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A Carnot heat engine receives heat from a reservoir at 900oC at a rate of 800 kJ/min and rejects the waste heat to the ambient a

ir at 27oC. The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at -5oC and transfers it to the same ambient air at 27oC. Determine (a) the maximum rate of heat removal from the refrigerated space and (b) the total rate of heat rejection to the ambient air.

1 answer:

Mrrafil [318]3 months ago

0 0

Answer:

a) $\dot Q_{L} = 4987.776\,\frac{kJ}{min}$ , b) $\dot Q = 5787.776\,\frac{kJ}{min}$

Explanation:

a) The Carnot heat engine's efficiency can be expressed as:

$\eta_{th} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%$

$\eta_{th} = \left(1-\frac{300.15\,K}{1173.15\,K} \right)\times 100\,\%$

$\eta_{th} = 74.415\,\%$

The energy utilized for operating the refrigerator is:

$\dot W = (0.744)\cdot \left(800\,\frac{kJ}{min}\right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)$

$\dot W = 9.92\,kW$

The refrigerator's coefficient of performance is determined by:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

$COP_{R} = \frac{268.15\,K}{300.15\,K-268.15\,K}$

$COP_{R} = 8.380$

The amount of heat extracted from the refrigerated area equals:

$\dot Q_{L} = \dot W\cdot COP_{R}$

$\dot Q_{L} = (9.92\,kW)\cdot \left(60\,\frac{s}{min} \right)\cdot (8.380)$

$\dot Q_{L} = 4987.776\,\frac{kJ}{min}$

b) The heat that the Carnot heat engine expels into the surrounding air amounts to:

$\dot Q_{L} = (1-0.744)\cdot \left(800\,\frac{kJ}{min} \right)$

$\dot Q_{L} = 204.8\,\frac{kJ}{min}$

The heat discharged from the refrigerator to the environment is:

$\dot Q_{H} = \dot W + \dot Q_{L}$

$\dot Q_{H} = (9.92\,kW)\cdot (60\,\frac{s}{min} ) + 4987.776\,\frac{kJ}{min}$

$\dot Q_{H} = 5582.976\,\frac{kJ}{min}$

The overall heat expulsion to the ambient air is then calculated as:

$\dot Q = 5582.976\,\frac{kJ}{min} + 204.8\,\frac{kJ}{min}$

$\dot Q = 5787.776\,\frac{kJ}{min}$

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