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clear;
clc;
disp('Example 4.19');
// aim : To determine the condition of the steam after
// (a) isothermal compression to half its initial volume,heat rejected
// (b) hyperbolic compression to half its initial volume
// Given values
V1 = .3951;// initial volume,[m^3]
P1 = 1.5;// initial pressure,[MN/m^2]
// solution
// (a)
// from steam table, at 1.5 MN/m^2
hf1 = 844.7;// [kJ/kg]
hfg1 = 1945.2;// [kJ/kg]
hg1 = 2789.9;// [kJ/kg]
vg1 = .1317;// [m^3/kg]
// calculation
m = V1/vg1;// mass of steam,[kg]
vg2b = vg1/2;// given,[m^3/kg](vg2b is actual specific volume before compression)
x1 = vg2b/vg1;// dryness fraction
h1 = m*(hf1+x1*hfg1);// [kJ]
Q = m*x1*hfg1;// heat loss,[kJ]
mprintf('\n (a) The Quantity of steam present is = %f kg \n',m);
mprintf('\n Dryness fraction is = %f \n',x1);
mprintf('\n The enthalpy is = %f kJ \n',h1);
mprintf('\n The heat loss is = %f kJ \n',Q);
// (b)
V2 = V1/2;
// Given compression is according to the law PV=Constant,so
P2 = P1*V1/V2;// [MN/m^2]
// from steam table at P2
hf2 = 1008.4;// [kJ/kg]
hfg2 = 1793.9;// [kJ/kg]
hg2 = 2802.3;// [kJ/kg]
vg2 = .0666;// [m^3/kg]
// calculation
x2 = vg2b/vg2;// dryness fraction
h2 = m*(hf2+x2*hfg2);// [kJ]
mprintf('\n (b) The dryness fraction is = %f \n',x2);
mprintf('\n The enthalpy is = %f kJ\n',h2);
// End
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