l33 - psychrometric properties of moist air
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Department of Mechanical Engineering
ME 322 – Mechanical Engineering Thermodynamics
Lecture 33
Psychrometric Properties of Moist Air
Air-Water Vapor Mixtures
• Atmospheric air– A binary mixture of dry air (a) + water vapor
(w)– The air in the mixture is treated as a pure
substance even though it is really a mixture itself
• Applications– Heating, ventilating, and air-conditioning
(HVAC)
• Analysis– HVAC – pressures are always low ~ Patm
• Ideal gas law can be used for both air and water vapor
2
Properties of Moist Air
Both air and water vapor are treated as ideal gases that obey Dalton’s Law of Partial Pressures.
28.97 lbm/lbmol 18.016 lbm/lbmol
0.06855 Btu/lbm-R 0.1102 Btu/lbm-Ra w
a w
M M
R R
Dry Air Water Vapor
Since moist air is a binary mixture,
1a wy y
Btu1.986
lbmol-RR Universal Gas Constant:
3
Properties of Moist Air
The field of psychrometrics (air-water vapor properties) has adopted other properties to represent the composition of the mixture rather than the mole fraction.
w
a
y
y w
a
m
m
These properties are related,
/ 28.971.608
/ 18.016w w a w
a w a a
n n m M m
n n M m m
4
Mole Fraction Ratio Humidity Ratio
Properties of Moist Air
Relative Humidity
,
w w w
w sat sat sat
y P PP
y P P P
T
s
wP
satP
T
T-s diagram of water
State of the water vapor in the mixture
Partial pressure of the water vapor in the mixture
Partial pressure of the water vapor in a saturated mixture
a wP P P
Partial pressure of the dry air
Total pressure of the mixture
dpT
5
Dew Point Temperature
Properties of Moist Air
All of these properties are related. For example,
/
/
/
/
18.016 lbm/lbmol
28.97 lbm/lbmol
w w w
a a a
a w a w
w a w a
w w w
a a a
m P V R T
m P V R T
R P R M P
R P R M P
M P P
M P P
6
0.622 0.622
0.622 0.622
w w
a w
sat sat
a w
P P
P P P
P P
P P P
Example
Given: Moist air at the following state
70 F 14 psia 60% 0.60T P
Find: Various psychrometric properties of the moist air
Solution: Partial pressure of the vapor 0.60 w
sat
P
P
The partial pressure of the water in a saturated mixture can be found from Table C.1a,
0.60 0.3632 psia
0.2179 psia
w sat
w
w
P P
P
P
7
satPT
Example
Dew Point Temperature The dew point temperature is the saturation temperature of the water vapor at its partial pressure. Using Table C.1a,
0.2179 psiawP
Interpolating ...
55.5 FdpT
8
T
s
0.2179 psiawP
0.3632 psiasatP
70 FT
55.5 FdpT
If the mixture drops below this temperature, the water vapor will start condensing.
Example
9
Humidity Ratio
0.2179 psia
0.622 0.62214 0.2179 psia
w w
a a
m P
m P
lbm 7000 grains grains0.009834 68.8
lbm lbm lbmw
a w a
Grains – A new unit! A ‘grain’ is an ancient Egyptian measure of the mass of one grain of barley (7000 grains/lbm). Since the humidity ratios are typically very small, the HVAC industry has adopted the use of ‘grains’ to represent humidity ratio ...
lbm0.009834 0.009834
lbmw
a
Example
Mole Fraction Ratio
1.608 1.608 0.009834 0.015813w
a
y
y
Mole Fraction of each Component
0.0158130.016
1 1 1 0.015813
1 1 10.984
1 1 0.015813
w ww
a w
w aa
a a
y yy
y y
y yy
y y
Notice that: 1v ay y
10
Intensive Moist Air Properties
Consider the enthalpy of the mixture …
a wH H H
Question: How can the specific enthalpy of the air-water vapor mixture be specified?
Answer: The total enthalpy must be divided by a mass value. Which mass value should be used?
11
a a w wH m h m h
a a w wH m h m h
Intensive Moist Air Properties
In air conditioning applications, the water vapor mass can vary due to condensation or evaporation (dehumidification or humidification). Thus, specific properties of the mixture are based on the dry air,
Units: Btu/lbma or kJ/kga
12
a wh h h
wa w
a a
mHh h h
m m
Intensive Moist Air Properties
, ,
a w
a w
p pa pw
v va vw
a a w w
u u T u T
h h T h T
c c T c T
c c T c T
s s T P s T P
Using ideal gas mixing for the components of moist air, the internal energy, enthalpy, heat capacities, and entropy of the mixture can be calculated by,
13
Department of Mechanical Engineering
ME 322 – Mechanical Engineering Thermodynamics
Example
Heating of a Moist Air Stream
14
Example
15
Given: Moist air flowing at 300 cfm enters a heating unit at 65°F, 14 psia with a relative humidity of 50%. The moist air leaves the heating unit at 110°F, 14 psia.
Find: (a) The heat transfer rate required (Btu/hr)(b) The relative humidity of the air leaving the heater
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
Example
16
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
The First Law applied to the heating system is,
2 2 2 2 1 1 2 1a a w w a a w wQ m h m h m h m h
The mass flow rate of the dry air does not change (in this case the water vapor mass flow does not change either … why?). Therefore,
1 2a a am m m
Example
17
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
2 12 2 1 1
w wa w a w
a a a
m mQh h h h
m m m
2 2 2 2 1 1 2 1a a w w a a w wQ m h m h m h m h
Rearranging the First Law,
The mass flow rate of the dry air is,
1
1a
a
Vm
v
2 2 2 1 1 1a w a wh h h h
Example
18
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
The specific volume of the dry air at state 1 is found using the ideal gas EOS with the partial pressure of the dry air,
11
1
aa
a
R Tv
P
The partial pressure of the dry air at state 1 is found knowing the relative humidity,
11
1
w
sat
P
P
1 1sat satP P T
1 1 1a wP P P
Example
19
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
The component enthalpy values can be found using the ideal gas model for each component,
The humidity ratio at state 1 can be found,
11
1
0.622 w
a
P
P
Example
20
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
No water vapor is added to or taken from the moist air from state 1 to 2. Therefore,
1 2
At this point, the problem can be solved for the heat transfer rate. We are also interested in the relative humidity at the exit of the heater. This can be found from the humidity ratio at 2,
22
2
w
sat
P
P
2 2sat satP P T 2 2 2a wP P P
22
2
0.622 w
a
P
P
Example
21
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
Solution (Key Variables):
Even though the humidity ratio stays constant in this process, the moist air leaving the heater will feel uncomfortably ‘dry’. This is a common problem encountered in heating processes. The ‘dryness’ can be alleviated by injecting water vapor into the moist air stream leaving the heater (humidification).
Example
22
Q
1
1
1
14 psia65 F0.50300 cfm
PT
V
2
2
14 psia110 F
PT
What would happen if the moisture content is neglected and the mixture is treated as dry air?
Since no water vapor is added or removed from the moist air in this process, neglecting the moisture results in a small error. However, neglecting the moisture does not reveal the relative humidity at the exit!