01 kern's method.

Kern's method for Tube and Shell Heat Exchanger Design

2. Designing Shell-and-Tube Heat Exchangers Using Softwares
Lecture 1: Kern’s Method
By:
MajidHayati
University of Kashan, Kashan, I.R. IRAN
2014

3. 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. 4
Objectives
This lecture on designing shell-and-tube HEsserves 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 inan
example

5. 5
Introduction to Kern’s method
Kern’s was based on experimental work on commercial
exchanger
Advantages:
Giving reasonably satisfactory prediction of the heat -transfer
coefficient for standard design
Simple to apply
Accurate 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 dropis less satisfactory, as pressure
drop is more affected by leakage and bypassing than heat transfer
The method does not take account of the bypassand leakage
streams

6. 6
Design procedure for shell-and-tube heat exchangers
(Kern’s method)
Start
Calculate tube number
Calculate shell diameter
Assume value of overall
coefficient U
o,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 parameters
Compare 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 duty
Make energy balance if needed
Calculate unspecified flow rates
Calculate ΔT
LMTD
and ΔT
M
Fig. 1: Algorithm of design procedure
Determine fouling factors
Step 5
Yes
No
NoYes
Start
Collect physical properties and
HE specifications
Step 1
Collect physical properties and
HE specifications
Step 1
Step 2
Define duty
Make energy balance if needed
Step 2
Define duty
Make energy balance if needed
Calculate unspecified flow rates
Calculate ΔT
LMTD
and ΔT
M
Assume value of overall
coefficient U
o,ass
Step 3
Assume value of overall
coefficient U
o,ass
Step 3
Calculate tube number
Calculate shell diameter
Step 4
Calculate tube number
Calculate shell diameter
Step 4
Determine fouling factors
Step 5
Determine fouling factors
Step 5
Estimate tube-and shell-side
heat transfer coefficient
Step 6
Estimate tube-and shell-side
heat transfer coefficient
Step 6
Estimate tube-and shell-side
pressure drop
Step 7
Estimate tube-and shell-side
pressure drop
Step 7
YesYes No
Determine overall heat transfer
coefficient
Step 8
Determine overall heat transfer
coefficient
Step 8
No
Yes
Accept all design parameters

7. 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. 8
Collect physical properties and HE specifications:
Physical properties
Solution: Step 1
WaterMethanolPhysical properties at
fluid mean temperature
4.22.84C
p
(Kj/Kg °C)
0.80.34μ(mNs/m
2
)
0.590.19k
f
(W/m °C)
995750ρ(Kg/m
3
)
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).

9. 9
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. 10
Tube Pattern Applications
The triangular and rotated square patterns give higher heat-
transferrates, but at the expense of a higher pressure drop
than 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. 11
Define duty, Make energy balance if needed
To 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×
3600
100000
= )T(TCm = Q = loadHeat
21ph
h
•
--
Fig. 3: Streams definitions.

12. 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 C
Q
= m=flow waterCooling
_
12cP
.
c _
Thewell-known “logarithmic mean”temperature difference (LMTD or lm) is
calculated by:
C31
25)(40
40)(95
ln
25)(4040)(95
tT
tT
ln
)t(T)t(T
ΔT
12
21
1221
LMTD











13. 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:
ΔT
m
= true temperature difference,
F
t
= the temperature correction factor.

14. 14
Temperature Correction Factor
The correction factor (F
t
) 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
tT
tt
S
tt
TT
R







15. 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. 16
Step 2 (Cont’d)
From Fig. 4, the correction factor (F
t
) is 0.85.
21.0
2595
2540
)t(T
)t(t
S
67.3
2540
4095
tt
TT
R
11
12
12
21














C 26 310.85ΔT FΔT
LMTDtm



17. 17
Assume value of overall coefficient U
o,ass
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/m
2
°C
Step 3

18. 18
Step 3 (Cont’d)
Table 2: Typical overall coefficients

19. 19
Step 3 (Cont’d)
Fig. 5: Overall coefficients (join process side duty to service side and read U from centre scale)

20. 20
Step 4
Calculate tube number, Calculate shell diameter
Provisional area:
So, the total outside surface area of tubes is 278 m
2
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 = 4.83x 20 x 10
-3
π= 0.303 m
2
2
3
M
m 278=
62× 600
10× 4340
=
TΔ U
Q
=A
ft 16 in.
4
3

918
0.303
278
tubeone of area surface Outside
area) al(Provision tubesof area surface outside Total
tubesof Numbers 

21. 21
Step 4 (Cont’d)
An estimate of the bundle diameter D
b
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 D
b
= bundle diameter in mm, do = tube outside diameter in
mm., N
t
= number of tubes.
As the shell-side fluid is relatively clean use 1.25 triangular pitch.
So, for this example:
1n
1
1
t
ob
)
K
N
(d D
mm 826 )
0.249
918
( 20 D diameter Bundle
2.207
1
b


22. 22
Step 4 (Cont’d)
Table 3: Constants K
1
and n
1

23. 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 (D
s
):
D
s
= 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)].

24. 24
Step 4 (Cont’d)
Fig. 6: Shell-bundle clearance

25. 25
Estimate tube-and shell-side heat transfer coefficient
Tube-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=
2
918
= pass per Tubes
222
3
avg
mm 201=16×
4
π
=D
4
π
= (a) area sectional-cross Tube
mkg 995=ρC 33=
2
25+40
= )(T etemperatur waterMean ⇒

26
m 0.092= )10×(201× 459= areaflow Total

26. 26
Step 6 (Cont’d)
Fig. 7: Equivalent diameter, cross-sectional areas and wetted perimeters.

27. 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 givenby
Eagle and Ferguson (1930):
where h
i
= inside coefficient, for water, W/m
2
°C,
t = water temperature, °C,
u
t
= water linear velocity, m/s,
d
i
= tube inside diameter, mm.
Step 6 (Cont’d)
2
m skg 749 =
0.092
68.9
=
areaflow Total
flow waterCooling
= velocity mass Water
sm 0.75 =
995
749
=
)ρ( density Water
)(G velocity mass Water
= )(u velocity linear Water
t
t
0.2
i
0.8
i
d
u 0.02t) + (1.35 4200
=h

28. 28
The equation can also be calculated using equation below; this is
done to illustrate use of this method.
where h
i
= inside coefficient, for water, W/m
2
°C,
d
i
= tube inside diameter, mm
k
f
= fluid thermal conductivity, W/m
2
°C
j
h
= heat transfer factor, dimensionless
Re = Reynolds number, dimensionless
Pr = Prandtlnumber, dimensionless
μ= viscosity of water, N s/m
2
μ
w
= viscosity of water at wall temperature, N s/m
2
Step 6 (Cont’d)
0.14
w
0.33
h
f
ii
)
μ
μ
( Pr Re j=
k
d h
C W/m3852=
16
0.75 33)×0.02+(1.35 4200
=
d
u 0.02t) + (1.35 4200
=h
2
0.2
0.8
0.2
i
0.8
i


29. 29
Neglect
Check reasonably the previously calculated value 3812 W/m
2
°C with
value calculated, 3852 W/m
2
°C.
Step 6 (Cont’d)
)
μ
μ
(
w
5.7=
0.59
10×0.8×10×4.2
=
k
μC
=Pr
14925=
10×8
10×16×0.75×995
=
μ
ρud
=Re
CmW0.59=1 Table from tyconductivi thermal Fluid
mmNs0.8=1 Table from )μ( waterof Viscosity
33
f
p
3
3
i
2
_
_

3
h
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 j
d
k
= h
20.140.333
3
0.14
w
0.33
h
i
f
i
_
_


30. 30
Step 6 (Cont’d)
Fig. 8: Tube-side heat-transfer factor

31. 31
Shell-side heat transfer coefficient:
Baffle spacing:The baffle spacingsused 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 A
s
for the hypothetical
row at the shell equator, given by:
Where p
t
= tube pitch (distance between the centers of two tubes, Fig. 7).
d
o
= tube outside diameter, m,
D
s
= shell inside diameter, m,
l
b
= 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
_
t
s
p
l)Dd(p
=A
t
o
_
t
p
)d(p

32. 32
Baffle spacing:
Choose baffle spacing = 0.2 D
s
=0.2 894 = 178 mm
Tube pitch:
P
t
= 1.25 d
o
= 1.25 20 = 25 mm
Cross-flow area:
Step 6 (Cont’d)
26
_
bs
t
o
_
t
s
m 0.032 = 10×178 × 894 ×
25
20)(25
=lD
p
)d(p
=A
_

33. 33
Shell-side mass velocity G
s
and the linear velocity u
t
:
Where W
s
= fluid flow-rate on the shell-side, kg/s,
ρ= shell-side fluid density, kg/m
3
.
Shell equivalent diameter (hydraulic diameter):calculate the shell-
side equivalent diameter, see Fig. 7. For an equilateral triangular
pitch arrangement:
Where d
e
= equivalent diameter, m.
Step 6 (Cont’d)
ρ
G
=u
A
W
=G
s
s
s
s
s
)d 0.917 (p
d
1.10
=
2
πd
)
4
d
π
2
1
0.87p×
2
p
( 4
=d
2
o
_2
t
oo
2
o_
t
t
e

34. 34
Shell-side mass velocity G
s
:
Shell equivalent diameter (hydraulic diameter):
Step 6 (Cont’d)
2
s
s
s
m s
kg
868 =
0.032
1
×
3600
100000
=
A
W
=G velocity, Mass
mm 14.4=)20 × 0.917 (25
20
1.1
= )d 0.917 (p
d
1.10
=d
2_22
o
_2
t
o
e

35. 35
Choose 25 per cent baffle cut, from Fig. 9
Step 6 (Cont’d)
5.1=
0.19
10×0.34×10×2.84
=
k
μC
=Pr
36762=
10×0.34
10×14.4×868
=
μ
dG
=
μ
dρu
=Re
Cm W0.19=1 Table tyconductivi Thermal
CkgkJ 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 =
2
40 + 95
= etemperatur side shell Mean
33
f
p
3
3
eses
2
3
_
_
-



3
h
_
10×3.3=j

36. 36
Step 6 (Cont’d)
Fig. 9: Shell-side heat-transfer factors, segmental baffles

37. 37
For the calculated Reynolds number, the read value of j
h
from Fig. 9
for 25 per cent baffle cut and the tube arrangement, we can now
calculate the shell-side heat transfer coefficient h
s
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 °C
μ= 0.37 mNs/m
2
Which shows that the correction for low-viscosity fluid is not significant.
Step 6 (Cont’d)
2740=
5.1×36762×10×3.3×
10 × 1.44
0.19
=h term) correction viscosity(without )
μ
μ
( Pr Rej =
k
dh
= Nu
313-
3-s
0.14
w
31
h
f
es
→
0.99=)
μ
μ
(
0.14
w
C 8=35×
2740
600
=ΔT×
h
U
=
o


38. 38
Pressure drop
Tube side: From Fig. 10, for Re = 14925
j
f
= 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 = 36762
j
f
= 4 10
-2
Neglect viscosity correction
Step 7 (Cont’d)
m/s 1.16 =
750
868
=
ρ
G
= velocity Linear
s
psi) (1.1 kPa 7.2=mN7211=
2
0.75×995
2.5)+ )
16
10×4.83
( 10×4.3×(8 2=
uρ
2.5]+)
μ
μ
)(
d
L
([8jN=ΔP
2
23
3-
2
tm-
wi
fpt

39. 39
Step 7 (Cont’d)
Fig. 10: Tube-side friction factors

40. 40
Step 7 (Cont’d)
Fig. 11: Shell-side friction factors, segmental baffles

41. 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
(h
o
Re
0.8
u
s
0.8
)
h
o
= 2740 (1/2)
0.8
= 1573 W/m
2
°C
This gives an overall coefficient of 615 W/m
2
°C –still above assumed value of
600 W/m
2
°C
Step 7 (Cont’d)
acceptable (10psi), kPa 68 =
4
272
= ΔP
s
high, too psi) (39 kPa 272=
mN272019=
2
1.16×750
)
178
10×4.83
( )
14.4
894
( 10×4×8=
2
uρ
)
L
L
)(
d
D
(8j =ΔP
2
23
2-
2
t
se
s
fs
∝ ∝

42. 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
3812
1
×
16
20
+
3000
1
×
16
20
+
50×2
16
20
ln10×20
+
5000
1
+
2740
1
=
U
1
h
1
×
d
d
+
h
1
×
d
d
+
K2
d
d
lnd
+
h
1
+
h
1
=
U
1
2
o
3-
o
ii
o
idi
o
w
i
o
o
odoo


43. 43
References
1.EAGLE, A. and FERGUSON, R. M. (1930) Proc. Roy. Soc.
A. 127, 540. On the coefficient of heat transfer fromthe
internal 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).

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

45. 45
Step 5
Table. 3: Fouling factors (coefficients), typical values