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clear;
clc;
disp('Example 5.12');
// aim : T0 determine
// (a) change in internal nergy of the air
// (b) work done
// (c) heat transfer
// Given values
m = .25;// mass, [kg]
P1 = 140;// initial pressure, [kN/m^2]
V1 = .15;// initial volume, [m^3]
P2 = 1400;// final volume, [m^3]
cp = 1.005;// [kJ/kg K]
cv = .718;// [kJ/kg K]
// solution
// (a)
// assuming ideal gas
R = cp-cv;// [kJ/kg K]
// also, P1*V1=m*R*T1,hence
T1 = P1*V1/(m*R);// [K]
// given that process is polytropic with
n = 1.25; // polytropic index
T2 = T1*(P2/P1)^((n-1)/n);// [K]
// Hence, change in internal energy is,
del_U = m*cv*(T2-T1);// [kJ]
mprintf('\n (a) The change in internal energy of the air is del_U = %f kJ',del_U);
if(del_U>0)
disp('since del_U>0, so it is gain of internal energy to the air')
else
disp('since del_U<0, so it is gain of internal energy to the surrounding')
end
// (b)
W = m*R*(T1-T2)/(n-1);// formula of work done for polytropic process,[kJ]
mprintf('\n (b) The work done is W = %f kJ',W);
if(W>0)
disp('since W>0, so the work is done by the air')
else
disp('since W<0, so the work is done on the air')
end
// (c)
Q = del_U+W;// using 1st law of thermodynamics,[kJ]
mprintf('\n (c) The heat transfer is Q = %f kJ',Q);
if(Q>0)
disp('since Q>0, so the heat is received by the air')
else
disp('since Q<0, so the heat is rejected by the air')
end
// End
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