Chapter15 Refrigeração e Ar Condicionado - Stoecker and Jones solution

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CHAPTER 15 - REFRIGERANTS 
 Page 1 of 4 
15-1. The machine room housing the compressor and condenser of a refrigerant 12 system has dimensions 5 by 4 
by 3 m. Calculate the mass of the refrigerant which would have to escape into the space to cause a toxic 
concentration for a 2-h exposure. 
 
Solution: 
 Section 15-7, Refrigerant 12 exposure for 2-h has 20 % by volume to become toxic. 
 Room volume = 5 x 4 x 3 m = 60 m
3
. 
 
 Volume of refrigerant 12. 
 = (0.20)(60) = 12 m
2
. 
 
 At atmospheric, 101.325 kPa, Table A-5. 
 ν
g
 = 158.1254 L/kg = 0.1581254 m
3
/kg 
 
 Mass of refrigerant 12. 
 = (12 m
2
) / (0.1581254 m
3
/kg) 
 = 76 kg - - - Ans. 
 
 
15-2. Using data from Table 15-4 for the standard vapor-compression cycle operating with an evaporating 
temperature of -15 C and a condensing temperature of 30 C, calculate the mass flow rate of refrigerant per 
kilowatt of refrigeration and the work of compression for (a) refrigerant 22 and (b) ammonia. 
 
Solution: 
 Table 15-4. 
 (a) Refrigerant 22. 
 
 Suction vapor flow per kW of refrigeration = 0.476 L/s 
 Table A-6, at -15 C evaporating temperature 
 ν
suc
 = 77.68375 L/kg 
 
 mass flow rate = (0.476 L/s) / (77.68375 L/kg) 
 = 0.0061274 kg/s - Ans. 
 
 Work of compression = (mass flow rate)(refrigerating effect) / COP 
 = (0.0061274 kg/s)(162.8 kJ/kg) / 4.66 
 = 0.2141 kW -- - Ans. 
 
 (b) Ammonia (717). 
 
 Suction vapor flow per kW of refrigeration = 0.476 L/s 
 Table A-3, at -15 C evaporating temperature 
 ν
suc
 = 508.013 L/kg 
 
 mass flow rate = (0.476 L/s) / (508.013 L/kg) 
 = 0.00090943 kg/s - Ans. 
 
 Work of compression = (mass flow rate)(refrigerating effect) / COP 
 = (0.00090943 kg/s)(1103.4 kJ/kg) / 4.76 
 = 0.2108 kW -- - Ans. 
 
15-3. A 20% ethylene glycol solution in water is gradually cooled/ 
(a) At what temperature does crystalluzation begin? 
(b) If the antifreeze is cooled to -25 C, what percent will have frozen into ice? 
CHAPTER 15 - REFRIGERANTS 
 Page 2 of 4 
 
Solution: 
 Figure 15-1 and Figure 15-2. 
 
 (a) At point B, 20 % Ethylene Glycol 
 Crystallization Temperature = -8.5 C 
 
 (b) If cooled to -25 C. 
 x
1
 = 0.20 
 x
2
 = 0.425 
 
( )100
xx
x
icePercent
21
1
+
=
 
 
( )100
0.4250.20
0.20
icePercent
+
=
 
 Percent ice = 32 % - - - Ans. 
 
 
15-4. A solution of ethylene glycol and water is to be prepared for a minimum temperature of -30 C. If the 
antifreeze is mixed at 15 C, what is the required specific gravity of the antifreeze solution at this 
temperature? 
 
Solution: 
 Fig. 15-1 and Fig. 15-2 at -30 C, point B 
 concentration = 46 % glycol 
 Figure. 15-3, at 15 C, 46 % glycol. 
 Specific gravity based on water = 1.063 - - - Ans. 
 
15-5. For a refrigeration capacity of 30 kW, how many liters per second of 30 % solution of ethylene glycol-water 
must be circulated if the antifreeze enters the liquid chiller at -5 C and leaves at -10 C? 
 
Solution 
Figure 15-6. 
 At -5 C, cp = specific heat = 3.75 kJ/kg.K 
 At -10 C, cp = specific heat = 3.75 kJ/kg.K 
 
 q = 30 kw = w (3.75 kJ/kg.K)(-5 C - (-10 C)) 
 
 w = 1.60 kg/s 
 
 Specific gravity at -7.5 C = 1.0475 
 
 Liters per second = (1.60 kg/s)(1 / 1.0475 kg/L) 
 Liters per second = 1.53 L/s - - - Ans. 
 
 
15-6. A manufacturer’s catalog gives the pressure drop through the tubes of a heat-exchanger as 70 kPa for a 
given flow rate of water at 15 C. If a 40 % ethylene glycol-water solution at -20 C flows through the heat 
exchanger at the same mass flow rate as the water, what will the pressure drop be? Assume turbulent flow. 
At 15 C the viscosity of water is 0.00116 Pa/.s. 
 
Solution: 
 Equation 15-3. 
CHAPTER 15 - REFRIGERANTS 
 Page 3 of 4 
 
w
w
w
w
w
a
a
a
a
a
w
a
2
2V
D
L
f
2
2V
D
L
f
p
p
ρ
ρ
=
∆
∆
 
 Equation 15-4. 
 
0.25Re
0.33
f =
 
 
 
µ
ρ
=
DV
Re
 
 ∆p
w
 = 70 kPa 
 µ
w
 = 0.0016 Pa.s 
 ρ
w
 = 0.99915 kg/L at 15 C 
 
 
wa
w
w
a
a DD,
D
L
D
L
==
 
 
0.25
aaw
wwa
0.25
a
0.25
w
w
a
V
V
Re
Re
f
f








ρµ
ρµ
==
 
 But: 
 a
a
w
w
A
w
V;
A
w
V
ρ
=
ρ
=
 
 Then: 
 
0.25
w
a
w
a
f
f






µ
µ
=
 
 Equation 15-3 then becomes, 
 
2
w
a
w
a
0.25
w
a
w
a
V
V
∆p
∆p












ρ
ρ






µ
µ
=
 
 
2
a
w
w
a
0.25
w
a
w
a
∆p
∆p








ρ
ρ






ρ
ρ






µ
µ
=
 
 








ρ
ρ






µ
µ
=
a
w
0.25
w
a
w
a
∆p
∆p
 
 For 40 % Ethylene Glycol, -20 C. 
 Fig. 15-3, Specific Gravity = 1.069 
 
 ρ
a
 = 1.069 kg/L 
 
 Fig. 15-5 
 
 µ
a
 = 0.01884 Pa.s 
 
 Substitute: 
 
 












=
1.069
0.99915
0.00116
0.01884
70
∆p
0.25
a
 
CHAPTER 15 - REFRIGERANTS 
 Page 4 of 4 
 ∆∆∆∆pa = 131 kPa - - - Ans. 
 
15-7. Compute the convection heat-transfer coefficient for liquid flowing through a 20-mm-ID tube when the 
velocity is 2.5 m/s if the liquid is (a) water at 15 C, which has a viscosity of 0.00116 Pa.s and a thermal 
conductivity of 0.584 W/m.K; (b) 40 % solution of ethylene glycol at -20 C. 
 
Solution: 
 Equation 15-5. 
 
0.4
p
0.8
k
cVD
D
k
0.023h 






 µ






µ
ρ
=
 
 (a) Water: 
 
 ρ = 0.99915 kg/L = 999.15 kg/m3 
 D = 0.020 m 
 µ = 0.00116 Pa.s 
 k = 0.584 W/m.K 
 c
p
 = 4190 J/kg.K 
 V = 2.5 m/s 
 
( )( )( ) ( )( ) 0.40.8
0.584
0.001164190
0.0016
999.150.0202.5
0.020
0.584
0.023h 

















=
 
 h = 6,177 W/m
2
.K - - - Ans. 
 
 (b) 40 % Solution, Ethylene Glycol at -20 C 
 
 ρ = 1.069 kg/L (Fig. 15-3) = 1069 kg/m3 
 D = 0.020 m 
 µ = 0.01884 Pa.s (Fig. 15-5) 
 k = 0.45 W/m.K (Fig. 15-4) 
 c
p
 = 3450 J/kg.K (Fig. 15-6) 
 V = 2.5 m/s 
 
( )( )( ) ( )( ) 0.40.8
0.450
0.018843450
0.01884
10690.0202.5
0.020
0.450
0.023h 

















=
 
 h = 2,188 W/m
2
.K - - - Ans. 
 
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