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clear;
clc;
disp('Example 5.16');
// aim : To determine the
// (a) initial partial pressure of the steam and air
// (b) final partial pressure of the steam and air
// (c) total pressure in the container after heating
// Given values
T1 = 273+39;// initial temperature,[K]
P1 = 100;// pressure, [MN/m^2]
T2 = 273+120.2;// final temperature,[K]
// solution
// (a)
// from the steam tables, the pressure of wet steam at 39 C is
Pw1 = 7;// partial pressure of wet steam,[kN/m^2]
// and by Dalton's law
Pa1 = P1-Pw1;// initial pressure of air, [kN/m^2]
mprintf('\n (a) The initial partial pressure of the steam is = %f kN/m^2',Pw1);
mprintf('\n The initial partial pressure of the air is = %f kN/m^2\n',Pa1);
// (b)
// again from steam table, at 120.2 C the pressure of wet steam is
Pw2 = 200;// [kN/m^2]
// now since volume is constant so assuming air to be ideal gas so for air P/T=contant, hence
Pa2 = Pa1*T2/T1 ;// [kN/m^2]
mprintf(' \n(b) The final partial pressure of the steam is = %f kN/m^2',Pw2);
mprintf('\n The final partial pressure of the air is = %f kN/m^2\n',Pa2);
// (c)
Pt = Pa2+Pw2;// using dalton's law, total pressure,[kN/m^2]
mprintf('\n (c) The total pressure after heating is = %f kN/m^2\n',Pt);
// End
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