Professor Modyn wants to power his refrigerator with a heat engine. A Carnot hea

Question

Professor Modyn wants to power his refrigerator with a heat engine. A Carnot heat engine receives heat from a reservoir at 493.0°C at a rate of 767 kJ/min and rejects heat to the ambient air at 29.1°C. The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at -3.23°C and transfers it to the same ambient air at 29.1°C. Note: The IUPAC sign conversion for work is used. Work into the system has a positive value.

a) Determine the maximum rate of heat removal from the refrigerated space (kW)

b) Determine the total rate of heat rejection to the ambient air. Heat rejection is a negative value. Account for both the heat engine and refrigerator. (kW)

Explanation / Answer

Let T1 be the temperature of the hot reservoir, in K. Let Q1 the rate at which heat is absorbed from this reservoir.

Let T2 be the temperature of the refrigerated space, in K. Let Q2 the rate at which heat is absorbed from this space.

Let T3 be the temperature of ambient air, in K. Let Q3 the rate at which heat is rejected to ambient air.

Now, we have, from the First Law,

Q1+Q2 = Q3.

Also, from the Second Law (the form of Clausius ineqality):

(Q1/T1)+(Q2/T2)-(Q3/T3) 'less than or equal to' 0 (zero).

For the limiting case (reversible limit):

Q1/T1)+(Q2/T2)-(Q3/T3) = 0

That is, (Q1/T1) + (Q2/T2) - ((Q1+Q2)/T3) = 0

Solve for Q2

(In this given problem we know Q1, T1, T2, and T3 are known

i.e

T1= 493oC+273 =766K

Q1 = 767 KJ/min

T2 = -3.23oC+273 = 269.77K

T3 = 29.1oC+273 = 302.1K).

therfore (767/766)+(Q2/269.77)-((767+Q2)/302.1) =0

Q2 = -3875 KJ/min

Then, Q3 = Q1 + Q2.

therfore Q3 = 767-3875 = -3108KJ/min

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