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A Carnot heat engine receives heat from a reservoir at 900∘C at a rate of 800 kJ/min and rejects the waste heat to the ambient air at 27∘C. The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at -5∘C and transfers it to the same ambient air at 27∘C. 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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a) The maximum rate of heat removal is 4987.776 kilojoules per minute.

b) The total rate of heat rejection is 5787.776 kilojoules per minute.

Now we proceed to describe the entire procedure:

a) Maximum rate of heat removal from refrigerated space:

The efficiency of the Carnot heat engine is:

[tex]\eta_{th} = 1 - \frac{T_{L}}{T_{H}}[/tex] (1)

Where:

  • [tex]T_{L}[/tex] - Temperature of the refrigerated space, in Kelvin.
  • [tex]T_{H}[/tex] - Temperature of the ambient air, in Kelvin.

[tex]\eta_{th} = 1 - \frac{300.5\,K}{1173.15\,K}[/tex]

[tex]\eta_{th} = 0.744[/tex]

The power needed to drive the refrigerator ([tex]\dot W[/tex]), in kilojoules per minute, is:

[tex]\dot W = \eta_{th} \cdot \dot Q_{H}[/tex] (2)

Where:

  • [tex]\eta_{th}[/tex] - Thermal efficiency, no unit.
  • [tex]Q_{H}[/tex] - Heat needed to drive the refrigerator, in kilojoules per minute.

[tex]\dot W = (0.744)\cdot (800\,\frac{kJ}{min} )[/tex]

[tex]\dot W = 595.2\,\frac{kJ}{min}[/tex]

The Coefficient of Performance of a Carnot Refrigerator ([tex]COP_{R}[/tex]), no unit, is:

[tex]COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}[/tex] (3)

[tex]COP_{R} = \frac{268.15\,K}{300.15\,K-268.15\,K}[/tex]

[tex]COP_{R} = 8.380[/tex]

The maximum rate of heat removal ([tex]\dot Q_{L}[/tex]), in kilojoules per minute, from the refrigerated space is:

[tex]\dot Q_{L} = COP_{R}\cdot \dot W[/tex] (4)

[tex]\dot Q_{L} = (8.380)\cdot (595.2\,\frac{kJ}{min} )[/tex]

[tex]\dot Q_{L} = 4987.776\,\frac{kJ}{min}[/tex]

The maximum rate of heat removal is 4987.776 kilojoules per minute.

b) Total rate of heat rejection to the ambient air:

This rate is the sum of the waste heat from heat engine ([tex]\dot Q_{L, ref}[/tex]), in kilojoules per minute, and heat transferred from refrigerator ([tex]\dot Q_{L, th}[/tex]), in kilojoules per minute. That is to say:

[tex]\dot Q_{net,out} = \dot Q_{L,th} + \dot Q_{H,ref}[/tex] (5)

The heat transferred from refrigerator is:

[tex]\dot Q_{H,ref} = 4987.776\,\frac{kJ}{min} + 595.2\,\frac{kJ}{min}[/tex]

[tex]\dot Q_{H,ref} = 5582.976\,\frac{kJ}{min}[/tex]

The waste heat from heat engine is:

[tex]\dot Q_{L,th} = 800\,\frac{kJ}{min} -595.2\,\frac{kJ}{min}[/tex]

[tex]\dot Q_{L,th} = 204.8\,\frac{kJ}{min}[/tex]

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

[tex]\dot Q_{net,out} = 5787.776\,\frac{kJ}{min}[/tex]

The total rate of heat rejection is 5787.776 kilojoules per minute.

To learn more on Carnot cycle, we kindly invite to check this verified question: https://brainly.com/question/25819144

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