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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1919/CH5/EX5.11/Ex5_11.sce | |
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diff --git a/1919/CH5/EX5.11/Ex5_11.sce b/1919/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 000000000..f3107212a --- /dev/null +++ b/1919/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,30 @@ +
+// 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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