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// Theory and Problems of Thermodynamics
// Chapter 5
//Second Law of Thermodynamcis
// Example 11
clear ;clc;
//Given data
T1 = 300 // Temperature of first reservoir in K
T2 = 400 // Temperature of second reservoir in K
T3 = 1200 // Temperature of third reservoir in K
Q3 = 1200 // heat abosrbed in third reservoir in kJ
QT = 400 // energy delivered from the heat engine in kJ
// Calculations
// First law of thermodynamics gives
// Q1 + Q2 + Q3 = QT => Q1 + Q2 = -800
// Clausis inequality gives
// Q1/T1 + Q2/T2 + Q3/T3 = 0 => 4Q1 + 3Q2 = -1200
// Solving the two equations by using AX = B
A = [1,1; 4,3]; // coefficent matrix of two equations
B = [-800; -1200] // costant matrix of two equations
X = A\B
// Output Results
mprintf('The amount of energy absorbed as heat by engine from reservoir at 300K = %4.0f kJ' ,X(1))
mprintf('\n The amount of energy rejected as heat by engine from reservoir at 400K = %4.0f kJ' ,-X(2))
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