Respuesta :
Answer:
(a) 185.1002 lbm/min
(b) 7125.12 Btu/min
(c) 0.42
Explanation:
Here we have;
[tex]\frac{T_2}{T_1} = \frac{T_4}{T_3}=\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}[/tex]
[tex]\frac{p_{2}}{p_{1}}[/tex] = 6
γ = 1.4 for air
T₁ = 460 R = 255.56 K
T₃ = 700 R = 388.89 K
1 ton of refrigeration capacity = 3.52 kWh = 0.98 W
∴ 15 ton = 15×3.52 = 52.8 kW/h
= 189910.0534716 kJ/h = 3165.1676 kJ/min
[tex]\therefore \frac{T_2}{460} = \left 6\right ^{\frac{1.4 -1}{1.4 }} = 1.669[/tex]
T₂ = 1.669 × 460 = 767.515 R = 426.4 K
Refrigeration effect per kg = 1 × [tex]c_p[/tex] × (T₃ - T₂) = 1.005 × (700 - 767.515) = 67.85 = 37.7 kJ/kg
[tex]Mass \, of \, air \, circulated = \frac{Refrigeration \ effect}{Refrigeration \ effect \, per \, kg}[/tex]
[tex]Mass \, of \, air \, circulated \ per \ minute = \frac{3165.1676 }{37.7 } = 83.96 \, kg/min[/tex]
83.96 kg/min = 185.1002 lbm/min
(b)[tex]W_{comp} = \frac{\gamma}{\gamma -1} mR(T_4 - T_3)[/tex]
T₄ = T₃×1.669 = 700×1.669 = 1167.96 °R = 648.9 K
[tex]= \frac{1.4}{1.4 -1} \times 83.96 \times 0.287 \times (648.9 - 388.89)[/tex]
= 21925.83 kJ/min
[tex]W_{exp} = \frac{\gamma}{\gamma -1} mR(T_1 - T_2)[/tex]
[tex]= \frac{1.4}{1.4 -1} \times 83.96 \times 0.287 \times (426.4 - 255.56)[/tex]
= 14408.4 kJ/min
[tex]W_{cycle} = W_{comp} - W_{exp}[/tex]
= 21925.83 - 14408.4 = 7517.43 kJ/min
[tex]Power \, required= \frac{7517.43}{60} \, kJ/s = 125.29 \, kW[/tex]
1 kW = 56.87 Btu/min
Therefore;
125.29 kW = 7125.12 Btu/min
(c) The coefficient of performance (COP), is given by the relation;
[tex]COP = \frac{3165.1676 }{7517.43} = 0.42[/tex]
