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//A Textbook of Chemical Engineering Thermodynamics
//Chapter 5
//Some Applications of the Laws of Thermodynamics
//Example 10
clear;
clc;
//Given:
T1= 290; //operating temperature (K)
W = 1000; //work (J)
tof = 3516.67; //ton of refrigeration (W)
//To determine COP, heat rejected and lowest temperature
//(a)
Q2 = tof;
COP = Q2/W; //coeffecient of performance
mprintf('(a). COP is %f',COP);
//(b)
Q1 = Q2+W; //heat rejected
mprintf('\n\n (b). Heat rejected is %f kW',Q1/1000);
//(c)
//Let T2 be the lowest temperature
T2 = T1*(Q2/Q1);
mprintf('\n\n (c). Lowest possible temperature in refrigerator is %f K',T2);
//end
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