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clear;
clc;
disp('Example 4.11');
// aim: To determine
// the change of internal energy
// Given values
m = 1.5;// mass of steam,[kg]
P1 = 1;// initial pressure, [MN/m^2]
t = 225;// temperature, [C]
P2 = .28;// final pressure, [MN/m^2]
x = .9;// dryness fraction of steam at P2
// solution
// from steam table at P1
h1 = 2886;// [kJ/kg]
v1 = .2198; // [m^3/kg]
// hence
u1 = h1-P1*v1*10^3;// internal energy [kJ/kg]
// at P2
hf2 = 551.4;// [kJ/kg]
hfg2 = 2170.1;// [kJ/kg]
vg2 = .646; // [m^3/kg]
// so
h2 = hf2+x*hfg2;// [kj/kg]
v2 = x*vg2;// [m^3/kg]
// now
u2 = h2-P2*v2*10^3;// [kJ/kg]
// hence change in specific internal energy is
del_u = u2-u1;// [kJ/kg]
del_u = m*del_u;// [kJ];
mprintf('\n The change in internal energy is = %f kJ \n',del_u);
// End
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