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clear;
clc;
disp(' Example 14.2');
// aim : To determine
// (a) the intermediate pressure
// (b) the total volume of each cylinder
// (c) the cycle power
// given values
v1 = .2;// air intake, [m^3/s]
P1 = .1;// intake pressure, [MN/m^2]
T1 = 273+16;// intake temperature, [K]
P3 = .7;// final pressure, [MN/m^2]
n = 1.25;// compression index
N = 10;// speed, [rev/s]
// solution
// (a)
P2 = sqrt(P1*P3);// intermediate pressure, [MN/m^2]
mprintf('\n (a) The intermediate pressure is = %f MN/m^2\n',P2);
// (b)
V1 = v1/N;// total volume,[m^3]
// since intercooling is perfect so 2 lie on the isothermal through1, P1*V1=P2*V2
V2 = P1*V1/P2;// volume, [m^3]
mprintf('\n (b) The total volume of the HP cylinder is = %f litres\n',V2*10^3);
// (c)
CP = 2*n/(n-1)*P1*v1*((P2/P1)^((n-1)/n)-1);// cycle power, [MW]
mprintf('\n (c) The cycle power is = %f MW\n',CP*10^3);
// there is calculation mistake in the book so answer is not matching
// End
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