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In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENTPowerPoint Presentation

In The Name OF ALLAH THE MOST MERCIFUL AND BENIFICIENT

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MUHAMMAD IMRAN 2006-CHEM-35

OSAMA AKRAM 2006-CHEM-39

WAQAS AHMAD 2006-CHEM-15

TAIMOOR RAI 2006-CHEM-81

SUMITTED TO:

Prof. Shah muhammad

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Main Parts

1.Tubes

2.Shell

3.Baffles

4.Tube Sheets

5.Head

6.Tube Bundle

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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 following
- topics would be presented:
- General Design Steps.
- Thermal Design.
- Hydraulic Design.

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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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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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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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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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- Prior entering to thermal
- design calculations;
- Tabulating the physical properties of
- Hot fluid (Ammonia)
- And cold fluid (Water)

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Physical Properties for Ammonia:

Properties: Values:

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

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Physical Properties for water:

Properties: Values:

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

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- An obvious answer would be from making
- an overall
- Energy And
- Material Balance.

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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.

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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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Calculating LMTD …

t2 = 95oF

t1 = 85oF

T1= 245oF

T2 = 95 oF

LMTD = [(T1- t2) - (T2-t1)]

[ln(T1- t2)/(T2-t1)]

LMTD = 51.7oF

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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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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

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1 – 1 Pass Arrangement calculation:

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

4

Flow 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.

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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

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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

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 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

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Summary of tube side Calculation...

Nre = 23302.51

Npr = 5.16

µ/µw = 1

Jh = 80

hi = 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 shell

which 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.

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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

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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

Flow Area

Mass Velocity of Ammonia

= 28950.45 lb/hr-ft2

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8.Calculating hydraulic diameter:

For Triangular Pitch hydraulic diameter can be calculated by using formula

Hydraulic Diameter

DH = 4(0.43×PT2-.39275×O.D2)/1.571×O.D

So;

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)=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

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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

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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 TransferCoefficient:

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 = 0.0625

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.

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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

=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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Pressure Drop Calculations across Tube side…

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.

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Pressure Drop Calculations across Tube side…

Return Losses Can Be Calculated As

∆Pr = 4n( V2) = 1.495 lbf/in2

2 g

Total 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 / in2

Nb = Tube Length - 1 = 16

Baffle Spacing

Note: This pressure drop is in allowable limit.

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