Chapter 6: The Second Law of Thermodynamics

Example 6-1 ,Page No.286

In [1]:
#given data
QH=80.0;#rate of heat transfer in MW
QL=50.0;#rate of waste heat rejection in MW

#calculations
Wnet=QH-QL;
print'net power output %i MW'%Wnet
nth=Wnet/QH;
print'the thermal efficiency is %f '%round(nth,3)

net power output 30 MW
the thermal efficiency is 0.375000

Example 6-2 ,Page No.286

In [2]:
#given data
Wnet=65;#power output in hp
nth=0.24;#effficiency of car
HV=19000;#heating value in Btu/lbm

#calculations
QH=Wnet/nth*2545;#factor of 2545 to convert to Btu/h
m=QH/HV;
print'the engine must burn at fuel rate %f lbm/h'%round(m,1)

the engine must burn at fuel rate 36.300000 lbm/h

Example 6-3 ,Page No.291

In [4]:
#given data
Wnet=2.0;#power input in kW
QL=360.0;#rate of heat rejection in kJ/min

#calculations
COPR=QL/Wnet/60;#factor of 60 to convert kW to kJ/min
print'coefficient of performance of refrigerator is %i'%COPR;
QH=QL+Wnet*60;#factor of 60 to convert kW to kJ/min
print'heat rejection rate %i kJ/min'%QH

coefficient of performance of refrigerator is 3
heat rejection rate 480 kJ/min

Example 6-4 ,Page No.291

In [5]:
#given data
COP=2.5;#COP in the given conditions of heat pump
QH=80000.0;#heat loss rate in kJ/hr

#calculations
Wnet=QH/COP;
print'the power consumed %i kJ/h'%Wnet
QL=QH-Wnet;
print'the rate at which heat is absorbed %i kJ/h'%QL

the power consumed 32000 kJ/h
the rate at which heat is absorbed 48000 kJ/h

Example 6-5 ,Page No.306

In [6]:
#given data
QH=500.0;#heat recieved per cycle in kJ
TL=30+273.0;#low temperature source in C
TH=652+273.0;#high temperature source in C

#calculations
nth=1-TL/TH;
print'the thermal efficiency of carnot engine is %f'%round(nth,3);
QL=TL*QH/TH;
print'the amount of heat rejected to the sink per cycle %i kJ'%round(QL)

the thermal efficiency of carnot engine is 0.672000
the amount of heat rejected to the sink per cycle 164 kJ

Example 6-6 ,Page No.310

In [9]:
#given data
COP=13.5;#COP of refrigerator
TH=75+460.0;#room temperature in R
TL=35+460.0;#refrigerated temp in R

#calculations
COPR=1/(TH/TL-1);
if COPR>=COP:
    print('claim is true');
else:
    print('claim is false')

claim is false

Example 6-7 ,Page No.310

In [2]:
#given data
TL=-5+273.0;#in C
TH=21+273.0;#house temp in C
QH=135000.0/3600;#heat loss rate in kW

#calculations
COPHP=1/(1-TL/TH);
Wnet=QH/COPHP;
print'minimum power required %f kW'%round(Wnet,2)

minimum power required 3.320000 kW

Example 6-8 ,Page No.314

In [3]:
#given data
Qrefrig=40.0;#heat load on referigerator in W
COPR=1.3;#COP of referigerator
Wlight=40.0;#rated power of bulb in W

#calculation
Wrefrig=Qrefrig/COPR;
Wt=Wrefrig+Wlight;
AnHr=365*24;#annual hours
NOH=20*30/3600.0*365;#normal operating hours
AOP=AnHr-NOH;#addtional operating hours
APC=Wt*AOP/1000;#additional power consumption; fator of 1000 to convert to kW
print'increase in power consumption %i kWh/yr'%round(APC);
print'increase in cost S%f/yr'%round((APC)*0.08,1)

increase in power consumption 616 kWh/yr
increase in cost S49.300000/yr