lecture 35
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Department of Mechanical EngineeringME 322 – Mechanical Engineering
Thermodynamics
Lecture 35
Analysis of Air Conditioning Processes
The Five HVAC Processes• Heating
– With and without humidification• Adiabatic humidification• Cooling
– With and without dehumidification• Adiabatic mixing of moist air streams• Evaporative cooling
2
Heating Without Humidification
3
HQ
1 2
1T 2T
12
#1h
#2h
12
1 2
# #2 1H aQ m h h
• Constant humidity ratio• Relative humidity
decreases• Outlet relative humidity
may be too dry to be comfortable
Heating with Humidification
4
HQ
1 2
water
1T 1'T
1
1'
#1h
#1'h
11'
1
1'
2
#2h
22 # #
2 1 2 1H
watera
Qh h h
m
# #1 2H a water water aQ m h m h m h
1 2
2 12 1
w water w
water w w
a a a
m m m
m m mm m m
2T
Adiabatic Humidification
5
1 2
water
1T
11 1
2b
# #2 1 2 1 0waterh h h
# #1 2a water water am h m h m h
1 2
2 12 1
w water w
water w w
a a a
m m m
m m mm m m
2c2a
The location of state 2 (2a, 2b, or 2c) depends on the state of the water being injected for humidification
Cooling Without Dehumidification
6
CQ
1 2
1T2T
12
#1h
#2h
12 1 2
# #1 2C aQ m h h
This process is not very common in HVAC systems. Often, the surface of the heat exchanger is below the dew point which causes water to condense
1'T
Cooling With Dehumidification
7
CQ
1 2
1T2T
12
#1h
#2h
121
If Twater is not given, it is common to assume that it is equal to T2
water
dpT
2
# #2 1c a water water aQ m h m h m h
2 1
1 21 2
w water w
water w w
a a a
m m m
m m mm m m
# #1 2 1 2
Cwater
a
Qh h h
m
Adiabatic Mixing
8
1 (cold)
2 (hot)
#1h
#2h
1
2
1
2
# # #1 1 2 2 3 3a a am h m h m h
3 (warm)
3
#3h
3
1 2 3a a am m m
# # #1 1 2 2 1 2 3
# #1 2 3
# #2 3 1
a a a a
a
a
m h m h m m hm h hm h h
1 1 2 2 1 2 3
1 2 3
2 3 1
a a a a
a
a
m m m mmm
1 1 2 2 3 3a a am m m
# #2 3 2 3# #3 1 3 1
h hh h
1-2-3 are on a straight line!
Evaporative Cooling
9
1 2
water
1T
1
1 1
# #2 1 2 1 0waterh h h
# #1 2a water water am h m h m h
1 2
2 12 1
w water w
water w w
a a a
m m m
m m mm m m
2
2
2
2T
This is the adiabatic humidification process when the water used for humidification is colder than T1
This system works very well in hot, dry climates. Notice that there can be a significant increase in the humidity levels (both relative humidity and humidity ratio).
Department of Mechanical EngineeringME 322 – Mechanical Engineering
Thermodynamics
Example
Combined Cooling and Heating Processes
Example
11
Given: In a combined cooling/heating system, moist air enters the cooling section at 90°F, = 50% at a volumetric flow rate of 5000 cfm. Saturated, moist air and liquid condensate leave the cooling section at a temperature that is 15 degrees below the dew point of the entering moist air. After leaving the cooling section, the saturated, moist air enters the heating section. After passing through the heater, the moist air leaves the heating section at 68°F. The pressure throughout the system can be assumed to be constant at normal sea-level pressure (29.921 inHg – consistent with the psychrometric chart).
Find:(a) The volumetric flow rate of the condensate (gpm)(b) The required refrigeration capacity of the cooling section (tons)(c) The relative humidity of the air leaving the heating section(d) The heat transfer rate required in the heating section (Btu/hr)
Example
12
A sketch of the system and a psychrometric chart showing the processes is shown below.
CQ
1 3
1T2T
12
#1h
#2h
121
water
dpT
2 3
HQ
2
3
#3h
3
3T
1
1
1
90 F50%5000 cfm
T
V
3 68 FT
2 1 15 RdpT T T
Properties from the Chart and Tables
13
1 90 FT
2 15°F 69 15 F 54°FdpT T
1 50% 2
#1 38.4 Btu/lbmah
#2 22.5 Btu/lbmah
121 0.0152
69 FdpT
2 3 0.0089 3
#3 26.0 Btu/lbmah
3 61%
3 68 FT
31 14.19 ft /lbmav
32 Table C.1aTable C.1a
54 F 22.1 Btu/lbm 0.01605ft / lbmwater water w wT T h v
Example
14
CQ
1 3
water
HQ
2
1
1
1
90 F50%5000 cfm
T
V
3 68 FT
2 1 15 RdpT T T Cooling section analysis …
# #1 2 1 2
Cwater
a
Qh h h
m
3
13
1
ft5000 lbm60 minmin 21141.6ft hr hr14.19
lbm
aa
a
Vmv
# #1 2 1 2
lbm lbmBtu Btu21141.6 38.4 22.5 0.0152 0.0089 22.1hr lbm lbm lbm
Btu333,208 27.8 tonshr
C a water
a wC
a a w
C
Q m h h h
Q
Q
1 90 FT
2 15°F 69 15 F 54°FdpT T
1 50% 2
#1 38.4 Btu/lbmah
#2 22.5 Btu/lbmah
121 0.0152
69 FdpT
2 3 0.0089 3
#3 26.0 Btu/lbmah
3 61%
3 68 FT
31 14.19 ft /lbmav
2Table C.1a
54 F 22.1 Btu/lbmwater water wT T h
Example
15
CQ
1 3
water
HQ
2
1
1
1
90 F50%5000 cfm
T
V
3 68 FT
2 1 15 RdpT T T
1 90 FT
2 15°F 69 15 F 54°FdpT T
1 50% 2
#1 38.4 Btu/lbmah
#2 22.5 Btu/lbmah
121 0.0152
69 FdpT
2 3 0.0089 3
#3 26.0 Btu/lbmah
3 61%
3 68 FT
31 14.19 ft /lbmav
2Table C.1a
54 F 22.1 Btu/lbmwater water wT T h
The condensate flow is determined by conservation of mass around the cooling section,
2 1
1 21 2
w water w
water w w
a a a
m m m
m m mm m m
1 2
lbm lbm lbm21141.6 0.0152 0.0089 133.2
hr lbm hr
water a
a w wwater
a
m m
m
3
3
Table C.1a
lbm ft hr gal133.2 0.01605 0.267 gpmhr lbm 60 min 0.13368 ft
wwater w wV m v
Example
16
CQ
1 3
water
HQ
2
1
1
1
90 F50%5000 cfm
T
V
3 68 FT
2 1 15 RdpT T T
1 90 FT
2 15°F 69 15 F 54°FdpT T
1 50% 2
#1 38.4 Btu/lbmah
#2 22.5 Btu/lbmah
121 0.0152
69 FdpT
2 3 0.0089 3
#3 26.0 Btu/lbmah
3 61%
3 68 FT
31 14.19 ft /lbmav
2Table C.1a
54 F 22.1 Btu/lbmwater water wT T h
The relative humidity leaving the heating section can be read from the psychrometric chart,
3 61%
Heating section analysis …
# #3 2H aQ m h h
lbm Btu Btu21141.6 26.0 22.5 73,996hr lbm hr
aH
a
Q
Example
17
CQ
1 3
water
HQ
2
1
1
1
90 F50%5000 cfm
T
V
3 68 FT
2 1 15 RdpT T T 1 90 FT
2 15°F 69 15 F 54°FdpT T
1 50% 2
#1 38.4 Btu/lbmah
#2 22.5 Btu/lbmah
121 0.0152
69 FdpT
2 3 0.0089 3
#3 26.0 Btu/lbmah
3 61%
3 68 FT
31 14.19 ft /lbmav
2Table C.1a
54 F 22.1 Btu/lbmwater water wT T h
EES Solution (Key Variables)
These are a bit different due to reading the psychrometric chart
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