Question

An ideal Brayton cycle refrigeration cycle has a compressor pressure ratio of 6. At the compressor inlet, the pressure and temperature of the entering air are 20 lbf/in2 and 460 R. The temperature at the inlet of the turbine is 700 R. For a refrigerating capacity of 15 tons, determine (a) the mass flow rate in lbm/min. (b) the net power input, in Btu/lbm. (c) the coefficient of performance."

Answer 1

Answer:

(a) 185.1002 lbm/min

(b) 7125.12 Btu/min

(c) 0.42

Explanation:

Here we have;

$\frac{T_2}{T_1} = \frac{T_4}{T_3}=\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}$

$\frac{p_{2}}{p_{1}}$ = 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  

$\therefore \frac{T_2}{460} = \left 6\right ^{\frac{1.4 -1}{1.4 }} = 1.669$

T₂ = 1.669 × 460 = 767.515 R = 426.4 K

Refrigeration effect per kg = 1 × $c_p$ × (T₃ - T₂) = 1.005 × (700 - 767.515) = 67.85 = 37.7 kJ/kg

$Mass \, of \, air \, circulated = \frac{Refrigeration \ effect}{Refrigeration \ effect \, per \, kg}$

$Mass \, of \, air \, circulated \ per \ minute = \frac{3165.1676 }{37.7 } = 83.96 \, kg/min$

83.96 kg/min = 185.1002 lbm/min

(b)$W_{comp} = \frac{\gamma}{\gamma -1} mR(T_4 - T_3)$

T₄ = T₃×1.669 = 700×1.669 = 1167.96 °R = 648.9 K

$= \frac{1.4}{1.4 -1} \times 83.96 \times 0.287 \times (648.9 - 388.89)$

= 21925.83 kJ/min

$W_{exp} = \frac{\gamma}{\gamma -1} mR(T_1 - T_2)$

$= \frac{1.4}{1.4 -1} \times 83.96 \times 0.287 \times (426.4 - 255.56)$

= 14408.4 kJ/min

$W_{cycle} = W_{comp} - W_{exp}$

= 21925.83 - 14408.4  = 7517.43 kJ/min

$Power \, required= \frac{7517.43}{60} \, kJ/s = 125.29 \, kW$

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;

$COP = \frac{3165.1676 }{7517.43} = 0.42$