Transcript Chapter 14: Gas-Vapor Mixtures and Air

Chapter 14
‫ بخار و تهویه مطبوع‬- ‫مخلوط هوا‬
Thermodynamics: An Engineering Approach, 7th edition
by Yunus A. Çengel and Michael A. Boles
‫• 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 بررس ی 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 ‪ -10oC‬تا ‪ 50oC‬است‪.‬‬
‫• در این شرایط هوا را می توان گاز آرمانی با گرمای ویژه ثابت در نظر گرفت‪.‬‬
‫)‪(Cpa=1.005 kJ/kgK‬‬
‫• در این صورت انتالپی هوای خشک با فرض مقدار انتالپی صفر در دمای صفر درجه سلسیوس به‬
‫عنوان نقطه مرجع چنین خواهد بود‪:‬‬
‫‪2‬‬
‫• فشار اشباع متناظر با دمای ‪ 50oC‬برابر با‬
‫‪12.3 kPa‬می 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.‬‬
‫‪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.‬‬
‫‪3‬‬
‫‪Pv  Psat @Tmix‬‬
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.
4
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.
Steam
375
325
275
T [C]
225
175
1.491 kPa
125
75
TDP
25
-25
0
2
4
6
s [kJ/kg-K]
8
10
12
5
Relative Humidity, ϕ

Mass of vapor in air
m
 v
Mass of in saturated air mg
Pv

Pg
Pv and Pg are shown on the following T-s diagram for the water-vapor alone.
Steam
125
T [C]
75
Tm
25
o Vapor State
Pg = 3.169 kPa
Tdp
Pv = 1.491 kPa
-25
0
2
4
6
8
10
12
s [kJ/kg-K]
Since
Pv 1491
.
kPa
Pg  Pv ,   1 or 100%,  

 0.47
Pg 3169
.
kPa
6
Absolute humidity or specific humidity (sometimes called humidity ratio), 

Mass of water vapor in air mv

Mass of dry air
ma
PVM
Pv M v
v
v / ( Ru T )


PVM
Pa M a
a
a / ( Ru T )
P
Pv
 0.622 v  0.622
Pa
P  Pv
Using the definition of the specific humidity, the relative humidity may be expressed
as
0.622Pg
P

and  
(0.622   ) Pg
P  Pg
Volume of mixture per mass of dry air, v
v
V
m R T /P
 m m m m
ma
ma
After several steps, we can show (you should try this)
v
V
RT
 va  a m
ma
Pa
7
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
mv
m  ma  mv  ma (1  )  ma (1   )
ma
m a
Mass flow rate of dry air,
Based on the volume flow rate of mixture at a given state, the mass flow rate of dry
air is
V
a 
m
v
m3 / s
kga

3
m / kga
s
Enthalpy of mixture per mass dry air, h
Hm Ha  Hv ma ha  mv hv
h


ma
ma
ma
 ha  hv
8
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.
Pv 2.548 kPa
 
 0.6 or 60%
Pg 4.247 kPa
9
2.548 kPa
kgv
  0.622
 0.01626
(100  2.548) kPa
kga
h  ha  hv
 C p , a T   (25013
.  182
. T)
kJ
kgv
kJ
o
o
 1005
.
(30 C )  0.01626
(25013
.  182
. (30 C ))
o
kga  C
kga
kgv
kJ
 7171
.
kga
10
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.
Pv  Pg  0.4(2.339 kPa )  0.936 kPa
w  0.622
Pv
0.936 kPa
 0.622
P  Pv
(100  0.936) kPa
 0.00588
kgv
kga
The change in mass of water per mass of dry air is
mv , 2  mv ,1
mv , 2  mv ,1
ma
ma
  2  1
 (0.00588  0.01626)
kgv
 0.01038
kga
kgv
kga
11
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
At
25oC,
Psat = 3.170 kPa, and with  1 = 50%
s
Pv ,1  1 Pg ,1  0.5(3.170 kPa )  1.585 kPa
Tdp ,1  Tsat @ Pv  13.8o C
12
w1  0.622
Pv ,1
P  Pv ,1
 0.01001
 0.622
15845
.
kPa
(100  15845
.
) kPa
kgv
kga
Therefore, when the mixture gets cooled to T2 = 10oC < Tdp,1, the mixture is saturated,
and  2 = 100%. Then Pv,2 = Pg,2 = 1.228 kPa.
w2  0.622
Pv ,2
P  Pv ,2
 0.622
1.228 kPa
(100  1.228) kPa
kgv
 0.00773
kg a
The change in mass of water per mass of dry air is
mv , 2  mv ,1
ma
  2  1
 (0.00773  0.01001)
kgv
 0.00228
kga
kgv
kga
13
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
Insulated flow duct
Atmospheric
air
T2 = 20oC
T1 = 30oC
P1 =100 kPa
 1 = 80%
3
V 1 = 17m /min
Condensate
at 20oC

P2 = 98 kPa
2=
95%
14
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?
Pv ,1  1 Pg ,1  0.8(4.247 kPa )  3.398 kPa
Tdp ,1  Tsat @ Pv  26.01o 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
2
2
V
V
 i (h 
 e (h 
Q net   m
 gz)i  Wnet   m
 gz) e
2
2
inlets
exits
Neglecting the kinetic and potential energies and noting that the work is zero, we get
 a1ha1  m
 v1hv1  m
 a 2ha 2  m
 v 2hv 2  m
 l 2hl 2
Q net  m
Conservation of mass for the steady-flow control volume is
 m   m
i
inlets
e
exits
15
For the dry air:
 a1  m
 a2  m
a
m
For the water vapor:
 v1  m
 v2  m
 l2
m
The mass of water that is condensed and leaves the control volume is
 l2  m
 v1  m
 v2
m
 a ( 1   2 )
m
 a , then
Divide the conservation of energy equation by m
Q net
 ha1   1hv1  ha 2   2 hv 2  ( 1   2 )hl 2
m a
Q net
 ha 2  ha1   2 hv 2   1hv1  ( 1   2 )hl 2
m a
Q net
 Cpa (T2  T1 )   2 hv 2   1hv1  ( 1   2 )hl 2
a
m
16
Now to find the 's and h's.
0.622 Pv1 0.622(3.398)
1 

P1  Pv1
100  3.398
kgv
 0.02188
kg a
Pv 2   2 Pg 2
 (0.95)(2.339 kPa )  2.222 kPa
0.622 Pv 2 0.622(2.222)
2 

P2  Pv 2
98  2.222
kgv
 0.01443
kga
17
Using the steam tables, the h's for the water are
hv1  2555.6
kJ
kg v
hv 2  2537.4
kJ
kg v
hl 2  83.91
kJ
kg v
The required heat transfer per unit mass of dry air becomes
Qnet
 C pa (T2  T1 )  2 hv 2  1hv1  (1  2 )hl 2
ma
 1.005
kg v
kJ
kJ
o
(20

30)
C

0.01443
(2537.4
)
o
kg a  C
kg a
kg v
 0.02188
 28.73
kg v
kg
kJ
kJ
(2555.6
)  (0.02188  0.01443) v (83.91
)
kg a
kg v
kg a
kg v
kJ
kg a
18
The heat transfer from the atmospheric air is
qout
Qnet
kJ

 28.73
ma
kga
The mass flow rate of dry air is given by
v1 
RaT1

Pa1
kJ
(30  273) K 3
kg a  K
m kPa
(100  3.398)kPa
kJ
0.287
m3
 0.90
kg a
Qout  ma qout  18.89
 9.04 kW
V1
a 
m
v1
m3
17
min  18.89 kg a
ma 
m3
min
0.90
kg a
kg a
kJ 1min 1kWs
(28.73
)
min
kg a 60s kJ
0.2843Tons
 2.57 Tons of refrigeration
kW
19
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.
20
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
2
2
V
V
 i (h 
 e (h 
Q net   m
 gz)i  Wnet   m
 gz) e
2
2
inlets
exits
Neglecting the kinetic and potential energies and noting that the heat transfer and
work are zero, we get
 a1ha1  m
 v1hv1  m
 l 2hl 2  m
 a 2ha 2  m
 v 2hv 2
m
Conservation of mass for the steady-flow control volume is
 m   m
i
inlets
e
exits
21
 a1  m
 a2  m
a
m
For the dry air:
For the water vapor:
 v1  m
 l2  m
 v2
m
The mass flow rate water that must be supplied to maintain steady-flow is,
 l2  m
 v2  m
 v1
m
 a ( 2   1 )
m
Divide the conservation of energy equation by m a , then
ha1   1hv1  ( 2   1 )hl 2  ha 2   2 hv 2
What are the knowns and unknowns in this equation?
22
Solving for 1
Since 1 is also defined by
We can solve for Pv1.
ha 2  ha1   2 (hv 2  hl 2 )
1 
(hv1  hl 2 )
C pa (T2  T1 )   2 h fg 2

(hg1  h f 2 )
Pv1
 1  0.622
P1  Pv1
Pv1 
Then, the relative humidity at state 1 is
 1 P1
0.622   1
Pv1
1 
Pg1
23
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.
24
Using the steam tables:
hf 2
kJ
 67.2
kg v
kJ
hv1  2544.7
kg v
h fg 2  2463.0
kJ
kg v
From the above analysis
1 
C pa (T2  T1 )   2 h fg 2
(hg1  h f 2 )
kg v
kJ
kJ
o
1.005 o 16  24  C  0.0115
(2463.0
)
kg a
kg v
kg a C

kJ
(2544.7  67.2)
kg v
 0.00822
kg v
kg a
25
We can solve for Pv1.
Pv1 
1 P1
0.622  1
0.00822(100 kPa)
0.622  0.00822
 1.3 kPa

Then the relative humidity at state 1 is
1 
Pv1
Pv1

Pg1 Psat @ 24o C
1.3 kPa

 0.433 or 43.3%
3.004 kPa
The enthalpy of the mixture at state 1 is
h1  ha1  1hv1
 C paT1  1hv1
 1.005
kg v
kJ
kJ
o
(24
C
)

0.00822
2544.7
kg a o C
kg a
kg v
 45.04
kJ
kg a
26
Wet-Bulb and Dry-Bulb Temperatures
In normal practice, the state of atmospheric air is specified by determining the wetbulb 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 wetbulb 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.
27
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:
28
29
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%
gv
kgv
  8.0
 0.008
kga
kga
h  46
kJ
kga
m3
v  0.853
kga
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
30
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.
31
The psychrometric diagram is
Psychrometric Diagram
0.050
0.045
Pressure = 101.3 [kPa]
0.040
Humidity Ratio
0.035
0.8
0.030
30 C
0.025
0.6
0.020
h3 =48 kJ/kga
0.015
0.010
 3 =0.0091kgv/kga
0.2
0C
-5
0.4
3
10 C
h1 =17 kJ/kga1
0.005
0.000
-10
20 C
h2 =37 kJ/kga
 1 = 2 =0.0049 kgv/kga
2
0
5
10
v1 =0.793 m^3/kga
15
20
25
30
35
40
T [C]
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   m
i
inlets
e
exits
32
For the dry air
 a1  m
 a2  m
a
m
For the water vapor (note: no water is added or condensed during simple heating)
 v1  m
 v2
m
Thus,
 2  1
Neglecting 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  ha  hv
we obtain
E in  E out
Q in  m a h1  m a h2
Q in  m a (h2  h1 )
33
 a and h's using the psychrometric chart.
Now to find the m
At T1 = 5oC, 1 = 90%, and T2 = 24oC:
The mass flow rate of dry air is given by
V1
a 
m
v1
34
m3
60
kga 1 min
kga
min
m a 
 75.66
 1261
.
3
m
min 60s
s
0.793
kga
The required heat transfer rate for the heating section is
kga
kJ 1kWs

Qin  1261
.
(37  17)
s
kga kJ
 25.22 kW
This 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
35
Apply conservation of mass and conservation of energy to process 2-3.
Conservation of mass for the steady-flow control volume is
 m   m
i
inlets
For the dry air
e
exits
 a2  m
 a3  m
a
m
For the water vapor (note: water is added during the humidification process)
m v 2  m s  m v 3
m s  m v 3  m v 2
m s  m a ( 3   2 )
kga
kgv
 1261
.
(0.0089  0.0049)
s
kga
kgv
 0.00504
s
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Neglecting the kinetic and potential energies and noting that the heat transfer and
work are zero, the conservation of energy yields
E in  E out
 a h2  m
 shs  m
 a h3
m
 shs  m
 a (h3  h2 )
m
Solving for the enthalpy of the steam,
m a ( 3   2 )hs  m a (h3  h2 )
h3  h2
hs 
3 2
37
hs 
kJ
(48  37)
kg a
kg v
(0.0089  0.0049)
kg a
kJ
 2750
kg v
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.
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