Heating of oil by air. a flow of 2200 lbm/h of hydrocarbon oil at 100°f enters a heat exchanger, where it is heated to 150°f by hot air. the hot air enters at 300°f and is to leave at 200°f. calculate the total lb mol air/h needed. the mean heat capacity of the oil is 0.45 btu/lbm · °f.

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:

› 2-

2. determine the proper pump operating conditions and bit nozzle sizes for - chegg

answer:

just reverse the digits.

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