tp - heat exchanger design basis

5/26/2018 TP - Heat Exchanger Design Basis

1/42

TP - heat exchanger design.ppt1

Transport ProcessesOverall heat transfer coefficient

From previous studies (CPP module):Q = U A TLM

oo

iioi

ii h

1

d

d

2k

)ddln(d

h

1

U

1

Some typical U values (all in W/m2K):steam/water: 6000 to 18000

water/water: 850 to 1700

steam condenser (water in tubes) 1000 to 6000

ammonia condenser (water in tubes) 800 to 1400

alcohol condenser (water in tubes) 250 to 700

finned tube (air outside, water inside) 25 to 50

deduce others from charts


2/42


Transport ProcessesOverall heat transfer coefficient

Contains manycombinations

May need to

transpose top

and bottom

fluids

Gives rather

conservativeestimates


3/42


Transport ProcessesChoosing right shell-and-tube type

Decision as toTEMA code

used depends on

fluids used

Shell& tube

exchangers

Severe thermal

exapansion stresses?

Are bellows

allowed?

Is chemical cleaning

possible?

High shellside

fouling > 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m2K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within

range for lantern ring?

Is F correction factor

< 0.75?Are there more than

2 passes?

Are F or multi shells

allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No

Is tubeside fouling >

0.00035 m2K/W?

Do we require tube access

without disturbing connections?

Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


4/42


Transport ProcessesLog Mean Temperature Difference

e.g. find TLMfor both co-current & counter-current flow

Fluid A Tin = 120 Tout = 90C

Fluid B tin = 20 tout = 80C

temperature

T1

T2

hot fluid

Tin

Tout

cold fluid

tout

tin

21

21LM

TlnTln

TTT


5/42



i.e. less driving force with co-current

Kdeg53.6ln40ln70

4070TLM

Kdeg39.1ln10ln100

10100TLM

120

809040

70

20

120

80

90

20

10010


6/42



i.e. more driving force than either of the first two

same value for both co and counter-current

Kdeg65.5ln40ln100

40100TLM

120

80

20

100 40

Now make fluid A condensing steam.What happens?


7/42



Why a log mean temperature difference ratherthan any other?

Consider point along heat exchanger tube:

T TdT

t + dt t

Area = dA

Heat = dQ

At this point: T = Tt

d(T) = dTdtalso dQ = -mhCphdT = mcCpcdt (sensible heat change)

/ / / /h h c c h h c c

dQ dQ 1 1d T dQ

m Cp m Cp m Cp m Cp


8/42



Hence:

/ /

h h c c

d T tdQ =

1 1

m Cp m Cp

= U(Tt).dA

/ /h h c c

d T t 1 1U.dAT t m Cp m Cp

assuming constant Cph& Cpc:

2 2 / /

1 1 h Ph c Pc

T t 1 1-ln UAT t m C m C

But/

h Ph

1 2

Qm C

T T

/

c Pc

2 1

Qm C

t t


9/42



Giving 2 2 1 2 2 11 1

T t T T t t-ln UAT t Q Q

2 2

1 1 2 21 1

T t UAln T t T t

T t Q

2 2 1 12 2

1 1

T t T tQ UA

T tlnT t

= UA.TLM

Counter-current derivation also true for co-current flow

Co-current flow rarely used in practice


10/42



Shell & tube exchanger NOT in true counter-

current flow if there is more than one tube-side pass

TLM< TLMfor pure counterflow

In this case, calculate TLMas if for counterflow.Multiply by correction factor F to give true value:

1R1R

1R1RS2ln1R1RS2ln

SR1lnS1lnF

2

22

tube(in)tube(out)

shell(out)shell(in)

TT

TTR

tube(in)shell(in)

tube(in)tube(out)

TT

TTS


11/42TP - heat exchanger design.ppt11


Alternatively,use charts to

evaluate F.

F should behigh (0.75 to

1.0) for

efficient

operation

If F > 0.75 inachievable, use single tube-side pass

F then becomes 1



Transport ProcessesDuties

For sensible heat (i.e. no boiling or condensing)QH= mHCPh(Tin - Tout)

QC= mCCPc(tout - tin)

For latent heat (boiling and/or condensing)Q = m fg

For perfect balance, QH= QC

i.e. heat lost by hot fluid = heat gained by cold

fluid

In reality, heat losses always occur



Transport ProcessesFouling

Standard formula for U assumes clean surfaces In reality, surface fouling increases thermal

resistanceExternal fouling layer

Internal fouling layer




Occurs for a number of reasons

Slimy film through microbial

activity in water

Precipitation of dissolved salts

Reaction of fluid alone (eg.

polymerisation of hydrocarbons)

Reaction of surface with fluid(eg. corrosion)

Freezing

Silt




Dynamic problem by nature

Fouling

resistance

Time

Can be held in check by

Regular cleaning High flow velocities

Low temperatures

Use of special devices and/or chemical additives

TEMA and others usually

quote this assymptotic value




Fouling resistances incorporated into formula:

Designers assume static Rfo& Rfi. A few examples:

FLUID Rf(m2K/W)

Seawater & treated boiler water (50C) 2 10-4

River water (



Transport ProcessesMechanical considerations of shell-and-tube heat exchanger

design

Tubes held in place by tube sheetswith drilled holes

Holes align the tubes in square or

triangular arrangement

Distance between centres of adjacent

tubes = tube pitch

Outer diameters:16, 20, 25, 30, 38, 50 mm, 2mm thick

Lengths:

1.83, 2.44, 3.66, 4.88, 6.10, 7.32 metres

T t P



Transport ProcessesMechanical considerations of shell-and-

tube heat exchanger design

Baffle spacing: minimum = Ds 5 (but > 5 cm)maximum = 74do

0.75(but < Ds)

Baffle cut (segment opening height Ds) = 0.25 to 0.40

eg. segmental baffle inside 1.00 m shell25% means segment 25cm high removed

Smaller cut leaves smaller hole

Higher shell-side film coefficientGreater shell side pressure drop

0.25 m

T P



Transport ProcessesFirst design of a shell-and-tube heat exchanger

Calculate duty Q (add 10% to include losses and

errors)

Find appropriate fouling resistances

Choose side for each fluid (based on fouling,corrosion and pressure)

Choose type of exchanger from TEMA tree

Calculate all temperatures TLM& F

Keep things simple to start with; assume 4.88m

tubes, do= 20 mm, 2 tube side passes (NP=2)

T P




Cool 5.0 kg/s of ethylene glycol from 370 to 330K with

cooling water from 283 to 323K

Ethylene glycol at 350K (average) has following

properties

k = 0.261 W/m.K Cp = 2637 J/kg.K

= 0.00342 Pa.s = 1079.0 kg/m3

Giving Pr = (26370.00342)/0.261 = 34.6

Anticipate fouling resistance of Rf= 0.00018 m2K/W

Duty is Q = 5.0 2637 (370330) = 527 400 Watts

Aim to transfer 580 140 W

T P




Water at 303K (average) has following properties

k = 0.618 W/m.K Cp = 4179 J/kg.K

= 0.000797 Pa.s = 995.6 kg/m3

Giving Pr = (41790.000797)/0.618 = 5.39 Anticipate fouling resistance of Rf= 0.0001 m

2K/W

Water fouls less and is on shell-side

We need water flowrate

inout ttCpQ

m

15506.3

8322331794

527400

3.16 kg/s water on shell-side

T P



Problemwe cannot calculate a log mean Solutiona log mean is just an average after all

What is average of 47 and 47?

?ln47ln47

7474TLM

370

32333047

47

283

1703303

233832R

4598.0703832

703330S

T = 47, F = 0.87


T P



Do we have

severe

expansion

stresses?

ie. are the

temperatures

greatly

different to

ambient?

Yes

Shell& tube

exchangers

Severe thermal


Are bellows

allowed?


possible?

High shellside

fouling > 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m2

K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within




2 passes?


allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No


0.00035 m2K/W?



Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


T P



Are bellows

allowed?

No reason

why not

Yes

Shell& tube

exchangers

Severe thermal


Are bellows

allowed?


possible?

High shellside

fouling > 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m2

K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within




2 passes?


allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No


0.00035 m2K/W?



Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


T P



High

shellsidefouling?

0.0001 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m2

K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within




2 passes?


allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No


0.00035 m2K/W?



Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


T t P



High

tubesidefouling?

0.00018 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m

2

K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within




2 passes?


allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No


0.00035 m2K/W?



Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


T t P



Is tube accessrequired

without

dismantling?

Unlikely

unless we

had solids or

other thingsthat may

block

No

Shell& tube

exchangers

Severe thermal


Are bellows

allowed?


possible?

High shellside

fouling > 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m

2

K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within




2 passes?


allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No


0.00035 m2K/W?



Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


T t P



BEM

exchanger

A fixed

tubesheet

design

Shell& tube

exchangers

Severe thermal


Are bellows

allowed?


possible?

High shellside

fouling > 0.00035

m2K/W?

High tubeside

fouling > 0.00035

m

2

K/W?


possible?

Removable

bundle design

Fixed

tubesheet

Is interstream

leakage allowed?

Are T & P within




2 passes?


allowed?

Frequency of bundle

removal

AEL

AEM

BEM

No NoYes

Yes No

AEU

AFU

AEU AFU

No NoYes

Yes No

AEP

BEP

No NoYes

Yes No

AEW

BEW

No NoYes

Yes No

AET

BET

No NoYes

Yes No

AES

BES

No NoYes

Yes No


0.00035 m2K/W?



Yes No

Yes

Yes

Yes

No

No

Yes Yes

No

Yes

No

Yes

No

Yes

No

Yes

No

LowHigh

Yes


T t P




T t P


30/42



Choose

best

case for

each

Usuggested

=500 W/m2K

2m.38824787.0500

580140A

T t P


31/42



Use NTto fix estimated coefficient as Uestimate

L = 4.88 m, do= 20 mm:

Area of one tube = 4.88 0.020 = 0.3066m2

Number of tubes needed = 28.38 0.3066 = 92.54

Obviously, should be an integer

Round up here, as 92 tubes means U>500

KW/m6.4974787.03066.093

580140U 2estimate

Aim to build exchanger with U = 497.6 W/m2K

T t P


32/42



Calculate tube side velocity

T

2

it

P/t

tNd

N4mu

Suggested ranges

Tubeside process liquids 1 to 2 m/s(up to 4 m/s if fouling risk)

Tubeside water 1.5 to 2.5 m/s

Vacuum gases/vapours 50 to 70 m/sAtmospheric pressure gases/vapours 10 to 30 m/s

High pressure gases/vapours 5 to 10 m/s

Note: di= 0.0202(0.002) = 0.016 m

m/s4956.0

39016.01079

25.042

T t P


33/42



Lower than the suggested 1 to 2 m/s

If tube side passes tripled to NP= 6, ut= 1.487 m/s7506

00342.0

016.0487.11079Ret

Use Nusselt turbulent correlation for forced

convection in tubes:

Nu = 0.036 (Re)

0.8

Pr

0.33

(di

L)

0.055

Nu = 0.036 (7506)0.8 (34.6)0.33 (0.016 4.88)0.055

Nu = 106.6 = hidik

hi= 106.60.261 0.016 = 1739 W/m2K

Transport Processes


34/42



Find tube bundle diameter DBthus:

assume tube pitch (pt)= 1.25 do

NP 1 2 4 6 8

K1 0.215 0.156 0.158 0.0402 0.0331

n1 2.207 2.291 2.263 2.617 2.643

1n1

1

ToB

K

NdD

NT= 93, NP = 6, pt= 1.25 0.020 = 0.025 m

m386.0

0402.0

93020.0

2.6171

So tube bundle is 0.386 m in diameter, but shell

needs to be wider still

Transport Processes


35/42



Use graph to find

clearance between

bundle and shell

diameter DS

12mm added so

DS= 0.386 + 0.012 =

0.398 m

Number of tubes atequator n = DB pt

tubes4.15

025.0

0.386n

Transport Processes


36/42



Find minimum baffle spacing

0.398 5 = 0.0796 m

Divide tube length by bminto find number of chambers

created by baffles

4.88 0.0796 = 61.3

Number of chambers should be integer i.e. round down

Actual baffle spacing b = tube length number ofchambers

b = 4.88 61 = 0.08 m

Transport Processes


37/42



Calculate equivalent diameter of shell-side fluid

(De)

perimeterwetted

areaflow4De

dopt

m0198.0020.0020.0

0.0254D

2

e

So for do

= 0.020 and pt

= 0.025

o

o

2

t

o

2o

2t

dd

4p

d

d4

p4

circle1ofncecircumfere

areacircleareasquare4

Transport Processes


38/42



Calculate cross-section for flow (S) for

hypothetical tube row mid-shell of n tubes

DS

b

pt do

S = b(DSndo)

= 0.08 [0.39815(0.02)]= 7.84103m2

Choose tube material

if stainless steel, k = 16 W/m.K

Transport Processes


39/42



Calculate shell side velocities

s

/s

sS

mu

Suggested ranges

Atmospheric pressure gases/vapours 10 to 30 m/sVacuum gases/vapours 50 to 70 m/s

High pressure gases/vapours 5 to 10 m/s

Shell-side liquids 0.3 to 1.0 m/s

Falls within accepted range

m/s4048.06.99500784.0

3.16

10013

000797.0

0198.04048.06.959Res

Transport Processes


40/42



Using Grimison correlation where C = 0.348 andn = 0.592

Nu = 1.130.348 (Re)0.592 Pr0.33

Nu = 0.39324 (10013)

0.592

(5.39)

0.33

Nu = 160.11 = hoDek

ho= 160.110.618 0.0198 = 4997 W/m2K

Now have all information needed for U-value

fo

oo

iioifi

iD

Rh

1

d

d

2k

ddlndR

h

1

U

1

Transport Processes


41/42



Inside

resistance K/Wm10550.700018.017391 24

K/Wm10116.1

162

0.0160.02ln0.016 24

K/Wm10401.20001.04997

1

0.02

0.016 24

Wall

resistance Outside

resistance Overall resistance(7.550 + 1.116 + 2.401)104= 1.1067103m2K/W

Overall heat transfer coefficient

1 (1.1067103) = 903.6 W/m2K

Transport Processes


42/42


Here, 903.6 497.6 W/m2K, over 81% out

Main resistance is tubeside, so ponder options

If UactualUestimate (30%) then do any of the

following:

A by reducing tube length (Uestimate)

A by increasing tube length/diameter ( Uestimate)

number of tube-side passes (Uactual)

number of shell-side baffles (Uactual)

If possible, alter the side where the MAIN

resistance lies

tp - heat exchanger design basis

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