Download - 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
-
5/26/2018 TP - Heat Exchanger Design Basis
2/42
TP - heat exchanger design.ppt2
Transport ProcessesOverall heat transfer coefficient
Contains manycombinations
May need to
transpose top
and bottom
fluids
Gives rather
conservativeestimates
-
5/26/2018 TP - Heat Exchanger Design Basis
3/42
TP - heat exchanger design.ppt3
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?
Is chemical cleaning
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
-
5/26/2018 TP - Heat Exchanger Design Basis
4/42
TP - heat exchanger design.ppt4
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
-
5/26/2018 TP - Heat Exchanger Design Basis
5/42
TP - heat exchanger design.ppt5
Transport ProcessesLog Mean Temperature Difference
i.e. less driving force with co-current
Kdeg53.6ln40ln70
4070TLM
Kdeg39.1ln10ln100
10100TLM
120
809040
70
20
120
80
90
20
10010
-
5/26/2018 TP - Heat Exchanger Design Basis
6/42
TP - heat exchanger design.ppt6
Transport ProcessesLog Mean Temperature Difference
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?
-
5/26/2018 TP - Heat Exchanger Design Basis
7/42
TP - heat exchanger design.ppt7
Transport ProcessesLog Mean Temperature Difference
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
-
5/26/2018 TP - Heat Exchanger Design Basis
8/42
TP - heat exchanger design.ppt8
Transport ProcessesLog Mean Temperature Difference
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
-
5/26/2018 TP - Heat Exchanger Design Basis
9/42
TP - heat exchanger design.ppt9
Transport ProcessesLog Mean Temperature Difference
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
-
5/26/2018 TP - Heat Exchanger Design Basis
10/42
TP - heat exchanger design.ppt10
Transport ProcessesLog Mean Temperature Difference
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
-
5/26/2018 TP - Heat Exchanger Design Basis
11/42TP - heat exchanger design.ppt11
Transport ProcessesLog Mean Temperature Difference
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
-
5/26/2018 TP - Heat Exchanger Design Basis
12/42TP - heat exchanger design.ppt12
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
-
5/26/2018 TP - Heat Exchanger Design Basis
13/42TP - heat exchanger design.ppt13
Transport ProcessesFouling
Standard formula for U assumes clean surfaces In reality, surface fouling increases thermal
resistanceExternal fouling layer
Internal fouling layer
-
5/26/2018 TP - Heat Exchanger Design Basis
14/42TP - heat exchanger design.ppt14
Transport ProcessesFouling
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
-
5/26/2018 TP - Heat Exchanger Design Basis
15/42TP - heat exchanger design.ppt15
Transport ProcessesFouling
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
-
5/26/2018 TP - Heat Exchanger Design Basis
16/42TP - heat exchanger design.ppt16
Transport ProcessesFouling
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 (
-
5/26/2018 TP - Heat Exchanger Design Basis
17/42TP - heat exchanger design.ppt17
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
-
5/26/2018 TP - Heat Exchanger Design Basis
18/42TP - heat exchanger design.ppt18
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
-
5/26/2018 TP - Heat Exchanger Design Basis
19/42TP - heat exchanger design.ppt19
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
-
5/26/2018 TP - Heat Exchanger Design Basis
20/42TP - heat exchanger design.ppt20
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
21/42TP - heat exchanger design.ppt21
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
22/42TP - heat exchanger design.ppt22
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T P
-
5/26/2018 TP - Heat Exchanger Design Basis
23/42TP - heat exchanger design.ppt23
Do we have
severe
expansion
stresses?
ie. are the
temperatures
greatly
different to
ambient?
Yes
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
m2
K/W?
Is chemical cleaning
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T P
-
5/26/2018 TP - Heat Exchanger Design Basis
24/42TP - heat exchanger design.ppt24
Are bellows
allowed?
No reason
why not
Yes
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
m2
K/W?
Is chemical cleaning
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T P
-
5/26/2018 TP - Heat Exchanger Design Basis
25/42TP - heat exchanger design.ppt25
High
shellsidefouling?
0.0001 0.00035
m2K/W?
High tubeside
fouling > 0.00035
m2
K/W?
Is chemical cleaning
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T t P
-
5/26/2018 TP - Heat Exchanger Design Basis
26/42TP - heat exchanger design.ppt26
High
tubesidefouling?
0.00018 0.00035
m2K/W?
High tubeside
fouling > 0.00035
m
2
K/W?
Is chemical cleaning
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T t P
-
5/26/2018 TP - Heat Exchanger Design Basis
27/42TP - heat exchanger design.ppt27
Is tube accessrequired
without
dismantling?
Unlikely
unless we
had solids or
other thingsthat may
block
No
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
m
2
K/W?
Is chemical cleaning
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T t P
-
5/26/2018 TP - Heat Exchanger Design Basis
28/42TP - heat exchanger design.ppt28
BEM
exchanger
A fixed
tubesheet
design
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
m
2
K/W?
Is chemical cleaning
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
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T t P
-
5/26/2018 TP - Heat Exchanger Design Basis
29/42TP - heat exchanger design.ppt29
Transport ProcessesFirst design of a shell-and-tube heat exchanger
T t P
-
5/26/2018 TP - Heat Exchanger Design Basis
30/42
TP - heat exchanger design.ppt30
Transport ProcessesFirst design of a shell-and-tube heat exchanger
Choose
best
case for
each
Usuggested
=500 W/m2K
2m.38824787.0500
580140A
T t P
-
5/26/2018 TP - Heat Exchanger Design Basis
31/42
TP - heat exchanger design.ppt31
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
32/42
TP - heat exchanger design.ppt32
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
33/42
TP - heat exchanger design.ppt33
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
34/42
TP - heat exchanger design.ppt34
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
35/42
TP - heat exchanger design.ppt35
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
36/42
TP - heat exchanger design.ppt36
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
37/42
TP - heat exchanger design.ppt37
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
38/42
TP - heat exchanger design.ppt38
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
39/42
TP - heat exchanger design.ppt39
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
40/42
TP - heat exchanger design.ppt40
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
41/42
TP - heat exchanger design.ppt41
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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
-
5/26/2018 TP - Heat Exchanger Design Basis
42/42
Transport ProcessesFirst design of a shell-and-tube heat exchanger
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