Heat Exchanger

In a 1-1 shell and tube heat exchanger, a fluid flowing through the tubes in turbulent flow, is being heated by means of steam condensing on the shell side. It is proposed to increase the tube side coefficient by one of the following methods:

1. Replace the existing tubes by the same number of tubes with half the original diameter but twice the length.
2. Increase the number of tube passes to 2.

Assuming that the fluid flow rate remains high enough to ensure a Reynolds number of over 10,000 in all cases, indicate the method you would select. Justify your selection in brief (in not more than five lines).

Calculations:

(a) Replace the existing tubes by the same number of tubes with half the original diameter but twice the length:

Original flow area = (π/4) D²

New flow area = (π/4) (D/2)² = 0.25 (π4) D²

For a given volumetric flow rate A v = Constant

If original A is taken as 1, and original v is taken as 1 then D = 1.1284

New diameter = 1.1284/2 = 0.5642

New velocity = 1/0.25 = 4

Under forced convection, heat transfer coefficient can be estimated from Dittus-Boelter equation, which can be written as,

Nu = 0.023 Re^0.8 Pr^0.4

i.e.,

hD/k = 0.023 (Dvρ/μ)^0.8 Pr^0.4

h = C D^-0.2 v^0.8

where C is a constant; D is the diameter and v is the velocity.

Original h = C x 1.1284^-0.2 x 1^0.8 = 0.976 C

New h = C x 0.5642^-0.2 x 4^0.8 = 3.399 C

Heat transfer coefficient increase by 100 x (3.399 - 0.976) / 0.976 = 248.3%.

(b) Increase the number of tube passes to 2.

Original flow area = n (π/4) D²

New flow area = 0.5 n (π/4) (D)²

New velocity = 1/0.5 = 2

New h = C x 1.1284^-0.2 x 2^0.8 = 1.6995 C

Heat transfer coefficient increases by 100 x (1.6995 - 0.976) / 0.976 = 74.1%

From the above calculations, it could be seen that Replacing the existing tubes by the same number of tubes with half the original diameter but twice the length will increase the heat transfer coefficient by a higher value than the other scheme. Hence the first method can be selected.