Chapter5 Refrigeração e Ar Condicionado - Stoecker and Jones solution

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CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 1 of 4 
5-1 A conditioned space that is maintained at 25 C and 50 percent relative humidity experience a sensible-heat 
of 80 kW and a latent-heat gain of 34 kW. At what temperature does the load-ratio line intersect the 
saturation line? 
 
Solution: 
 
 LS
S
qq
q
ratioLoad
+
=−
 
 q
S
 = 80 kW 
 q
L
 = 34 kW 
 
 3480 +
=−
80
ratioLoad
 
 
 Load-ratio = 0.7018 
 
 But, 
 
 
( )
ratioLoad
h-h
t-tc
ic
icp
−=
 
 
 At 25 C, 50 percent relative humidity 
 h
c
 = 50.5 kJ/kg 
 
 Try t
i
 = 15 C 
 
 
( )
0.7018
h-h
t-tc
ic
icp
=
 
 
 
( )( )
0.7018
h-50.5
15-251.0
i
=
 
 
 Connecting the two-points gives the load-ratio line which intersects the saturation line at 9.75 C with hi = 
28.76 kJ/kg. 
 
 Ans. 9.75 C. 
 
 
5-2. A conditioned space receives warm, humidified air during winter air conditioning in order to maintain 20 C 
and 30 percent relative humidity. The space experiences an infiltration rate of 0.3 kg/s of outdoor air and an 
additional sensible-heat loss of 25 kW. The outdoor air is saturated at a temperature of -20 C (see Table A-
2). If conditioned air is supplied at 40 C dry-buld, what must be the wet-bulb temperature of supply air be in 
order to maintain the space conditions? 
 
Solution: 
 
 At -20 saturated, 
 h
1
 = -18.546 kJ/kg 
 m
1
 = 0.3 kg/s 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 2 of 4 
 
 Additional heat loss = 25 kW 
 
 At 20 C and 30 percent relative humidity, 
 h
2
 = 31 kJ/kg 
 
 t
3
 = 40 C 
 
 Equations: 
 
 Sensible Heat Balance: 
 
 m
2
(t
3
 - t
2
) + m
1
(t
1
 - t
2
) = 25 kW 
 m
2
(40 - 20) + (0.3)(-20 - 20) = 25 
 m
2
 = 1.85 kg/s 
 
 Total Heat Balance: 
 m
2
(h
3
 - h
2
) + m
1
(h
1
 - h
2
) = 25 kW 
 (1.85)(h
3
 - 31) + (0.3)(-18.546 - 31) = 25 
 h
3
 = 52.55 kJ/kg 
 
 Then at 40 C and 52.55 kJ/kg. 
 
 Wet-Bulb Temperature = 18.8 C - - - Ans. 
 
5-3. A laboratory space to be maintained at 24 C and 50 percent relative humidity experiences a sensible-cooling 
load of 42 kW and a latent load of 18 kW. Because the latent load is heavy, the air-conditioning system is 
equipped for reheating the air leaving the cooling coil. The cooling coil has been selected to provide outlet air 
at 9.0 C and 95 percent relative humidity. What is (a) the temperature of supply air and (b) the airflow rate? 
 
Solution: 
 
 q
S
 = 42 kW 
 q
L
 = 18 kw 
 
 At 24 C , 50 percent relative humidity 
 h
i
 = 47.5 kJ/kg 
 
 At 9.0 C, 95 percent relative humidity 
 h
c
 = 26 kJ/kg 
 
 
( )
ic
icp
hh
ttc
lineratioloadCoil
−
−
=−
 
 
( )( ) 70.024 =−=−
47.5-26
91.0
lineratioloadCoil
 
 
0.70
1842
42
qq
q
lineratioloadCoil
LS
S
=
+
=
+
=−
 
 
 (a) Since 9 C < 13 C minimum. 
 Temperature of supply air = 13 C - - - Ans. 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 3 of 4 
 
 (b) 
 
( ) ( )( )13241.0
42
tt1.0
q
m
21
S
−
=
−
=
 
 m = 3.82 kg/s - - - - Ans. 
 
5-4. In discussing outdoor-air control Sec. 5-3 explained that with outdoor conditions in the X and Y regions on 
the psychrometric chart in Fig. 5-5 enthalpy control is more energy-efficient. We now explore some 
limitations of that statement with respect to the Y region. Suppose that the temperature setting of the outlet 
air form the cooling coil is 10 C and that the outlet air is essentially saturated when dehumidification occurs in 
the coil. If the condition of return air is 24 C and 40 percent relative humidity and the outddor conditions are 
26 C and 30 percent relative humidity, would return air or outside air be the preferred choice? Explain why. 
 
Solution: 
 
 See Fig. 5-5 and Sec. 5-3. 
 
 Outside Air: At 26 C, 30 percent relativw humidity 
 h
o
 = 42 kJ/kg 
 Coil outlet = 10 C saturated 
 q = 42 kJ/kg - 29.348 kJ/kg 
 q = 12.652 kJ/kg 
 
 Recirculated air: At 24 C, 40 percent relative humidity 
 h
i
 = 43 kJ/kg 
 With 10% outdoor air. 
 h
m
 = (0.10)(42) + (0.90)(43) = 42.9 kJ/kg 
 q = 42.9 kJ/kg - 29.348 kJ/kg 
 q = 13.552 kJ/kg > 12.652 kJ/kg. 
 
Ans. Outside air is preferred due to lower cooling required. 
 
 
5-5. A terminal reheat system (Fig. 5-9) has a flow rate of supply air of 18 kg/s and currently is operating with 3 
kg/s of outside air at 28 C and 30 percent relative humidity. The combined sensible load in the spaces is 140 
kw, and the latent load is negligible. The temperature of the supply air is constant at 13 C. An accountant of 
the firm occupying the building was shocked by the utility bill and ordered all space thermostat be set up from 
24 to 25 C. What is the rate of heat removal in the cooling coil before and after the change and (b) the rate of 
heat supplied at the reheat coils before and after change? Assume that the space sensible load remains at 
140 kw? 
 
Solution: See Fig. 5-9. 
 
 Outside air at 28 C and 30 percent relative humidity 
 h
o
 = 46 kJ/kg 
 
 At 24 C Set-Up. 
 Coil entering temperature, t
m
 
 t
m
 = [(3)(28) + (18 - 3)(24)] / 18 = 24.667 C 
 
 Coil supply temperature = 13 C constant 
 Cooling rate = (18)(24.667 - 13) = 210 kw 
 
CHAPTER 5 - AIR CONDITIONING SYSTEMS 
Page 4 of 4 
 Space sensible load = 140 kw constant 
 
 Reheat supply temperature, t
s
. 
 t
s
 = 24 - 140 / 18 = 16.222 C 
 Heating Rate = (18)(16.222 - 13) 
 Heating Rate = 58 kw 
 
 At 25 C Set-Up. 
 Coil entering temperature, t
m
 
 t
m
 = [(3)(28) + (18 - 3)(25)] / 18 = 25.5 C 
 
 Coil supply temperature = 13 C constant 
 Cooling rate = (18)(25.5 - 13) = 225 kw 
 
 Space sensible load = 140 kw constant 
 
 Reheat supply temperature, t
s
. 
 t
s
 = 25 - 140 / 18 = 17.222 C 
 Heating Rate = (18)(17.222 - 13) 
 Heating Rate = 76 kw 
 
Answer: 
 
 (a) Before = 210 kw 
 After = 225 kw 
 15 kw increase in cooling rate. 
 (b) Before = 58 kw 
 After = 76 kw 
 18 kw increase in heating rate 
 
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