Answer:
∆S1 = 0.5166kJ/K
∆S2 = 0.51826kJ/K
Explanation:
Check attachment for solution
Answer:
curly brackets are missing
Explanation:
The body of the main() function need to be enclosed in curly brackets. Try this:
int P = 3000;
int main( ) {
for (int t = 0; t < 10; t++) {
cout << P;
}
}
Answer:A. 40% B.50% C. 60% Od 70%
Explanation:A. True B. False
Answer:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
