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+// Theory and Problems of Thermodynamics

+// Chapter 4

+// Energy Analysis of Process

+// Example 2

+

+clear ;clc;

+

+//Given data

+V = 0.3             // Volume of container in m^3

+P1 = 0.2            // Initial Pressure in MPa

+power = 200         // Power of electric motor in watts

+t = 15              // electric motor run time in min

+// at 0.2 MPa

+h_1 = 2706.7        // Specific volume of vapor in kJ/kg

+v_1 = 0.8857        // Specific volume of vapor in m^3/kg

+

+// Units conversion

+t = t*60            // From min to seconds

+P1 = P1*1e3        // From MPa to kPa

+

+// Calculations

+W = power*t         // work done in joules

+W = W*1e-3          // units conversion J to kJ

+

+// at 0.2 MPa

+u_1 = h_1 - P1*v_1  // internal energy of stream in kJ/kg

+

+m = V/v_1          // Mass of stream in the tank in kg

+

+del_U = W         // Amount of heat energy transferred 

+

+u_2 = del_U/m + u_1

+v_2 = v_1

+

+// Final state of stream 

+P = 0.4        // Assumed temperature for trail an error method

+    // read value of Temperature by interpolation of P assumed

+T = 496.92     // From super heated steam tables at P = 0.4MPa, v = 0.8857m^3/kg

+h = 3478.39    // at P = 0.4MPa and T = 496.92 C

+

+u = h - P*T     // calculated internal energy from assumed value

+

+// Output Results

+mprintf('The final state of steam is: Superheated steam')

+

+

+