ME 441
Homework Solutions
Set 3
E6
3-11 Air (assumed to be at standard atmospheric pressure) is cooled from 80 F db and 67 F wb until it is saturated at
55 F. Using the psychrometric chart, find (a) the moisture removed per pound of dry air, (b) the heat removed to
condense the moisture, (c) the sensible heat removed, and (d) the total heat removed.
Solution: The solution utilizes the ASHRAE sea level psychrometric chart. The initial and final states of
the air are plotted along with an intermediate point “a” which represents a state with the same dry bulb
temperature as “1” and the same humidity ratio as “2”. Properties are read from a full-size psychrometric
chart (the chart below is only used to illustrate the graphical procedure).
(a) The pounds of moisture removed per pound of dry air is given by ω1 - ω2 = 0.0112-0.0092
∆ω = 0.0020 lbmv/lbma
(b) The heat removed to condense the moisture is the latent heat. This is the energy required to move from
point “1” to point “a”. On a pound of dry air basis this is the change in enthalpy between “1” and “a”.
Latent heat transfer = h1 – ha = 31.47 – 29.31 = 2.16 Btu/lbma
(c) The sensible heat removed is the heat transfer that causes a change in the air temperature. This is the
energy required to move from point “a” to point“2”. On a pound of dry air basis this is the change in
enthalpy between “a” and “2”
Sensible heat transfer = ha – h2 = 29.31 – 23.19 = 6.12 Btu/lbma
(d) The total heat removed is the energy required to move from point “1” to point “2” and represents the
sum of the latent and sensible heat transfer.
Total heat transfer = h1 – h2 = 31.47 – 23.19 = 8.28 Btu/bma
The sensible heat factor (or ratio) is the sensible heat transfer divided by the total heat transfer. This value
can be obtained analytically from the above results or by translating the line 12 to the protractor scale as
illustrated below.
1
2
a
ME 441
Homework Solutions
Set 3
E6
3-13 Complete the table using the ASHRAE psychrometric chart for (a) sea level and (b) 5,000
feet.
(a) Sea level
Dry
Bulb
(F)
85
75
74.7
88.6
100
Wet
Bulb
(F)
60
59.6
65.1
70
85.7
Dew
Point
(F)
40.6
49.1
59.9
61
81.6
Humidity
Ratio
(lbmv/lbma)
0.0053
0.0074
0.0110
0.01143
0.0235
Enthalpy
(Btu/lbma)
Wet
Bulb
(F)
60
58.6
61.5
70
81.1
Dew
Point
(F)
44.9
49.1
56.7
55.8
76.0
Humidity
Ratio
(lbmv/lbma)
0.0076
0.0089
0.0118
0.01143
0.0235
Enthalpy
(Btu/lbma)
26.26
26.08
30
33.88
50
Relative
Humidity
(%)
20.9
40
60
39.56
56.22
Specific
Volume
(ft3/lbma)
13.84
13.63
13.71
14.07
14.64
Relative
Humidity
(%)
24.7
40
60
21.4
46.8
Specific
Volume
(ft3/lbma)
16.70
16.43
16.39
17.35
17.59
(b) 5,000 feet
Dry
Bulb
(F)
85
75
71.3
102.8
100
28.74
27.73
30
37.34
50
ME 441
Homework Solutions
Set 3
E6
3-15 Air enters a cooling coil at the rate of 5,000 cfm at 80 F db, 68 F wb and sea-level pressure. The air leaves the
coil at 55 F db and 54 F wb. (a) Determine the coil SHF and apparatus dew point temperature (b) compute the total
and sensible heat transfer rates for the air.
Solution: (a) Plot the entering and leaving air conditions on the sea-level psychrometric chart and connect
the two points with a straight line. Translating the line to the protractor scale provides a graphical method
of determining the SHF for the coil (SHF=0.63). The SHF for the coil can also be determined by
decomposing the coil process line into the sensible and latent components. These are shown as line
segments “2a” and “1a”, respectively on the chart. The value of coil SHF can then be determined using
enthalpy values for the three points.
SHF =
ha − h2 28.65 − 22.58
=
= 0.626
h1 − h2 32.28 − 22.58
By extending the line connecting the two points to the saturation curve yields the apparatus dew point
temperature. While the chart below illustrates the procedure, an actual chart should be used to obtain the
numerical answer. Apparatus dew point = 52 F
1
a
2
(b) In order to calculate the heat transfer rates, the specific volume must be determined. Since the
volumetric airflow rate is given for the coil inlet conditions, the specific volume of the air at the inlet
conditions must be used. v1=13.86 ft3/lbma.
The total heat transfer rate is given by
ft 3
5,000
Q
min ( 22.58−32.28 ) Btu = −3, 499 Btu
q = ( h2 − h1 ) =
t v1
lbma
min
ft 3
13.86
lbma
The sensible heat transfer rate is given by
ft 3
5,000
Q
min ( 22.58− 28.65 ) Btu = −2,190 Btu
q = ( h2 − ha ) =
t v1
lbma
min
ft 3
13.86
lbma
ME 441
Homework Solutions
The negative signs indicate heat is removed from the air stream.
Set 3
E6
ME 441
Homework Solutions
Set 3
E6
3.32 Air enters a refrigeration coil at 90 F db and 75 F wb at a rate of 1,400 cfm. The apparatus dew point
temperature of the coil is 55 F. If 5 tons of refrigeration are produced, what is the dry-bulb temperature of the air
leaving the coil?
Solution: The state of the air entering the cooling coil (1) and the saturated air state based on the apparatus
dew point temperature (adp) are plotted on the psychrometric chart. The state of the air leaving the cooling
coil lies on the line connecting states (1) and (adp). An energy balance on the cooling coil is used to
determine the enthalpy of the air leaving the coil. The intersection of the leaving air enthalpy line with the
line segment 1(adp) is the state of the air leaving the coil. The leaving dry bulb temperature is read from
the chart.
i=28.26
1
adp
The energy balance for the cooling coil is
q=
Q1
( i2 − i1 )
v1
Solving for the enthalpy of the air leaving the coil and substituting values yields:
Btu ⎞ ⎛
ft 3 ⎞
⎛
( −5tons ) ⎜ 200
⎟⎜14.19
min-ton ⎠ ⎝
lbma ⎟⎠
Btu
Btu
qv1
⎝
+ 38.40
= 28.26
i2 = + i1 =
3
ft
lbma
lbma
Q
1,400
min
The air temperature leaving the cooling coil is approximately 67 F db.
ME 441
Homework Solutions
Set 3
E6
3-21 Air at sea level pressure, 100 F dry bulb and 65 F wet bulb is humidified adiabatically with steam. The steam
supplied contains 20 percent moisture (80% quality) at 14.7 psia. If the air is humidified to 60% relative humidity,
what is the dry bulb temperature of the humidified air?
1
db=100 F
wb=65 F
2
RH=60%
Steam: x=.8, P=14.7psia
Solution: The solution presented here is graphical and makes use of the ASHRAE sea level psychrometric
chart. The chart below is used only to illustrate the procedure. A full-size chart should be used to solve the
problem.
For adiabatic humidification, the energy balance yields:
∆i
= iw
∆ω
The enthalpy of the steam used in the humidification process is obtained from the steam tables.
iw = 956.22
Btu
lbmv
Utilizing this value along with the protractor scale on the psychrometric chart, a process line can be
constructed passing through state 1. State 2 must be along this line and is found where the line intercepts
the relative humidity line of 60%. The corresponding dry-bulb temperature is read directly from the chart
and equals 91.5 F.
1
ME 441
Homework Solutions
Set 3
E6
3-22 Air at sea level pressure, 84 F dry bulb and 60 F wet bulb is humidified adiabatically while the dry-bulb
temperature remains constant. Wet steam is supplied for humidification at 20 psia. What quality must the steam
have (a) to provide saturated air and (b) to provide air at 70% relative humidity?
1
db=84 F
wb=60 F
2
db=84 F
Wet steam at 20 psia
Solution: This problem can be solved analytically, computationally and graphically. The solution
presented here is graphical and makes use of the ASHRAE sea level psychrometric chart. The chart below
is used only to illustrate the procedure. A full-size chart should be used to solve the problem.
First construct a line of constant dry-bulb temperature that passes through state 1. Next, translate this line
onto the protractor scale and read the value of ∆i/∆w (=1085 Btu/lbmv). For adiabatic humidification, the
energy balance yields:
∆i
= iw
∆ω
Thus, the enthalpy of the steam used in the humidification process is read directly from the protractor scale.
From the steam table at a pressure of 20 psia, it is seen that the value of 1085 Btu/lbm falls within the 2phase region (if = 196.35 and ifg = 959.82). The quality of the “wet” steam is found from the relation
i = i f + x ⋅ i fg
Substituting values, x = .93 Note that the answer is the same for both parts (a) and (b) since the process
line of constant dry-bulb temperature defines a single value of ∆i/∆w.
2b
1