Process Design Process Equipments Solved Sample Problems

Problem solving – Heat exchanger shellside pressure drop calculation

Problem Statement -

Calculate the shell side pressure drop for the following heat exchanger specification,

Process fluid = water
Inlet pressure = 4 barg
Inlet temperature = 500C
Outlet temperature = 300C
Tubeside flowrate = 50000 kg/hr
Shell diameter =22 inches
Number of baffles = 32
Baffle spacing = 6 inches
Tube diameter = 1 inch
Number of tubes = 10
Pitch = 1.25 inches, triangular pitch
Shellside roughness = 0.06 mm

Solution -

Calculation of shell side pressure loss is not as straightforward as the tubeside pressure drop calculation. Presence of baffles in the shell poses a challenge in calculation of the pressure drop on shell side. The shell side has to be approximated as a series of tube banks connected by window zones, for estimating the exchanger shellside pressure drop. Exchanger pressure drop has to be a summation of pressure drop estimated for crossflow across the tube banks and pressure drop estimated for window zones. These pressure drop estimation values have to be corrected by accounting for leakage through baffles and bypassing of tube bundles.

The sample problem presented here using a simplistic approximation which holds quite accurate for large tube bundles. The equations used in the solution presented here also govern the calculations performed by EnggCyclopedia's Heat Exchanger Shell side Pressure Drop Calculator. Problem solving is performed in following 4 basic steps.

Step 1.

First step of problem solving requires determination the important physical properties of given fluid (water) at given temperature and pressure conditions. Since, water density will be lowest at inlet temperature (500C), which corresponds to highest volumetric flow. Hence for conservative pressure drop estimate, physical properties of water are calculated at the inlet conditions. Using EnggCyclopedia's Liquid Density Calculator,

water density at 500C = 988.0 kg/m3 Using EnggCyclopedia's Liquid Viscosity Calculator,

water viscosity at 500C =0.53 cP

Step 2.

Next the effective area for the crossflow across tubes between baffles is calculated using following  equation,

Effective Area =  Ae = Ds × Bs × (P-Dt)  ⁄  P

where, Ds = shell diameter
Bs = Baffle spacing
P = pitch (distance between center axes of two adjacent tubes)
Dt = Tube diameter

For our case, Ae = 0.0341 m2

Velocity of the crossflow then becomes,

V = Mass flow/(ρ×Ae×3600)
V = 50000/(988×0.0341×3600) m/s
V = 0.4127 m/s

Step3

Next, the effective diameter of fluid path is determined using following approximation,

De = 4 × (P2 - (π Dt2/4) ) / π Dt
De = 0.0484 m

And the factor fk is then calculated as a function of Reynold's number,
fk = 1.79×(ρ×V×D/μ)-1.9
fk = 0.2424 3.7×10-9

Step4

The shell side pressure drop is finally calculated using the following equation,

ΔP = ( (N+1)×fk×Ds×ρV2 ) / ( 2×De )
where, N = number of baffles

ΔP = 0.0777 bar

Alternative Solution

Another alternative is to directly use EnggCyclopedia’s Heat Exchanger Shell side Pressure Drop Calculator. All the inputs given in the sample problem statements are given to the calculator and pressure drop across the heat exchanger shell side is calculated as output. This calculator uses the same basic steps discussed above and hence the answer also matches with the figure above (0.0778 bar) . The following image is a snapshot of this direct calculation of tubeside pressure drop.

 
 
 

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