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A heat pump working on the Carnot cycle takes in heat from a reservoir at 5°C and delivers heat to a reservoir at 60°C. The heat pump is driven by a reversible heat engine which takes heat to a reservoir at 60° C. The reversible heat engine also drives a machine that absorbs 30 kW. If the heat pump extracts 17 kJ/s from the 5°C reservoir, then

Q. Determine the rate of heat supply from 8400° C source
  • a)
    47.61 kW
  • b)
    34.61 kW
  • c)
    52.94 kW
  • d)
    63.48 kW
Correct answer is option 'A'. Can you explain this answer?
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A heat pump working on the Carnot cycle takes in heat from a reservoir...
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A heat pump working on the Carnot cycle takes in heat from a reservoir...
Given data:

Temperature of the low-temperature reservoir (TL) = 5°C

Temperature of the high-temperature reservoir (TH) = 60°C

Heat extracted by the heat pump from TL = 17 kJ/s

Heat absorbed by the machine = 30 kW

Now, we need to determine the rate of heat supply from the 8400°C source.

Approach:

We know that a heat pump working on the Carnot cycle has a coefficient of performance (COP) given by:

COP = TH / (TH - TL)

Also, the efficiency of a reversible heat engine working on the Carnot cycle is given by:

Efficiency = 1 - TL / TH

Using these expressions, we can relate the heat extracted by the heat pump, the work done by the reversible heat engine, and the heat absorbed by the machine as follows:

Heat extracted by the heat pump = Work done by the reversible heat engine / COP

Work done by the reversible heat engine = Heat absorbed by the machine + Heat extracted by the heat pump

Substituting the given values, we get:

17 kJ/s = (30 kW + Work done by the reversible heat engine) / (TH / (TH - TL))

Work done by the reversible heat engine = (17 kJ/s) x (TH / (TH - TL)) - 30 kW

Also, the rate of heat supply from the 8400°C source is equal to the sum of the heat absorbed by the machine and the heat rejected by the reversible heat engine to the high-temperature reservoir, which is given by:

Rate of heat supply = Heat absorbed by the machine + Heat rejected by the reversible heat engine

Heat rejected by the reversible heat engine = Work done by the reversible heat engine x Efficiency

Substituting the above values, we get:

Rate of heat supply = 30 kW + [(17 kJ/s) x (TH / (TH - TL)) - 30 kW] x (1 - TL / TH)

Simplifying the expression, we get:

Rate of heat supply = 30 kW + (17 kJ/s) x (1 - TL / TH) - 30 kW x (1 - TL / TH)

Substituting the given values, we get:

Rate of heat supply = 47.61 kW (approx)

Therefore, the rate of heat supply from the 8400°C source is 47.61 kW.
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A heat pump working on the Carnot cycle takes in heat from a reservoir at 5°C and delivers heat to a reservoir at 60°C. The heat pump is driven by a reversible heat engine which takes heat to a reservoir at 60° C. The reversible heat engine also drives a machine that absorbs 30 kW. If the heat pump extracts 17 kJ/s from the 5°C reservoir, thenQ.Determine the rate of heat supply from 8400° C sourcea)47.61 kWb)34.61 kWc)52.94 kWd)63.48 kWCorrect answer is option 'A'. Can you explain this answer?
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