Question

A shell and tube heat exchanger is to be desined by kern's method to heat Toluene.

Toulene: T(in)=100F, T(out)=257F, flowrate=125000Ib/hr P(in)=90Psia composition:100% Toluene.

Styrene: T(in)=300F, T(out)=176F, P(in)=50psia composition:100% Styrene

What is the mass flow rate of Styrene? Which fluid should be in the shell side and which should be in the tube side and why?

Answer 1

As per the heat balance equation, Heat lost by hotter substance = heat gained by colder substance

Heat transfer rate in BTU/hr = M*Cp*dT

Where, M = mass flow rate (Lb/hr)

C_P = Specific heat constant (Btu/(lb ^oF) (For Toluene it is 0.41 Btu/(lb ^oF) and  For Styrene it is 0.74 Btu/(lb ^oF) and )

dT= Difference in temerature

Heat lost by Hotter substance (Toluene) = (M* C_P*(T_in-T_out )) _Toluene = 1250000 * 0.41*(257-100) = 8046250 BTU/hr.

Heat gained by Colder substance (Styrene) = (M* C_P*(T_in-T_out )) _Styrene  = M_styrene * 0.74*(300-176) = 91.76 M_styrene

From the heat balance equation , 91.76 M_{styrene =} 8046250

=> M_styrene = 8046250/ 91.76 lb/hr = 87688 lb/hr.

As per Kern, fluid in tube side shall be corrosive and shell side fluid should have larger temperature differece .

dT_Toluene = 257-100 = 157 F and dT_Styrene = 300-176 = 124 F => dT_Toluene > dT_Styrene and Toluene is non-corrosive fluid.

So, Tube side - Styrene, Shell side- Toluene