01 kern's method

Designing Shell-and-Tube Heat Exchangers Using Softwares

Lecture 1: Kern’s Method

By:

Majid Hayati

University of Kashan, Kashan, I.R. IRAN

2014

3

Schedule – Kern’s Method

Kern’s method

Introduction to Kern’s method

Algorithm of design procedure for shell-and-tube heat exchangers

Design procedure steps along with an example

4

Objectives

This lecture on designing shell-and-tube HEs serves as an introduction lecture to the subject, and covers:

Introduction to “Kern’s method” definition along with its advantages and disadvantages

Developing an algorithm for the design of shell-and-tube exchangers

Finally, following up the procedure set out in the algorithm in an example

5

Introduction to Kern’s method

Kern’s was based on experimental work on commercial exchangerAdvantages:

Giving reasonably satisfactory prediction of the heat-transfer coefficient for standard design Simple to applyAccurate enough for preliminary design calculations Accurate enough for designs when uncertainty in other design

parameter is such that the use of more elaborate method is not justified

Disadvantage: The prediction of pressure drop is less satisfactory, as pressure drop is more affected by leakage and bypassing than heat transferThe method does not take account of the bypass and leakagestreams

6

Design procedure for shell-and-tube heat exchangers(Kern’s method)

Start

Calculate tube numberCalculate shell diameter

Assume value of overall coefficient Uo,ass

Collect physical properties and HE specifications

End

Estimate tube- and shell-side heat transfer coefficient

Estimate tube- and shell-side pressure drop

Question: Are pressure drops within specification?

Estimate tube- and shell-side heat transfer coefficient-go to step 3

Accept all design parametersCompare to estimated overall heat transfer coefficient

Determine overall heat transfer coefficient

Start from step 3

Step 1

Step 2

Step 3

Step 4

Step 6

Step 7

Step 8

Define dutyMake energy balance if needed

Calculate unspecified flow ratesCalculate ΔTLMTD and ΔTM

Fig. 1: Algorithm of design procedure

Determine fouling factors

Step 5

YesNo

NoYes

Start


Step 1


Step 1

Step 2


Step 2


Calculate unspecified flow ratesCalculate ΔTLMTD and ΔTM


Step 3


Step 3

Calculate tube numberCalculate shell diameter

Step 4Calculate tube numberCalculate shell diameter

Step 4


Step 5


Step 5


Step 6


Step 6

Estimate tube- and shell-side pressure drop

Step 7Estimate tube- and shell-side pressure drop

Step 7

YesYes No


Step 8


Step 8

NoYes

Accept all design parameters

7

Kern’s Method Design Example

Design an exchanger to sub-cool condensate from a methanol condenser from 95 °C to 40 °C

Flow-rate of methanol 100,000 kg/h

Brackish water (seawater) will be used as the coolant, with a temperature rise from 25° to 40 °C

8

Collect physical properties and HE specifications: Physical properties

Solution: Step 1

WaterMethanolPhysical properties at fluid mean temperature

4.22.84Cp (Kj/Kg °C)0.80.34μ (mNs/m2)

0.590.19kf (W/m °C)995750ρ (Kg/m3)

Table 1

HE specifications: Coolant (brackish water) is corrosive, so assign to tube-side.Use one shell pass and two tube passes.At shell side, fluid (methanol) is relatively clean. So, use 1.25 triangular pitch

(pitch: distance between tube centers).

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Tube Arrangements

The tubes in an exchanger are usually arranged in an equilateral triangular, square, or rotated square pattern (Fig. 2)

Fig. 2: Tube patterns

10

Tube Pattern Applications

The triangular and rotated square patterns give higher heat-transfer rates, but at the expense of a higher pressure dropthan the square pattern.

A square, or rotated square arrangement, is used for heavily fouling fluids, where it is necessary to mechanically clean the outside of the tubes.

The recommended tube pitch is 1.25 times the tube outside diameter; and this will normally be used unless process requirements dictate otherwise.

11

Define duty, Make energy balance if neededTo start step 2, the duty (heat transfer rate) of methanol (the hot stream or

water, the cold stream) needed to be calculated.

Step 2

kW 4340=40)(95 2.84×3600100000

= )T(TCm = Q = loadHeat 21phh

•--

Fig. 3: Streams definitions.

12

Step 2 (Cont’d)

The cold and the hot stream heat loads are equal. So, cooling water flow rate is calculated as follow:

kg/s 68.9 =25)(40 4.2

4340=

)t(t CQ

= m=flow waterCooling _12cP

.c _

The well-known “logarithmic mean” temperature difference (LMTD or lm) is calculated by:

C31

25)(4040)(95ln

25)(4040)(95

tTtTln

)t(T)t(TΔT

12

21

1221LMTD

13

Mean Temperature Difference

The usual practice in the design of shell and tube exchangers is to estimate the “true temperature difference”from the logarithmic mean temperature by applying a correction factor to allow for the departure from true counter-current flow:

LMTDtm ΔT FΔT

Where:

ΔTm = true temperature difference,

Ft = the temperature correction factor.

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Temperature Correction Factor

The correction factor (Ft) is a function of the shell and tube fluid temperatures, and the number of tube and shell passes.

It is normally correlated as a function of two dimensionless temperature ratios:

11

12

12

21

tTttS

ttTTR

15

For a 1 shell : 2 tube pass exchanger, the correction factor is plotted in Fig. 4.

Step 2

)TEMAre even tube passes (available in Fig. 4: Temperature correction factor: one shell pass; two or mo

16

Step 2 (Cont’d)

From Fig. 4, the correction factor (Ft) is 0.85.

21.025952540

)t(T)t(tS

67.325404095

ttTTR

11

12

12

21

C 26 310.85ΔT FΔT LMTDtm

17


Typical values of the overall heat-transfer coefficient for various types of heat exchanger are given in Table 1.

Fig. 5 can be used to estimate the overall coefficient for tubular exchangers (shell and tube).

The film coefficients given in Fig. 5 include an allowance for fouling.

The values given in Table 1 and Fig. 5 can be used for the preliminary sizing of equipment for process evaluation, and as trial values for starting a detailed thermal design.

From Table 2 or Fig. 5: U=600 W/m2°C

Step 3

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Step 3 (Cont’d)

Table 2: Typical overall coefficients

19

Step 3 (Cont’d)

Fig. 5: Overall coefficients (join process side duty to service side and read U from centre scale)

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Step 4

Calculate tube number, Calculate shell diameterProvisional area:

So, the total outside surface area of tubes is 278 m2

Choose 20 mm o.d. (outside diameter), 16 mm i.d. (inside diameter), 4.88-m-long tubes ( ), cupro-nickel.Allowing for tube-sheet thickness, take tube length: L= 4.83 m

Surface area of one tube: A = πDL = 4.83 x 20 x 10-3π = 0.303 m2

23

Mm 278=62× 600

10× 4340=TΔ U

Q=A

ft 16 in.43

9180.303278

tubeone of area surface Outsidearea) al(Provision tubesof area surface outside Total tubesof Numbers

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Step 4 (Cont’d)

An estimate of the bundle diameter Db can be obtained from equation below which is an empirical equation based on standard tube layouts. The constants for use in this equation, for triangular and square patterns, are given in Table 3.

where Db = bundle diameter in mm, do = tube outside diameter in mm., Nt = number of tubes.As the shell-side fluid is relatively clean use 1.25 triangular pitch.So, for this example:

1n1

1

tob )

KN(d D

mm 826 )0.249918( 20 D diameter Bundle 2.207

1b

22

Step 4 (Cont’d)

Table 3: Constants K1 and n1

Nav

Rectangle

23

Step 4 (Cont’d)

Use a split-ring floating head type for Fig. 6.

From Fig. 6, bundle diametrical clearance is 68 mm.

Shell diameter (Ds):Ds= Bundle diameter + Clearance = 826 + 68 = 894 mm.

Note 1: nearest standard pipe size are 863.6 or 914.4 mm.

Note 2: Shell size could be read from standard tube count tables [Kern (1950), Ludwig (2001), Perry et al. (1997), and Saunders (1988)].

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Step 4 (Cont’d)

Fig. 6: Shell-bundle clearance

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Estimate tube- and shell-side heat transfer coefficientTube-side heat transfer coefficient:

Since we have two tubes pass, we divide the total numbers of tubes by two to find the numbers of tubes per pass, that is:

Total flow area is equal to numbers of tubes per pass multiply by tube cross sectional area:

Step 6

459=2918

= pass per Tubes

222

3avg

mm 201=16×4π

=D 4π

= (a) area sectional-cross Tube

mkg 995=ρC 33=225+40

= )(T etemperatur waterMean ⇒

26 m 0.092= )10×(201× 459= areaflow Total

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Step 6 (Cont’d)

Fig. 7: Equivalent diameter, cross-sectional areas and wetted perimeters.

27

Coefficients for water: a more accurate estimate can be made by using equations developed specifically for water. The physical properties are conveniently incorporated into the correlation. The equation below has been adapted from data given by Eagle and Ferguson (1930):

where hi = inside coefficient, for water, W/m2 °C,t = water temperature, °C,ut = water linear velocity, m/s,di = tube inside diameter, mm.

Step 6 (Cont’d)

2m skg 749 = 0.09268.9

=areaflow Totalflow waterCooling

= velocity mass Water

sm 0.75 = 995749

=)ρ( density Water)(G velocity mass Water

= )(u velocity linear Water tt

0.2i

0.8

i d u 0.02t) + (1.35 4200

=h

Nav

Rectangle

28

The equation can also be calculated using equation below; this is done to illustrate use of this method.

where hi = inside coefficient, for water, W/m2 °C,di = tube inside diameter, mmkf = fluid thermal conductivity, W/m2 °Cjh = heat transfer factor, dimensionless Re = Reynolds number, dimensionless Pr = Prandtl number, dimensionless μ = viscosity of water, N s/m2

μw = viscosity of water at wall temperature, N s/m2

Step 6 (Cont’d)

0.14

w

0.33h

f

ii )μμ

( Pr Re j=kd h

C W/m3852=16

0.75 33)×0.02+(1.35 4200=

d u 0.02t) + (1.35 4200

=h 20.2

0.8

0.2 i

0.8

i

29

Neglect

Check reasonably the previously calculated value 3812 W/m2°C with value calculated, 3852 W/m2°C.

Step 6 (Cont’d)

)μμ

(w

5.7=0.5910×0.8×10×4.2

=kμC

=Pr

14925=10×8

10×16×0.75×995=μ

ρud=Re

CmW0.59=1 Table from tyconductivi thermal FluidmmNs0.8=1 Table from )μ( waterof Viscosity

33

f

p

3

3i

2

_

_

3h

3

i

_10×3.9=j8, Fig. From302=16

10×4.83=d

L⇒

CmW 3812= 1 × 5.7 × 14925×10 × 3.9 × 10×16

0.59 = )μ

μ( Pr Re jd

k = h 20.140.333

30.14

w

0.33h

i

fi

__

30

Step 6 (Cont’d)

Fig. 8: Tube-side heat-transfer factor

31

Shell-side heat transfer coefficient:Baffle spacing: The baffle spacings used range from 0.2 to 1.0 shell

diameters.A close baffle spacing will give higher heat transfer coefficients but at the

expense of higher pressure drop.Area for cross-flow: calculate the area for cross-flow As for the hypothetical

row at the shell equator, given by:

Where pt = tube pitch (distance between the centers of two tubes, Fig. 7).do = tube outside diameter, m,Ds = shell inside diameter, m,lb = baffle spacing, m.Note: the term is the ratio of the clearance between tubes and the total distance between tube centers.

Step 6 (Cont’d)

t

bso_

ts p

l)Dd(p=A

to

_t

p)d(p

Nav

Rectangle

32

Baffle spacing:Choose baffle spacing = 0.2 Ds=0.2 894 = 178 mm

Tube pitch:Pt = 1.25 do= 1.25 20 = 25 mm

Cross-flow area:

Step 6 (Cont’d)

26_

bst

o_

ts m 0.032 = 10×178 × 894 × 25

20)(25=lDp

)d(p=A

_

33

Shell-side mass velocity Gs and the linear velocity ut:

Where Ws = fluid flow-rate on the shell-side, kg/s,ρ = shell-side fluid density, kg/m3.

Shell equivalent diameter (hydraulic diameter): calculate the shell-side equivalent diameter, see Fig. 7. For an equilateral triangular pitch arrangement:

Where de = equivalent diameter, m.

Step 6 (Cont’d)

ρG

=u

AW

=G

ss

s

ss

)d 0.917 (p d1.10

=

2πd

)4d

π21

0.87p×2p

( 4 =d 2

o_2

too

2o_

tt

e

34

Shell-side mass velocity Gs:

Shell equivalent diameter (hydraulic diameter):

Step 6 (Cont’d)

2s

ss m s

kg 868 =0.0321

×3600100000

=AW

=G velocity, Mass

mm 14.4=)20 × 0.917 (25 201.1 = )d 0.917 (p d

1.10=d 2_22

o_2

to

e

35

Choose 25 per cent baffle cut, from Fig. 9

Step 6 (Cont’d)

5.1=0.1910×0.34×10×2.84

=kμC

=Pr

36762=10×0.34

10×14.4×868=μ

dG=μ

dρu=Re

Cm W0.19=1 Table tyconductivi ThermalCkgkJ 2.84=1 Table from capacityHeat

)mmNs 0.34=1 Table from (μ methanol of Viscosity)mkg 750 =1 Table from )ρ( density Methanol

C68 =240 + 95

= etemperatur side shell Mean

33

f

p

3

3eses

23

_

_

-

3h

_10×3.3=j

36

Step 6 (Cont’d)

Fig. 9: Shell-side heat-transfer factors, segmental baffles

37

For the calculated Reynolds number, the read value of jh from Fig. 9 for 25 per cent baffle cut and the tube arrangement, we can now calculate the shell-side heat transfer coefficient hs from:

The tube wall temperature can be estimated using the following method:Mean temperature difference across all resistance: 68 -33 =35 °C

across methanol film

Mean wall temperature = 68 – 8 = 60 °Cμ = 0.37 mNs/m2

Which shows that the correction for low-viscosity fluid is not significant.

Step 6 (Cont’d)

2740= 5.1×36762×10×3.3×

10 × 1.440.19

=h term) correction viscosity(without )μμ

( Pr Rej = kdh

= Nu 313-3-s

0.14w

31 h

fes

→

0.99=)μμ

( 0.14

w

C 8=35×2740600

=ΔT×hU

=o

38

Pressure dropTube side: From Fig. 10, for Re = 14925

jf = 4.3 10-3

Neglecting the viscosity correction term:

low, could consider increasing the number of tube passes. Shell side

From Fig. 11, for Re = 36762jf = 4 10-2

Neglect viscosity correction

Step 7 (Cont’d)

m/s 1.16 =750868

= ρG

= velocity Linear s

psi) (1.1 kPa 7.2=mN7211= 20.75×995

2.5)+ )1610×4.83

( 10×4.3×(8 2=

uρ 2.5]+)μ

μ)(d

L([8jN=ΔP

2

233-

2tm-

wifpt

39

Step 7 (Cont’d)

Fig. 10: Tube-side friction factors

40

Step 7 (Cont’d)

Fig. 11: Shell-side friction factors, segmental baffles

41

could be reduced by increasing the baffle pitch. Doubling the pitch halves the shell side velocity, which reduces the pressure drop by a factor of approximately (1/2)2

This will reduce the shell-side heat-transfer coefficient by a factor of (1/2)0.8(ho Re0.8 us

0.8)ho = 2740 (1/2)0.8 = 1573 W/m2°C

This gives an overall coefficient of 615 W/m2°C – still above assumed value of 600 W/m2°C

Step 7 (Cont’d)

acceptable (10psi), kPa 68 = 4272

= ΔPs

high, too psi) (39 kPa 272= mN272019=

21.16×750

)17810×4.83

( )14.4894

( 10×4×8=2uρ

)LL

)(dD

(8j =ΔP2

232-

2t

se

sfs

∝ ∝

42

Take the thermal conductivity of cupro-nickel alloys from Table 1, 50 W/m°C, the fouling coefficients from Table 3; methanol (light organic) 5000 Wm-2°C-1, brackish water (sea water), take as highest value, 3000 Wm-2°C-1

Well above assumed value of 600 Wm-2°C

Step 8 (Cont’d)

CmW 738=U

38121

×1620

+30001

×1620

+50×21620

ln10×20+5000

1+2740

1=U

1

h1

×dd

+h1

×dd

+K2dd

lnd+h

1+h

1=U

1

2o

3-

o

ii

o

idi

o

w

i

oo

odoo

43

References

1. EAGLE, A. and FERGUSON, R. M. (1930) Proc. Roy. Soc. A. 127, 540. On the coefficient of heat transfer fromtheinternal surfaces of tube walls.

2. KERN, D. Q. (1950) Process Heat Transfer (McGraw-Hill).

3. LUDWIG, E. E. (2001) Applied Process Design for Chemical and Petroleum Plants, Vol. 3, 3rd edn (Gulf).

4. PERRY, R. H., GREEN, D.W. and MALONEY, J. O. (1997) Perry’s Chemical Engineers Handbook, 7th edn (McGraw-Hill).

5. SAUNDERS, E. A. D. (1988) Heat Exchangers (Longmans).

442/1/2009 8:11:44 AM 44

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Step 5

Table. 3: Fouling factors (coefficients), typical values

01 kern's method

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