Question

A heat engine operating between two reservoirs at 1000 K and 300 K is used to drive a heat pump which extracts heat from the reservoir at 300 K at a rate twice than at which the engine rejects heat to it. If the efficiency of the engine is 40% of the maximum possible and the COP of the heat pump is 50% of the maximum possible, what is the temperature of the reservoir to which the heat pump rejects heat? What is the rate of heat rejection from the heat pump if the rate of heat supply to the engine is 50 kW?​

Answer 1

Answer:

Ans. 326.5 K, 86 kW

Explanation:

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Answer 2

Concept:

A spherical mirror is a mirror that resembles a portion that has been removed from a spherical surface.

The usable heating or cooling delivered to work (energy) required ratio, also known as the coefficient of performance, or COP, of a heat pump, refrigerator, or air conditioning system.

Given:

The temperature of the heat engine source is 1000 K.

The temperature of the second reservoir is 300 K.

The heat extracted by a pump from the reservoir at 300 K is $2Q_2$.

The rate of heat supply to the engine is 50 kW.

Find:

The rate of rejection from the heat pump.

Solution:

The temperature of the first reservoir, $T_1=1000K$

The temperature of the second reservoir, $T_2=300K$

The efficiency of the heat engine, $\eta _{actual}=40%(\eta _{max})$

Now,

$\eta _{max}=\frac{T_1-T_2}{T_1}=\frac{Q_1-Q_2}{Q_1}$

$\eta _{max}=(1-\frac{T_2}{T_1})=(1-\frac{300}{1000})=0.7$

So,

$\eta _{actual}=0.4\times 0.7=0.28$

Therefore,

$(COP)_{actual}=(COP)_{max}\times0.5$

But,

$\eta _{actual}=\frac{W_E}{Q_1}=0.28$

The rate of heat supply to the engine, $Q_1=50kW$

Therefore, $W_E=0.28\times50=14kW$

We know,

$W_E=Q_1-Q_2\\0.28Q_1=Q_1-Q_2\\Q_2=0.72Q_1$

Now,

$(COP)_{actual}=0.5(\frac{T_p}{T_p-T_2} )$

$(COP)_{max}=[\frac{Q_3}{Q_3-2Q_2} ]$

Therefore,

$(COP)_{actual}=0.5(\frac{Q_3}{Q_3-2Q_2} )$

Now,

$Q_3=2Q_2+W_E\\Q_3=2(0.72Q_1)+0.28(Q_1)=1.72Q_1$

So,

$(COP)_{actual}=0.5(\frac{1.72Q_1}{0.28Q_1})=6.1428$

Now,

$6.1428=0.5(\frac{T_p}{T_p-T_2})$

$(11.2857)T_p=12.2857(300)=3685.71\\T_p=326.58K$

So, the heat rejection from the heat pump is:

$Q_3=1.72Q_1\\Q_3=1.72(50)=86kW$

Hence, the temperature of the reservoir is 326.58K and the heat rejection from the heat pump is 86kW.

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