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Gnom [1K]
3 years ago
15

"A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively.

Mathematics
1 answer:
klemol [59]3 years ago
6 0

Answer:

Step-by-step explanation:

In order to answer all this questions, you need to know that the the coefficient of performance is the amount of heat removed by the system divided by the work required by the system. Using first law of thermodynamics, the work W can be calculated with the expression Q_H-Q_C, where Q_H is the heat transferred by the hot reservoir and Q_C  the heat collected by the cold reservoir. Then, the expression to calculate the coefficient of performance of a refrigeration cycle would be:

COP_{refrig}=\frac{Q_C}{Q_H - Q_C} =\frac{T_c}{T_H-T_c}

Important: The temperatures must be in absolute units (Rankine or Kelvin)

a) T_H = \frac{T_C + T_C*COP}{COP} = \frac{(-40+459.67)R + (-40+459.67)R*6.8}{6.8}=481.386 R - 459.67 = 21.72°F

b)COP = \frac{(-15+273.15)K}{(30+273.15)K -(-15+273.14)K } = 5.74

c) \frac{Q_C}{Q_H - Q_C} =\frac{T_c}{T_H-T_c}

T_H=  T_C*\frac{Q_H - Q_C}{Q_C} + T_C = (20 +459.67)R * \frac{600 Btu - 500Btu}{500 Btu} + (20 +459.67)R  = 575.604 R = 115.934 F

d) COP = \frac{T_c}{T_H-T_c} = \frac{(10+459.67)F}{(100+459.67)-(10+459.67)} = 5.22

e) T_H = \frac{T_C}{COP} + T_C =  \frac{(-5 + 273.15)K}{2} +  (-5 + 273.15)K = 402.225K = 129.075 C

f) T_H = \frac{T_C}{COP} + T_C =  \frac{(-40 + 459.67)R}{6.8} +  (-40 + 459.67)R = 481.39R = 21.72 F

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