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clear;
clc;
disp('Example 2.4');
// Given values
m_dot = 4; // fluid flow rate, [kg/s]
Q = -40; // Heat loss to the surrounding, [kJ/kg]
// At inlet
P1 = 600; // pressure ,[kn/m^2]
C1 = 220; // velocity ,[m/s]
u1 = 2200; // internal energy, [kJ/kg]
v1 = .42; // specific volume, [m^3/kg]
// At outlet
P2 = 150; // pressure, [kN/m^2]
C2 = 145; // velocity, [m/s]
u2 = 1650; // internal energy, [kJ/kg]
v2 = 1.5; // specific volume, [m^3/kg]
// solution
// for steady flow energy equation for the open system is given by
// u1+P1*v1+C1^2/2+Q=u2+P2*v2+C2^2/2+W
// hence
W = (u1-u2)+(P1*v1-P2*v2)+(C1^2/2-C2^2/2)*10^-3+Q; // [kJ/kg]
mprintf('\n workdone is, W = %f kJ/kg ',W);
if(W>0)
disp('Since W>0, so Power is output from the system')
else
disp('Since <0, so Power is input to the system')
end
// Hence
P_out = W*m_dot; // power out put from the system, [kW]
mprintf('\n The power output from the system is = %f kW \n',P_out);
// End
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