Design Heatexchanger

7/21/2019 Design Heatexchanger

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PAPER PRESENTATION ON

OPTIMUM THERMAL DESIGN OF ETHYL ALCOHOL HEAT EXCHANGER

USING SOFTWARE

SUBMITTED BY:-CHIRAG MEHTA (B.E. MECH.)SALMAN DABIR (B.E. MECH)

GUIDED BY:-

PROF. S.S. PAWAR

(ASST. PROF. MECH. ENGG.)

DEPARTMENT OF MECHANICAL ENGINEERING

LOKMANYA TILAK COLLEGE OF ENGINEERING

NAVI-MUMBAI.

2005- 2006


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ABSTRACT

Optimum design of heat exchanger is more effective in thermal design stage as compared to

mechanical design. The thermal design is a trial and error method of design i.e. iterative. The

application of computer in such design method gives very fast and optimum design by comparing all

the possible resource.

The objective of this project is to do optimum design of ethyl alcohol heat

exchanger using C++ language for thermal design and for checking parameters like pressure drop,

velocity drop, dirt factor etc by using area based and required flow condition.

At last the conclusive remark and results are discussed by comparing required output.


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CONTENTS

1. INTRODUCTION

2. OBJECTIVE AND NECESSITY OF OPTIMUM THERMAL DESIGN

3. PROBLEM

4. NOMENCLATURE

5. RESULTS OF 1-4 PASS - FIXED TUBESHEET SHELL AND TUBE HEAT EXCHANGER

6. OPTIMUM THERMAL DESIGN BASED ON MINIMUM AREA

PASS 1-1

PASS 1-2

PASS 1-4

7. RESULTS AND CONCLUSIONS

8. REFERENCES


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INTRODUCTION:-

A device used for transferring or exchanging energy in the form of heat between two fluids is

known as heat exchanger. This heat exchanger can be either by direct or indirect method. In direct

method heat exchanger heat transfer between two fluids takes place without mixing. Two fluids areseparated by a surface metallic or non- metallic through which heat transfer takes place in the direction

perpendicular to the flow direction.

TYPES OF HEAT EXCHANGER:-1. fixed tube sheet2. outside packed floating head.3. internal floating head4. U- tube type.5. Reboiler (with internal floating head or U- tube type).

COMPONENTS OF HEATEXCHANGER:-1. Shell2. Shell cover3. Tubes4. tube sheet5. tie rods or spacer.6. baffles7. channel8. channel cover9. pass partition10.nozzles11.flanges12.support

type of heat exchanger we have selected for given specification is a fixed tube sheet shell and tube dueto following reasons.:-

to prevent corrosion in the shell, water is taken inside the tube. There is a possibility of foulingand hence clogging of tubes. This will reduce heat transfer capability of heat exchanger. To avoid

clogging it is recommended that velocity of fluid flowing on the tube side should be more then 3 fps.

Since there is low temperature difference on shell side and tube side and material ofconstruction is also same, there will be practically no thermal expansion, therefore u tube heatexchanger will not be suitable choice.

Among the heat exchanger configuration 1-1,1-2, 1-4 pass , we have selected 1-4 pass heatexchanger, which will satisfy the above criteria of increase in tube side fluid velocity

Also we have selected fixed tube sheet configuration, because it will be suitable for thisapplication as there is no problem of thermal stresses and there are no internal joints in this, thus

eliminating leakage of one fluid into another.


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OBJECTIVE:-

The application of the principle of heat transfer to the design of equipment to accomplish acertain engg. Objective is of extreme importance. For applying the principles to the design, individual

is working towards the important goals of product development for economic gain.

The objective of this project is to do optimum design of ethyl alcohol heat exchanger using C++language for thermal design and for checking parameters like pressure drop, velocity drop, dirt factor

etc by using area based and required flow condition.

NECESSITY OF OPTIMUM THERMAL DESIGN:-

Economics play a key role in the design and selection of heat exchanger equipment. The weight

and size of heat exchanger used in space or aeronautical applications are very important parameters.

The weight and size are important cost factor in the overall application in the chemical process

industries and thus may still be considered as economic variables.

Optimum design of heat exchanger is more effective in thermal design stage as compared to

mechanical design. The thermal design is a trial and error method of design i.e. iterative. Theapplication of computer in such design method gives very fast and optimum design by comparing all

the possible resource.


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PROBLEM:-

Ethyl alcohol is manufactured by m/s Krishna Sahakari Sakar Karakhana Ltd. Rethare BK. Tal.. Karad

at the rate of 30kl/day. The ethyl alcohol is condensed in condenser operating at atmospheric pressure,

which gives condensing temp. as 78 c . the hot ethyl alcohol is to be cooled from 78 c to 45 c comingwith pressure 3.5 kg/cm2 using cooling water available at 30 c and pressure 3.5 kg/cm2 with 5 c as

approach

Design suitable cooler (heat exchanger) with dirt coefficient = 0.002 (hr)(ft2)(F)/Btu and

allowable pressure drop as 10 psi(g) on both sides of the cooler .

SOLUTION:

This problem is solved by theoretical calculation which is very long lengthy , tedious and time

consuming. The various formulas used for calculation are as follows:

LMTD = (t2t1)

Ln [t2/t1]

R = (T1T2) / (t2t1) and S = (t2t1) / (T1t1)

True Temperature Difference,t = LMTD x FT

a) For viscosity at Taand ta, referring from graph from Donald Q.kern

Viscosity of water at 90.5 F is = 0.80 cp

Viscosity of ethyl alchohol at 143.0 F is al= 0.62 cp

b) For Thermal conductivities, refer table 4 from Donald Q.kern

Kalat 122F = 0.087

Kalat 68F = 0.105

c) Prendtal number :- (c/ K)al

Surface Area, A = Q / (UDxt)

Mass velocity, Gt= Mt/ at

Water velocity V = Gt/ (3600 x w)

Reynold number, Ret = D x Gt / w

hi= JHx (k / D) x (c/ K)1/3

x t


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hio/ t = (hi/ t ) x (ID / OD)

pt = (f x Gt2x L x n) / (5.22 x 10

10x D x s x t)

Return pressure loss :-

pr = (4 x n/s) x (V

2

/ 2 x g)

Total pressure loss from tube side:PT = (pt +pr)

Shell side calculation :

Flow area as= (ID x C x B) / (144 x PT )

Mass velocity Gs = W / as

Shell side Reynold number: Res= Dex Gs / al

ho= JHx (K / De) x (c / K)1/3

x s

Number of crosses,N + 1 = 12 x L/B

Ps= (f x Gs

2x Dsx (N + 1)

(5.22 x 1010

x Dex s x s)

Clean overall coefficient Uc:Uc= hiox ho

hio+ ho

Dirt factor Rd:

Rd= Uc- Ud

Ucx Ud


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NOMENCLATURE:

A = heat transfer surfacea = flow area, ft2

a = external surface per linear foot, ft

B = baffle spacing, in.C = clearance between tubes, in.

C = specific heat of fluid, Btu/ (lb) (F)

D = Inside diameter of tube, ft.Ds= Inside diameter of shell, ft

De, De1= equivalent diameter for heat transfer and pressure drop, ft

d = inside diameter of tubes, in.

FT= temperature- difference factor, dimensionlessf = friction factor, ft

2/ in

2.

G = mass velocity, lb / (hr) (ft2)

g = acceleration of gravity, ft / hr2

G= acceleration of gravity, ft / sec2

.H, hi, ho= heat transfer coefficient in general, for inside fluid, and for outside

Fluid respectively, Btu / hr (ft2) (F / ft).

hio= value of hi when referred to tube outside diameter, Btu / hr (ft2) (F / ft)

jH= factor for heat transfer, dimensionless.

k = thermal conductivity, Btu / hr (ft2) (F / ft)

L = tube length, ft.

LMTD = Log mean temperature difference, F.

N = number of shell side baffles.

Nt= number of tubes.n = number of tube passes

Pt= tube pitch, in.Q = heat flow, Btu / hr.R = TEMPERATURE GROUP (T1-T2)/ (t2-t1), dimensionless.

Rd= dirt factor, (hr) (ft2) (F) / Btu.

Re= Reynolds no for heat transfer,), dimensionlessS = temperature group, (t2-t1) / (T1t1),), dimensionless.T1, T2= inlet and outlet temperature of hot fluid, F.

Ta= average temperature of hot fluid, F

ta= average temperature of cold fluid, FUc, Ud= clean and design overall coefficient of heat transfer, Btu / hr (ft

2) (F / ft)

V = Velocity, fps

W = weight flow of hot fluid, lb / hrMt= weight flow of cold fluid, lb / hr

s = shell

t = tube.


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As it is seen that the thermal design is a trial and error i.e. iterative method of design, therefore it isvery difficult, tedious and time consuming method to get optimum design. Hence it is proposed to

prepare general program in C++ language to get quick, accurate and optimum thermal design

parameters to reduce the weight to achieve cost effectiveness. So we solve the above problem usingC++ software

We obtain the following results as shown below for all the three passes (C++ program is not shown

here as it is lengthy)

RESULTS OF 1-4 PASS

FIXED TUBESHEET SHELL AND TUBE HEAT EXCHANGER

INSTRUCTION: - in this program you feed the data such that, T1>T2, t2>t1 otherwise it willgive wrong results.

Now enter the values of following parameters:

Enter tube od in inches

.75

Enter triangular pitch in inches

1

Enter clearance in inches

.25

Enter surface per linear feet value ft square from kern book

.1963

Enter flow area per tube value inch sqr

.302

Enter inner diameter of tube di in ft

.502

Enter mass flow rate on shell side (lb / hr)

2200

Enter shell side inlet temp (deg F) T1

173

Shell side outlet temperature (deg F) t2113


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Enter tube side inlet temperature t1 (deg F)86

Tube side outlet temperature t2 (deg F)95

Enter shell side fluid specific gravity SG_s

.81

Enter tube side fluid specific gravity SG_t

1

Enter shell side fluid specific heat (Btu/Lb deg F)

.78

Enter tube side fluid sp. Heat (Btu/Lb deg F)

11

Enter viscosity of shell side fluid

1.5

Enter tube side fluid viscosity

1.936

Enter shell side fluid thermal conductivity

0.08

Enter tube side fluid thermal conductivity

0.358

Enter the term fay_s

1

Enter the term fay_t

1

Enter lower (limit of Ud-1) in British units so that it will start from lower limit

74

Enter upper limit of Ud

150

Enter final limit of tube length, ft

6


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ENTER THE LOWER LIMIT OF TUBE SIDE PASSES AS FOLLOWS

For np=1, enter 0

For np=2, enter 1

For np=4, enter 22

Now enter upper limit of no of passes which you want

But remember this upper limit >= lower limit4

Enter the minimum tube side velocity which you want (ft/sec)1.5

Enter lower limit of dirt factor in British units0.002

Enter upper limit of dirt factor in British units0.002

Heat transferred =102960

Mt=11440LMTD=46.0198

Ud lower limit=74

Enter tube length lower limit (i.e. suppose 816 then enter 8)6

Lower limit of tube length selected=6No of passes before beginning of loop =2

Press any key!

RESULTS:-

/ Rd / Ps / Pt / TU / TBLN / B / SH_ID / Ud / N_P / cnt / EPC / EPD // 0.002264 / 0.31 / 3.46 / 24 / 6.0 / 1.60 / 8.0 / 75 / 4 / 1 /0.62 / 0.53/

/ 0.002264 / 0.31 / 3.46 / 24 / 6.0 / 1.60 / 8.0 / 76 / 4 / 1 /0.62 / 0.53 /

/ 0.002264 / 0.31 / 3.46 / 24 / 6.0 / 1.60 / 8.0 / 77 / 4 / 1 /0.62 / 0.53 /

/ 0.002264 / 0.31 / 3.46 / 24 / 6.0 / 1.60 / 8.0 / 78 / 4 / 1 /0.62 / 0.53 /

hence optimum thermal design based on minimum area for the passes 1-1, 1-2 and 1-4 is as

shown below:-


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OPTIMUM THERMAL DESIGN BASED ON MINIMUM AREA

FOR 1-1 PASS

PARAMETERS MAGNITUDE UNITS

Area 43.5 ft sqr

UD 51.34 Btu/hr ftsqr deg FLMTD 46.02 deg F

TUBE LENGTH 6.00 ft

TUBES 37.0 number

SHELL ID 8.00 inchesAs 0.0319 ft sqr

Gs 68869.58 Lb/hr ft sqr

Re_s 2755.42 dimensionless

Friction Factor 0.002868 dimensionlessOn shell side

Pressure drop 0.11 psiOn shell side

At 0.0776 ft sqr

Gt 147427.97 Lb/hr ft sqrTube velocity 0.66 ft/sec

Re_t 3959.84 dimensionless

Friction factor 0.000345 dimensionless

On tube sidePressure drop 0.03 psi

On tube sideJho 27.41Ho 89.34 Btu/hr ftsqr deg F

Hio 159.24 Btu/hr ftsqr deg F

Uc 57.23 Btu/hr ftsqr deg FRd 0.002006 hr ft sqr deg F/BtuNo of passes 1.0 number

EPC_Uc 0.58

EPC_Ud 0.53


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FOR 1-2 PASS


Area 35.33 ft sqr

UD 63.32 Btu/hr ftsqr deg F

LMTD 46.02 deg F

TUBE LENGTH 6.00 ftTUBES 30.0 number

SHELL ID 8.00 inches

As 0.0319 ft sqrGs 68869.58 Lb/hr ft sqr

Re_s 2755.42 dimensionlessFriction Factor 0.002868 dimensionlessOn shell side

Pressure drop 0.1100 psi

On shell sideAt 0.0315 ft sqr

Gt 363655.62 Lb/hr ft sqr

Tube velocity 1.62 ft/sec

Re_t 9767.61 dimensionlessFriction factor 0.000271 dimensionless

On tube side

Pressure drop 0.31 psiOn tube side

Jho 27.41

Ho 89.35 Btu/hr ftsqr deg FHio 392.80 Btu/hr ftsqr deg F

Uc 72.79 Btu/hr ftsqr deg F

Rd 0.002055 hr ft sqr deg F/Btu

No of passes 2.0 numberEPC_Uc 0.59

EPC_Ud 0.53


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FOR 1-4 PASS


Area 28.27 ft sqr

UD 79.15 Btu/hr ftsqr deg F

LMTD 46.02 deg F

TUBE LENGTH 6.00 ftTUBES 24.0 number

SHELL ID 8.00 inches

As 0.0200 ft sqrGs 99000.0 Lb/hr ft sqr

Re_s 3960.92 dimensionlessFriction Factor 0.002672 dimensionlessOn shell side

Pressure drop 0.31 psi

On shell sideAt 0.0126 ft sqr

Gt 909139.06 Lb/hr ft sqr

Tube velocity 4.04 ft/sec

Re_t 24419.03 dimensionlessFriction factor 0.000211 dimensionless

On tube side

Pressure drop 3.46 psiOn tube side

Jho 33.43

Ho 108.98 Btu/hr ftsqr deg FHio 837.03 Btu/hr ftsqr deg F

Uc 96.42 Btu/hr ftsqr deg F

Rd 0.002264 hr ft sqr deg F/Btu

No of passes 4.0 numberEPC_Uc 0.62

EPC_Ud 0.53


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RESULTS AND CONCLUSION:-

The thermal design of the heat exchanger is carried out as per D.Q. Kern. The manual designprocedure is found tedious and time consuming. Therefore computer program in C++ language is

prepared and results obtained are compared for different pass arrangements. The optimum thermal

design parameters are selected on the basis of minimum area. The key parameters are tabulated as

shown below:-

From the result mention in table it can be concluded that 1-4 pass arrangement is the best optimumdesign. Further 1-6 pass arrangement is not possible because pressure drop from tube side increases

eight times which crosses the limit of allowable pressure drop 10psi. also the tube velocity and dirt

factor crosses the allowable limit for next pass arrangement.

REFERENCES:-

Donald Q Kern. process heat transfer Tata McGraw Hill.

Let us C ++ by Yashwant Kanitkar.

Passarrangement Area

No. ofTubes Sheet ID Uc UD Rd

TubeVelocity

Total pressuredrop

1-1 43.58 37 8 57.23 51.34 0.002006 0.66 0.14

1-2 35.33 30 8 72.79 63.32 0.002055 1.62 0.42

1-4 28.27 24 8 96.42 79.15 0.00226 4.04 3.77

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