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diff --git a/647/CH2/EX2.2/Example2_2.sce b/647/CH2/EX2.2/Example2_2.sce
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+clear;
+clc;
+
+// Example: 2.2
+// Page: 40
+
+printf("Example: 2.2 - Page: 40\n\n");
+
+// Solution
+
+//*****Data*****//
+U1 = 1000;// [kJ]
+Q = -600; // [kJ]
+W = -100;// [kJ]
+//************//
+
+// The system is considered to be a closed system. No mass transfer takes place across the system. The tank is rigid. 
+// So, the kinetic and the potential energies is zero.
+// Therefore:
+// delta_E = delta_U + delta_PE + delta_KE
+// delta_E = delta_U
+// From the first law of thermodynamics:
+// Q = delta_U + W
+// delta_U = Q - W
+// U2 - U1 = Q - W
+U2 = U1 + Q - W;// [kJ]
+printf("The final internal energy of the fluid is %d kJ\n",U2);
+\ No newline at end of file