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In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT
In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT
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TOPIC OF PRESENTATIONTOPIC OF PRESENTATION
DESIGN OF SHELL AND TUBE HEAT EXCHANGER…..
DESIGN OF SHELL AND TUBE HEAT EXCHANGER…..
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GROUP MEMBERSMUHAMMAD IMRAN 2006-CHEM-35OSAMA AKRAM 2006-CHEM-39WAQAS AHMAD 2006-CHEM-15TAIMOOR RAI 2006-CHEM-81
SUMITTED TO: Prof. Shah muhammad
GROUP MEMBERSMUHAMMAD IMRAN 2006-CHEM-35OSAMA AKRAM 2006-CHEM-39WAQAS AHMAD 2006-CHEM-15TAIMOOR RAI 2006-CHEM-81
SUMITTED TO: Prof. Shah muhammad
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Shell And Tube Heat Exchanger
Main Parts1.Tubes2.Shell3.Baffles4.Tube Sheets5.Head6.Tube Bundle
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Problem Statement: Problem Statement:
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.
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.
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Design of a Shell And Tube Exchanger:
In this presentation followingtopics would be presented: General Design Steps. Thermal Design. Hydraulic Design.
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Part (A)
General Design Steps of Shell and Tube Heat exchanger:
General Design Steps of Shell and Tube Heat exchanger:
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General Design Steps:
Step 1. Perform the energy balance and calculate heat exchanger duty.Step 2.Obtain the necessary thermo physical properties of hot and cold fluid streams at their mean temperature.
Step 3.Select the tentative number of shell and tube passes.Step 4.Calculate the LMTD and the correction factor FT .
Step 5.Assume a reasonable value of the overall coefficient on outside tube area designated as Udo .
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General Design Steps:
Step 6.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.
Step 7.Select the tube pitch. Select the shell diameter that can accommodate the required number of tubes.
Step 8.Select the type, size (e.g. percentage cut) number and spacing of baffles.
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General Design Steps: Step 9.
Estimate the shell side and tube side heat transfer coefficients.
Step 10.Select the dirt factors Rd applicable to the system.Step11.Calculate the overall coefficient Udo
Step 12.Calculate the area based on this Udo.
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General Design Steps: Step13.
Compare the Udo and A values with those assumed
Step 14.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
Step 15. Calculate the shell side and tube side pressure drops
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Part (B)
Thermal Design of Shell and Tube Heat exchanger:
Thermal Design of Shell and Tube Heat exchanger:
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Physical Properties:
Prior entering to thermal design calculations; Tabulating the physical properties ofHot fluid (Ammonia) And cold fluid (Water)
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Physical Properties for Ammonia:
Properties: Values:Mass Flow Rate mh 9872 lb/hrEntering Temperature T1 245 oFLeaving Temperature T2 95 oFAverage temperature Tavg 170 oFSpecific Heat Capacity Cph 0.53 Btu/lb-oFViscosity Of Ammonia µh 0.027 lb/ft-hrThermal Conductivity kh 0.0176Btu/hr-ft-oFDensity ρh 0.02079lb/ft3
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Physical Properties for water:
Properties: Values:Mass Flow Rate mc 78482.4 lb/hrEntering Temperature t1 85 oFLeaving Temperature t2 95oFAverage temperature tavg 90oFSpecific Heat Capacity Cpc 1 Btu/lb-oFViscosity Of water µc 1.846 lb/ft-hrThermal Conductivity kc 0.358 Btu/hr-ft-oFDensity ρc 62.11 lb/ft3
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How should I start ?How should I start ?
An obvious answer would be from making an overall Energy And Material Balance.
An obvious answer would be from making an overall Energy And Material Balance.
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HOW TO CALCULATE HEAT DUTY?
Using Values Of m, Cp and ∆T from ammonia Q = 784824 Btu/hr.
By using Formula Q = m Cp ∆T
From Which Mass Flow Rate of Water would be calculated using Q =m Cp ∆T Yields m = 78482.4 lb/hr.
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Calculation of Volumetric Flow Rate of Water:
Volumetric Flow Rate of Water q = mc /ρc
= 1263.6 ft3/hr.
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t1 = 85oF t2 = 95oF
T1= 245oFT2 = 95 oF
LMTD = 51.7oF
Calculating LMTD …Calculating LMTD …
LMTD = [(T1- t2) - (T2-t1)]
[ln(T1- t2)/(T2-t1)]
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Calculation Of Heat Transfer Area :
Assume Udo=27 Btu/hr-ft2-oF
Using the formula Q = Udo × A× LMTD
Calculate Heat Transfer Area
A = 562.235 ft2
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Select A Suitable Size Tube:
A Iterative Selection is3/4" and 16 BWGSelect A suitable length L = 16 ft
Outside Diameter of Tube = O.D = 0.0625 ftInside Diameter of Tube = I.D = 0.0516 ftWall Thickness = xw = 0.005416 ft
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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 TubeNumber of Tubes Required = 178.94Flow Area = π× (I.D) 2×Number of Tubes 4Flow Area = 0.3742 ft2
Linear Velocity within the Tubes = Volumetric flow Rate/Flow Area Linear Velocity within the Tubes = 0.937 ft/sec
Deduction: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.
1 – 1 Pass Arrangement calculation:
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1 – 4 Pass Arrangement calculation: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 4Flow 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
Deduction:This velocity is in the allowable range so we proceed further with 1-4 Pass Exchanger.
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1 – 4 Pass Arrangement calculation:
Calculation of Prandtl No for water:Npr = (Cp×µ)/kNpr = 5.16Calculation of L/Di = (16 ft / 0.0516 ft) =310
Calculation Of jh factor Which from chart comes to be = 80Using value of jh to calculate hi by using correlation jh = [hi (Di × Npr
1/3)] / khi = 959.115
Calculation of reynolds no for water:NRe = (density×velocity*Di)/µNRe = 23302.51
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Summary of tube side Calculation...
Nre = 23302.51 Npr = 5.16µ/µw = 1Jh = 80hi = 959.11 (Btu/hr- ft2-oF)
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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 shellwhich can accommodate 180 tubes have I.D =17.25 inches.Inside Diameter of Shell=Ds=1.4375ft Pitch=PT=0.0833ft
3. Selecting Baffle: Select 25% cut segmental baffles.
SHELL SIDE CALCULATIONS:
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SHELL SIDE CALCULATIONS:
4. Selecting Baffle Spacing:According to the TEMA standards the allowed baffle spacing is 0.2Ds--------Ds we considerBaffle Spacing B = 0.66×DsBaffle Spacing = 0.95ft
5. Calculating tube clearance:Tube Clearance=Pitch - O.DTube Clearance= PD = 0.0208ft
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6. Calculating flow area:Flow Area = PD × B ×Ds
PT
Flow Area = 0.3409 ft2
7. Calculating mass velocity for ammonia:Mass Velocity of Ammonia Gs
= Mass Flow Rate of Ammonia
Flow AreaMass Velocity of Ammonia = 28950.45 lb/hr-ft2
SHELL SIDE CALCULATIONS:
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SHELL SIDE CALCULATIONS:
8.Calculating hydraulic diameter:For Triangular Pitch hydraulic diameter can be calculated by using formula Hydraulic Diameter DH = 4(0.43×PT
2-.39275×O.D2)/1.571×O.DSo;Hydraulic Diameter = 0.059 ft
9. Calculating shell side reynold's number:Shell Side Reynold’s Number= DH×Gs/µh = 63262.09
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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)=140As 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
SHELL SIDE CALCULATIONS:
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SHELL SIDE CALCULATIONS:
12. Deducing inside and outside heat transfer Dirt co-efficient from data tables:Now from the tables we see that for ammonia water systemsInside Heat Transfer Dirt Coefficient=hi,d=1007.516Btu/hr-ft2-oFOutside Heat Transfer Coefficient=ho,d=3526.306 Btu/hr-ft2-oF
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SHELL SIDE CALCULATIONS:
13. Selecting compatible Materials Of Construction Material selected Carbon steel.Thermal Conductivity of Carbon Steel = kw =29.913Btu/hr-ft-oF
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Calculation Of Overall Heat Transfer Coefficient: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
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Calculation of LMTD Correction Factor
As R = (T1-T2)/(t2-t1) R = 15 P = (t2-t1)/(T1-t1) P = 0.0625Now there is a well known equation for LMTD correction factorF= (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 = 0.838The value of F is acceptable as it is above 0.75.
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AREA VERIFICATION:Area Required AR= Q/Udo × F × LMTDArea Required = 517.487ft2
Area Available AA
= 3.142×O.D×Length×Number Of tubesArea Available = 565.56ft2
Percentage excess area: % excess area=100× (AA-AR)/AR
% excess area = 9.29%Note:Since an excess area of 10% is allowed so our design is acceptable
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Part (C)
Hydraulic Design involves:
Pressure Drop calculation across:
Tube Side
Shell Side
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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 Where:f = friction factor.Gt = Mass velocity of tube side fluid.
n = no. of tube passes. φt = Dimensionless Viscosity ratio.
Pressure Drop Calculations across Tube side…
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Pressure Drop Calculations across Tube side…
Return Losses Can Be Calculated As
∆Pr = 4n( V2) = 1.495 lbf/in2
2 gTotal Pressure Losses:ΔPT = ΔPt + ΔPr = 1.958 lbf/in2 Note: Pressure Drop is in allowable limit.
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Pressure Drop Calculations across shell side…
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 / in2Nb = Tube Length - 1 = 16 Baffle SpacingNote: This pressure drop is in allowable limit.
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THANK’S …
THANK’S …
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Any Question ? Any Question ?
