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clear;
clc;
disp('Example 2.3');
// Given values
Q = -150; // Heat transferred out of the system, [kJ/kg]
del_u = -400; // Internal energy decreased ,[kJ/kg]
// solution
// using equation [3],the non flow energy equation
// Q=del_u+W
W = Q-del_u; // [kJ/kg]
mprintf('\n The Work done is, W = %f kJ/kg \n',W);
if(W>0)
disp('Since W>0, so Work done by the engine per kilogram of working substance')
else
disp('Since <0, so Work done on the engine per kilogram of working substance')
end
// End
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