Table of Contents1.0 Introduction...............................................................................................................................3

2.0 Selection of Heat Exchanger.....................................................................................................4

3.0 Specification..............................................................................................................................5

4.0 Physical Properties of Fluid.......................................................................................................6

5.0 Exchanger Type and Dimensions.............................................................................................6

6.0 Heat Transfer Area....................................................................................................................7

6.1 Overall Coefficient.................................................................................................................7

7.0 Material Selection, Sizing and Layout for Tube.........................................................................8

7.1 Material Selection..................................................................................................................8

7.2 Tube and Shell Fluid..............................................................................................................8

7.3 Tube Dimension....................................................................................................................8

7.3.1 Tube size........................................................................................................................9

7.3.2 Tube Thickness..............................................................................................................9

7.4 Tube Arrangement...............................................................................................................10

7.5 Number of Tubes.................................................................................................................11

8.0 Shell Selection and Sizing.......................................................................................................12

9.0 Heat Transfer Coefficient........................................................................................................13

9.1 Tube-side Heat Transfer Coefficient....................................................................................13

9.2 Shell-side Heat Transfer Coefficient....................................................................................14

9.3 Overall Heat Transfer Coefficient........................................................................................16

10.0 Pressure Drop.........................................................................................................................17

10.1 Tube-side Pressure Drop....................................................................................................17

10.2 Shell-side Pressure Drop.....................................................................................................17

11.0 Mechanical Design..................................................................................................................19

11.1 Heat Exchanger Tubes........................................................................................................19

11.2 Heat Exchanger Shell..........................................................................................................19

11.3 Tube Sheets........................................................................................................................19

11.4 Vent Point............................................................................................................................19

1

11.5 Baffle number and distance.................................................................................................20

11.6 Weight Loads.......................................................................................................................20

12.0 Schematic Diagram of Heat Exchanger..................................................................................21

13.0 Mechanical Drawing of Key Item.............................................................................................22

14.0 Piping and Instrumental..........................................................................................................23

15.0 Costing....................................................................................................................................24

16.0 Auxiliary Equipment, pump and piping....................................................................................25

17.0 Mass Balance and Safety Review...........................................................................................26

17.1 Mass Balance Review.........................................................................................................26

17.2 Safety Review......................................................................................................................26

18.0 Conclusion..............................................................................................................................27

19.0 References..............................................................................................................................28

Appendix A Nomenclature...........................................................................................................29

Appendix B Process Flow Diagram For Plant Design..................................................................30

Appendix C Mass Balance For Plant Design...............................................................................31

Appendix D Pipe Sizes – ANSI/ASME B36.19M - 1985...............................................................32

Appendix E TEMA Shell and Tube Nomenclature.......................................................................33

2

1.0 IntroductionThe aim of this project is to design Heat Exchanger as one of the selected major equipment in the

Plant Design project. The values used in the design of this equipment will be based on the revised

mass balance which can be found in Appendix C with reference to Appendix B. Equations used in

the calculation are presented in the Appendix A with the nomenclature right under the equation.

It should be noted that there is no best heat exchanger unit in this world as any unit of heat

exchanger need to be custom made and design to suit the process. There are myriads of

combination and types of heat exchanger proposed in the literature. The selection of heat

exchanger type, the number of tubing, the number of plate, the size of tube, the amount of pass in

tube and shell, the use of baffle, the type of material used, all are part of design consideration when

making decision.

The design criteria of this equipment are as follow:

I. Safe to operate

II. Low operating cost

III. Low maintenance cost

IV. Low fabrication cost

The safety aspect of the equipment will be placed before any other criteria when designing the

equipment.

3

2.0 Selection of Heat ExchangerThere are many types of heat exchanger to choose from before starting the design of the

equipment. Among the heat exchanger type that is widely discussed in the books and literature are

Shell and Tube Heat Exchanger, Double Pipe Heat Exchanger, and Gasketed Plate Heat

Exchanger.

It is known that each type of heat exchanger have its pros and cons, after weighing both

pros and cons was the decision to use Shell and Tube Heat Exchanger in the design project One of

the major concern in the selection of the Heat Exchanger type is whether it can sustain high

pressure while maintaining good heat transfer efficiency and low cost.

The following data need to be considered when choosing a heat exchanger.

I. Can withstand 30bar of pressure

II. Can withstand up to 150 ℃III. Can be cleaned or maintained in case of fouling

IV. Can perform well while support load up to 200,000 kg/h

Ethylene oxide has been found to undergo polymerization with iron oxide and strong alkali. It

has also been reported that chlorine and alkali contamination can also cause polymerization. Since

the plant would be running using commercially available water, it is almost impossible to avoid any

contamination of water that could potentially cause the polymerization of ethylene oxide. A periodic

cleaning or maintenance will have to be carried out to prevent any safety hazard from happening.

The Gasketed-Plate Heat Exchanger has been reported to exhibit high heat transfer

coefficient, but it cannot hold high pressured fluid. Also, the use of gasket is not appropriate with the

high reactive fluid such as Ethylene Oxide, there have been accounts of report whereby Ethylene

Oxide would attack the gasket and cause leakage. While Welded-Plate Heat Exchanger can hold

high pressure fluid while exhibit high heat transfer efficiency and prevent leakage, the heat

exchanger cannot be cleaned, or by any means maintained.

Double pipe heat exchanger is one of the high cost equipment that is usually used for

sensible heating and cooling of relatively small area. It can withstand high pressure and

temperature but the cost of the equipment out weight the advantage it can offer and will not be

considered in the design.

4

Shell and Tube heat exchanger will be chosen in this design project as it can withstand the

pressure of the cold fluid, it is relatively cheaper than double pipe heat exchanger and it can be

cleaned with a proper design.

5

3.0 SpecificationThe equipment will have the following specification as illustrated in the diagram below:

Figure 2.1 Specification of Equipment

The values in the diagram above are extracted from Appendix C. The general process flow diagram

can be found in Appendix B for reference.

The heat transfer rate (duty) is part of the limiting specification which can be calculated from the

inlet using the formula:

Q=mC p∆T (3.1)

Q=mC p∆T

Q=(148798 kgh−1)(4113.35Jk g−1 K−1)(50K )

Q=30602939400 J h−1

Q=8500.823 kW

6

148798 kg/hr(4.85% mol EO & 95.15% Water)20 bar100℃

54625 kg/hr(100% Saturated Steam & Water)1.0135 bar100℃

54625 kg/hr(100% Saturated Steam)1.0135 bar100℃

148798 kg/hr(4.85% mol EO & 95.15% Water)20 bar50℃

4.0 Physical Properties of FluidThe physical properties of the hot fluid involved in the design of the heat exchanger are tabulated in

the table below.

Steam Inlet Outlet Unit

Temperature 100 100 ℃Specific heat 1.988 1.988 Jk g−1K−1

Latent heat of vaporization 2259.44 2259.44 kJk g−1

Thermal conductivity 0.016 0.016 W /(m∙ K )

Density (1bar) 0.59 0.59 kg m−3

Viscosity 0.00001297 0.00001297 Pa∙ sTable 3.1 Physical Properties of Hot Fluid

It should be noted that the cold fluid is a mixture of Ethylene Oxide and Water, the physical

properties of the mixture at 75℃ is calculated and tabulated in the following table.

Physical Properties Water Ethylene Oxide Mixture Unit

Specific heat 4181.3 2782 4113.35 Jk g−1K−1

Thermal conductivity 0.58 0.14 0.161365 W m−1K−1

Density 0.97755 0.785175 0.960636 gmL−1

Viscosity 0.00018 0.0003797 0.00037 Pa.s

Table 3.2 Physical Properties of Cold Fluid Mixture

5.0 Exchanger Type and DimensionsThe heat exchanger type and dimension discussed here will focus on the number of tube passes,

number of shell passes. As the heat exchanger would have to be cleaned with periodic

maintenance, the number of passes in both shell and tube will be limited to one only. Counter flow

arrangement is preferable over parallel flow, as the mean temperature difference between the two

fluids over the length of heat exchanger is maximized as reported in many literatures.

The heat exchanger will be used as a condenser with the saturated steam from the

evaporator used to heat up the Ethylene Oxide and Water mixture. A simple energy balance

calculation is performed to check the outlet temperature of the hot fluid.

7

Latent Heat of Vaporization = 2259.44 kJ/kg

Steam Flow rate = 54625 kg/h

Calculation,

Total energy released upon condensation = (2259.44)(54625)

= 12342.21 MJ/h

= 34283.96 kW

Comparing,

Energy required to heat up the cold fluid to 100 ℃ = 8500.823 kW

From the calculation above, we can see that the hot fluid have more than enough heat

energy to warm up the cold fluid. Thus, it can be assumed that the hot fluid, despite undergoing

condensation will remain at 100 ℃ with 24.8% of the hot fluid condensate into liquid. The hot fluid is

assumed to absorb all the heat released by the condensation and reach 100 ℃

The log mean temperature cannot be used in this situation as there are phase change occurs during

the process with excess energy provided in the hot fluid. The temperature difference used in the

calculation will only be based on the temperature difference of the cold fluid, that is 50 ℃.

6.0 Heat Transfer AreaThe heat transfer area calculated here is for the preliminary analysis which later will be changed

again after the coefficient of heat transfer for each part of the heat exchanger is obtained in the later

part.

6.1 Overall CoefficientFor Tube and Shell Heat Exchanger, the overall coefficient will be in the range of 490 – 1000 for

fluids of steam to light organics. Taking the design conservatively 490 W/m2 ℃ will be used in the

preliminary calculation.

The heat transfer area can be calculated using the equation

Q=UA ∆T m (6.1)Rearranging we have,

A=Q /U ∆Tm (6.2)

Substituting the values,

8

A=8500.823×1000490×50

=346.9724m2

A preliminary estimation of the area needed for heat transfer is roughly 347m2.

7.0 Material Selection, Sizing and Layout for Tube7.1 Material SelectionThere are many cases of run off reaction concerning Ethylene Oxide, one of them is the

polymerization of Ethylene Oxide. It is reported that rust, or iron oxide will catalyzed and initiate

polymerization. The presence of loose quantity of rust could pose a significant safety hazard. The

condition of metal surfaces is extremely important in determining the rate of ethylene oxide polymer

formation. It has been reported that even clean carbon steel catalyzes polymerization, although at a

much slower rate than rusty steel. The polymerization of ethylene oxide usually cause line and

equipment plugging and off-specification product.

Stainless steel would be the best choice for materials of construction, especially when the

surface to volume ratio is high. The polymerization reaction has not been found to be auto-catalytic.

That is, the presence of polymer does not accelerate the polymerization process. The type of

stainless steel that will be used in the fabrication of this equipment would be Stainless Steel 316.

7.2 Tube and Shell FluidThe cold fluid which is the mixture of ethylene oxide and water has higher pressure than the hot

fluid. Higher pressure fluid should be contained in the tube while the lower pressure steam will be

put in the shell side.

7.3 Tube DimensionIn order to maximize the heat transfer area and increase the heat transfer efficiency, a good

selection of tube dimension is important. While the calculation will be perform to check the best tube

size for the equipment, it is usually not easy to get one. There are many commercially available tube

size on the market which may be able to serve the need of the equipment. Hence, the next best

available commercial tube that is close to the one we need will be used.

9

7.3.1 Tube size

The tube selected will be based on the ASME B36.19 standard which can be found in the Appendix

D. The reason for choosing the tube size from the ASME B36.19 standard is due to the convenient

of obtaining the material for fabrication while getting a fair price from the market.

The nominal pipe size of ¼ Schedule 10S will be used in the calculation. The size of the pipe

is subjected to change after checking the pressure drop and heat transfer rate. A 358, Electric

Fusion Welded stainless steel will be used.

According to the ANSI/ASME B36.19 standard the ¼ Schedule 10S have the following dimensions

Dimension Value

Outer diameter 13.7 mm

Thickness 1.65mm

Table 6.1 Dimension of ¼ Schedule 10S

7.3.2 Tube Thickness

Before the selection of tube, a pressure test on the tube will need to be done. The following

equation will be used to examine the minimum thickness of the tube required for the working

pressure.

The minimum wall thickness of a tube is given by the equation

tm=t p+c (7.1)

t p=Pd

2(SE+Pγ) (7.2)

The data from the table below is used for the calculation.

Properties Value

Internal design gauge pressure,P(N /m2) 5000000*

Pipe outside diameter,d (m) 0.0137

Basic allowable stress for pipe material, S(N /m2) 115142446.576

Casting Quality,P(N /m2) 1

Temperature coefficient,P(N /m2) 0.4

Table 6.1 Properties of Stainless steel 316 for ¼ Schedule 10S

10

* Taking the calculation conservatively, an extra 250% of internal pressure is allocated for safety

purposes.

The calculation is worked out as follow

t p=5000000×0.0137

2 [ (115142446.6×1 )+(5000000×0.4 )]t p=0. 000292379m

t p=0.292379mm

As calculated above, the pressure design thickness has to be at least 0.15mm to withhold the

internal pressure without failing. The schedule 10S has a thickness of 1.65mm which is more than

enough to hold the internal pressure and give allowance for the corrosion and erosion.

7.3.3 Tube Length

A longer tube length is usually favourable over shorter one as longer tube length would result in less

number of tubes and smaller diameter of the shell. The decision of having longer tube could be

justified as the cost of larger diameter of shell is usually high. A standard 6 meter pipe length is

usually readily available on the market and will be used in the design.

7.4 Tube ArrangementTriangular tube arrangement will be used in this design project. It is reported that triangular pitch

gives greater turbulence than square pitch, under comparable conditions of flow and tube size, the

heat transfer coefficient for triangular pitch are roughly 25% greater than that of square pitch.

As the heat transfer process includes condensation, condensate on the tube will flows by

gravity onto lower tubes in bundle. The condensate thickness around the lower tube will increase

and hence decrease the heat transfer coefficient. It was found out by theory and calculation that

triangular, staggered arrangement of tubes generally yields an average heat transfer coefficient that

is generally larger than that for a square in-line arrangement.

The tube bundle will also be inclined by 5 degree in the design as it is found out by Shklover

and Buevich that inclination of the tube bundle increases the average heat transfer coefficient as

much as 25%.

11

7.5 Number of TubesThe number of tube will be calculated based on the data specified above which is summarized in

the table below.

Dimension Value

Outer diameter of tube 13.7mm

Length of tube 6.0 m

Table 8.1 Dimension of Tube

Area of one tube ¿ π×13.7×10−3×6=0.258239m2

Number of tubes ¿Total areaneeded for heat transfer /areaof one tube

¿346.9724 /0.258239=1343.61tubes

The number of tubes will be rounded up to 1350.

The tube-side velocity can be calculated using the formula

ut=m

ρ× A ID×NT(7.3)

Substituting the values we have,

ut=148798 kg /h

960.64 kg /m3×0.000114m2×1350=0.28m /s

This velocity of the fluid should be fast enough to achieve turbulent flow, a check on the Reynolds’

number will be carried out to determine the fluid flow profile. The Reynolds number of the flow can

be calculated using the formula

ℜ=ρv DH

μ(7.4)

Substituting the values, we have

ℜ=960.64 kg /m3×0.28m /s ×0.001205m0.00037

=8743.36

As the Reynolds number is more than 4000, the fluid flow in tube is considered as turbulent and

hence is acceptable by most standards. The purpose of achieving turbulent flow is to have higher

heat transfer coefficient, low fouling rate, corrosion rate and low deposition rate.

12

13

8.0 Shell Selection and SizingAs the heat transfer process involves condensation of steam, it is therefore good to use J type

shells, which have two inlet for vapour phase and one central outlet for condensate. It has been

reported that J-Shell has approximately 1/8 pressure drop of a comparable E-Shell. The Shell type

and its alphabetical character is standardized by TEMA and can be seen in Appendix F. In addition

to that a type B front-end stationary head will be used along with type M rear-end stationary head in

the design for the ease of cleaning and maintenance.

8.1 Bundle and Shell Diameter

The bundle diameter can be calculated using the equation

Db=do( N t

K1)1/n1

(8.1)

For triangular pitch with one pass,

K 1=0.319

n1=2.142

Substituting values into the equation we have,

Db=13.7( 13500.319 )1 /2.142

=675.747mm

For an Outside packed head exchanger, the typical shell clearance is 38mm, so the shell inside

diameter is

Ds=675.747+38=713.747mm

As the shell diameter is 714mm, which fall under category of the nominal shell range 610-740, for

alloy steel the minimum thickness is 4.8mm. An extra 0.5mm will be allocated for fouling, corrosion

and erosion of the shell wall. Hence the shell outside diameter would be, Dso=720mm, after

rounding up.

A check up on the dimension of Shell diameter and Tube length is done by referring to a figure

illustrating a graph of tube outside surface area as a function of shell inside diameter and effective

tube length (Bell, 1998). For tube heat transfer area of 350m2 and tube length of 6m, the appropriate

shell inside diameter is 737mm, very close to our calculation.

14

15

9.0 Heat Transfer Coefficient9.1 Tube-side Heat Transfer CoefficientNumerous Nusselt numbers correlation where proposed in the literature. In order to choose a

suitable correlation that fits the flow profile of the design, it is important to calculate the Prandtl

numbers and Reynolds number.

The Reynolds number is as calculated before which is 8743.36

Prandtl number can be calculated using the following equation

Pr=C p μk f

(9.1)

Substituting the values into the equation, we have

Pr=4113.35×0.000370.16136

=9.431

Since the Re is more than 2300 and below 5000000, with Pr more than 0.5, Gnielinski’s correlation

will be used in this case. Gnielinski’s correlation is as follow

Nu= (f /2 ) ( ℜ−1000 )Pr1+12.7 (f /2 )1 /2 (Pr2/3−1 )

(9.2)

Where, f=(1.58 lnℜ−3.28 )−2 (9.3)

Substituting the values, we have

f=(1.58 ln (8743.36)−3.28 )−2=0.008174

Substituting the f values into Gnielinski, we have

Nu= (0.008174 /2 ) (8743.36−1000 )×9.4311+12.7 (0.008174/2 )1/ 2 (9.4312/3−1 )

=78.297

From the Nusselt numbers we can calculate the heat transfer coefficient with the equation

hi=Nu kd i

(9.4)

16

Substituting the values, we have

hi=78.2971×0.1613

0.01205=1048.5W /m2∙ K

9.2 Shell-side Heat Transfer CoefficientAs the shell-side heat transfer involves condensation, a different heat transfer correlation must be

used in this case. For partial condenser, the effect of film condensation must be taken into account

when calculating the heat transfer coefficient. Two types of correlation can be used in determining

the heat transfer coefficient, laminar film condensation and forced convection.

A check on the velocity of the fluid flow in the shell,

ug=Volumetric flow rate of steam /effectivecross sectional area of shel l

Volumetric flow rate=54625 kg /h3600

× 10.590kg /m3

=25 .7179m3 /s

Effective cross sectional area = cross sectional area of shell – cross sectional area of tube bundle

Effective cross sectional area=[ π4 ×( 715.21000 )2]−[ π4 ×( 13.71000 )

2

×1350]=0.202734m2

Fluid velocity ,ug=25.71790.202734

=126.855m / s

This fluid velocity is clearly too high for a pipe flow, it is also known that the steam feed is in excess,

one way to reduce the velocity of the steam is to reduce the feed. Since from the calculation above

that only 24.8% of the total gas is condensed into fluid, a reduction of feed by 75% would

tremendously reduce the gas velocity in the shell. Hence, by limiting the steam feed to

13656.25kg/h. we get

Volumetric flow rate=13656.25 kg /h3600

× 10.590kg /m3

=6.4295m3/s

Hence,

Fluid velocity ,ug=6.42950.202734

=31.714m /s

17

A gas velocity of 32m/s is more reasonable compared to the previous calculation. A high velocity

could cause serious erosion in the shell side.

To calculate the forced convection on the tube and the effect of condensation, a modified version of

Shekriladze and Gomelauri equation on the surface shear convection by Butterworth using

interpolation formula will be used in the calculation. This formula satisfies gravity-controlled and

shear-controlled condensation and is considered a more conservative approach compared to other

approach.

The Butterworth equation is as follow

Num~R e1 /2

=0.416 (1+(1+9.47 F )1/2 )1 /2 (9.5)

Where F is given by

F=gd μl i gug2 k l∆T

(9.6)

The two-phase Reynolds number is

~R e=ρlugdμl

(9.7)

Calculating the F values gives

F=9.81×0.0137×0.00001297×2.2011×106

(31.714 )❑2 ×0.16×25

=0.009537

Calculating the two-phase Reynolds Number gives

~R e=958.4×31.714×0.01370.000282

=1.476×106

Rearranging and substituting both ~R e and F into equation (9.5) gives

N um=0.416 (1+(1+9.47 F )1 /2 )1 /2×~Re1/2

N um=0.416 (1+(1+9.47 (0.009537 ) )1/2)1/2× (1.476×106 )1 /2

N um=722.74 8

From the equation (9.4) the mean heat transfer coefficient can be calculated

18

hm=Nu k l

d

hm=722.728×0.16×0.3268

0.0137=17240W /m2∙ K

The heat transfer coefficient is appropriate for a condensing steam.

9.3 Overall Heat Transfer CoefficientThe overall heat transfer coefficient can be calculated using the equation

1U o

=1ho

+1hod

+

do ln( d o

d i)

2k w+do

d i× 1

hid+do

d i× 1h i

(9.8)

The condensing steam from the evaporator is assumed to have traces of light hydrocarbon in it, a

fouling coefficient of 5000 W/m2℃ will be used in the calculation.

There are insufficient amount of data for the fouling coefficient for the cold fluid mixture of

ethylene oxide and water. Taking that the fouling factor for water and ethylene oxide is addictive,

that is, independent of each other, a combine fouling factor of 10000 W/m2℃ would be taken for

conservative calculation.

The following data will be used to calculate the overall heat transfer coefficient.

Coefficient type Values (W/m2K

¿

ho 17240

hod 5000

hid 10000

hi 1048.5

Table 9.1 Heat transfer coefficient for heat exchanger

Substituting the values into the equation (9.8) we have

1U o

= 117240

+ 15000

+0.0137 ln( 13.712.05 )

2 (16.3)+ 13.712.05

× 110000

+ 13.712.05

× 11048.5

19

1U o

=0.001454572

U o=687.487W /m2℃

20

10.0 Pressure Drop10.1 Tube-side Pressure DropTube side pressure drop can be referred from the Tube-side friction factors in the Sinnot & Towler

book. The Reynolds number of cold fluid in tube is 8743.4. From the figure, this correspond to

friction factor of 5×10−3.

The pressure drop can be calculated using the equation

∆ Pt=N p[8 if ( Ldi )+2.5] ρ ut2

2 (10.1)

Substituting the known values into the equation (10.1) we have

∆ Pt=1350[8×0.005( 60.01205 )+2.5] 960.64×0.2795❑

2

2

∆ Pt=840.97 N /m2

The pressure drop is very low and is way below the recommended pressure drop guideline outlined

by Sinnot & Towler, hence it is acceptable.

10.2 Shell-side Pressure DropAs the shell-side involves condensation which reduces volume dramatically, the pressure drop

should be inspected. The pressure drop for shell-side associated with condensation will be

calculated using Kern’s method to make an approximate estimate.

Area for cross-flow is given by

A s=(Pt−do)D s lB

Pt(10.2)

Using a baffle spacing of that equal to the internal diameter of Shell with 45% cut lined vertically.

Since Pt=1.25do , the equation can be further simplified to

A s=DslB5

(10.3)

Substituting the values into equation (10.3) we have

A s=(0.715)2

5=0.1023m2

21

Mass flow rate over area = (13656.25/3600)/0.1023

= 37.08kg/s m2

Calculating the shell-side equivalent diameter, for a triangular arrangement, the following equation

can be used.

De=1.10do

(P t2−0.917do

2 ) (10.4)

Substituting the values, we have

De=1.100.0137

(0.017125−0.917×0.0137 )=0.009728m

The Reynolds Number is,

ℜ=148.321×0.0097280.00001297

=111243.2

Using the friction factor diagram found in Sinnot and Tawlor

if=2.2×10−2

Fluid speed in the shell-side is

us=G s

ρv(10.5)

Where

us=148.3210.59

=62.85m /s

Take pressure drop as 50% of that calculated using the inlet flow, neglect viscosity correction. We

have the equation

∆ P s=8 if ( D s

de )( ρus2

2 )( LlB ) (10.6)

Substituting the values, we have

∆ P s=8×0.022( 0.71520.009728 )( 0.59×62.852

2 )( 60.715 )=63245.6N /m2

∆ P s=63.2kPa

22

The amount of pressure drop in the shell-side is small compared to the minimum pressure drop

value recommended of 0.5 atm. Hence the design is acceptable

23

11.0 Mechanical Design11.1 Heat Exchanger TubesAs mentioned above during the calculation, the heat exchanger tube will be made with the following

material, specification and dimension

Material Stainless Steel, 316

Nominal Size ¼

Schedule 10S

Internal Diameter 13.7mm

Thickness 1.65mm

Table 11.1 Material, specification and dimension of tubes

These tubes are to be welded to the fixed-tube sheet and rolled in the tube sheets. As the tube-side

liquid is high pressured, special caution has been taken to measure the minimum allowable

thickness of the tube. The calculation can be referred above. To minimize the vibrational damage,

wide grid bars will be resting on the tube wall. The baffles in the shell will also act as a vibrational

cushion to the tubes. The tubes should also be inclined 5 degree as mentioned above to facilitate

the film condensate drop flow and increase the overall heat transfer coefficient.

11.2 Heat Exchanger ShellThe heat exchanger shell will have the following material, specification and dimension

Material Stainless Steel, 316

Internal Diameter 715.2mm

Thickness 4.8mm

Table 11.2 Material, specification and dimension of shells

11.3 Tube SheetsThe tube sheet is extremely important in this design as it separate two pressure sections. The

thickness of the tube sheet has to be constant. The tubes will be fastened in a tube sheet by

welding and roller expansion. The tube must be fixed before welding to prevent eccentric weld

joints.

11.4 Vent PointAs condensation process is in the shell-side, a special attention is needed to vent the system

properly, the steam will be drawn off at vent point but screened off to prevent the non-cooled

heating steam from taking the direct path to the vent system.

24

11.5 Baffle number and distanceEach baffle will have a spacing of 0.715m, for a rough estimate on the number of baffle needed.

Total spacing = 6m/0.715m

= 8.39

Number of baffle needed = 8 – 1

= 7

Rounding down the numbers, a total of 7 baffles will be used.

11.6 Weight LoadsThe dead weight of the heat exchanger consist of several parts as described below

I. Shell

II. Shell Cover

III. Tubes

IV. Baffle

V. Tube Sheets

Approximate weight of the heat exchanger will be calculated by assuming it as a cylindrical vessel

with domed ends and uniform thickness. The following equation can be used

W v=Cw π ρmDmg (H v+0.8Dm ) t ×10−3 (11.1)Substituting the values,

W v=1.15× π×8000×0.720 (6.72+0.8×0.720 )5.8×10−3

W v=8638.77 kg

The weight of the tube can be calculated as follow,

weight of tube=den sity of stainless steel ×effective volume of tub e

weight of tube=2162.35kg

Adding up the tube and shell, we have an approximate weight of 10801.12kg.

11.7 Nozzle

The nozzle of the heat exchanger will have the internal diameter of 5 inch with 8 bolts and nuts for

tightening. Special gasket will be used as a seal when the nozzle is connected to a pipe.

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12.0 Schematic Diagram of Heat Exchanger

26

13.0 Mechanical Drawing of Key Item

27

14.0 Piping and Instrumental

28

15.0 CostingThe equipment costing can be approximate using correlation data by Sinnot and Tawlor. For the

estimation of heat exchanger, the U-tube shell and tube heat exchanger correlation data will be

used.

The data of correlation is as follow

Equipment Unit for size Slower Supper a b n

U-tube shell and tube Area, m2 10 1000 24000 4

6

1.2

Table 15.1 Equipment cost correlation for U-Tube Shell and Tube

Using the equation

C e=a+b Sn (15.1)

The area of the heat exchanger is 346.97m2 , substituting the values into the equation, we have

C e=24000+46 (346.97 )1.2

C e=75417

Since stainless steel 316 is used instead of carbon steel, the estimated cost need to be multiplied

by 1.3 for stainless steel factor. This would total up the estimation cost to

C e=75417×1.3=96872.1

This again should be coupled with escalation of cost since the estimated cost is for the year 2007. A

cost index is referred in the Sinnot and Towler design book with the current cost index extrapolated

to be 2500

A correction to the cost estimation with cost index gives

C e=96872.1×25002059.1

=$117614.6

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16.0 Auxiliary Equipment, pump and pipingThere is no auxiliary equipment needed for the operation of the heat exchanger. There is, however,

a need for pump and compressor in order for the heat exchanger to work.

It is calculated that the tube side pressure drop is ∆ Pt=840.97N /m2 which is negligible. The pipe

that carry the feed into heat exchanger would have to be stainless steel due to the nature of fouling.

The heat exchanger will be operating from the gound. The internal diameter of the pipe used will

have 5 inch to match the nozzle of the heat exchanger. Assume that the compressor is situated 20

meter away.

Cross-sectional area of pipe = π4

(0.127 )2=0.01267m2

Fluid velocity ¿ 148798960.63

× 13600

× 10.01267

=3.3965m /s

Reynolds Number ¿(960.63×0.127×3.3965)/0.00037=1119942

Absolute roughness of pipe =0.000015m

Relative roughness = 0.000118

From the diagram of pipe friction versus Reynolds number and relative roughness, the friction factor

is 0.001625.

Friction loss in pipeline, ∆ P f=8×0.00018×( 200.127 )×( 960.63×3.396522 )=1256.5453N /m2

Since the pressure before the compressor is atmospheric pressure, the pressure difference is,

∆ P❑=20−1atm=1898675Pa

The energy balance is

(1898675/960.63)+(1256.5453/960.63) = 1977.8J/kg

The required power

Power = 109965.53W

Taking the compressor efficiency as 70%,

Power = 157093.624kW

This would require a heavy duty compressor as the duty is as high as 157kW.

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17.0 Mass Balance and Safety Review17.1 Mass Balance ReviewThroughout the design process, the mass balance has been reviewed again and again by out

group. While performing energy balance to determine the duty of the heat exchanger, it was found

out that there are excessive steam energy from the evaporator and only 24.8% of them is fully

utilized.

Additional steam has to be diverted to other stream due to the limitation in the designing of

the equipment such as keeping the shell-side velocity within desirable range. In that regard, the

mass balance will probably need to be adjusted again for better thermal efficiency.

17.2 Safety ReviewThe usage of gasket and O-ring for pump and heat exchanger is usually inevitable. There have

been reports of ethylene oxide chemically attack the gasket and O-ring, especially those made from

asbestos or PTFE. This would usually resulted in leakage and in other serious case, a hazard.

Special O-ring and Gasket need to be used when handling ethylene oxide, the item listed below

have high resistance and stable against ethylene oxide. Only this items and this item only should be

used, other

Gaske

t

Polycarbon Sigraflex BTCSS Flexible Compressed Graphite – Laminated on Stainless

Steel Tang Sheet

O-ring Chemraz 505

Kalrez 2035

Kalrez 6375

Parker EPDM-740-75

Parker EPDM-962-90

Parker E-515-8-EPM

Table 17.1 Item compatible with Ethylene Oxide

As the ethylene oxide is a hazardous substance, the use of pressure relief valve along the pipes

that contain the liquid is not encouraged. Appropriate ethylene oxide leak detecting system should

be employed, online pressure monitoring probe should be install where ethylene oxide is present in

the pipe line. For insulated flanges, it is useful to install leak detection tubes. Sealed stainless steel

bands installed around the flange can help prevent leakage of liquid along the piping.

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18.0 ConclusionThis equipment was design with the priority of safety over the other factors and many design

consideration was made by making calculation conservatively. The heat exchanger can withstand

high pressure load even if there is a sudden fluctuation of pressure during the operation. The

equipment has both the capacity to heat up the feed as well as low operating cost due to the low

pressure drop over the entire unit.

The use of Stainless steel would ensure low fouling rate in both the tube and the shell, while

maintenance should be carried out periodically. Overall, I think the design of this heat exchanger

would satisfy the need of the plant to operate without much concern about time lost as well as

saving the cost in maintenance.

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19.0 References

Bott T.R. 1995. Chemical Engineering Monographs 26: Fouling of Heat Exchanger. Amsterdam: Elsevier Science

Kakac S. & Liu H.T. 2002. Heat Exchangers Selection, Rating, and Thermal Design 2nd Ed. Florida: CRC Press

Kuppan T. 2000. Heat Exchanger Design Handbook. Basel: Marcel Dekker

Ramesh K.S & Dusan P.S. 2003. Fundamentals of Heat Exchanger Design. New Jersey: Wiley & Son

Sinnot.R & Towler. C. 2009. Chemica Engineering Design 5th Ed. China: Butterworth – Heinemann

Podhorsky M. & Krips H. 1998. Heat Exchanger: A Practical Approach to Mechanical Construction, Design, and Calculation. New York: Begell House

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Appendix A Nomenclature

hi Tube-side heat transfer coefficient

hid Tube-side fouling

ho Shell-side heat transfer coefficient

hod Shell-side fouling

A ID Area of Internal Diameter

C p Specific heat at constant pressure, J/kg·KDb Bundle Diameter

Gs Specific gravity

K 1 constant

NT Number of tubes

U o Overall Heat transfer coefficient

do Outside diameter

ig Enthalpy of gas

k f Thermal conductivity

m Fluid mass flow rate, kg/sn1 constant

tm Minimum thickness

t p Pressure thickness

∆ P f Pressure loss in pipe friction, N/m2

∆ P s Pressure difference in Shell-side, N/m2

∆ Pt Pressure difference in Tube-side, N/m2

∆T m Mean temperature

∆ P Pressure difference, N/m2

∆T Local temperature difference between two fluids, ℃, KA Area, m2

U Heat transfer coefficientE Casting quality factorQ Heat transfer rate, WS Basic allowable stressc Sum of mechanical allowanceg Gravityv Velocityγ Temperature coefficientμ Viscosityρ Density

35

Appendix B Process Flow Diagram For Plant Design

36

Appendix C Mass Balance For Plant Design

37

Appendix D Pipe Sizes – ANSI/ASME B36.19M - 1985

Dimensions and Weights per metre – stainless steel pipe

38

Appendix E TEMA Shell and Tube Nomenclature

39