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// Theory and Problems of Thermodynamics
// Chapter 5
// Second law of Thermodynamics
// Example 8
clear ;clc;
//Given data
Q = 100 // rate of energy losses as heat from roof and walls of auditorium in kW per degree celsius
TD = 298.15 // desired temperature to be maintained in K
TW = 273.15 // outside temperature during winter in K
TS = 314.15 // outside temperature during summer in K
// Winter: Device is used as heat pump
TL = TW // outside temperature
TH = TD // Desired temperature
QH = Q*(TH-TL) // total heat loss to surroundings
// COP_HP = QH/W = TH/(TH-TL)
W = QH * (TH-TL)/TH // Power required in kW
mprintf('The power required to operate the device in winter = %5.2f kW', W)
// Summer: Device is used as refrigerator
TL = TD // Desired temperature
TH = TS // Outside temperature
QL = Q*(TH-TL) // Total heat loss to surroundings
// COP_HP = QH/W = TL/(TH-TL)
W = QL * (TH-TL)/TL // Power required in kW
mprintf('\n The power required to operate the device in summer = %5.2f kW', W)
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