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Geo777

Geo777

05.05.2020 •

Engineering

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 air 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.

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Ответ:

alexisozl9253

26.01.2022

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

Explanation:

a) The efficiency of the Carnot heat engine is:

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

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

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

The power output used to drive 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 coefficient of performance of the refrigerator is:

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

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

$COP_{R} = 8.380$

The heat removal from the refrigerated space is:

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

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

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

b) The heat rejected from the Carnot heat engine to the ambient air is:

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

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

Now, the heat rejected from the refrigerator to the ambient air is:

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

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

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

Finally, the total rate of heat rejection to the ambient air is:

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

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

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Ответ:

spicee68003

03.02.2020

255

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