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+// Example 4_4

+clc;funcprot(0);

+// Given data

+P=400;// kPa

+V_1=2;// m^3

+T_2=200;// °C

+Q=3500;// The amount of heat added in kJ

+

+// Calculation

+// Using the steam tables

+v_1=0.5342;// m^3/kg

+u_1=2674;// kJ/kg

+m=V_1/v_1;// kg

+// V_2=m*v_2

+// Q-(P*(V_2-V_1))=(u_2-u_1)*m---->(a)

+// This requires the trial and error process.

+// For example,guess

+v_2=1.0;// m^3/kg

+u_2=((Q-(P*((m*v_2)-V_1)))/m)+u_1;// kJ/kg

+// From the steam tables at P=0.4 MPa

+T_2=654;// °C

+// The v_2 gives 

+T_2=600;// °C

+// Guess

+v_2=1.06;// m^3/kg

+u_2=((Q-(P*((m*v_2)-V_1)))/m)+u_1;// kJ/kg

+// The tables are interpolated to give 

+T_2=640;// °C

+// The v_2 gives 

+T_2=647;// °C

+// The final temperature being approximately

+T_2=644;// °C

+printf("\nThe final temperature being approximately,T_2=%3.0f°C",T_2);