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clear;
clc;
disp('Example 5.18');
// aim : To determine the
// (a) final pressure
// (b) final dryness fraction of the steam
// Given values
P1 = 130;// initial pressure, [kN/m^2]
T1 = 273+75.9;// initial temperature, [K]
x1 = .92;// initial dryness fraction
T2 = 273+120.2;// final temperature, [K]
// solution
// (a)
// from steam table, at 75.9 C
Pws = 40;// partial pressure of wet steam[kN/m^2]
Pa = P1-Pws;// partial pressure of air, [kN/m^2]
vg = 3.99// specific volume of the wet steam, [m^3/kg]
// hence
V1 = x1*vg;// [m^3/kg]
V2 = V1/5;// [m^3/kg]
// for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so
P2 = Pa*V1*T2/(V2*T1);// final pressure,[kN/m^2]
// now for steam at 120.2 C
Ps = 200;// final partial pressure of steam,[kN/m^2]
// so by dalton's law total pressure in cylindert is
Pt = P2+Ps;// [kN/m^2]
mprintf('\n (a) The final pressure in the cylinder is = %f kN/m^2\n',Pt);
// (b)
// from steam table at 200 kN/m^2
vg = .885;// [m^3/kg]
// hence
x2 = V2/vg;// final dryness fraction of the steam
mprintf('\n (b) The final dryness fraction of the steam is = %f\n ',x2);
// End
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