Question

a) A reversible heat engine operates between two reservoirs at temperature 700°C and 50°C. The engine drives a reversible refrigerator which operates between reservoirs at temperatures of 50°C and -25°C. The heat transfer to the engine is 2500 kJ and the net work output of the combined engine refrigerator plant is 400 kJ. (i) Determine the heat transfer to the refrigerant and the net heat transfer to the reservoir at 50°C; (ii) Reconsider (i) given that the efficiency of the heat engine and the COP of the refrigerator are each 45% of the maximum possible values.

Answer 1

Explanation:

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Answer 2

Given,

A reversible heat engine operates at temperatures 700°C and 50°C.

The engine drives a reversible refrigerator at temperatures of 50°C and -25°C.

Heat transfer to the engine = 2500 kJ

The work output of the combined engine refrigerator plant = 400 kJ.

To find,

(i) Heat transfer to the refrigerant and the net heat transfer to the reservoir at 50°C

(ii) Reconsider (i) given that the efficiency of the heat engine and the COP of the refrigerator are each 45% of the maximum possible values.

Solution,

We may easily answer the numerical problem by following the steps below.

We are aware of this.

T₁ = 973K

T₂ = 323K

T₃ = 248K

Q₁ = 2500 kJ

W = W₁ - W₂ = 400kJ

Now,

(i) Maximum efficiency of the heat engine cycle:

η(max) = $1- \frac{T2}{T1}$ = 1 - $\frac{323}{973}$ = 0.668

Again, $\frac{W1}{Q1} = 0.668$

           W₁ = 1670 kJ

(C.O.P)max = $\frac{T3}{T2-T3}$ = $\frac{248}{323-248}$ = 3.306

Furthermore, COP = $\frac{Q4}{W2}$ = 3.306

Since,

W₁ - W₂ = W = 400kJ

W₂ = W₁ - W = 1670 - 400 = 1270 kJ

Q₄ = 3.306 * 1270 = 4198.6 kJ

Q₃ = Q₄ + W₂ = 4198.6 + 1270 = 5468.6 kJ

Q₂ = Q₁ - W₁ = 830kJ

As a consequence, the heat rejection to the 50°C reservoirs = Q₂ + Q₃ = 6298.6 kJ

(ii) Efficiency of actual heat engine cycle;

η = 0.45 * η(max) = 0.45 * 0.668 = 0.3

W₁ = η* Q₁ = 0.3 * 2500 = 750kJ

W₂ = 350 kJ

C.O.P of the actual refrigerator cycle;

COP = $\frac{Q4}{W2}$ = 0.45 * 3.306 = 1.48

Q₄ = 350 * 1.48 = 518 kJ

Q₃ = 518 + 350 = 868 kJ

Q₂ = 2500 - 750 = 1750 kJ

As a consequence, Heat rejected to the 50°C reservoirs = Q₂ + Q₃  = 2618 kJ