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Assignment of Process Heat Transfer

This document summarizes the analysis of heat transfer through a shell and tube heat exchanger using iso-butane as the fluid. Key parameters and equations are provided to calculate the heat duty, mass flow rates, log mean temperature difference, heat transfer coefficients, pressure drops, and clean and dirty overall heat transfer coefficients. The analysis finds the tube side pressure drop to be very high, making this configuration of tubes not feasible for the shell and tube exchanger.

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Atif Mehfooz
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1K views6 pages

Assignment of Process Heat Transfer

This document summarizes the analysis of heat transfer through a shell and tube heat exchanger using iso-butane as the fluid. Key parameters and equations are provided to calculate the heat duty, mass flow rates, log mean temperature difference, heat transfer coefficients, pressure drops, and clean and dirty overall heat transfer coefficients. The analysis finds the tube side pressure drop to be very high, making this configuration of tubes not feasible for the shell and tube exchanger.

Uploaded by

Atif Mehfooz
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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ASSIGNMENT OF PROCESS

HEAT TRANSFER

Submitted to :

Muhammad Sulaiman
Submitted by:
M Shahbaz 2014-CH-220
Aitazaz Ahsan Noor 2014-CH-221
Hussain Kazmi 2014-CH-219
  
Problem 7.10
Given data:

Mass flow rate of iso-butane =m’=78359lb/hr

No of tubes =nT=178

Shell passes = 1

Tube passes = 4

Length of tubes=l=12ft

Solution:

Physical properties of solution :


Specific heat capacity of iso-butane=cP=0.689 Btu/lb.0F

Specific heat capacity of n-butane=cP=0.631 Btu/lb.0F

Viscosity=µn-butanne=µiso-butane=0.2904lb/ft.hr

Thermal conductivity =kn-butane=0.01203 Btu/hr.ft2.oF/ft

Thermal conductivity =kiso-butane=0.0139 Btu/hr.ft2.oF/ft


Heat Duty:
We know that

Q0t=m’.cP.∆ T

Q0t=23×0.689×78359

Q0t=1241755.1 But/hr

m’n-butane= 1241755.1/23×0.2904

m’n-butane= 185913.7472lb/hr

LMTD:
LMTD=(T1-t2)-(T2-t1)/ln((T1-t2)/(T2-t1))

LMTD=(203-177)-(180-154)/ln(49/3)

LMTD=0

As LMTD=0 which means this heat exchanger is possible but infinite area would be require for
the transformation of heat . But here we are given with the data so we are going to calculate
pressure drops and dirt factor.

Tube side calculations:


As we know that

Re=G×Di/µ

As,

G=m’/aT

While aT=No. of tubes×a1/np

aT= 178×0.268/4×144=0.0828ft2

And

G=78359/0.0828=946364.73 lb/hr.ft 2
Re=Di × G/µ

=0.584×946364.73/12×0.2904

Re= 1903157.7

Here we don’t need to calculate L/D because its huge Re .

From graph,

JH=1000

We know that

JH=(hi×di/k)(cP×µ/k)-1/3(µ/µW)-0.14

hi=(1000×0.01323×12/0.584)(0.2904×0.689/0.584) 1/3

hi=190.2 Btu/hr.ft2.oF

hio=di×hi/do=0.584×190.2/0.75=148.1024 Btu/hr.ft 2.oF

Tube side pressure drop:


∆Pt = fG2ln/5.22×1010×ds∅ t

s=0.6

f=0.00009 ft2/in2

∆Pt = 0.00009×(946364.73)2×12×4/(5.22×1010×0.6×1×0.04867)

∆Pt = 2.5381 psi

Now,

∆PT=∆Pt.∆Pr ------> (1)

∆Pr= (4n/s)(V2/2g)

∆Pr= 16/0.6×1

∆Pr= 26.66psi
∆PT= 2.5381+26.66=29.1981psi

It’s very huge pressure drop so this kind of tubes is not feasible for shell and tube heat
exchanger.

Clean overall coefficient:


Uc=hi×hio/hi+hio

=190.2×148.10/190.2+148.10

=669.131N

And

Ud=Q/A×∆ T

Ad=N×l×π

=178×12×.1963= 419.29 ft2

Now

m.Cp∆ T = Q

Q=23×0.689×78359

=1241755.1 BTU/hr

Ud=α

So,

Rd=1/Ud-1/Uc =1/α-1/669.131 = 0

Shell side calculation:


As = iD×c’×B/Pt

C’ = Pt-do=1-0.75= 0.25

As = 17.25×0.25×6/1×144

= 0.1797 ft2

G = m/as
=185913.74/0.1797

=1034578.45

De = 4×(1/2Pt×0.86Pt-1/2×πdo2/4)/1/2πdo

= 0.711 in

We know that

Re = De×G/µ

= 0.05925×1034578.45/0.2904

Re = 211083.93

Shell side Pressure drop:


∆Ps = fGi×Di(N+1)/5.22×1010De×S×φs

f= .00119 S= 0.6 N+1 = l/B

N+1 = 12×12/6

= 24

Now

∆Ps = 0.00119×1034578.45×17.25×24×12/5.22×10 10×0.711×0.6

∆Ps = 0.00033 Psia

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