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+// Example 10_1

+clc;funcprot(0);

+// Given data

+p_0=0.101;// MPa

+T=10;// °C

+T_0=20+273;// K

+L=0.150;// m

+D=0.0700;//  m

+R=D/2;// m

+rho=1000;// kg/m^3

+Z=0.762;// m

+g=9.81;// m/s^2

+g_c=1;// The gravitational constant

+

+// Calculation

+m=%pi*R^2*((3/4)*L)*rho;// kg

+// From Table C.1b of Thermodynamic Tables to accompany Modern Engineering Thermodynamics 

+u=42.0;// kJ/kg

+u_0=83.9;// kJ/kg

+v=0.001000;// m^3/kg

+v_0=0.001002;// m^3/kg

+s=0.1510;// kJ/kg.K

+s_0=0.2965;// kJ/kg.K

+A=m*[(u-u_0)+((p_0*10^3)*(v-v_0))-(T_0*(s-s_0))+0+((g*Z)/g_c)];// kJ

+printf("\nThe total availability of the water in the glass relative to the floor,A=%1.2f kJ",A);