Vertical thermosyphon reboiler recirculation ratio and total pressure

[hswang2]

Dear Sir/ Madam
I am sorry to bother you this trivial again. I have tried to get the minimum thermosyphon reboiler height through writing a program in excel.

But, I don’t understand how to get this total pressure drop formula (Please see the attachment), How does this formula come from?

The other question is the recirculation ratio (liquid divided to vapor) is increasing, it may cause the resistance smaller than head. And, how is the frictional pressure drop may be reduced by the square of the mass velocity if the tubes are made shorter?

Could anyone kindly explain the reason behind this?

Thank you so much.

 

[25362]

IMHO:

a. The formula is the conventional one used for [Δ]P_f using "english" units. In this case the excerpt refers to a no-slip condition, meaning vapor and liquid flow at the same velocity. The factor 5.22[×]10¹⁰ is easily obtained when trying to get consistency in the units. Suggest paying a visit to the old Ludwig's Vol. 3, "Applied process design for chemical and petrochemical plants" - Gulf.

b. It seems to me the word "and", before "if the tubes are made shorter", is missing.

 

[ione]

You have already received valid input and reliable reference from 25362.

The pressure drop formulation is that typical for tube-side in Shell&Tube heat exchangers with consistent imperial units.

?P =(f*G^2*L*n)/(2g*rho*Di*fi) (Kern)

Where
f = friction factor
G = mass velocity kg/m^2/s
L = tube length, m
n = number of tube passes
g = gravitational acceleration, 9.8 m/s^2
rho = density of the tube fluid, kg/m^3
Di = inside diameter, m
fi = dimensionless viscosity ratio
?P = pressure drop, kg/m^2


Further to your second question, I would interpret the text this way: reducing the length L of the tubes has an impact on the pressure drop since pressure drop is directly proportional to the square of the mass velocity (G expressed in lb/hr/ft^2), via the length of tubes.