Two powerful analytical tools of the HVAC design engineer are the conservation of energy or energy balance, and the conservation of mass or mass balance. These con- servation laws are the basis for the analysis of moist air processes. In actual practice the properties may not be uniform across the flow area, especially at the outlet, and a considerable length may be necessary for complete mixing. It is customary to analyze these processes by using the bulk average properties at the inlet and outlet of the device being studied.
In this section we will consider the basic processes that are a part of the analysis of most systems.
Heating or Cooling of Moist Air
When air is heated or cooled without the loss or gain of moisture, the process yields a straight horizontal line on the psychrometric chart, because the humidity ratio is con- stant. Such processes may occur when moist air flows through a heat exchanger. In cooling, however, if part of the surface of the heat exchanger is below the dew point of the air, condensation and the consequent dehumidification will occur. Figure 3-2 shows a schematic of a device used to heat or cool air. For steady-flow-steady-state heating the energy balance becomes
(3-22) However, the direction of the heat transfer may be implied by the terms heatingand cooling, with the heating process going from left to right and cooling from right to left in Fig. 3-3. The enthalpy of the moist air, per unit mass of dry air, at sections 1 and 2 is given by
(3-23) and
(3-24) i2 =ia2 +W iv2 2
i1 =ia1+W i1 1v
˙ ˙ ˙
m ia2 + =q m ia1
3-5 Classic Moist Air Processes 57
Figure 3-2 Schematic of a heating or cooling device.
Heating or cooling medium
W1 i1
q
ma W2 –W1
i2 m•a •
1 2
Figure 3-3 Sensible heating and cooling process.
0.030
0.024 0.022 0.020 0.018 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.026 0.028
Humidity ratio (W),pounds of moisture per pound of dry air
60
55
50
45
40
35
120 30
115
110
105
100
95
90
85
80
75
70
65
60
55
50
45
4035
25 12.5 30 35
45
50 55
60
10 15 20
2 1
25
40 Dry bulb,F
Dry bulb temperature, F 20%
75 80
85
70 F Wet bulb 14.5 volume, ft
per pound of dry air3 15.0
40%
14.0
13.5
13.0
60% Relative humidity 80%
80 85
35 40
45 50
55 60
65 70
75
15 20
25 30
35 40
45 50
Saturation temperature, F Enthalpy, Btu per pound of dry air
Se nsible heat =SHF
T otal heat–0 .1
–0.3 –0.5
–1 .0 –2.0
–4.0 8.02.0
1.0 1.0
0.8 5000
3000 2000
1500 1000
0.6 0.5 0.4
0.3 0.2 0.1
i1
i2
t2 t1
W1= W2
2 1
Alternatively i1and i2 may be obtained directly from the psychrometric chart. The convenience of the chart is evident. Because the moist air has been assumed to be a perfect gas, Eq. 3-22 may be arranged and written
(heating) (3-25a)
or
(cooling) (3-25b)
where
(3-26) In the temperature range of interest, cpa = 0.240 Btu/(lbma-F) or 1.0 kJ/(kga-C), cpv = 0.444 Btu/(lbmv-F) or 1.86 kJ/(kgv-C), and W is the order of 0.01. Then cp is about 0.244 Btu/(lbma-F) or 1.02 kJ/(kga-C).
EXAMPLE 3-4
Find the heat transfer rate required to warm 1500 cfm (ft3/min) of air at 60 F and 90 percent relative humidity to 110 F without the addition of moisture.
SOLUTION
Equations 3-22 or 3-25 may be used to find the required heat transfer rate. First it is necessary to find the mass flow rate of the dry air:
(3-27) The specific volume is read from Chart 1aat t1=60 F and φ=90 percent as 13.33 ft3/lbma:
Also from Chart 1a, i1 = 25.1 Btu/lbma and i2 = 37.4 Btu/lbma. Then by using Eq. 3-22, we get
or if we had chosen to use Eq. 3-25,
Agreement between the two methods is within 1 percent.
We can see that the relative humidity decreases when the moist air is heated. The reverse process of cooling results in an increase in relative humidity but the humidity ratio is constant.
Cooling and Dehumidifying of Moist Air
When moist air is passed over a surface so that a part of the stream is cooled to a tem- perature below its dew point, some of the water vapor will condense and may leave
˙ ( . ) ( ) ,
q =6752 0 244 110−60 =82 374 Btu/ hr
˙ ( . . ) ,
q =6752 37 4−25 1 =83 050 Btu/ hr
˙ ( )
ma =1500 60. =
13 33 6752 lbma / hr
˙ ˙
ma = V A1 1v = Qv
1 1 1
cp =cpa +Wcpv
˙ ˙ ( )
qs =m c ta p 2 −t1
˙ ˙ ( )
qs = m c ta p 2 −t1
the airstream. Figure 3-4 shows a schematic of a cooling and dehumidifying device, and Fig. 3-5 shows the process on the psychrometric chart. Although the actual process path may vary considerably depending on the type of surface, surface tem- perature, and flow conditions, the net heat and mass transfer can be expressed in terms of the initial and final states, neither of which has to be at saturation conditions. By referring to Fig. 3-4, we see that the energy balance gives
(3-28) and the mass flow balance for the water in the air is
(3-29) Combining Eqs. 3-28 and 3-29 yields
(3-30) Equation 3-30 gives the total rate of heat transfer from the moist air. The last term on the right-hand side of Eq. 3-30 is usually small compared to the others and is often neglected. Example 3-5 illustrates this point.
EXAMPLE 3-5
Moist air at 80 F db and 67 F wb is cooled to 58 F db and 80 percent relative humid- ity. The volume flow rate is 2000 cfm, and the condensate leaves at 60 F. Find the heat transfer rate.
˙ ˙ ( ) ˙ ( )
q =m ia 1−i2 −m Wa 1−W i2 w
˙ ˙ ˙
m Wa 1 = mw +m Wa 2
˙ ˙ ˙ ˙
m ia1 = +q m ia2 +m iw w
3-5 Classic Moist Air Processes 59
Figure 3-4 Schematic of a cooling and dehumidifying device.
ma W1 i1
q
ma W2 i2
mw iw
1 2
Refrigerant
•
•
•
•
Figure 3-5 Cooling and dehumidifying process.
0.024 0.022 0.020 0.018 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.026 0.028
Humidity ratio (W),pounds of moisture per pound of dry air
60
55
50
45
35
120 30
115
110
105
100
95
90
85
80
75
70
65
60
55
50
45
4035
25 12.5 30 35
45
50 55
60
10 15 20
3 1 2
25
40 Dry bulb,F
Dry bulb temperature, F 20%
75 80
85
70 F Wet bulb 14.5 volume, ft
per pound of dry air3 15.0
40%
14.0
13.5
13.0
60% Relative humidity 80%
80 85
35 40
45 50
55 60
65 70
75
15 20
25 30
35 40
45 50
Saturation temperature, F Enthalpy, Btu per pound of dry air
Se nsib
le heat =SHF T otal heat–0
.1 –0.3
–0.5 –1 .0 –2.0
–48.0.02.0 1.0 1.0
0.8 5000
3000 2000
1500 1000
0.6 0.5 0.4
0.3 0.2 0.1
i1
i3 i2
t2 t1,t3
W2 W1
2
1
0.030
SOLUTION
Equation 3-30 applies to this process, which is shown in Fig. 3-5. The following prop- erties are read from Chart 1a:v1=13.85 ft3lbma, i1=31.4 Btu/lbma,W1=0.0112 lbmv/lbma,i2=22.8 Btu/lbma,W2=0.0082 lbmv/lbma. The enthalpy of the conden- sate is obtained from Table A-1a, iw = 28.08 Btu/lbmw. The mass flow rate ma is obtained from Eq. 3-27:
Then
The last term, which represents the energy of the condensate, is seen to be small.
Neglecting the condensate term,q=74,356 Btu/hr =6.2 tons.
The cooling and dehumidifying process involves both sensible and latent heat transfer; the sensible heat transfer rate is associated with the decrease in dry bulb tem- perature, and the latent heat transfer rate is associated with the decrease in humidity ratio. These quantities may be expressed as
(3-31) and
(3-32) By referring to Fig. 3-5 we may also express the latent heat transfer rate as
(3-33) and the sensible heat transfer rate is given by
(3-34) The energy of the condensate has been neglected. Obviously
(3-35) The sensible heat factor(SHF) is defined as qs/q.This parameter is shown on the semi- circular scale of Fig. 3-5. Note that the SHF can be negative. If we use the standard sign convention that sensible or latent heat transfer to the system is positive and trans- fer from the system is negative, the proper sign will result. For example, with the cool- ing and dehumidifying process above, both sensible and latent heat transfer are away from the air,qsand qlare both negative, and the SHF is positive. In a situation where air is being cooled sensibly but a large latent heat gain is present, the SHF will be neg- ative if the absolute value of qlis greater than qs. The use of this feature of the chart is shown later.
Heating and Humidifying Moist Air
A device to heat and humidify moist air is shown schematically in Fig. 3-6. This process is generally required to maintain comfort during the cold months of the year.
An energy balance on the device yields
(3-36)
˙ ˙ ˙ ˙
m ia1+ +q m iw w =m ia2
˙ ˙ ˙
q =qs +ql
˙ ˙ ( )
qs =m ia 2 −i3
˙ ˙ ( )
ql = m ia 3 −i1
˙ ˙ ( )
ql = m Wa 2 −W i1 fg
˙ ˙ ( )
qs = m c ta p 2 −t1
˙ ( . . ) ( . . ) .
˙ ( . ) ( . )
q q
= [ − − − ]
=8646 31 4[ − 22 8 ] 0 0112 0 0082 28 8 8646 8 6 0 084
˙ ( )
ma = 2000 60. =
13 88 8646 lbma / hr
and a mass balance on the water gives
(3-37) Equations 3-36 and 3-37 may be combined to obtain
(3-38a) or
(3-38b) Equations 3-38a and 3-38b describe a straight line that connects the initial and final states on the psychrometric chart. Figure 3-7 shows a combined heating and humidi- fying process, states 1–2.
A graphical procedure makes use of the semicircular scale on Chart 1ato locate the process line. The ratio of the change in enthalpy to the change in humidity ratio is (3-39) Figure 3-7 shows the procedure where a straight line is laid out parallel to the line on the protractor through state 1. Although the process may be represented by one line
∆
∆ i W
i i
W W
q
m i
w
= − w
− = +
2 1
2 1
˙
˙ i i
W W
q
m i
w w
2 1
2 1
−
− = ˙ +
˙ i i
W W
q
m W W i
a
w
2 1
2 1 2 1
−
− =
− +
˙
˙ ( )
˙ ˙ ˙
m Wa 1+mw =m Wa 2
3-5 Classic Moist Air Processes 61
Figure 3-6 Schematic of a heating and humidifying device.
ma W1
i1 ma
W2 i2
mw q
iw
•
• •
1 χ 2
Heating medium
Figure 3-7 Combined heating and humidifying process.
0.024 0.022 0.020 0.018 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.026 0.028
Humidity ratio (W),pounds of moisture per pound of dry air
60
55
50
45
40
35
120 30
115
110
105
10095
90
85
80
75
70
65
60
55
50
45
4035
25 12.5 30 35
45
50 55
60
10 15 20
1
2
x
25
40 Dry bulb,F
Dry bulb temperature, F 20%
80 85
70 F Wet bulb
15.0
40%
14.0
13.5
13.0
80%
80 85
35 40
45 50
55 60
65 70
75
15 20
25 30
35 40
45 50
Saturation temperature, F Enthalpy, Btu per pound of dry air
Se nsible heat =SHF
T otal heat–0 .1–0.3
–0.5 –1 .0 –2.0
–4.0 8.02.0
1.0
Parallel 1.0
0.8 5000
3000 2000
1500 1000
0.6 0.5 0.4
0.3 0.2 0.1
Parallel
i1
i2
ix
tx
t1 t2
W1 W2
0.030
60% Relative humidity 14.5 volume, ft
per pound of dry air3
75
from state 1 to state 2, it is not practical to perform it in that way. The heating and humidification processes are usually carried out separately, shown in Figs. 3-6 and 3-7 as processes 1 −χand χ−2.
Adiabatic Humidification of Moist Air
When moisture is added to moist air without the addition of heat, Eq. 3-38b becomes (3-40) Close examination of the protractor on Chart 1areveals that ∆i/∆Wcan vary from pos- itive infinity on the left to negative infinity on the right. Therefore, in theory, the adi- abatic humidification process can take many different paths depending on the condition of the water used. In practice the water will vary from a liquid at about 50 F (10 C) to a saturated vapor at about 250 F (120 C). The practical range of ∆i/∆Wis shown on the chart and protractor of Fig. 3-8.
EXAMPLE 3-6
Moist air at 60 F db and 20 percent relative humidity enters a heater and humidifier at the rate of 1600 cfm. Heating of the air is followed by adiabatic humidification so that it leaves at 115 F db and a relative humidity of 30 percent. Saturated water vapor at 212 F is injected. Determine the required heat transfer rate and mass flow rate of water vapor.
SOLUTION
Figure 3-6 is a schematic of the apparatus. Locate the states as shown in Fig. 3-7 from the given information and Eq. 3-40 using the protractor feature of the psychrometric chart. Process 1 − χ is sensible heating; therefore, a horizontal line to the right of state 1 is constructed. Process χ−2 is determined from Eq. 3-40 and the protractor:
i i
W W i i
w W
2 1
2 1
−
− = = ∆
∆
Figure 3-8 Practical range of adiabatic humidifying processes.
0.030
0.024 0.022 0.020 0.018 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.026 0.028
Humidity ratio (W),pounds of moisture per pound of dry air
60
55
50
45
40
35
120 30
115
110
105
10095
90
85
80
7570
6560
55
50
45
4035
25 12.5 30 35
45 50
55 60
10 15 20
1 2b
2a
x
25
40 Dry bulb,F
Dry bulb temperature, F 20%
75 80
85
70 F Wet bulb 14.5 volume, ft
per pound of dry air3 15.0
40%
14.0
13.5
13.0
60% Relative humidity 80%
80 85
35 40
45 50
55 60
65 70
75
15 20
25 30
35 40
45 50
Saturation temperature, F Enthalpy, Btu per pound of dry air
Sen sible heat =SHF
T otal heat –0
.1 –0.3
–0.5 –1.0
–2.0 –48.0.02.0
1.0
Practical range 1.0
0.8 5000
3000
2000
1500 1000
0.6 0.5 0.4
0.3 0.2 0.1
where iwis read from Table A-1a. A parallel line is drawn from state 2 as shown in Fig. 3-7. State χis determined by the intersection on lines 1 −χand χ−2. The heat transfer rate is then given by
where
and i1and ix, read from Chart 1a, are 16.8 and 29.2 Btu/lbma, respectively. Then
The mass flow rate of the water vapor is given by
where W2and W1are read from Chart 1aas 0.0193 and 0.0022 lbmv/lbma, respectively.
Then
Adiabatic Mixing of Two Streams of Moist Air
The mixing of airstreams is quite common in air-conditioning systems. The mixing usually occurs under steady, adiabatic flow conditions. Figure 3-9 illustrates the mix- ing of two airstreams. An energy balance gives
(3-41) The mass balance on the dry air is
(3-42) and the mass balance on the water vapor is
(3-43) Combining Eqs. 3-41, 3-42, and 3-43 and eliminating ma3yields
(3-44) i i
i i
W W
W W
m m
a a
2 3
3 1
2 3
3 1
1 2
−
− = −
− = ˙
˙
˙ ˙ ˙
m Wa1 1+m Wa2 2 = m Wa3 3
˙ ˙ ˙
ma1+ma2 = ma3
˙ ˙ ˙
m ia1 1 +m ia2 2 =m ia3 3
˙ ( . . )
mv =7296 0 0193−0 0022 =125 lbmv/ hr
˙ ˙ ( )
mv = m Wa 2 −W1
˙ ( . . ) ,
q =7296 29 2−16 8 =90 500 Btu/ hr
˙ ˙ ( )
m Q .
a = v60 = 1600 =
13 1660 7296
1
lbma / hr
˙ ˙ ( )
q =m ia x −i1
∆∆i
W = iw =1150 4. Btu/ lbm
3-5 Classic Moist Air Processes 63
Figure 3-9 Schematic of the adiabatic mixing of two airstreams.
1
2 Adiabatic 3 i3
, i2
W3
, W2
m⋅a3
i1 W1 m⋅a1
m⋅a2
The state of the mixed streams lies on a straight line between states 1 and 2 (Fig. 3-10).
From Eq. 3-44 the lengths of the various line segments are proportional to the masses of dry air mixed:
(3-45) This is most easily shown by solving Eq. 3-44 for i3and W3:
(3-44a)
(3-44b)
Clearly for given states 1 and 2, a straight line will be generated when any constant value of ma1/ma2is used and the result plotted on the psychrometric chart. It is also clear that the location of state 3 on the line is dependent on ma1/ma2. This provides a very convenient graphical procedure for solving mixing problems in contrast to the use of Eqs. 3-44a and 3-44b.
Although the mass flow rate is used when the graphical procedure is employed, the volume flow rates may be used to obtain good approximate results.
EXAMPLE 3-7
Two thousand cubic feet per minute (cfm) of air at 100 F db and 75 F wb are mixed with 1000 cfm of air at 60 F db and 50 F wb. The process is adiabatic, at a steady flow rate and at standard sea-level pressure. Find the condition of the mixed streams.
W m
m W W
m m
a a
a a 3
1 2
1 2
1 2
1
=
+ +
˙
˙
˙
˙ i
m m i i
m m
a a
a a 3
1 2
1 2
1 2
1
=
+ +
˙
˙
˙
˙
˙
˙ , ˙
˙ , ˙
˙ m
m
m m
m m
a a
a a
a a 1
2
1 3
2 3
32 13
32 12
13
= = = 12
Figure 3-10 Adiabatic mixing process.
0.030
0.024 0.022 0.020 0.018 0.016 0.014 0.012 0.010 0.008 0.006 0.004 0.002 0.026 0.028
Humidity ratio (W),pounds of moisture per pound of dry air
60
55
50
45
40
35
120 30
115
110
105
100
95
90
85
80
75
70
65
60
55
50
45
4035
25 12.5 30 35
45
50 55
60
10 15 20
1
3
2
25
40 Dry bulb,F
Dry bulb temperature, F 20%
75 80
85
70 F Wet bulb 14.5 volume, ft
per pound of dry air3 15.0
40%
14.0
13.5
13.0
60% Relative humidity 80%
80 85
35 40
45 50
55 60
65 70
75
15 20
25 30
35 40
45 50
Saturation temperature, F Enthalpy, Btu per pound of dry air
Se nsible heat =SHF
T otal heat–0 .1
–0.3 –0.5
–1.0 –2.0
–48.0.02.0 1.0 1.0
0.8 5000
3000 2000
1500 1000
0.6 0.5 0.4
0.3 0.2 0.1
i3 i2
i1
t1 t3 t2
W1 W2 W3
SOLUTION
A combination graphical and analytical solution is first obtained. The initial states are first located on Chart 1aas illustrated in Fig. 3-10 and connected with a straight line.
Using Eq. 3-44b or another form of Eqs. 3-42 and 3-43, we obtain
(3-46) Using the property values from Chart 1a, we obtain
The intersection of W3 with the line connecting states 1 and 2 gives the mixture state 3. The resulting dry bulb temperature is 86 F, and the wet bulb temperature is 68 F.
Equation 3-44a could have also been solved for i3to locate the mixture state 3.
The complete graphical procedure could also be used, where
The lengths of line segments 12— and 13—
depend on the scale of the psychrometric chart used. However, when the length 13—
is laid out along 12—
from state 1, state 3 is accu- rately determined. An excellent approximate solution for Example 3-7 may be obtained by neglecting the effect of density and using the volume flow rates to find state 3.
A computer program named PSYCH is given on the website for this text. The pro- gram carries out all of the processes presented so far, allowing for the variation of barometric pressure and determination of other properties.