1.4. Design problem

The above design procedure is elaborated through the calculation of the following example

Problem Statement:

150000 lb per hour of kerosene will be heated from 75 to 120°F by cooling a gasoline stream from 160 to 120°F. Inlet pressure will be 50 psia for each stream and the maximum pressure drop of 7 psi for gasoline and 10 psi for kerosene are permissible. Published fouling factors for oil refinery streams should be used for this application. Design a shell and tube heat exchanger for this service.

Part 1: Thermal design :

( Part 2: Mechanical design provided in module #4 )

Given data :

Hot fluid inlet temperature (T₁)= 160°F

Hot fluid outlet temperature (T₂) = 120°F

Cold fluid inlet temperature (t₁) = 75°F

Cold fluid outlet temperature (t₂) = 120°F

Fouling factor of hot fluid (R_dg) = 0.0005 (for gasoline)

Fouling factor of cold fluid (R_dk) = 0.001 (for kerosene)

P_inlet (for hot fluid) = 50 psia

P_inlet (for cold fluid) = 50 psia

ΔP_max (for cold fluid) = 7 psia

ΔP_max (for cold fluid) = 10 psia

Mass flow rate of cold fluid ( ) = 150000 lb.h^-1

(Subscripts ‘k' for kerosene and ‘g ' for gasoline)

 

I. Calculation of calorific temperature

For the calculation of calorific temperature please refer [3] (page 827) .

° API of hot fluid=76° ; Therefore K_c = 1; F_c = 0.456

(The calorific temperature factor, F_c with ° API as a function K_c is available in reference [3] (page 827) .

Calorific temperature of the hot fluid,

    =120+0.455×(160-120)

    =138.2°F

Calorific temperature of the cold fluid,

    =75+0.455×(120-75)

    =95.475°F