A liquid toluene stream flowing at 100 kmolhr is to be coole

A liquid toluene stream flowing at 100 kmol/hr is to be cooled in a heat exchanger from 100 degree C to 30 degree C using cooling water. The cooling water enters at 20 degree C and is to leave at 40 degree C. At what rate must cooling water be fed to the heat exchanger in kg/hr?

Solution

Liquid stream flowing rate = 100 Kmol/hr

1 mol water = 18 gm

1 Kmol water = 18 Kg

Therefore rate of flow = 18 Kg/hr

Let the amount of cool water required per hour = X kg/hr

and let the specific heat = S

Heat loss for hot liquid stream = 18 x S x (100 - 30)

Heat gain for cool water = X x S x (40 - 20)

Therefore 18 x S x 70 = X x S x 20

X = 18 x 70 / 20 = 63 Kg/hr