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In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT. TOPIC OF PRESENTATION. DESIGN OF SHELL AND TUBE HEAT EXCHANGER…. GROUP MEMBERS MUHAMMAD IMRAN 2006-CHEM-35 OSAMA AKRAM 2006-CHEM-39 WAQAS AHMAD 2006-CHEM-15
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THE MOST MERCIFUL
DESIGN OF SHELL AND TUBE HEAT EXCHANGER…..
MUHAMMAD IMRAN 2006-CHEM-35
OSAMA AKRAM 2006-CHEM-39
WAQAS AHMAD 2006-CHEM-15
TAIMOOR RAI 2006-CHEM-81
Prof. Shah muhammad
Dry ammonia gas at 83 psia and
at a rate of 9872 lb/hr is to be
cooled from 245oF to 95oF using
cooling water from 85oF to 95oF.
Design a shell and tube
heat exchanger to perform
the above duty.
General Design Steps of Shell and Tube Heat exchanger:
Perform the energy balance and calculate heat
Obtain the necessary thermo physical properties of hot and cold fluid streams at their mean temperature.
Select the tentative number of shell and tube passes.
Calculate the LMTD and the correction factor FT .
Assume a reasonable value of the overall coefficient on outside tube area designated as Udo .
Select the tube diameter, it\'s wall thickness (In Terms Of BWG) and the tube length. Calculate the number of tubes required to provide the area calculated above.
Select the tube pitch. Select the shell diameter that can accommodate the required number of tubes.
Select the type, size (e.g. percentage cut) number and spacing of baffles.
Estimate the shell side and tube side heat transfer coefficients.
Select the dirt factors Rd applicable to the system.
Calculate the overall coefficient Udo
Calculate the area based on this Udo.
Compare the Udo and A values with those assumed
If area is in excess of 10% of that calculated then the design is acceptable. This excess area is sometimes required and sometimes not. If area calculations do not agree assume a new value of Udo and proceed in a similar way
Calculate the shell side and tube side pressure drops
Thermal Design of Shell and Tube Heat exchanger:
Mass Flow Rate mh 9872 lb/hr
Entering Temperature T1 245 oF
Leaving Temperature T2 95 oF
Average temperature Tavg 170 oF
Specific Heat Capacity Cph 0.53 Btu/lb-oF
Viscosity Of Ammonia µh 0.027 lb/ft-hr
Thermal Conductivity kh 0.0176Btu/hr-ft-oF
Density ρh 0.02079lb/ft3
Mass Flow Rate mc 78482.4 lb/hr
Entering Temperature t1 85 oF
Leaving Temperature t2 95oF
Average temperature tavg 90oF
Specific Heat Capacity Cpc 1 Btu/lb-oF
Viscosity Of water µc 1.846 lb/ft-hr
Thermal Conductivity kc 0.358 Btu/hr-ft-oF
Density ρc 62.11 lb/ft3
By using Formula Q = m Cp ∆T
Using Values Of m, Cp and ∆T from ammonia Q = 784824 Btu/hr.
From Which Mass Flow Rate of Water would be calculated using
Q =m Cp ∆T
Yields m = 78482.4 lb/hr.
Volumetric Flow Rate of Water
q = mc /ρc
= 1263.6 ft3/hr.
Calculating LMTD …
t2 = 95oF
t1 = 85oF
T2 = 95 oF
LMTD = [(T1- t2) - (T2-t1)]
LMTD = 51.7oF
Assume Udo=27 Btu/hr-ft2-oF
Using the formula Q = Udo × A× LMTD
Calculate Heat Transfer Area
A = 562.235 ft2
A Iterative Selection is
3/4" and 16 BWG
Select A suitable length
L = 16 ft
Outside Diameter of Tube = O.D = 0.0625 ft
Inside Diameter of Tube = I.D = 0.0516 ft
Wall Thickness = xw = 0.005416 ft
Outer Surface Area of Tube = 3.142×O.D×LOuter Surface Area of Tube = 3.142 ft2
No. of Tubes Required = Area/ Outside Surface Area of One Tube
Number of Tubes Required = 178.94
Flow Area = π× (I.D) 2×Number of Tubes
Flow Area = 0.3742 ft2
Linear Velocity within the Tubes = Volumetric flow Rate/Flow Area
Linear Velocity within the Tubes = 0.937 ft/sec
According to rule of thumbs and conventions it is well known that the velocity in the tubes should be between 3-10 ft/sec.
So 1-1 pass is rejected and we go for a 1-4 pass exchanger that might bring the velocity in the range.
Take 180 tubes Number of tubes per pass = 180/4Number of tubes per pass = 45
Flow Area Per Pass
= π× (I.D)2×Number Of Tubes Per Pass
Flow Area per Pass = 0.0941ft2
Linear Velocity within the Tube = Volumetric flow Rate/Flow Area
Linear Velocity within the Tubes = u = 3.73 ft/sec
This velocity is in the allowable range so we proceed further with 1-4 Pass Exchanger.
Calculation of reynolds no for water:
NRe = (density×velocity*Di)/µ
NRe = 23302.51
Calculation of Prandtl No for water:
Npr = (Cp×µ)/k
Npr = 5.16
Calculation of L/Di = (16 ft / 0.0516 ft) =310
Calculation Of jh factor
Which from chart comes to be = 80
Using value of jh to calculate hi by using correlation
jh = [hi (Di× Npr1/3)] / k
hi = 959.115
Nre = 23302.51
Npr = 5.16
µ/µw = 1
Jh = 80
hi = 959.11 (Btu/hr-
1. Selecting Tube Arrangement:Triangular (0.75 inch O.D tubes on 1 inch triangular pitch)
2. Shell Diameter:
From the tables we see that the shell
which can accommodate 180 tubes
have I.D =17.25 inches.
Inside Diameter of Shell=Ds=1.4375ft
3. Selecting Baffle:
Select 25% cut segmental baffles.
4. Selecting Baffle Spacing:
According to the TEMA standards the allowed baffle spacing is 0.2Ds--------Ds we consider
Baffle Spacing B = 0.66×Ds
Baffle Spacing = 0.95ft
5. Calculating tube clearance:
Tube Clearance=Pitch - O.D
Tube Clearance= PD = 0.0208ft
6. Calculating flow area:Flow Area = PD× B ×DsPTFlow Area = 0.3409 ft2
7. Calculating mass velocity for ammonia:
Mass Velocity of Ammonia Gs
= Mass Flow Rate of Ammonia
Mass Velocity of Ammonia
= 28950.45 lb/hr-ft2
8.Calculating hydraulic diameter:
For Triangular Pitch hydraulic diameter can be calculated by using formula
DH = 4(0.43×PT2-.39275×O.D2)/1.571×O.D
Hydraulic Diameter = 0.059 ft
9. Calculating shell side reynold\'s number:
Shell Side Reynold’s Number=
DH×Gs/µh = 63262.09
10.Calculating shell side Prandtl number:Shell Side Prandtl Number = Cph×µh/kh = 0.81
11. Evaluating Colburn\'s Factor from chart:
Colburn’s Factor (jh)=140
As jh= ho× (Pr) -0.33 ×DH/kh
Thus ho= jh/(Pr) -0.33 ×DH/kh
Outside Heat Transfer Coefficient
= 39.063 Btu/hr-ft-oF
12. Deducing inside and outside heat transfer Dirt co-efficient from data tables:
Now from the tables we see that for ammonia water systems
Inside Heat Transfer Dirt Coefficient=hi,d=1007.516Btu/hr-ft2-oF
Outside Heat Transfer Coefficient=ho,d=3526.306 Btu/hr-ft2-oF
13. Selecting compatible Materials Of Construction
Material selected Carbon steel.
Thermal Conductivity of Carbon Steel
= kw =29.913Btu/hr-ft-oF
The overall heat transfer coefficient can be
calculated from the following formula
1 = O.D + O.D .
Udo ((hi ×I.D) (hi.d × I.D)
+ O.D × Xw × ln(O.D/I.D) + 1 + 1 .
[kw× (O.D-I.D)] ho ho.d
Overall Heat Transfer Dirt Coefficient
Udo = 35.03 Btu/hr-ft2-oF
As R = (T1-T2)/(t2-t1)
R = 15
P = (t2-t1)/(T1-t1)
P = 0.0625
Now there is a well known equation
for LMTD correction factor
F= (R2+1)0.5ln (1-P/1-RP) .
(R-1) × ln [2-P(R+1-(R2+1)0.5]
2- P[R+1+ (R2+1)0.5]
F = 0.838
The value of F is acceptable as it is above 0.75.
Area Required AR= Q/Udo × F × LMTD
Area Required = 517.487ft2
Area Available AA
= 3.142×O.D×Length×Number Of tubes
Area Available = 565.56ft2
Percentage excess area:
% excess area
% excess area = 9.29%
Since an excess area of 10% is allowed
so our design is acceptable
Hydraulic Design involves:
Using The Relation
∆Pt = f x G2t x L x n .
2 x g x p x Di x φt
ΔPt = 3194.77 N/m2 = 0.463 lbf/in2
f = friction factor.
Gt = Mass velocity of tube side fluid.
n = no. of tube passes.
φt = Dimensionless Viscosity ratio.
Return Losses Can Be Calculated As
∆Pr = 4n( V2) = 1.495 lbf/in2
Total Pressure Losses:
ΔPT = ΔPt + ΔPr = 1.958 lbf/in2
Note:Pressure Drop is in allowable limit.
Using The Relation
∆Ps = f sx G2s x Ds x (N b+1) .
2 x g x ps x DHx φs
∆Ps = 352.46 Kg/m2 = 0.5 lbf / in2
Nb = Tube Length - 1 = 16
Note: This pressure drop is in allowable limit.