Overall heat transfer co-efficient (U_o,cal ):

Fouling factor, R_dk =0.001 h ft² °F Btu ^-1 for kerosene and R_dg = 0.0005 h ft ² °F Btu ^-1 for gasoline is taken for this service.

                                           (1.6)

Let select, Admirality brass as tube material with thermal conductivity, k_w =70 Btu h ^-1 ft^-1 °F^-1 .

U_o,cal = 53.5 Btu h ^-1 ft ^-2 °F^-1

 

Now,

 

Therefore, the calculated overall heat transfer co-efficient is well within the design criteria.

VI. Pressure drop calculation

VI.1. Tube side pressure drop:

Friction factor f = 0.00028 x 144 for Re = 0.04032 ft² /ft² for Re=11571.4 ( [3] page 836 ]

a_t = (no. of tubes)×(flow area per tube)/(no. of passes)

= 0.232 ft ²

Tube side mass velocity:

=646552 lb. h^-1.ft^-2

Frictional pressure drop:

= 5.81 psi

Return loss ΔP_rt : (due to change in flow direction of the tube side fluid)

= 0.73 psi

Total tube side drop neglecting nozzle loss:

                                                                            (1.8)

=5.81+0.73

=6.54 psi<10 psi

 

Therefore the tube side pressure drop is within the maximum allowable pressure drop of 10 psi.

 

VI.2. Shell side pressure drop calculation

Tube clearance, C =0.25″

Spacing, B =15.5″

a_s = 0.444 ft²

Mass velocity, G_S = 210526 lb. h^-1.ft^-2

Re = 35668

No of baffles,

Friction factor, f = 0.0017 X 144= 0.2448 ft² / ft² with 25% cut segmental baffles ( [3] page 839 )

 

Shell side frictional pressure drop ΔP_s :

=1.4 psi <7 psi

ΔP_rs = 0 (in case of single shell pass flow)

 

Total shell side drop neglecting nozzle loss:

= 1.4 psi                                                                            (1.9)

Therefore the shell side pressure drop is within the maximum allowable pressure drop of 7 psi.

 

VII. Over surface and over design

Over surface =

The clean overall heat fransfer co-efficient:

= 141.3×0.834=117.8 Btu h^-1 ft^-2 °F^-1

U_c = 66.98 Btu hr ^-1 ft^-2 °F^-1

% Over surface =

                         =20% (acceptable)

Over design:

The design area of heat transfer in the exchanger, ( n_t =318) :

=π× ×24×368=2312 ft²

The required heat transfer area (where, n_t =335):

= π× ×24×335=2105 ft²

% Overdesign=9.8% which is within the acceptable limit.

Refer module # 2 for the mechanical design of shell and tube heat exchanger .