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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. νg = 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 νsuc = 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. = 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 νsuc = 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. = 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?
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. x 2 = 0.
x x
x Percentice 1 2
1
Percent ice
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.
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.
∆∆∆∆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.
p
k
VD c D
k h 0.023
μ
μ
ρ
(a) Water:
ρ = 0.99915 kg/L = 999.15 kg/m
3
D = 0.020 m μ = 0.00116 Pa.s k = 0.584 W/m.K cp = 4190 J/kg.K V = 2.5 m/s
0.8 0.
h 0.023
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/m^3 D = 0.020 m μ = 0.01884 Pa.s (Fig. 15-5) k = 0.45 W/m.K (Fig. 15-4) cp = 3450 J/kg.K (Fig. 15-6) V = 2.5 m/s
h 0.023
h = 2,188 W/m
2 .K - - - Ans.
