Chapter 14

مخلوط هوا - بخار و تهویه مطبوع

Thermodynamics: An Engineering Approach, 7th editionby Yunus A. Çengel and Michael A. Boles

2

در بحث مخلوط گازها در فصل گذشته با هیچ گونه فرایند تبدیل فاز •روبرو نبودیم. در این فصل مخلوط هوای خشک و بخار آب مورد

بررسی قرار می گیرد که ممکن است با چگالش بخار و تبدیل آن به مایع روبرو باشیم. به مخلوط هوا و بخار آب، هوای اتمسفری می

گویند.- تغییرات دمای هوای اتمسفری در مسایل تهویه مطبوع معموال از •

10oC 50 تاoC .است در این شرایط هوا را می توان گاز آرمانی با گرمای ویژه ثابت در •

نظر گرفت. (Cpa=1.005 kJ/kgK)

در این صورت انتالپی هوای خشک با فرض مقدار انتالپی صفر در •دمای صفر درجه سلسیوس به عنوان نقطه مرجع چنین خواهد بود:

3

Note: For the dry air-water vapor mixture, the partial pressure of the water vapor in the mixture is less that its saturation pressure at the temperature.

P Pv sat Tmix @

فشار اشباع متناظر با دمای •50oC 12.3 برابر با kPa می

باشد. در این محدوده فشار، همان طور که از شکل روبرو

پیداست، در دمای ثابت، انتالپی ثابت می ماند. بنابراین، انتالپی

بخار آب را می توان معادل انتالپی بخار اشباع در دمای مخلوط هوا و بخار گرفت.

انتالپی بخار اشباع بنابراین به •صورت تابعی از دما به شکل

زیر قابل بیان خواهد بود.

4

Consider increasing the total pressure of an air-water vapor mixture while the temperature of the mixture is held constant. See if you can sketch the process on the P-v diagram relative to the saturation lines for the water alone given below. Assume that the water vapor is initially superheated.

P

v

When the mixture pressure is increased while keeping the mixture temperature constant, the vapor partial pressure increases up to the vapor saturation pressure at the mixture temperature and condensation begins. Therefore, the partial pressure of the water vapor can never be greater than its saturation pressure corresponding to the temperature of the mixture.

5

Definitions

Dew Point, Tdp

The dew point is the temperature at which vapor condenses or solidifies when cooled at constant pressure.

Consider cooling an air-water vapor mixture while the mixture total pressure is held constant. When the mixture is cooled to a temperature equal to the saturation temperature for the water-vapor partial pressure, condensation begins.

When an atmospheric air-vapor mixture is cooled at constant pressure such that the partial pressure of the water vapor is 1.491 kPa, then the dew point temperature of that mixture is 12.95oC.

0 2 4 6 8 10 1212-25

25

75

125

175

225

275

325

375

s [kJ/kg-K]

T [

C]

1.491 kPa

Steam

TDP

6

Relative Humidity, ϕ

Mass of vapor in air

Mass of in saturated air

m

m

P

P

v

g

v

g

Pv and Pg are shown on the following T-s diagram for the water-vapor alone.

0 2 4 6 8 10 1212-25

25

75

125

s [kJ/kg-K]

T [

C]

Pv = 1.491 kPa

Pg = 3.169 kPa

Steam

o

Tdp

Tm Vapor State

Since P P or

P

P

kPa

kPag vv

g

,.

.. 1 100%,

1491

31690 47

7

Absolute humidity or specific humidity (sometimes called humidity ratio),

Mass of water vapor in air

Mass of dry air

m

m

PVM R T

PVM R T

P M

P M

P

P

P

P P

v

a

v v u

a a u

v v

a a

v

a

v

v

/ ( )

/ ( )

. .0 622 0 622

Using the definition of the specific humidity, the relative humidity may be expressed as

P

Pand

P

P Pg

g

g( . )

.

0 622

0 622

Volume of mixture per mass of dry air, v

vV

m

m R T P

ma

m m m m

a

/

After several steps, we can show (you should try this)

vV

mv

R T

Paa

a m

a

8

So the volume of the mixture per unit mass of dry air is the specific volume of the dry air calculated at the mixture temperature and the partial pressure of the dry air.

Mass of mixture

m m m mm

mma v a

v

aa ( ) ( )1 1

Mass flow rate of dry air, ma

Based on the volume flow rate of mixture at a given state, the mass flow rate of dry air is

/

/m

V

v

m s

m kg

kg

saa

a 3

3

Enthalpy of mixture per mass dry air, h

hH

m

H H

m

m h m h

m

h h

m

a

a v

a

a a v v

a

a v

9

Example 14-1

Atmospheric air at 30oC, 100 kPa, has a dew point of 21.3oC. Find the relative humidity, humidity ratio, and h of the mixture per mass of dry air.

2.5480.6 60%

4.247v

g

P kPaor

P kPa

10

h h h

C T T

kJ

kg CC

kg

kgC

kJ

kg

kJ

kg

a v

p a

ao

o v

a

o

v

a

, ( . . )

. ( ) . ( . . ( ))

.

25013 182

1005 30 0 01626 25013 182 30

7171

0 6222 548

100 2 5480 01626.

.

( . ).

kPa

kPa

kg

kgv

a

11

Example 14-2

If the atmospheric air in the last example is conditioned to 20oC, 40 percent relative humidity, what mass of water is added or removed per unit mass of dry air?

At 20oC, Pg = 2.339 kPa.

P P kPa kPa

wP

P P

kPa

kPa

kg

kg

v g

v

v

v

a

0 4 2 339 0 936

0 622 0 6220 936

100 0 936

0 00588

. ( . ) .

. ..

( . )

.

The change in mass of water per mass of dry air is

m m

mv v

a

, ,2 12 1

m m

m

kg

kg

kg

kg

v v

a

v

a

v

a

, , ( . . )

.

2 1 0 00588 0 01626

0 01038

12

Or, as the mixture changes from state 1 to state 2, 0.01038 kg of water vapor is condensed for each kg of dry air. Example 14-3

Atmospheric air is at 25oC, 0.1 MPa, 50 percent relative humidity. If the mixture is cooled at constant pressure to 10oC, find the amount of water removed per mass of dry air.

Sketch the water-vapor states relative to the saturation lines on the following T-s diagram.

T

sAt 25oC, Psat = 3.170 kPa, and with = 50%1

,1 1 ,1

,1 @

0.5(3.170 ) 1.585

13.8v

v g

odp sat P

P P kPa kPa

T T C

13

wP

P P

kPa

kPa

kg

kg

v

v

v

a

11

1

0 622 0 62215845

100 15845

0 01001

. ..

( . )

.

,

,

Therefore, when the mixture gets cooled to T2 = 10oC < Tdp,1, the mixture is saturated,

and = 100%. Then Pv,2 = Pg,2 = 1.228 kPa. 2

,22

,2

1.2280.622 0.622

(100 1.228)

0.00773

v

v

v

a

P kPaw

P P kPa

kg

kg

The change in mass of water per mass of dry air is

m m

m

kg

kg

kg

kg

v v

a

v

a

v

a

, ,

( . . )

.

2 12 1

0 00773 0 01001

0 00228

14

Or as the mixture changes from state 1 to state 2, 0.00228 kg of water vapor is condensed for each kg of dry air.

Steady-Flow Analysis Applied to Gas-Vapor Mixtures

We will review the conservation of mass and conservation of energy principles as they apply to gas-vapor mixtures in the following example.

Example 14-3

Given the inlet and exit conditions to an air conditioner shown below. What is the heat transfer to be removed per kg dry air flowing through the device? If the volume flow rate of the inlet atmospheric air is 17 m3/min, determine the required rate of heat transfer. Cooling fluid

In Out

Atmospheric air

T1 = 30oC

P1 =100 kPa

1 = 80%

1 = 17m3/min

Condensateat 20oC

Insulated flow duct

T2 = 20oC

P2 = 98 kPa

2 = 95%

V

15

Before we apply the steady-flow conservation of mass and energy, we need to decide if any water is condensed in the process. Is the mixture cooled below the dew point for state 1?

,1 1 ,1

,1 @

0.8(4.247 ) 3.398

26.01v

v g

odp sat P

P P kPa kPa

T T C

So for T2 = 20oC < Tdp, 1, some water-vapor will condense. Let's assume that the condensed water leaves the air conditioner at 20oC. Some say the water leaves at the average of 26 and 20oC; however, 20oC is adequate for our use here.

Apply the conservation of energy to the steady-flow control volume

( ) ( )Q m hV

gz W m hV

gznet iinlets

i net eexits

e

2 2

2 2

Neglecting the kinetic and potential energies and noting that the work is zero, we get

Q m h m h m h m h m hnet a a v v a a v v l l 1 1 1 1 2 2 2 2 2 2

Conservation of mass for the steady-flow control volume is

m miinlets

eexits

16

For the dry air: m m ma a a1 2

For the water vapor: m m mv v l1 2 2

The mass of water that is condensed and leaves the control volume is

( )

m m m

ml v v

a

2 1 2

1 2

Divide the conservation of energy equation by , then ma

( )

( )

Q

mh h h h h

Q

mh h h h h

net

aa v a v l

net

aa a v v l

1 1 1 2 2 2 1 2 2

2 1 2 2 1 1 1 2 2

( ) ( )

Q

mC T T h h hnet

apa v v l 2 1 2 2 1 1 1 2 2

17

Now to find the 's and h's.

11

1 1

0.622 0.622(3.398)

100 3.398

0.02188

v

v

v

a

P

P P

kg

kg

P P

kPa kPa

P

P P

kg

kg

v g

v

v

v

a

2 2 2

22

2 2

0 95 2 339 2 222

0 622 0 622 2 222

98 2 222

0 01443

( . )( . ) .

. . ( . )

.

.

18

Using the steam tables, the h's for the water are

1

2

2

2555.6

2537.4

83.91

vv

vv

lv

kJh

kg

kJh

kg

kJh

kg

The required heat transfer per unit mass of dry air becomes

2 1 2 2 1 1 1 2 2( ) ( )

1.005 (20 30) 0.01443 (2537.4 )

0.02188 (2555.6 ) (0.02188 0.01443) (83.91 )

28.73

netpa v v l

a

o vo

a a v

v v

a v a v

a

QC T T h h h

m

kgkJ kJC

kg C kg kg

kg kgkJ kJ

kg kg kg kg

kJ

kg

19

The heat transfer from the atmospheric air is

28.73netout

a a

Q kJq

m kg

The mass flow rate of dry air is given by

m

V

va 1

131

11

3

0.287 (30 273)

(100 3.398)

0.90

a a

a

a

kJK

R T kg K m kPav

P kPa kJ

m

kg

3

3

17min 18.89

min0.90

aa

a

mkg

mmkg

1min 118.89 (28.73 )

min 60

0.28439.04 2.57

aout a out

a

kg kJ kWsQ m q

kg s kJ

TonskW Tons of refrigeration

kW

20

The Adiabatic Saturation Process Air having a relative humidity less than 100 percent flows over water contained in a well-insulated duct. Since the air has < 100 percent, some of the water will evaporate and the temperature of the air-vapor mixture will decrease.

21

If the mixture leaving the duct is saturated and if the process is adiabatic, the temperature of the mixture on leaving the device is known as the adiabatic saturation temperature.

For this to be a steady-flow process, makeup water at the adiabatic saturation temperature is added at the same rate at which water is evaporated.

We assume that the total pressure is constant during the process.

Apply the conservation of energy to the steady-flow control volume

( ) ( )Q m hV

gz W m hV

gznet iinlets

i net eexits

e

2 2

2 2

Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero, we get

m h m h m h m h m ha a v v l l a a v v1 1 1 1 2 2 2 2 2 2

Conservation of mass for the steady-flow control volume is

m miinlets

eexits

22

For the dry air: m m ma a a1 2

For the water vapor: m m mv l v1 2 2

The mass flow rate water that must be supplied to maintain steady-flow is,

( )

m m m

ml v v

a

2 2 1

2 1

Divide the conservation of energy equation by , thenma

h h h h ha v l a v1 1 1 2 1 2 2 2 2 ( )

What are the knowns and unknowns in this equation?

23

Since 1 is also defined by 11

1 1

0 622

.P

P Pv

v

We can solve for Pv1. PP

v11 1

10 622

.

Then, the relative humidity at state 1 is 11

1

P

Pv

g

Solving for 1

12 1 2 2 2

1 2

2 1 2 2

1 2

h h h h

h h

C T T h

h h

a a v l

v l

pa fg

g f

( )

( )

( )

( )

24

Example 14-4

For the adiabatic saturation process shown below, determine the relative humidity, humidity ratio (specific humidity), and enthalpy of the atmospheric air per mass of dry air at state 1.

25

Using the steam tables:

2

1

2

67.2

2544.7

2463.0

fv

vv

fgv

kJh

kg

kJh

kg

kJh

kg

From the above analysis

2 1 2 21

1 2

( )

( )

1.005 16 24 0.0115 (2463.0 )

(2544.7 67.2)

0.00822

pa fg

g f

o vo

a va

v

v

a

C T T h

h h

kgkJ kJC

kg kgkg CkJkg

kg

kg

26

We can solve for Pv1.

1 11

10.622

0.00822(100 )

0.622 0.008221.3

v

PP

kPa

kPa

Then the relative humidity at state 1 is

1 11

1 @24

1.30.433 43.3%

3.004

o

v v

g sat C

P P

P P

kPaor

kPa

The enthalpy of the mixture at state 1 is

1 1 1 1

1 1 1

1.005 (24 ) 0.00822 2544.7

45.04

a v

pa v

o vo

a a v

a

h h h

C T h

kgkJ kJC

kg C kg kg

kJ

kg

27

Wet-Bulb and Dry-Bulb Temperatures

In normal practice, the state of atmospheric air is specified by determining the wet-bulb and dry-bulb temperatures. These temperatures are measured by using a device called a psychrometer. The psychrometer is composed of two thermometers mounted on a sling. One thermometer is fitted with a wet gauze and reads the wet-bulb temperature. The other thermometer reads the dry-bulb, or ordinary, temperature. As the psychrometer is slung through the air, water vaporizes from the wet gauze, resulting in a lower temperature to be registered by the thermometer. The dryer the atmospheric air, the lower the wet-bulb temperature will be. When the relative humidity of the air is near 100 percent, there will be little difference between the wet-bulb and dry-bulb temperatures. The wet-bulb temperature is approximately equal to the adiabatic saturation temperature. The wet-bulb and dry-bulb temperatures and the atmospheric pressure uniquely determine the state of the atmospheric air.

28

The Psychrometric Chart

For a given, fixed, total air-vapor pressure, the properties of the mixture are given in graphical form on a psychrometric chart.

The air-conditioning processes:

29

30

Example 14-5 Determine the relative humidity, humidity ratio (specific humidity), enthalpy of the atmospheric air per mass of dry air, and the specific volume of the mixture per mass of dry air at a state where the dry-bulb temperature is 24oC, the wet-bulb temperature is 16oC, and atmospheric pressure is 100 kPa.

From the psychrometric chart read

44%

8 0 0 008

46

08533

. .

.

g

kg

kg

kg

hkJ

kg

vm

kg

v

a

v

a

a

a

NOTE: THE ENTHALPY READ FROM THE PSYCHROMETRIC CHART IS THE TOTAL ENTHALPY OF THE AIR-VAPOR MIXTURE PER UNIT MASS OF DRY AIR. h= H/ma = ha + ωhv

31

Example 14-6

For the air-conditioning system shown below in which atmospheric air is first heated and then humidified with a steam spray, determine the required heat transfer rate in the heating section and the required steam temperature in the humidification section when the steam pressure is 1 MPa.

32

The psychrometric diagram is

-10 -5 0 5 10 15 20 25 30 35 400.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

0.040

0.045

0.050

T [C]

Hu

mid

ity

Ra

tio

Pressure = 101.3 [kPa]

0.2

0.4

0.6

0.8

0 C

10 C

20 C

30 C

Psychrometric Diagram

1

2

3

1=

2=0.0049 kgv/kga

3=0.0091kgv/kga

h2=37 kJ/kga

h1=17 kJ/kga

h3=48 kJ/kga

v1=0.793 m^3/kga

Apply conservation of mass and conservation of energy for steady-flow to process 1-2.

Conservation of mass for the steady-flow control volume is

m miinlets

eexits

33

For the dry air m m ma a a1 2

For the water vapor (note: no water is added or condensed during simple heating)

m mv v1 2Thus,

2 1Neglecting the kinetic and potential energies and noting that the work is zero, and letting the enthalpy of the mixture per unit mass of air h be defined as

h h ha v we obtain

( )

E E

Q m h m h

Q m h h

in out

in a a

in a

1 2

2 1

34

Now to find the and h's using the psychrometric chart.ma

At T1 = 5oC, 1 = 90%, and T2 = 24oC:

The mass flow rate of dry air is given by

mV

va 1

1

35

min

..

min

min.m

m

mkg

kg

s

kg

sa

a

a a 60

0 79375 66

1

601261

3

3

The required heat transfer rate for the heating section is

. ( )

.

Qkg

s

kJ

kg

kWs

kJ

kW

ina

a

1261 37 171

2522This is the required heat transfer to the atmospheric air. List some ways in which this amount of heat can be supplied.

At the exit, state 3, T3 = 25oC and 3 = 45%. The psychrometric chart gives

36

Apply conservation of mass and conservation of energy to process 2-3.

Conservation of mass for the steady-flow control volume is

m miinlets

eexits

For the dry air m m ma a a2 3

For the water vapor (note: water is added during the humidification process)

( )

. ( . . )

.

m m m

m m m

m m

kg

s

kg

kg

kg

s

v s v

s v v

s a

a v

a

v

2 3

3 2

3 2

1261 0 0089 0 0049

0 00504

37

Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero, the conservation of energy yields

( )

E E

m h m h m h

m h m h h

in out

a s s a

s s a

2 3

3 2

Solving for the enthalpy of the steam,

( ) ( )m h m h h

hh h

a s a

s

3 2 3 2

3 2

3 2

38

At Ps = 1 MPa and hs = 2750 kJ/kgv, Ts = 179.88oC and the quality xs = 0.985.

See the text for applications involving cooling with dehumidification, evaporative cooling, adiabatic mixing of airstreams, and wet cooling towers.

(48 37)

(0.0089 0.0049)

2750

as

v

a

v

kJkg

hkgkg

kJ

kg