ω,

4. finally, able to determine the φ.

� This approach is simple, but not quite practical.

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diagram of water.

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Temperatures

� A more practical approach is to use a thermometer whose bulb is covered with a cotton wick saturated water and to blow air over the wick.

� The temperature measured in this manner is called the wet-bulb temperature, Twb

and it is commonly used in air-conditioning applications.

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A simple arrangement to

measure the wet-bulb

temperature


Temperatures

� This method is related to an

adiabatic saturation process,

shown schematically and on

a T-s.

� Where it can be analyzed as

a steady-flow process, which

involves no heat and work

interactions, and kinetic and

potential energy changes

are neglected.

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The adiabatic saturation

process & its representation on

a T-s diagram of water

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Temperatures

Thus, the ω and φ of air can be determined from

equations below,

( )

2

2

21

2

2

2

212

1

622.0

where,

g

g

fg

fgp

PP

P

hh

hTTc

−=

−

+−=

ω

ωω

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Temperatures

22

22

22

11

@

@

2

@

@

22

1

1

2

o

pressure catmospheri

at humidity specific

air ofhumidity specific

e)temperatur(ordinary re temperatubulb-dry

re temperatubulb-wet

C)kJ/kg. (1.005heat specific pressure-constant

Where,

Tfgfg

Tsatg

Tff

Tgg

p

hh

PP

P

hh

hh

T

T

T

c

=

=

=

=

=

=

=

=

=

=

ω

ω

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Temperatures

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Sling psychrometer

Example Example 44

The air in a room has a dry-bulb temperature of

22oC and a wet-bulb temperature of 16oC.

Assuming a pressure of 100 kPa, determine,


b) the relative humidity;

c) the dew-point temperature.

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The Psycrometric Chart

� Psychrometric charts

present the moist air data

in the form of easily

readable charts,

calculated from the

previous relations.

� They are used extensively

in air-conditioning

applications.

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Schematic for a psyhcrometricchart

Example 5Example 5

A room contains air at 1 atm, 26oC, and 70 %

relative humidity. Using the psychrometric

chart, determine,


b) the enthalpy;

c) the wet-bulb temperature;

d) the dew-point temperature;

e) the specific volume of the air.

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Air-Conditioning Processes

� Air-conditioning processes are processes to

maintain a living space or an industrial facility

at the desired temperature and humidity.

� These processes include,

i. Simple heating (raising the temperature)

ii. Simple cooling (lowering the temperature)

iii. Humidifying (adding moisture)

iv. Dehumidifying (removing moisture)

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� Sometimes two or more of

these processes are needed to

bring the air to a desired

temperature and humidity

level.

� Various air-conditioning

processes can be illustrated

on the psychrometric chart.

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Various air-conditioning

processes

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Simple heating and cooling

processes appear as horizontal

lines on this chart since the

moisture content of the air remains

constant (ω = constant) during

these processes.

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processes


Air commonly heated and

humidified in winter.

And it commonly cooled

and dehumidified in

summer.

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processes

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Most air-conditioning processes can be

modelled as steady-flow processes, and thus

the mass balance relation can be

expressed for dry air and water as,outin mm

••

=

∑∑••

==out

a

in

a mmairdryforbalanceMass

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∑∑

∑∑

••

••

=

==

out

a

in

a

out

w

in

w

mm

mmwaterforbalanceMass

ωω

or

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Disregarding the kinetic and potential energy changes, the

steady-flow energy balance relation can be expressed in

this case as,

The work term usually consists of fan work input, which is

small relative to other terms in energy balance relation.

∑ ∑••••••

++=++in out

outoutininhmWQhmWQ

outin EE••

=

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Simple Heating and Cooling (ω = constant)

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During simple heating, specific

humidity remains constant, but

relative humidity decreases.

During simple cooling, specific


relative humidity increases.

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Simple Heating and Cooling (ω = constant)

During simple heating, specific humidity remains

constant, but relative humidity decreases.

Many residential heating systems consist of a stove, a heat pump, or an electric

resistance heater. The air in these systems is heated by circulating it through a duct

that contains the tubing for the hot gases or the electric resistance wires.

Cooling can be accomplished by passing the air over some coils through which a

refrigerant or chilled water flows.

Heating and cooling appear as a horizontal line since no moisture is added to or

removed from the air.

Dry air mass balance

Water mass balance

Energy balance

During simple cooling, specific


relative humidity increases.

Example Example 66

A heating section consists of a 38 cm-diameter

duct that houses a 4-kW electric resistance

heater. Air enters the heating section at 1 atm,

10oC, and 40%rh at a velocity of 8 m/s.

Determine,

a) the exit temperature,

b) the relative humidity of the air,

c) the exit velocity.

40

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1 2

10oC 13.5oC

40% 31.5%

Heating with Humidification

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Heating with Humidification

Problems with the low relative humidity resulting from simple heating can be

eliminated by humidifying the heated air. This is accomplished by passing the air

first through a heating section and then through a humidifying section.

Example Example 77

Air at 1 atm, 15oC, and 60%rh is first heated to 20oC in a

heating section and then humidified by introducing water

vapor. The air leaves the humidifying section at 25oC

and 65%rh. Determine,

a) the amount of steam added to the air [kg w.v/kg d.a],

b) the amount of heat transfer to the air in the heating

section [kJ/kg d.a].

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15oC 20oC

Cooling with Dehumidification

� The specific humidity of air remains constant

during a simple cooling process, but its relative

humidity increases.

� If the φ reaches undesirably high levels, it may be

necessary to remove some moisture from the air,

that is, to dehumidify it.

� This requires cooling the air below its dew point

temperature.

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Cooling with Humidification

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Example of schematic and psychrometric chart

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Cooling with DehumidificationThe specific humidity of air remains constant during a simple cooling process, but

its relative humidity increases. If the relative humidity reaches undesirably high

levels, it may be necessary to remove some moisture from the air, that is, to

dehumidify it. This requires cooling the air below its dew-point temperature.

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Example 8Example 8

Air enters a window air conditioner at 1 atm, 30oC,

and 80% relative humidity at a rate of 10 m3/min,

and it leaves at saturated air at 14oC. Part of the

moisture in the air that condenses during the

process is also removed at 14oC. Determine the rates

of heat and moisture removal from the air.

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Example Example 99

2000 m3/hr of atmospheric air at 28oC with a dew point of

25oC flows into an air conditioner that uses chilled water as

the working fluid. The atmospheric air is to be cooled to

18oC. Determine mass flow rate of the condensate water, if

any, leaving the air conditioner, in kg/hr. If the cooling

water has a 10oC temperature rise while flowing through

the air conditioner, determine the volume flow rate of

chilled water supplied to the air conditioner heat

exchanger, m3/min. The air conditioning process takes

place at 100 kPa.

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Date post:	19-Jan-2015
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Lecture note psycrometry air-conditioning

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