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+// Example 8_11

+clc;funcprot(0);

+// Given data

+// from Example 5.7

+D=0.100;// m

+T_1=200;// K

+p_1=0.140;// MPa

+h=3.50;// W/(m^2.K)

+T_infinitive=15.0;// K

+c_v=3.123;// kJ/kg.K

+R=2.077;// kJ/kg.K

+t=5.00;// seconds

+

+// Calculation

+// (a)

+V=(%pi/6)*D^3;// m^3

+A=%pi*D^2;// m^2

+m=((p_1*10^3)*V)/(R*(T_1+273.15));// kg

+hAbymc_v=(h*A)/(m*c_v*1000);// s^-1

+T_2=((T_1-T_infinitive)*exp((-(h*A)/(m*c_v*1000))*t))+T_infinitive;// °C

+// (b)

+delU=m*c_v*(T_2-T_1);// kJ

+// (c)

+// Let s_2-s_1=ds

+ds=(c_v*log((T_2+273.15)/(T_1+273.15)))+0;// kJ/(kg.K)

+dQbyT_b=-1.35*10^-4;// kJ/K

+S_P=((m*ds)-(dQbyT_b));// kJ/K

+S_P=S_P*10^3;// J/K

+printf("\n(a)The final temperature of the helium,T_2=%2.1f°C \n(b)The change in total internal energy of the helium,U_2-U_1=%0.3f kJ \n(c)The total entropy production in the helium,S_P=%0.4f J/K",T_2,delU,S_P);