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Concentric tube heat exchanger
Group A5
Done by
P. Neha CH10B044
A. Harika CH10B079Shilochana CH10B091
Shivani Patel CH10B101
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Problem Statement
1800 kg/h of ethylene glycol is to be cooled from 1000Cto 600C by water available at 150C. The maximumtemperature to which water can be heated is 420C.Ethylene glycol is circulated through the tubes while
water flows through the annulus of a concentric tubeheat exchanger. Inside tube is of copper while outsideis of steel. Inside and outside diameters of copper tubeare 12.5 and 14.5 mm respectively. Inside diameter ofthe outer steel tube is 22 mm. Fouling resistances and
metal wall resistances can be neglected. Suggest asuitable design of a concentric tube heat exchanger.Will the heat exchanger does the duty if fluids streamsare swapped?
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Data Provided
Flow rate of ethylene glycol = 1800 kg/h
Initial temperature of ethylene glycol = 1000C
Desired final temperature of ethylene glycol = 600C
Initial temperature of water = 15
0
C Acceptable final temperature of water = 420C
Inside diameter of copper tube = 12.5mm
Outside diameter of copper tube = 14.5mm
Inside diameter of steel tube = 22mm
Density of ethylene glycol = 1078 kg/m3
Viscosity of ethylene glycol = 3.2 x 10-3 N-s/m2
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Data Provided (Continued.)
Specific heat of ethylene glycol = 2650 J/kgK
Thermal conductivity of ethylene glycol = 0.261 W/mK
Density of water = 995 kg/m3
Viscosity of water = 0.853 x 10
-3
N-s/m2
Specific heat of water = 4180 J/kgK
Thermal conductivity of water = 0.614 W/mK
Fouling resistances and metal wall resistances are considered
to be negligible.
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Introduction
Concentric tube heat exchangers create a
temperature driving force by passing fluid
streams of different temperatures parallel to
each other, separated by a physical boundary
in the form of a pipe.
They are generally used in industries for
purposes such as material processing, foodpreparation and air-conditioning
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Typical concentric tube heat exchanger
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Solution
According to conservation of energy
Heat lost by ethylene glycol = Heat gained by water
1800*2650*(100-60) = m*4180*(42-15)Hence m = 1690.59kg/h
For the outer tube:
According to correlation
V = m/(*(/4)*(D^2 D^2))Now substituting the corresponding values of all the variables
and solving for velocity
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V = 1690.6*4*10^6/(995*3.14*(22^2-14.5^2)*3600)
= 2.19 m/s
Therefore velocity of fluid is found to be 2.19 m/s
Calculation of Reynolds number:Re = (*v*Dh/)
Where Dh is the hydraulic diameter = D0 Di
Dh = 22-14.5
= 7.5mm
Substituting all the values of variables to determine Reynolds
number
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Re = (995*2.19*7.5*10^-3)/0.853*10^-3
= 19159.29
As the value of Reynolds number is found to be very high the
Flow is considered to be turbulent.Calculation of Nusselts number:
Correlation of nusselts number for turbulent flow is
Nu = 0.027*Re^(4/5)*Pr^(1/3) where Pr = k/(*Cp)
= 0.027*19159^0.8*6.808^0.33
= 135.56
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As we know that Nu = h*Dh/k
Where h is the convective heat transfer coefficient
Dh is hydraulic diameter
k is the thermal conductivity of waterNow we know that Nu = 135.56
h*Dh/k = 135.56
h = 135.56*0.614/7*10^-3
= 11097.8
Therefore the convective heat transfer coefficient is found to be
11097.8
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For the inner tube:
V = m/(*(/4)*(D^2 D^2))
Plugging in all the necessary values of variables
= 1800*4/(3600*1078*3.14*12.5^2*10^-6)= 3.78 m/s
Velocity of ethylene glycol flowing inside the tubes is 3.78 m/s
Calculation of Reynolds number:
Re = (*v*Dh/)
Plugging in all the necessary values of variables
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Re = (1078*3.78*12.5*10^-3)/3.2*10^-3
Re = 15917
As the value is very high in this case also the flow inside the tube
is considered to be turbulent.Calculation of Nusselts number:
Correlation of nusselts number for turbulent flow is
Nu = 0.027*Re^(4/5)*Pr^(1/3) where Pr = k/(*Cp)
= 0.027*15917^0.8*32.49^0.33
= 195
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As we know that Nu = h*Dh/k
Where h is the convective heat transfer coefficient
Dh is the diameter of tube
k is the thermal conductivity of ethylene glycolNow that we know Nu = 195
h*D/k = 195
h = 195*0.261/12.5*10^-3
= 4071.6
Therefore the convective heat transfer coefficient is found to be
4071.6
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Calculation of overall heat transfer coefficient:
= 1/11097 + 2/401* + (1/4071)*
= 1/11097 + (2*10^-3/401)*1/ln(22/14.5) + (1/4071)*(22/14.5)
= 4.7477*10^-4
Hence Uo = 2106
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Calculation of length of concentric tube heat exchanger:
As we know that
Where is the logarithmic mean temperature difference
Plugging in all the values and appropriate expressions to
calculate the length of the heat exchanger
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Plugging in all the values of variables
(1800*7650*40)/3600 = 2106*3014*22*10^-3*L*((100-42)-(60-15)
ln((100-42)/(60-15)
53000 = 7452.4*L
L = 53000/7452.4
L = 7.118m
Therefore the length of the concentric tube heat exchanger is
Obtained to be 7.118m
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Calculation of heat duty:
The amount of heat exchanged from a fluid at a higher
temperature to the fluid at lower temperature in unit time by a
heat exchanger is called its dutyHeat duty Q = UATlmtd
= 2106*3.14*22*7.118*((100-42)-(60-15)
ln((100-42)/(60-15)
= 2106*3.14*22*10^-3*7.118*51.23
= 53046.135 J
= 53.05 KJ
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References
Fundamentals of Heat and Mass Transfer, Fifth
Edition by Frank P. Incropera and David P.
Dewitt
Wikipedia