The complete air-conditioning system may involve two or more of the processes just considered. For example, in the air conditioning of space during the summer, the air supplied must have a sufficiently low temperature and moisture content to absorb the total cooling load of the space. As the air flows through the space, it is heated and humidified. Some outdoor air is usually mixed with the return air and sent to the conditioning equipment, where it is cooled and dehumidified and supplied to the space again. During the winter months the same general processes occur, but in reverse. Systems described in Chapter 2 carry out these conditioning processes with some variations.
13 12
2000
2000 1000 0 67 13 0 67 12
2 3
≈ =
+ = =
˙
˙ . . ( )
Q
Q and
13 12
8332
8332 4542 0 65 13 0 65 12
2 3
= =
+ = =
˙
˙ . . ( )
m m
a a
or
˙ ( )
.
˙ ( )
.
. ( . . )
. m m W W
a
a 1
2 3 3
1000 60
13 21 4542 2000 60
14 4 8332
0 0054 8332
4542 8332 0 013 0 0054 0 0103
= =
= =
= +
+
−
=
lbma/hr lbma/hr
lbmv/ lbma
W W m
ma W W
a
3 1 2
3
2 1
= + ˙ −
˙ ( )
3-6 Space Air Conditioning—Design Conditions 65
Sensible Heat Factor
The sensible heat factor(SHF) was defined in Sec. 3-5 as the ratio of the sensible heat transfer to the total heat transfer for a process:
(3-47) If we recall Eqs. 3-33 and 3-34 and refer to Chart 1a, it is evident that the SHF is related to the parameter ∆i/∆W. The SHF is plotted on the inside scale of the protrac- tor on Chart 1a. The following examples will demonstrate the usefulness of the SHF.
EXAMPLE 3-8
Conditioned air is supplied to a space at 54 F db and 90 percent RH at the rate of 1500 cfm. The sensible heat factor for the space is 0.80, and the space is to be maintained at 75 F db. Determine the sensible and latent cooling loads for the space.
SOLUTION
Chart 1acan be used to solve this problem conveniently. A line is drawn on the pro- tractor through a value of 0.8 on the SHF scale. A parallel line is then drawn from the initial state, 54 F db and 90 percent RH, to the intersection of the 75 F db line, which defines the final state. Figure 3-11 illustrates the procedure. The total heat transfer rate for the process is given by
and the sensible heat transfer rate is given by
and the mass flow rate of dry air is given by
˙ ( ) ˙
qs = SHF q
˙ ˙ ( )
q = m ia 2 −i1 SHF=
+ =
˙
˙ ˙
˙
˙ q
q q q
q
s
s l
s
Figure 3-11 The condition line for the space in Example 3-8.
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
d
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
Parallel
i1 i2
t2 t1
where v1=13.11 ft3/lbma is read from Chart 1a. Also from Chart 1a,i1=21.6 Btu/lbm dry air and i2=27.8 Btu/lbm dry air. Then
and
The process 1–2 with its extension to the left is called the condition linefor the space. Assuming that state 2, the space condition, is fixed, air supplied at any state on the condition line will satisfy the load requirements. However, as that state is changed, different quantities of air must be supplied to the space. The closer point 1 is to point 2, the more air is required; the converse is also true.
We will now consider several examples of single-path, constant-flow systems. Heat losses from and gains to the ducts and fan power will be neglected for the time being.
EXAMPLE 3-9
A given space is to be maintained at 78 F db and 65 F wb. The total heat gain to the space has been determined to be 60,000 Btu/hr, of which 42,000 Btu/hr is sensible heat transfer. The outdoor air requirement of the occupants is 500 cfm. The outdoor air has a temperature and relative humidity of 90 F and 55 percent, respectively. Deter- mine the quantity and the state of the air supplied to the space and the required capac- ity of the cooling and dehumidifying equipment.
SOLUTION
A simplified schematic is shown in Fig. 3-12. The given quantities are shown and sta- tions are numbered for reference. By Eq. 3-47 the sensible heat factor for the condi- tioned space is
SHF = 42 000 = 60 000, 0 7
, .
˙ ˙ ˙
ql = −q qs =8500 Btu/ hr
˙ ( . . ) ,
˙ ˙( ) , ( . ) ,
q
qs q SHF
= − =
= = =
6865 27 8 21 6 42 600 42 600 0 8 34 100
Btu/ hr Btu/hr
˙ ˙ ( )
m Q .
a = v = =
1
1500 60
13 11 6865 lbma /hr
3-6 Space Air Conditioning—Design Conditions 67
Figure 3-12 Single-line sketch of cooling and dehumidifying system for Example 3-9.
2 3 5
1' 1
4
0 Mixing
box
Exhaust Return fan
Supply fan
Conditioned space t3 = 78 F t3wb = 65 F t0 = 90 F
0 = 55%
Q0 = 500 cfm q = 60,000 Btu/hr
qs = 42,000 Btu/hr Cooling and
dehumidifying unit
State 3 is located as shown in Fig. 3-13, where a line is drawn from point 3 and par- allel to the SHF =0.7 line on the protractor. State 2, which may be any point on that line, fixes the quantity of air supplied to the space. Its location is determined by the operating characteristics of the equipment, desired indoor air quality, and what will be comfortable for the occupants. These aspects of the problem will be developed later.
For now assume that the dry bulb temperature of the entering air t2is 20 F less than the space temperature t3. Then t2=58 F, which fixes state 2. The air quantity required may now be found from an energy balance on the space:
or
and
From Chart 1a,i3=30 Btu/lbma,i2=23 Btu/lbma, and
Also from Chart 1a,v2=13.21 ft3/lbma and the air volume flow rate required is
Before attention is directed to the cooling and dehumidifying process, state 1 must be determined. A mass balance on the mixing section yields
˙ ˙ ˙ ˙
˙
˙
, .
m m m m
m Q
v v
a a a a
a
0 4 1 2
0 0
0
0 14 23
+ = =
= = ft / lbma3
˙ ˙ ( . )
Q2 m va2 2 8570 13 21
60 1885 1890
= = = or cfm
˙ ˙ ,
ma2 ma3 60 000
30 23 8570
= =
− = lbma/hr
˙ ˙
m q
i i
a2
3 2
= −
˙ ˙ ( )
q =ma2 3i −i2
˙ ˙ ˙
m ia2 2 + =q m ia3 3
Figure 3-13 Psychrometric processes for Example 3-9.
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
3 1
0
d
2 d
25
40 Dry bulb,F
Dry bulb temperature, F 20%
75 80
85
70 F Wet bulb 14.5 volume, ft
3 per pound of dry air 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 Space, 0.7
Coil, 0.6
i1 i0
i3 i2
t2 t3
td t1 t0
W0
W3 W1 W2
0.030
Then the recirculated air is
By using the graphical technique discussed in Example 3-7 and referring to Fig. 3-13, we see that
State 1 is located at 81 F db and 68 F wb. A line constructed from state 1 to state 2 on Chart 1a then represents the process for the cooling coil. An energy balance gives
Solving for the rate at which energy is removed in the cooling coil From Chart 1a,i1=32.4 Btu/lbma and
The SHF for the cooling coil is found to be 0.6 using the protractor of Chart 1a (Fig. 3-13). Then
and
The sum of qcsand qc1is known as the coil refrigeration load. Notice that because of outdoor air cooling the coil refrigeration load it is different from the space cooling load. Problems of this type may be solved using the program PSYCH given on the website.
An alternate approach to the analysis of the cooling coil in Example 3-9 uses the so-called coil bypass factor. Note that when line 1–2 of Fig. 3-13 is extended, it inter- sects the saturation curve at point d. This point represents the apparatus dew point(td) of the cooling coil. The coil cannot cool all of the air passing through it to the coil sur- face temperature. This fact makes the coil perform in a manner similar to what would happen if a portion of the air were brought to saturation at the coil temperature and the remainder bypassed the coil unchanged. Using Eq. 3-44 and the concept of mix- ing described in the previous section, the resulting mixture is unsaturated air at point 2. In terms of the length of the line d–1, the length d–2 is proportional to the mass of air bypassed, and the length 1–2 is proportional to the mass of air not bypassed.
Because dry bulb lines are not parallel, are inclined, and the line 1–2–d is not hori- zontal, it is only approximately true that
(3-48) b t t
t t
d d
= −
−
2 1
˙ , , ,
qcl =80 600−48 400 =32 200 Btu/hr
˙ . ( , ) ,
qcs =0 6 80 600 = 48 400 Btu/hr
˙ ( . ) , .
qc =8570 32 4−23 =80 600Btu/ hr =6 7tons
˙ ˙ ( )
qc = ma1 1i −i2
˙ ˙ ˙
m ia1 1 = qc +m ia2 2 31
30
2108
8570 0 246 31 0 246 30
0 1
= = =
=
˙
˙ .
. ( )
m m
a a
˙ ˙ ˙
ma4 = ma2 −ma0 =8570−2108=6462lbma/hr
˙ ( )
ma0 500 60.
14 23 2108
= = lbma /hr
3-6 Space Air Conditioning—Design Conditions 69
and
(3-49) where bis the fraction of air bypassed, or the coil bypass factor, expressed as a deci- mal, and where the temperatures are dry bulb values. The coil sensible heat transfer rate is
(3-50a) or
(3-50b) The bypass factor is not used extensively for analysis. The ability to model coils with a computer (Chapter 14) makes the procedure unnecessary. However, some manufac- turers still use the concept in catalog data, where the bypass factor is determined from simulation and experiment.
In an actual system fans are required to move the air, and some energy may be gained from this. Referring to Fig. 3-12, the supply fan is located just downstream of the cooling unit and the return fan is just upstream of the exhaust duct. All of the power input to the fans is manifested as a sensible energy input to the air, just as if heat were transferred. Heat may also be gained in the supply and return ducts. The power input to the supply air fan and the heat gain to the supply air duct may be summed as shown on Chart 1a, Fig. 3-14, as process 1′–2. It is assumed that all of the supply fan power input is transformed to internal energy by the time the air reaches the space, state 2. Likewise, heat is gained from point 3 to point 4, where the return fan power also occurs, as shown in Fig. 3-14. The condition line for the space, 2–3, is the same as it was before when the fans and heat gain were neglected. However, the requirements of the cooling unit have changed. Process 1–1′ now shows that the capacity of the coil must be greater to offset the fan power input and duct heat gain.
Example WS3-1 given on the website is similar to Example 3-9 and includes the sup- ply and return fans with both IP and SI units.
˙ ˙ ( )( )
qcs = m c ta1 p 1−td 1−b
˙ ˙ ( )
qcs = m c ta1 p 1−t2
1 1 2
1
− = − b t −t
t td
Figure 3-14 Psychrometric processes for Example 3-9, showing the effect of fans and heat gain.
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
3 4
1 0
d
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
–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
W0
Space, 0.7 Coil, 0.65
i2 i3
i4 i0
i1
t2
t1 t3 t4 t0
W3= W4 W1 = W2 W1
0.030
In Example 3-9 the outdoor air was hot and humid. This is not always the case, and state 0 (outdoor air) can be almost anywhere on Chart 1a. For example, the south- western part of the United States is hot and dry during the summer, and evaporative cooling can often be used to advantage under these conditions. A simple system of this type is shown in Fig. 3-15. The dry outdoor air flows through an adiabatic spray chamber and is cooled and humidified. An energy balance on the spray chamber will show that the enthalpies i0and i1are equal; therefore, the process is as shown in Fig.
3-16. Ideally the cooling process terminates at the space condition line. The air then flows through the space and is exhausted. Large quantities of air are required, and this system is not satisfactory where the outdoor relative humidity is high. If W0 is too high, the process 0–1 cannot intersect the condition line.
Evaporative cooling can be combined with a conventional system as shown in Fig.
3-17 when outdoor conditions are suitable. There are a number of possibilities. First, 3-6 Space Air Conditioning—Design Conditions 71
Figure 3-15 A simple evaporative cooling system.
0 1 2
Conditioned space
Figure 3-16 Psychrometric diagram for the evaporative cooling system of Fig. 3-15.
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
0
d 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 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 == i0
W0 W2
0.030
Figure 3-17 Combination evaporative and regular cooling system.
2 3 5
1 0′
4 0
Evaporative cooler
Exhaust
Conditioned space
q Cooling
coil
qc
if the outdoor air is just mixed with return air without evaporative cooling, the ideal result will be state 1 in Fig. 3-18. The air will require only sensible cooling to state 2 on the condition line. The outdoor air could ideally be evaporatively cooled all the way to state 1′. This would require the least power for sensible cooling, and the air sup- plied to the space would be 100 percent outdoor air.
EXAMPLE 3-10
A space is to be maintained at 75 F and 50 percent relative humidity. Heat losses from the space are 225,000 Btu/hr sensible and 56,250 Btu/hr latent. The latent heat trans- fer is due to the infiltration of cold, dry air. The outdoor air required is 1000 cfm at 35 F and 80 percent relative humidity. Determine the quantity of air supplied at 120 F, the state of the supply air, the size of the furnace or heating coil, and the humidifier characteristics.
SOLUTION
Figure 3-19 is a schematic for the problem; it contains the given information and ref- erence points. First consider the conditioned space:
Figure 3-18 Psychrometric diagram for Fig. 3-17.
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
0 2 1
1′ 3
25
40 Dry bulb,F
Dry bulb temperature, F 20%
75 80
85
70 F Wet bulb 14.5 volume, ft
3 per pound of dry air 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
i2 i3
t2 t1't3 t1 t0
W0
0.030
Figure 3-19 The heating and humidifying system for Example 3-10.
2 3 5
1 x 4 0 Exhaust
Return
Humidifier
Furnace Conditioned
space t3 = 75 F
3 = 50%
t2 = 120 F t0 = 35 F
0 = 80%
qs = 225,000 Btu/hr Q0 = 1000 cfm
q1x mv
ql = 56,250 Btu/hr
The state of the supply air lies on a line drawn through state point 3 parallel to the SHF =0.8 line on the protractor of Chart 1a. Figure 3-20 shows this construction.
State 2 is located at 120 F dry bulb and the intersection of this line. An energy bal- ance on the space gives
or
From Chart 1a,i2=42 Btu/lbma,i3=28.2 Btu/lbma, and
From Chart 1a,v2=14.89 ft3/lbma, and
To find the conditions at state 1, the mixing process must be considered. A mass bal- ance on the mixing section yields
or
˙ ˙ ˙
˙ ˙
.
m m m
m Q
v and v
a a a
a
4 2 0
0 0
0
0 12 54
= −
= = ft /lbma3
˙ ˙ ˙ ˙
ma0 +ma4 = ma1 = ma2
˙ ,
. Q2 20 400
60 14 89 5060
= × = cfm
˙ ˙ ,
. ,
m q
i i
a2
2 3
281250
42 28 2 20 400
= − =
− = lbma/ hr
˙ ˙ ( )
q =ma2 i2 −i3
˙ ˙ ˙
m ia2 2 = +q m ia3 3 SHF =
+ =
225 000
225 000, 56 250 0 80
, , .
3-6 Space Air Conditioning—Design Conditions 73
Figure 3-20 Psychrometric processes for Example 3-10.
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
6560
55
50
45
4035
25 12.5 30 35
45 50
55 60
10 15 20
d
0
1
3 2
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
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 1146
1000 0.6
0.5 0.4
0.3 0.2 0.1 Space, 0.8
i3
i2
t1 t3 tx
t0
W2 W1= Wx
0.030
Using the graphical technique and referring to Fig. 3-20, we obtain
State 1 is then located at 65.5 F db and 57 F wb. A line 12—
could be constructed on Chart 1a, Fig. 3-20, representing the combination heating and humidifying process that must take place in the heating and humidifying unit. However, in practice the processes must be carried out separately. Assume that saturated vapor at 200 F is used in the humidifier.
Then iw=1145.8 Btu/lbm from Table A-1a. The required sensible heating is
The amount of water vapor supplied to the humidifier is given by
where W2=0.012 lbv/lba and W1=0.0078 lbv/lba from Chart 1a, so that
It is usually necessary to use a preheat coil to heat the outdoor air to a temperature above the dew point of the air in the equipment room so that condensation will not form on the air ducts upstream of the regular heating coil. Figure 3-21 shows this arrangement. The outdoor air is heated to state 0′, where it is mixed with return air, resulting in state 1. The mixed air is then heated to state x, where it is humidified to state 2 on the condition line for supply to the space. Figure 3-22 shows the states on Chart 1a.
Example CD3-5 illustrates a system with preheat of outdoor air. Examples of other single-path systems such as VAV or multizone could be presented here; however, under the full-flow design condition, these systems operate the same as the simple sys- tem of Figs. 3-12 and 3-13. They will be discussed further in the following section on part-load operation.
˙ , ( . . )
mv =20 400 0 012−0 0078 =86 lbv/hr
˙ ˙ ( )
mv = m Wa 2 −W1
˙ ˙ ˙ ( )
˙ , ( . )( . ) ,
q q m c t t
q
x s a pa x
s
1 1
20 400 0 244 119 65 5 266 000
= = −
= − = Btu/hr
31 30 4800
20 40030 0 235 30
0 1
= ˙ = =
˙ , . ( )
m m
a a
˙ ( )
.
˙ , ,
m m
a a
0 4
1000 60
12 54 4800 20 400 4800 15 600
= =
= − =
lbma/hr lbma/hr
Figure 3-21 Heating system with preheat of outdoor air.
2 3 5
1 x 4
0 0′
Humidifier Heating
coil
Preheat coil
Conditioned space
qs q1x mv
ql