Question #196249

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 25°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.

1
Expert's answer
2021-05-24T15:56:03-0400

Heat engine

TL=27°c=27+273=300KT_L=27°c=27+273=300K

TH=900°c=900+273=1173KT_H=900°c=900+273=1173K

QH=800KJ/minQ_H=800KJ/min


The power output is obtained from the different efficiency relation

W=QH(1TLTH)W=Q_H(1-\frac{T_L}{TH})

Put value


W=800(13001173)KJ/minW=800(1-\frac{300}{1173})KJ/min

W=595kJ/min 

The rate of heat rejected to the ambient air QL=QHWQ_L=Q_H-W

Put value

QL=800595=205KJ/minQ_L=800-595=205 KJ/min

Refrigerator 

W=595kJ/minW=595kJ/min

TH=27°c=300KT_H=27°c=300K

TL=25°c=273+25=298kT_L=25°c=273+25=298k

The rate of heat removed from the refrigerator is obtained by manipulating the different efficiency relation

QL=WTHTL1,Q_L=\frac{W}{\frac{T_H}{T_L}-1,}

Put values

QL=5953002981Q_L=\frac{595}{\frac{300}{298}-1} =88655KJ/min

The rate of heat related of the ambient air is

QH=W+QLQ_H=W+Q_L

QH=(88655+595)KJ/minQ_H=(88655 +595)KJ/min

QH=89250KJ/minQ_H=89250KJ/min

(b)

Total rate of heat rejected to the ambient air is

Qamb=QH+QLQ_{amb}=Q_H+Q_L

Q=(89250+205)KJ/minQ=(89250+205)KJ/min

Q=89455KJ/minQ=89455KJ/min



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Comments

Theophilus Joseph Attah
04.05.23, 12:13

Thank you.

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