Heat engine
TL=27°c=27+273=300K
TH=900°c=900+273=1173K
QH=800KJ/min
The power output is obtained from the different efficiency relation
W=QH(1−THTL)
Put value
W=800(1−1173300)KJ/min
W=595kJ/min
The rate of heat rejected to the ambient air QL=QH−W
Put value
QL=800−595=205KJ/min
Refrigerator
W=595kJ/min
TH=27°c=300K
TL=25°c=273+25=298k
The rate of heat removed from the refrigerator is obtained by manipulating the different efficiency relation
QL=TLTH−1,W
Put values
QL=298300−1595 =88655KJ/min
The rate of heat related of the ambient air is
QH=W+QL
QH=(88655+595)KJ/min
QH=89250KJ/min
(b)
Total rate of heat rejected to the ambient air is
Qamb=QH+QL
Q=(89250+205)KJ/min
Q=89455KJ/min
Comments
Thank you.