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Department of Mechanical Engineering
ME 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
1
2#
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 1
H
water
a
Qh h h
m
# #
1 2H a water water aQ m h m h m h
1 2
2 1
2 1
w water w
water w w
a a a
m m m
m m m
m m m
2T
Adiabatic Humidification
5
1 2
water
1T
1
1 1
2b
# #
2 1 2 1 0waterh h h
# #
1 2a water water am h m h m h
1 2
2 1
2 1
w water w
water w w
a a a
m m m
m m m
m 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
1
2
#
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
1
2
#
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 2
1 2
w water w
water w w
a a a
m m m
m m m
m m m
# #
1 2 1 2
C
water
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 h
m h h
m h h
1 1 2 2 1 2 3
1 2 3
2 3 1
a a a a
a
a
m m m m
m
m
1 1 2 2 3 3a a am m m
# #
2 3 2 3
# #
3 1 3 1
h h
h h
1-2-3 are on a straight line!
Evaporative Cooling
9
1 2
water
1T
1
11
# #
2 1 2 1 0waterh h h
# #
1 2a water water am h m h m h
1 2
2 1
2 1
w water w
water w w
a a a
m m m
m m m
m 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 Engineering
ME 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
1
2
#
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
3
1 14.19 ft /lbmav
3
2Table 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
C
water
a
Qh h h
m
3
1
3
1
ft5000
lbm60 minmin 21141.6ft hr hr
14.19lbm
aa
a
Vm
v
# #
1 2 1 2
lbm lbmBtu Btu21141.6 38.4 22.5 0.0152 0.0089 22.1
hr lbm lbm lbm
Btu333,208 27.8 tons
hr
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
3
1 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
3
1 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 2
1 2
w water w
water w w
a a a
m m m
m m m
m 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 gpm
hr 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
3
1 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 Btu
21141.6 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
3
1 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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