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
 
        
             
        
        
        
Answer:
a) 0.489
b) 54.42 kg/s
c) 247.36 kW/s
Explanation:
Note that all the initial enthalpy and entropy values were gotten from the tables.
See the attachment for calculations
 
        
             
        
        
        
B because thermal has to do with temperature and it’s the amount of kinetic and potential energy in and object