applications of the principles of heat transfer to design of heat exchangers

1
Momentum and Heat Transfer
Applications of the principles of heat transfer to design of
heat exchangers
Dr. Anand V. Patwardhan
Professor of Chemical Engineering
Institute of Chemical Technology
Nathalal M. Parikh Road
Matunga (East), Mumbai-400019
av.patwardhan@ictmumbai.edu.in; avpuict@gmail.com; avpiitkgp@gmail.com

2
Temperature driving force (log-mean ΔT)
Consider a Double Pipe Heat Exchanger (DPHE):
Cold stream OUT
T = Tc2
Cold stream IN
T = Tc1
Hot stream IN
T = Th1
Hot stream OUT
T = Th2
dll
Th
Tc
Tc
L

3
Macroscopic (OVERALL) steady state energy balance
for each stream is given by (for negligible changes in
kinetic and potential energy):
( )
( )Q
Q
m H H
h
m H
h
H
c c c
h1
1 c2
h2
= −
= −
For no heat loss to surroundings, Qh = Qc.
For incompressible liquids, ideal gases, and for
constant cP:
( )
( )Q m c T T
h h
Q m
P
c T T
c c Pc c
h h2 1
c
h
1 2
Q
c
= −
== −

4
Differential steady state heat balance for hot stream:
( ) ( )( )lU 2m c dT T
h P
R d
h
T
o o ch h
= π −
RO = outer radius of the inner pipe
UO = overall heat transfer coefficient based on RO
Rearranging the above differential balance:
( )
( ) ( )
( )
( ) ( )
l
For
l
For
cold
hot stream
stream:
: ...
...
T
c
U 2 R dd
U 2 R ddT
T o
o oh 1
T m
o
c
h h
c 2
T m c
c
P
T
h c
h
Pc
−
π
π
=
−
=

5
Subtracting Equation (2) from Equation (1) gives the
relation between (Th–Tc) as a function of l (length of
heat exchanger):
( )
( ) ( ) l
T 1h
T m c
d T 1c
h h Ph
U 2 R d
o oT m c
c c Pc
−
−
⎛ ⎞
= π⎜ ⎟⎜ ⎟−
⎝ ⎠
Assuming UO independent of length l, and integrating
between length zero (∆T1=Th1–Tc2) to L (∆T2=Th2–Tc1) :
( )
T 1h2
T m c
T 1c1ln U 2 R L
o oT m c
c2 c Pch1 h Ph
− ⎛ ⎞
= π⎜ ⎟⎜ ⎟−
⎝ ⎠
−

6
( ) ( )
( )
( ) ( )
( )
( )
Q
hQ m c T T m c
h h P
Q
cQ m c T T m c
c
h h2 h1 h Ph
c Pc c1 c2 c Pc
Now,
T T
c1 c2
T T1 c
T T
h2 h1
T T1 h2 h1m c
h Ph T T m c
h2 h1 h Ph
1 c2
m c Q
c Pc c
Q
c
Q
c
= − ⇒ =
= − ⇒
−
−
= =
−
⇒ ⇒
=
−
−
⇒ =

7
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
T TT
h2 h1h2
T
h1
T T
h2 h1
T TT c1 c2c1ln U 2 R L
o oT Q Q
c2 c c
T T
c1 c2U 2 R L
o o Q
c
T TT T
h2 h1
T
h2
T
h1
c1 c2Q U 2 R L
c o o T
c1ln
T
c2
⎛ ⎞−−
⎜ ⎟= π
⎜ ⎟−
⎝ ⎠
⎛ ⎞− − −
⎜
−
−
⎟= π
⎜ ⎟
⎝ ⎠
⎧ ⎫− − −
⎪ ⎪= π ⎪ ⎪−⎨ ⎬
⎪ ⎪−⎪ ⎪⎩ ⎭
⇒
⇒

8
( )
( ) ( )
( )
( ) ( )
( )( )
( )( )
log-mean
Similarly, log-mean
T T
2 1
T
2ln
T
1
T T
1 2
T
1ln
U 2 R L
o o
U 2 R L
o o
Q U 2 R L
c o o
Q 2 L
c
U
i
T
T
TR
i
2
⎧ ⎫Δ − Δ
⎪ ⎪⎪ ⎪Δ⎛ ⎞⎨ ⎬
⎜ ⎟⎪ ⎪⎜ ⎟Δ
⎪ ⎪⎝ ⎠⎩ ⎭
⎧ ⎫Δ − Δ
⎪ ⎪⎪ ⎪Δ⎛ ⎞⎨ ⎬
⎜ ⎟⎪ ⎪⎜ ⎟Δ
⎪ ⎪⎝ ⎠⎩ ⎭
⇒ Δ
Δ
= π
= π
= π
= π

9
Temperature profile for co-current DPHE
Th1
Th2
Tc1
Tc2
ΔT1
ΔT2
Th
Tc
Length coordinate →
Generally, less corrosive liquid flows through the outer
pipe (through the annulus). Why?

10
Temperature profile for counter-current DPHE
Th2
Th1
Tc1
Tc2ΔT1
ΔT2
Th
Tc
Length coordinate →

11
Shell and tube heat exchangers (STHE)
Most widely used heat transfer equipment.
Large heat transfer area in a relatively small volume
Fabricated from alloy steels to resist corrosion and used
for heating, cooling, and for condensing wide range of
vapours
Various types of construction of STHE, and a typical
tube bundle is shown in the Figures on the following
slides.

12
1,4-STHE
(1 shell-side pass, 4 tube-side passes)
STHE with fixed tube sheet
Exchanger
support Shell-side
inlet
Shell-side
outlet
Baffles Tube-side
outlet
Tube-side
inletFixed
tube sheet

13
The simplest type of STHE is “fixed tube sheet type”:
Fixed tube sheets at both ends into which the tubes
are welded and their ends are expanded (flared).
Tubes can be connected so that the internal fluid
can make several passes, which results into a high
fluid velocity for a given heat transfer area and fluid
flow rate.
Shell-side fluid is made to flow in a ZIG-ZAG
manner across the tube bundle by fitting a series of
baffles along the tube length.

14
Baffles can be segmental with ≈ 25% cutaway (see
Figure) to provide some free space to increase fluid
velocity across the tubes, resulting into higher heat
transfer rates for a given shell-side fluid flow rate.
Limitations of fixed tube sheet type STHE:
⌧ Tube bundle cannot be removed for cleaning
⌧ No provision exists for differential expansion
between the tubes and the shell (an expansion joint
may be fitted to shell but this results into higher
fabrications cost).

15
Segmental baffle
(≈ 25% area is cutaway)

17
STHE with floating head
If we want to allow for tube bundle removal and for
tubes’ expansion (thermal expansion): floating head
exchanger is used (see Figure).
One tube sheet is fixed, but the second is bolted to a
floating head cover so that the tube bundle can move
relative to the shell (in the case of thermal expansion).
The floating tube sheet is clamped between the floating
head and a split backing flange in such a way that the
tube bundle can be taken out by breaking the flanges.

18
The shell cover at the floating head end is larger than
that at the other end.
Therefore, the tubes can be placed as near as possible
to the edge of the fixed tube sheet, thus utilising the
space to the maximum.

19
Shell-side
inlet
Shell-side
outlet
Tube-side
outlet
Tube-side
inlet
Fixed
tube sheet
Floating
head
Split
ring
Tube
support
Baffles
Exchanger
support
Tube-pass
partition
Floating
tube sheet
STHE with floating tube sheet and head

21
STHE with hairpin tubes
This arrangement also provides for tubes’ expansion.
This involves the use of hairpin tubes (see Figure).
This design is very commonly used for the reboiler of
distillation columns where steam is condensed inside
the tubes to provide for the latent heat of vaporisation.

22
STHE with hairpin tubes
Shell-side
inlet
Shell-side
outlet
Tube-side
outlet
Tube-side
inlet
Tube
sheet
Saddle
support
Tube supports
and baffles
Tube-pass
partition
Gaskets
Shell
vent Tubes
Tie-rod
Spacers

23
Sometimes, it is advantageous to have two or more
shell-side passes, although this increases the difficulty
of construction.

24
Design considerations for STHE
The HE should be reliable with the desired capacity.
Use of standard components and fittings and making
the design as simple as possible.
Minimum overall cost.
Balance between depreciation cost and operating cost.

25
For example, in a vapour condenser:
Higher water velocity in tubes ⇒ higher Reynolds
number Re ⇒ higher heat transfer coefficient on
TUBE SIDE ⇒ higher OVERALL transfer
coefficient ⇒ smaller exchanger (lower area).
However, pumping cost increases rapidly with
increase in velocity (kinetic head increases).
Economic optimum is required to be calculated (see
Figure).

26
Effect of water velocity on
annual operating cost of condenser
Water velocity →
Cost→
Total overall cost
Depreciation
Operating cost

27
Classification of STHE
Basis for classification of STHE: “standard” published
by Tubular Exchanger Manufacturer’s Association
(TEMA), 8th Edition, 1998. Supplements pressure
vessel codes like ASME and BS 5500.
Sets out constructional details, recommended tube
sizes, allowable clearances, terminology etc.
Provides basis for contracts.
Tends to be followed rigidly even when not strictly
necessary.
Many users have their own additions to the standard
which suppliers must follow.

28
TEMA terminology
• Letters given for the front end, shell and rear end types
• Exchanger given three letter designation
ShellFront end
stationary
head type
Rear end
head type

29
Front head type
• A-type is standard for dirty tube side
• B-type for clean tube side duties. Use if possible since
cheap and simple.
B
Channel and removable cover Bonnet (integral cover)
A

30
More front-end head types
• C-type with removable shell for hazardous tube-side
fluids, heavy bundles or services that need frequent
shell-side cleaning
• N-type for fixed for hazardous fluids on shell side
• D-type or welded to tube sheet bonnet for high pressure
(over 150 bar)
C N D

31
TEMA shell types for STHE
E
F
G
H
J
K
X
One-pass shell Split flow Divided flow
Two-pass shell
with
longitudinal
baffle
Double split flow Kettle type
reboiler
Cross flow

32
TEMA E-type shell for STHE
The simplest possible construction.
Entry and exit nozzles at opposite ends of a single pass
exchanger.
Most design methods are based on TEMA E-type shell,
although these methods may be adapted for other shell
types by allowing for the resulting velocity changes.
One-pass shell

33
TEMA F-type shell for STHE
Longitudinal baffle gives two shell passes (alternative
to the use of two shells for a close temperature
approach or low shell-side flow rates).
ΔP for two shells (instead of F-type) is ≈ 8× that for E-
type design (pumping cost).
Limitation: probable leakage between longitudinal
baffle and shell may restrict application range.
Two-pass
shell with
longitudinal
baffle

34
TEMA G-type shell for STHE
Performance is superior to E-type although ΔP is
similar to the E-type.
Used mainly for reboiler and only occasionally for
systems without phase.
Split flow

35
TEMA J-type shell for STHE
“Divided-flow” type
One inlet and two outlet nozzles for shell
ΔP ≈ one-eighth of the E-type, and hence,
Gas coolers and condensers operating at low pressures.
Divided flow

36
TEMA X-type shell for STHE
No cross baffles and hence the shell-side fluid is in
counter-flow giving extremely low ΔP, hence,
Gas coolers and condensers operating at low pressures.
Cross flow

37
MOC of shell of STHE: carbon steel (C.S.); standard
pipes for smaller sizes and rolled welded plate for
larger sizes (> 0.4-1.0 m).
Except for high pressure, calculated wall thickness is
usually < minimum recommended values, although a
corrosion allowance of 3.2 mm is added for C.S.
Shell thickness: calculated using Equation for thin-
walled cylinders (minimum thickness = 9.5 mm for
shells > 0.33 m o.d. and 11.1 mm for shells > 0.9 m o.d.)
Thickness is determined more by rigidity requirements
than by internal pressure.

38
Minimum shell thickness for various materials is given
in BS-3274 (and many international standards).
Shell diameter should be such that the tube bundle
should fit very closely ⇒ reduces bypassing of fluid
outside the tube bundle.
Typical values for the clearance between the outer
tubes in the bundle and the inside diameter of the shell
are available for various types of HEs (see Figure).

39
Shell – tube bundle clearance
Pull-through floating head
Split-ring floating head
Outside packed head
Fixed head and U-tube
Tube bundle diameter, m →
Clearance=
(shellID–tubebundlediameter)→

40
Tube bundle design takes into account shell-side and
tube-side pressures since these affect any potential
leakage between tube bundle and shell which cannot be
tolerated where high purity or uncontaminated
materials are required.
In general, tube bundles make use of a fixed tube sheet,
a floating-head or U-tubes.

41
Thickness of fixed tube-sheet is obtained from a
relationship of form:
0.25 P
d d
t G f
d
G
P
f
d
t
gasket diameter, m
2design pressure, MN m
2allowable working stress, MN m
tube sh
floating head tube
eet thickness, m
Thickness of is usually
calculated as:
sheet
d2
t
=
=
=
=
=

42
Tube DIAMETER selection
Smaller tubes give a larger heat transfer area for a
given shell diameter (16 mm o.d. tubes are minimum
size to permit adequate cleaning).
Smaller diameters ⇒ shorter tubes ⇒ more holes to be
drilled in tube sheet ⇒ adds to construction cost and
increases tube vibration.
Heat exchanger tubes size range = 16 mm (⅝ inch) to
50 mm (2 in) o.d.

43
Smaller diameter tubes are preferred ⇒ more compact
and hence cheaper units.
Larger tubes are easier to clean by mechanical methods
and are hence widely used for heavily fouling fluids.
The tube thickness should withstand internal pressure
and should provide adequate corrosion allowance.
Details of steel tubes used in heat exchangers are given
in BS-3606, and standards for other materials are given
in BS-3274.

44
Tube LENGTH selection
As tube length ↑ cost ↓: for a given surface area
because of smaller shell diameter, thinner tube sheets
and flanges, smaller number of holes in tube sheets.
Preferred tube lengths: 1.83 m (6 ft), 2.44 m (8 ft), 3.88
m (12 ft) and 4.88 m (16 ft).
Longer tubes: for low tube-side flow rate.
For given number of tubes per pass for the required
fluid velocity, the total length of tubes per tube-side
pass is determined by heat transfer surface required.

45
Then, the tubes are fitted into a suitable shell to give
the desired shell-side velocity.
With long tubes and relatively few tubes, it may be
difficult to arrange sufficient baffles for sufficient
support to the tubes.
For good all-round performance, the ratio of tube
length to shell diameter is typically in the range 5-10.

46
In-line layout, Rectangular pitch
PTY
PTX
C
C = clearance
PTX = pitch
PTY = pitch
Tube layout and pitch

47
PTY
PTX
C
C = clearance
PTX = pitch
PTY = pitch
Staggered layout, Rectangular pitch

48
Equilateral triangular pitch
Φ = 300
CV = 0.5 PT
CH = 0.866 PT
CV
CH
PT
Φ
Y

49
In-line layout, Square pitch
Φ = 900
CV = PT
CH = PT
PT
Φ
PT
CH
CV

50
Staggered layout, Square pitch
CV
CH
PT Φ
P
T
Φ = 450
CV = 0.707 PT
CH = 0.707 PT

51
Tube layout and pitch: equilateral triangular, square
and staggered square arrays.
Triangular layout: robust tube sheet.
Square layout: simplifies maintenance and shell side
cleaning.
Minimum pitch: 1.25 × tube diameter.
Clean fluids: smallest pitch (triangular 30° layout) is
used for clean fluids in both laminar and turbulent
flow.

52
Fluid with probable scaling: 90° or 45° layout with a
6.4 mm clearance to facilitate mechanical cleaning.
Tube bundle diameter (db): estimated from an
empirical equation based on standard tube layouts:
b
d
bNumber of tubes N a
t d
o
⎛ ⎞
= = ⎜ ⎟
⎜ ⎟
⎝ ⎠
The values of a and b are available for different
exchanger types for ▲ and ■ pitch, for different
number of tube-side passes (see Table on next slide).

53
Tube-side
passes ⇒
1 2 4 6 8
▲ pitch
(= 1.25 dO) a 0.319 0.249 0.175 0.0743 0.0365
▲ pitch
(= 1.25 dO) b 2.412 2.207 2.285 2.499 2.675
■ pitch
(= 1.25 dO) a 0.215 0.156 0.158 0.0402 0.0331
■ pitch
(= 1.25 dO) b 2.207 2.291 2.263 1.617 2.643
( )Number of tube
b
as N d d
t b o
= =

54
Baffle (cross-baffle) designs
Baffle: designed to direct shell-side flow across the tube
bundle and to support the tubes against sagging and
possible vibration:
Segmental baffle: most common type is the which
provides a baffle window.
Ratio (baffle spacing/baffle cut): decides the
maximum ratio of heat transfer rate to ΔP.
Double segmental (disc and doughnut) baffles: to
reduce ΔP by about 60%.
Triple segmental baffles: all tubes are supported by
all baffles ⇒ low ΔP and minimum tube vibration.

55
Baffle (cross-baffle) designs
Baffle spacing: TEMA recommendation:
Segmental baffles spacing ≥ 20% shell ID or 50 mm
whichever is greater.
It may be noted that the majority of failures due to
vibration occur when the unsupported tube length is
in excess of 80 per cent of the TEMA maximum; the
best solution is to avoid having tubes in the baffle
window.

56
Baffles
Shell flange
Tube sheet
(stationary) Channel
flange

57
Segmental baffle
Drillings
Shell
Tubes
Baffles

58
Disk and doughnut baffle
Shell Doughnut
Disk
Doughnut Disk

59
Orifice baffle
Orifice
Baffle
OD of tubes
Tube
Tube

60
Correction for LMTD for 1,2-STHE
θ1
θ2
T1
T2
( )
( )
T T
1 2Y
2 1
−
=
θ − θ
( ) ( )X T
2 1 1 1
= θ − θ − θ →
CorrectionfactorF→
0.5
0.6
0.7
0.8
0.9
1.0
1.00.90.1 0.3 0.5 0.7

61
( ) ( )X T
2 1 1 1
= θ − θ − θ →
CorrectionfactorF→
0.5
0.6
0.7
0.8
0.9
1.0
1.00.90.1 0.3 0.5 0.7
( )
( )
T T
1 2Y
2 1
−
=
θ − θ
T1
T2
θ1
θ2

62

63

64
(ΔT)MEAN in multipass STHE
Multipass STHE (having more tube-side passes then
shell-side passes): flow is countercurrent in some
sections and cocurrent in other sections.
The LMTD does not apply in this case.
Correction factor F: when F is multiplied by LMTD for
countercurrent flow, the product is true average
temperature driving force.

65
The assumptions involved are:
The shell fluid temperature is uniform over the
cross-section in a pass.
Equal heat transfer area in each pass.
Overall heat transfer coefficient U is constant
throughout the exchanger.
Heat capacities of the two fluids are constant over
the temperature range involved.
No change in phase of either fluid.
Heat losses from the unit are negligible.

66
Then, Q = UA {F(ΔT)MEAN} = F UA(ΔT)MEAN
F is expressed as a function of 2 parameters, X and Y:
T T
tubeOUT tubeIN shellIN shellOUTX ; Y
T
shellIN tubeIN tubeOUT tubeIN
T T
2 1 1 2X ; Y
T
1 1 2 1
θ − θ −
= =
− θ θ − θ
θ − θ −
= =
− θ θ − θ

67
Physical significance of X and Y
X = ratio of heat actually transferred to cold fluid to
heat which would be transferred if the same fluid were
to be heated to hot fluid inlet temperature =
temperature effectiveness of HE on cold fluid side.
Y = ratio of McP value of cold fluid McP values of hot
fluid = heat capacity rate ratio.
( )
( )
( )
( )
X P in some books
T T
h,I
T T
c,OUT c,IN
T
c,IN
Mc
P cold
T T
c,
T
h,I
N h,OUT
Y R in some books
N
M
OUT c,IN
c
P hot
=
−
−
=
−
=
−

68
Temperature profiles in 1,2-STHE
T1
T2
θ1
θ2
T1
T2θ1
θ2
T1
T2
θ1
θ2
T1
T2
θ1
θ2
F is the same in both cases.
(a)
(b)

69
There may be some point where the temperature of the
cold fluid is greater than θ2. Beyond this point the
stream will be cooled rather than heated.
This situation is avoided by INCREASING the number
of shell passes.
If a temperature cross occurs in a 1 shell-side pass
STHE, 2 shell-side passes should be used.
The general form of the temperature profile for a two
shell-side unit is as shown in the Figure on next slide.

70
Temperature profiles in 2,4-STHE
T1
T2
θ1
θ2
T2
T1
θ1
θ2
(c)

71
Shell-side (longitudinal) baffles: difficult to fit and
serious chance of leakage between two shell sides ⇒ use
two exchangers in series, one below the other.
On very large installations it is necessary to link up a
number of exchangers in SERIES (Figure on next
slide).
For A ≥ 250 m2, consider using multiple smaller units
in series; initial cost is higher.

72
3 × 1,2-STHE in series
Effectively: 3,6-STHE in series
θ1
θ2
T1
T1

73
For example, a STHE is to operate as following:
T 455 K, T 372 K
1 2
283 K, 388 K
1 2
388 2832 1X 0.61
T 455 283
1 1
T T 455 3721 2Y 0.79
388 283
2 1
⇒
= =
θ = θ =
θ − θ −
= = =
− θ −
− −
= = =
θ − θ −
⇒
For 1,2-STHE: F ≈ 0.65 (from graph)
For 2,4-STHE: F ≈ 0.95 (from graph)

74
For maximum heat recovery from the hot fluid, θ2
should be as high as possible.
The difference (T2–θ2) is known as the approach
temperature OR temperature approach.
If θ2 > T2: temperature cross (F decreases very rapidly
when there is only 1 shell-side pass) ⇒ in parts of HE,
heat is transferred in the wrong direction.
Consider an example (1,2-STHE) where equal
ranges of temperature are considered:

75
Case
↓
T1 T2 θ1 θ2
Approach
(T2–θ2)
X Y F
1 613 513 363 463 50 0.4 1 0.92
2 573 473 373 473 0 0.5 1 0.80
3 543 443 363 463
–20
(cross of 20)
0.55 1 0.66

76
There may be a number of process streams, some to be
heated and some to be cooled.
Overall heat balance: indicates whether there is a net
PLUS or MINUS heat available.
The most effective match of the hot and cold streams in
the heat exchanger network: reduces the heating and
cooling duties to a minimum.
This is achieved by making the best use of the
temperature driving forces.

77
There is always a point where the temperature
difference between the hot and cold streams is a
minimum and this is referred to as the pinch.
Lower temperature difference at the pinch point means
lower demand on utilities.
However, a greater area (and hence cost) is involved
and an economic balance must be made.

78
Fully Developed Forced Convection Heat Transfer
0.8 0.4Nu 0.023 Re Pr
0.40.8 cd vh d pi avei i 0.023
... Dittus-Boelter equation
0.7 Pr 160
4... Re 10
L
1
i
k k
0
d
=
μ⎛ ⎞ρ⎛ ⎞⎛ ⎞
⎜ ⎟= ⎜ ⎟⎜
< <⎧
⎪ >⎪
⎨
>
⎟
μ⎝ ⎠ ⎝ ⎠⎝ ⎠
⎪
⎪⎩
Dittus-Boelter equation is less accurate for liquids with
high Pr, and following equation is recommended:
( )20.795 0.495Nu 0.0225 Re Pr 0.0225 Pexp ln
L4 6... 0.3 Pr 300; 4 10
r
Re 10 ; 10
d
i
⎡ ⎤= −⎣ ⎦
< < × < < >

79
Effect of variation of fluid properties with temperature on
turbulent convective heat transfer
The previous equations can be used without correction
for variation of physical properties with temperature
provided that the driving force (Tbulk – Twall) is small,
that is, less than ≈ 15% of absolute temperature of fluid.
If a gas is being cooled (Tbulk > Twall), error analysis
shows that no correction is necessary even for large ΔT.
If gas is being heated (Tbulk < Twall), then heat transfer is
reduced and a correction must be applied. The
recommended form of correction is:

80
( ) ( ) ( )
( ) ( )
for
where function of gas propertie
nh Ti,corrected b
s
N ; air ; He ;
2
ulk
h , T
i T T wall
bulk wall
n
0.44 0.40 0.38
0.37 H ; 0.1 steam ; etc.
2
8
⎛ ⎞
= ⎜ ⎟
⎜
= ⎠
=
⎟
⎝
=
If gas is being heated (Tbulk < Twall), then heat transfer is
reduced and a correction must be applied. The
recommended form of correction is:

81
In liquids, μ decreases with T, hence effect of
temperature is the opposite to that in gases and heat
transfer is increased in case of heating (Tbulk < Twall).
Therefore, in liquids, (cooling and heating), a correction
is applied if Tbulk and Twall are significantly different.
The correction factor recommended is (Sieder and Tate,
1936) as:
0.14h
i,corrected bulk
h , T T
i bulk wall wall
μ⎛ ⎞
= ⎜ ⎟
⎜ ⎟≠ μ
⎝ ⎠

82
Non-circular pipes and ducts
All the corrections (for fully developed turbulent flow)
are directly applicable, provided equivalent diameter De
replaces diameter d.
4 cross-sectional area
D
e wetted perimeter
×
=

83
EXAMPLE (Effect of fluid properties on turbulent
convective heat transfer):
Compare the heat transfer rates for air, water, and oil
flowing in a pipe of 2.5 cm diameter at a Reynolds
number of 105. The internal surface temperature of
pipe is 99 0C and fluid mean temperature is 55 0C. How
would the pressure drop vary for the three cases?
Properties at the mean film temperature (Tbulk+Twall)/2
are as given in the following table:

84
Properties ↓ Air Water Oil
Density, ρ (kg/m3) 0.955 974 854
Kinematic viscosity,
μ/ρ = ν (m2/s)
2.09×10–5 3.75×10–7 4.17×10–5
Thermal conductivity,
k [W/(m K)]
0.030 0.668 0.138
Prandtl number Pr 0.70 2.29 546
Mean film temperature = (Tbulk+Twall)/2 = (372+328)/2
= 350 K. Properties of the three fluids at 350 K are as:
Dittus-Boelter equation is used to compute the heat
transfer coefficient hi.
... Dit0.8 0.4Nu 0.02 tus-Boelter3 Re P equ nr atio=

85
Therefore, for given Reynolds number (Re), kinematic
viscosity (μ/ρ = ν), and pipe diameter (di), the average
fluid velocity will be different for each fluid. That is,
( )
d v d v
i ave i ave
d v
i aveRe
ρ
= = =
μ μ ρ ν
( )Re
d
i
Re
v
ave d
i
μ ρ
= =
ν
For the given Re (105), and given three Prandtl
numbers Pr (0.70, 2.29, 548), compute Nusselt number
Nu for the three fluids. This will give the heat transfer
coefficient hi.
Nu k
h
i d
i
=

86
From hi, compute the convective heat transfer flux q:
( ) ( ) ( )q h T T h 372 328 h 44
i wall bulk,ave i i
= − = − =
Friction factor is same in all the three cases because
Reynolds number Re = 10000:
( ) 0.20.20.046 Re 0.046 10000 3f 4.6 10−−= ×= −=
Now, compute the pressure drop per unit length of the
pipe (ΔP/L).
22 f vP ave
L d
i
ρΔ
=
Computation summary is presented on the next slide:

87
Fluid Nu
hi
J/(s m2
K)
q
J/(s m2
)
vave
m/s
ΔP/L
(N/m2
)/m
Air 199 239 10529 84 2456
Water 320 8817 387939 2 806
Oil 2862 15796 695019 167 8743751
ΔP/L is an indication of pressure loss in pipe.
Pressure loss in case of oil is very high ⇒ velocity of
167 m/s is impractical.

88
If average oil velocity is brought down to 2 m/s, then
(ΔP/L)oil will reduce to 707 (N/m2)/m, but (Re)oil will
also reduce from 105 to ≈ 1199.
That will bring down (hi)oil drastically.
This means: heat transfer rates equivalent to water can
not be obtained with oil for commensurate pressure
loss. WHY???
Please note: Dittus-Boelter equation is no longer valid
for Re of 1199, that is, laminar flow regime; some other
appropriate equation will have to be used.

89
Heat transfer
and
pressure loss calculations
for
Shell and Tube Heat Exchangers

90
Calculations for heat transfer and pressure loss for
fluids flowing inside tubes is relatively straightforward.
Heat transfer and pressure loss calculations within the
shell of the exchanger are not straightforward, because
of the complex flow conditions.
The calculation procedure has evolved over the
decades.
Initially, methods (correlations) were developed for
computing shell-side pressure drop and heat transfer
coefficient based on experimental data for “typical”
exchangers.

91
Kern (1950) method: correlation of data for standard
exchangers by a simple equation analogous to
equations for flow inside tubes. (Kern, Donald Q. 1950,
“Process Heat Transfer”, McGraw-Hill).
Limitations:
Restricted to a fixed baffle-cut (25%), can not
account for effect of other baffle configurations,
Can not adequately account for leakages through
gaps between tubes and baffles, and between baffles
and shell,
Can not account for bypassing of the flow around
the gap between tube bundle and shell.

92
Nevertheless, Kern method: very simple and rapid for
the calculation of shell-side heat transfer coefficients
and pressure losses.
Based on data from industrial heat transfer operations
and for a fixed baffle cut of 25%, Kern gives the
equation:
0.55 1 3h d d c 0.14
shell eq eq p
0.36
k k
wall
0.55 1 3h d
v
shell
m
sh
d c 0.14
shell eq eq p
0
e
.36
k k
wa
l
l
l
l
μ⎛ ⎞ ⎛ ⎞ μ⎛ ⎞⎜ ⎟ ⎜ ⎟= ⎜ ⎟ ⎜ ⎟⎜ ⎟μ μ⎝ ⎠⎝ ⎠ ⎝ ⎠
μ⎛ ⎞ ⎛ ⎞ μ⎛ ⎞⎜ ⎟ ⎜ ⎟= ⎜ ⎟ ⎜ ⎟⎜ ⎟μ μ⎝ ⎠⎝ ⎠ ⎝ ⎠
ρ

93
shell-side heat transfer coefficient
equivalent diameter of shell-side flow
thermal conductivity of shell side fluid
mass velocity on the shell-side
tota
h
shell
d
eq
k
m
sh
l mass f
el
low rate of f
l
=
=
=
=
=
luid on shell-side
shell cross-flow area
total mass flow rate on shell-side
S
S
specific heat of shell-side fluid
at the diameter
= viscosity of shell-side flu
c
id
of the s
at bulk
h
p
ell
flμ
⎛ ⎞
⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
uid temperature
= viscosity of shell-side fluid at wall tempera
wa l
e
l
turμ

94
PT
C
C = clearance
PT = pitch
SQUARE pitch tube layout

95
( )
( )
( )
If shell inside diameter
tube pitch tube centre-to-centre distance
tube clearance space between tubes
baffle spacing distance between baffles
Then, Number of tube o
d
sh
n a
ell
P
T
shels l diamat
C
L
B
=
=
=
=
, and
Flow area associated with each tube betwe
d
sh
en
ellN
baffl
e
e
T P
L
r
T
B
s
C
=
=
=

97
( )
( )
Therefore, shell cross-flow area
Number of tubes
Flow area associated with each tube between baffles
N
on a
C
at diamete
L
r of shell
S
S
d
shell C L
BT B
The
shell diam
equivalent diameter i
t
s
a er
P
T
= = ×
= × =
4
d
eq
2 2 2 24 P d 4 P d
T o4 flow area T o4 4
dwetted perimeter d
o o
defined as,
where d outside diameter of tubes
o
4
4
4
⎡ ⎤π⎛ ⎞ π⎡ ⎤− −⎜ ⎟⎢ ⎥ ⎢ ⎥× ×⎝ ⎠⎣ ⎦ ⎣ ⎦= = =
π π⎛ ⎞
⎜ ⎟
⎝ ⎠
=

98
Flow
area
PT
C
TRIANGULAR pitch tube layout

99
( )
( )
( )
( )
22 dP 3
oT
4 8triangular pitch d
eq d 2
o
shell-side pressure loss
22 N 1 f d m
shell shellP
shell 0.14
d
e
For tube layout:
Kern 1950 correlation for :
where, shell inside diame
q wall
d
shell
f
ter
friction factor;
π
−
=
π
+
Δ =
μ
=
=
ρ μ
[ ]
Number of
= number of across the
Does not account
N baffles
B
N
for "leak
tube bund
ages" bet
+
ween baffle spaces
1 fluid pas leses
B
=
⇒

100
Shell-side Reynolds number Re →
Shell-sidefrictionfactorf→

101
EXAMPLE: A shell and tube heat exchanger has the
following geometry:
Shell ID = dshell = 0.5398 m
No. of tubes = NT = 158
Tube OD = do = 2.54 cm; Tube ID = di = 2.0574 cm
Tube pitch (square) = PT = 3.175 cm
Baffle spacing = LB = 12.70 cm
Shell length = LS = 4.8768 m

102
Tube-to-baffle diametrical clearance = ΔTB = 0.8 mm
Shell-to-baffle diametrical clearance = ΔSB = 5 mm
Bundle-to-shell diametrical clearance = ΔBuS = 35 mm
Split backing floating head = assumed
No. of sealing strips per cross-flow row = NSS/NC = 1/5
Baffle thickness = TB = 5 mm
No. of tube-side passes = n = 4

103
Use the Kern method to calculate shell-side heat transfer
coefficient and pressure drop for flow of a light
hydrocarbon with following specification (at Tbulk):
Total mass flow rate = MT = 5.5188 kg/s
Density = ρ = 730 kg/m3
Thermal conductivity = k = 0.1324 W/(m K)
Specific heat capacity = cP = 2.470 kJ/(kg K)
Viscosity = μ = 401 (μN s)/m2
Assume no change in viscosity from bulk to wall.

104
Kern method is to be followed. Therefore, compute:
Cross-flow area at the shell diameter (shell centre-
line), mass flux (mass velocity), equivalent shell
diameter,
Shell Reynolds no. (Reshell) and Shell Prandtl no.
(Prshell),
Heat transfer coefficient and pressure drop (loss).
Tube-to-tube clearance C:
Gap between tubes = clearance
= C = PT – do = 0.03175 – 0.0254 = 0.00635 m

105
( ) ( )
( )
2 2
4 0.
2 24
031
P d
T o4d
eq d
o
0.
75 0.0254
4 02513 m
0.0254
π⎡ ⎤
−
π⎡ ⎤
−⎢ ⎥
⎢ ⎥⎣
⎦
=
⎣
π
⎦ =
π
=
Equivalent shell diameter deq:
( )( )0.5398
0.00635 0.1270
0.031
d
shellS C L
S BP
T
20.0137 m
75
1
= =
=
Cross-flow area at shell diameter (shell centre-line) SS:

106
total mass flow rate
m
shell at diameter of shell
M kgT
shell cross-flow area
5.5188
0.01371
402.5
2S s mS
=
= ==
Mass flux (mass velocity) mshell:
v m
ave she
d d
eq eq
Re 25224
0.025l 1
shel
.l
l
3 402 5×
= =
μ
=
μ μ
ρ
=
Reynolds number Re:
( )( )3 62.470 10 40
c
p
Pr
1 10
0.1324
7.481
k
−
= ==
×
μ
×
Prandtl number Pr:

107
Shell-side heat transfer coefficient hshell:
( ) ( )
( )
( ) ( )
h d
shell eq 0.55 1 3
Nu 0.36 Re Pr
shell shell shellk
0.55 1 3d m c
0.36 k eq shel
0.36 0.13
l p
h
shell d k
eq
W
977.9
24 0.55 1 325224 7.481
0.
4
025
2m
1
K
3
= ≈
μ⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟≈ ⎜ ⎟μ ⎝ ⎠⎝
=
=
⎠

108
Calculation of pressure drop (loss) ΔPshell:
Calculate number of baffles in the shell (N):
L
S
L T
B
1 1
1 1
shell length
N
B
B
4.8768 4.8768
35.95
0
baffle spacing + baffle thi
.1270 0.005 0.1320
ckness
36
=
+
=
− −
− − = =
+
=
=
Estimated friction factor from the GRAPH ≈ 0.063
( )
( )
( )
( )( )( )( )
( )( )
2 22 N 1 f d m 2 N 1 f d m
B shell shell B shel
22 3
l shellP
shell 0.14 dd e
6 1 0.063 0.5
qeq wal
398 402
l
222
.5
730 0.02513
24 Pa
+
+ +
Δ
≈
=
ρμ
≈
≈
ρ μ

109
Bell Delaware method
(Bell, K.J. 1963, University of Delaware, U.S.A.)

110
Bell (1963) developed a method in which correction
factors were introduced for the following:
Leakage through gaps between tubes and baffles,
and baffles and shell, respectively,
Bypassing of flow around the gap between tube
bundle and shell,
Effect of baffle configuration (recognition: only a
fraction of the tubes are in pure cross-flow),
Effect of adverse temperature gradient on heat
transfer in laminar flow.

111
Flow Stream-Analysis method
(Tinker, T. 1951; Wills, M.J.N. and Johnston, D. 1984)
Tinker (1950) suggested “stream-analysis” method for
calculation of shell-side flow and heat transfer.
Formed the basis of modern computer codes for shell-
side prediction.
Wills and Johnston (1984) developed a simplified
version suitable for hand calculations.
Simultaneously adopted by Engineering Sciences Data
Unit (ESDU, 1983), and provides a useful tool for
realistic checks on “black box” computer calculations.

112
Shell
baffle
A
B
b
c
t
s
w
Flow Stream-Analysis method
(Tinker, T. 1951; Wills, M.J.N. and Johnston, D. 1984)

113
Refer to the Figure: Fluid flows from A to B via
various routes, b, c, t, s, and w.
Leakage between tubes and baffle (t),
Leakage between baffle and shell (s),
Partly, flow passes over the tubes in cross-flow (c),
Partly, flow bypasses the tube bundle (b),
Streams b and c combine to form stream w, that
passes through the baffle-window zone.

114
This method also depends on empirically based
resistance coefficients for the respective streams.
This problem may be partly overcome by using
Computational Fluid Dynamics (CFD) techniques.

115
Heat Exchanger Performance
Effectiveness of heat exchanger E: defined as ratio of
actual heat transfer rate Q to the thermodynamically
maximum possible heat transfer rate Qmax:
Q
E
Q
max
=
Qmax = heat transfer rate that would be achieved if the
outlet temperature of the fluid with lower heat
capacity rate was brought equal to the inlet
temperature the other fluid.

116
T1
T2
T1
T1
T2 T2
M1
M2
Temperature→
Distance →
Condenser-Evaporator
(isothermal hot fluid; isothermal cold fluid)

117
Condenser-Heater
(isothermal hot fluid; non-isothermal cold fluid)
T1
T2,1
T1
T1
T2,1 T2,2
M1
M2
Temperature→
Distance →
T2,2
T1

118
Cooler-Evaporator
(non-isothermal hot fluid; isothermal cold fluid)
T2
T1,1
T1,2
T2 T2
M1
M2
Temperature→
Distance →
T1,2
T1,1
T2

119
Cooler-Heater (co-current flow)
(non-isothermal hot fluid; non-isothermal cold fluid)
T2,1
T1,1
T1,2
T2,1 T2,2
M1
M2
Temperature→
Distance →
T1,2
T1,1
T2,2

120
Cooler-Heater (counter-current flow)
(non-isothermal hot fluid; non-isothermal cold fluid)
T1,2
T1,1
T1,2
T2,1
T2,2
M1
M2
Temperature→
Distance →
T2,2
T2,1
T1,1

121
Assuming “fluid 1” as having lower value of (McP),
(see Figures on the next few slides):
( )1 1
Q M c T T
max P ,1 1 ,12
= −
Over heat balance gives:
( ) ( )Q M c T T M c T T
P ,1 ,21 1 1 1 2 2 2P ,2 2,1
= − = −
Therefore, effectiveness (based on fluid 1) is given by:
( )
( )
( )
( )
M c T T T T
P ,1 ,2 ,1 ,2
E
M c T T T T
P ,1
1 1 1 1 1 1
1 1 1 2 1,1 ,1 2,1
− −
= =
− −

122
Similarly, effectiveness (based on fluid 2) is given by:
( )
( )
M c T T
P ,2 ,1
E
M c T T
P
2 2 2 2
1 1 1,1 ,12
−
=
−
In calculating temperature differences, the positive
value is considered.

123
Example: A flow of 1 kg/s of an organic liquid of heat
capacity 2.0 kJ/(kg K) is cooled from 350 K to 330 K
by a stream of water flowing counter-currently
through a DPHE. Estimate effectiveness of heat
exchanger if water enters DPHE at 290 K and leaves at
320 K.
Solution:
MORG = 1 kg/s
cP,ORG = 2.0 kJ/(kg K) = 2000 J/(kg K)
ΔTORG = TORG,OUT – TORG,IN = 350 – 330 = 20 K

124
Therefore, heat duty (heat load)
= Q = MORGcP,ORG ΔTORG = 1 × 2000 × 20 = 40,000 J/s
(McP)ORG = (1 × 2000) = 2000 J/(s K)
The mass flow rate of water is calculated using overall
heat balance:
( )
40000 kg
M 0.3185
Water 4186.8 320 290 s
= =
× −
⇒ (McP)WATER = (0.3185 × 4186.8) = 1333.3 J/(s K)
⇒ (McP)MIN = (McP)WATER = 1333.3 J/(s K)

125
( ) ( )
( )
Q
E
Q
max
Actual heat lo
Actual heat
ad
MC T maxP m
load
Maximum he
in
40000
1333.
at load
3 350 290
0.5
=
=
=
Δ
−
=
⇒
=

126
Heat Transfer Units
Number of heat transfer units (NHTU) is defined by:
( )
( ) ( ) ( )
U A
NHTU
Mc
P min
Mc M c M c
P 1 P1 2 P2min
where of aLO ndWER
=
=
NHTU: heat transfer rate for a unit temperature
driving force divided by heat taken (given) by fluid
when its temperature is changed by 1 K.
NHTU: measure of amount of heat which the heat
exchanger can transfer.

127
The relation for effectiveness of heat exchanger in
terms of heat capacity rates of fluids can be DERIVED
for a number of flow conditions.
We will consider two cases:
(I) Co-current flow double-pipe heat exchanger
(without phase change)
(II) Counter-current flow double-pipe heat exchanger
(without phase change)

128
CASE I: Co-current flow
( )dQ U A T T
1 2
= −
T2,1
T1,1
T1,2
T2,1 T2,2
M1
M2
Temperature→
Distance →
T1,2
T1,1
T2,2
T1
T2
θ

129
For a differential area dA of heat exchanger, heat
transfer rate is given by:
( )
( )
where, and = temperatures of two streams, and
local temperature
dQ U dA T T U dA
1 2
T T
1 2
T T
1 2
dQ M c dT M c dT
2 P2 2 1 P1 1
dQ dQ
dT dT
2 1M c M c
2
difference
Also,
and
P2 1 P1
dT dT d T T
1 2 1 2
= − = θ
θ
= = −
−
= =
− =
= −
⇒ −
=
⇒

130
( )
1 1
dQ U dA d U dA
M c M c
1 P1 2 P2
d 1 1
U dA
M c M c
1 P1 2 P2
1 12 U A
M c M c
1 1 P1 2 P2
T T
1 11,2 2,2
U A
T T M c M c
1,1 2,
1 1
d dQ
M c M c
1 P1
1 1 P
2 P
But
1 2 P2
ln
ln
2
⇒
⇒
⎡ ⎤
= θ θ = − θ
⎡ ⎤
θ
+⎢ ⎥
⎣ ⎦
θ ⎡ ⎤
= − +⎢ ⎥θ
⎣ ⎦
θ ⎡ ⎤
= − +⎢ ⎥θ
⎣ ⎦
−
⎡ ⎤
=
= − +⎢ ⎥
⎣ ⎦
− +⎢ ⎥−
⎣
⇒
⇒
⎦
⇒

131
( )
( )
M c M c M c M c
1 P1 2 P2 1 P1 1 P1min
T T
1 11,2 2,2
U A
T T M c M c
1,1 2,1 1 P1 2 P2
U A U A
NHTU
Mc M c
P 1 P1min
T T M cU A1,2 2,2 1 P11
T T M c M c
1,1 2,1 1 P1 2 P2
M c
1 P1NHTU 1
M c
If
ln
ln
T T
1,2 2,2
T T
1,1 2,1 2 P
exp
2
< =
−
⎡ ⎤
= − +⎢ ⎥−
⎣ ⎦
= =
− ⎡ ⎤−
= +⎢ ⎥
− ⎢
⇒
⇒
⇒
⎥⎣ ⎦
⎡ ⎤
= − +⎢ ⎥
⎢⎣
−
−
⎦
⇒
⎥
⎧ ⎫⎪ ⎪
⎨ ⎬
⎪ ⎪⎩ ⎭

132
( )
( )
( )
( )
( )
M c T TT T 2 P2 2,2 2,11,1 1,2
E E
T T M c T T
1,1 2,1 1 P1 1,1 2,1
T T E T T
1,1 1,2 1,1 2,1
M c
1 P1T T E T T
2,2 2,1 1,1 2,1M c
2 P2
M c
1 P1T T E 1
1,2 2
Since and
and
Adding:
T T T T
1,1 2,1 1,1 2,1,2 M c
2 P
Dividing both sides by
2
T
1 1,
−−
= =
− −
⇒ − = −
− = −
⎡ ⎤
− + = +⎢ ⎥
⎢ ⎥⎣
− −
⎦
( )T
2,1
− ⇒

133
M c
1 P11 E 1
M c
2 P2
M c M c
1 P1 1 P11 exp NHTU 1 E 1
M
T T
1,2 2,2
T T
1,1 2,1
M c
1 P11 exp NHTU 1
M c
2 P2
E
M c
1 P11
M c
2 P
c M c
2 P2
2
2 P2
⎡ ⎤
⇒ − = +⎢ ⎥
⎢ ⎥⎣ ⎦
⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪
⇒ − − + =
−
−
⎧ ⎫⎡ ⎤⎪ ⎪
− − +⎢ ⎥⎨ ⎬
⎢ ⎥⎪
+⎢ ⎥ ⎢ ⎥⎨ ⎬
⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦
⎪⎣ ⎦⎩ ⎭=
⎡ ⎤
+⎢ ⎥
⎢⎣
⎦⎩ ⎭
⎦
⎣
⎥

134
( ){ }
[ ]
{ }[ ]
( )
{ }
[ ]
[ ]
1 exp NHTU 1 1
E
1 1
0.5 1 exp 2
Special case: M c M c
1 P1 2 P2
Special case: NHTU very hig
NHTU
1 exp
E
2
1 0
0.5
2
h
=
⇒
→
− − +
=
+
= − −
− −∞
=
−
= =
∞
⇒

135
T1,2
T1,1
T1,2
T2,1
T2,2
M1
M2
Temperature→
Distance →
T2,2
T2,1
T1,1
CASE II: Counter-current flow

136
Similar to co-current flow, equation can derived for
effectiveness of heat exchanger for counter-current
flow.
Please note: in case of counter-current heat exchanger,
θ1 = T2,2–T1,1; θ2 = T2,1–T1,2.
Equation for effectiveness E is given by:
M c
1 P11 exp NHTU 1
M c
2 P2
E
M c M c
1 P1 1 P11 exp NHTU 1
M c M c
2 P2 2 P2
⎧ ⎫⎡ ⎤⎪ ⎪
− − −⎢ ⎥⎨ ⎬
⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭=
⎧ ⎫⎡ ⎤⎪ ⎪
− − −⎢ ⎥⎨ ⎬
⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

137
[ ]{ }1 exp NHTU 1 1
E
M c M c
1 P1 1 P11 exp NHTU 1
M c M c
2 P2 2 P
Special case: M c M c
1
INDETERM
P1
INAT
2 P2
2
1 1 0
1
E
1 0
− − −
=
⎧ ⎫⎡ ⎤⎪ ⎪
− − −⎢ ⎥⎨ ⎬
⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
−
= =
−
=
=
Therefore, expanding the exponential term gives:
NHTU
E
1 NHTU
=
+
Special case: NHTU → ∞ (very high) ⇒ E = 1.

138
If one component is undergoing a ONLY PHASE
CHANGE at constant temperature, M1cP1 = 0.
In that case, both cases lead to the following equation
for heat exchanger effectiveness:
( )E 1 exp NHTU= − −

139
Counter-current flow
NHTU →
EffectivenessE→
0.0
0.2
0.4
0.6
0.8
1.0
0 1 2 3 4 5
M c M c 0
1 P1 2 P2
=
0.25
0.50
0.75
1.0

140
1,2-STHE
NHTU →
EffectivenessE→
0.0
0.2
0.4
0.6
0.8
1.0
0 1 2 3 4 5
M c M c 0
1 P1 2 P2
=
0.25
0.50
0.75
1.0

141
Example: A process requires a flow of 4 kg/s of purified
water at 340 K to be heated from 320 K by 8 kg/s of
untreated water which can be available at 380, 370, 360
or 350 K. Estimate the heat transfer surfaces of 1,2-
STHE suitable for these duties. In all cases, the mean
heat capacity of the water streams is 4.18 kJ/(kg K) and
the overall coefficient of heat transfer is 1.5 kW/(m2 K).
Solution:
For the untreated water (hot fluid):
(McP)HOT = 8 × 4180 = 33,440 J/(s K)

142
For the purified water (cold fluid):
(McP)COLD = 4 × 4180 = 16,720 J/(s K)
Therefore, (McP)MIN = (McP)COLD = 16,720 J/(s K)
( )
( ) ( )
( )
( )
Actual heating load
Maximum hea
M c 167201 P1 0.5
M c 33440
2 P2
Q
E
Q
max
M c 340 320 M c 340 320
1 P1 1 P1
MC T 320 M c T 320
P hot,1
t load
1 P1 hot,1min
= =
= =
− −
= =
− −
⇒

143
( )
( ) ( )
Therefore,
for T : E
hot,1
for T K :
340 320 20
E
T 320 T 32
E
hot,1
for T K : E
hot
0
hot,1 hot,1
380 K 0.3333
370 0.4
36
,1
for T K : E
hot
0 0.5
350 0.66
,1
67
⇒
= =
= =
= =
= =
−
= =
− −

144
( )P MIN
From graph for 1,2-STHE, is found as:
T : NHTU 0.45
hot,1
T K : NHTU 0.6
hot,1
T K : NHTU 0.9
hot,1
T K : NHTU 1.7
hot,1
The area is calculat
NHTU
for 380 K
for 370
for 360
for 350
A NHTU(Mc )ed as:
for T
h t,1
U
o
= =
= =
= =
= =
=
= : A
for T K : A
hot,1
for T K : A
ho
2380 K 5.02 m
2370 6.69 m
2360 10
t,1
for T K : A
hot
.03 m
2350 18.95 m
,1
=
= =
= =
= =

145
Plate heat exchangers
First developed by APV, and then by Alfa-Laval.
Series of parallel plates held firmly together between
substantial head frames.
Plates: one-piece pressings (usually SS), and are spaced
by rubber sealing gaskets cemented into a channel
around the edge of each plate.
Plates have troughs pressed out perpendicular to flow
direction and arranged to interlink neighbouring plates
to form a channel of constantly changing direction.

146
Generally, the gap between the plates is 1.3-1.5 mm.
Each liquid flows in alternate spaces and a large
surface can be obtained in a small volume.
Because of shape of the plates, the developed area of
surface is appreciably greater than the projected area.
High degree of turbulence is obtained even at low flow
rates and the high heat transfer coefficients obtained.
The high transfer coefficient enables these exchangers
to be operated with very small temperature differences,
so that a high heat recovery is obtained.

147
Gasketed-plate heat exchanger

148
Plate Heat Exchanger
Plate

149
Because of shape of
the plates, the
developed area of
surface is
appreciably greater
than the projected
area.

150
Plate and Gasket

151
Plate and Gasket
Plate of
suitable metal
Gasket of
suitable polymer

152
Two-pass plate and frame flow arrangement

156
Plate Heat Exchanger (PHE)

157

158

159

161

162

163

164
Schematic

165
Schematic

166
Schematic

167
Functioning

168
details

169
These units have been particularly successful in the
dairy and brewing industries, where the low liquid
capacity and the close control of temperature have been
valuable features.
A further advantage is that they are easily dismantled
for inspection of the whole plate.
The necessity for the long gasket is an inherent
weakness, but the exchangers have been worked
successfully up to 423 K and at pressures of 930 kN/m2.
They are now being used in the processing and gas
industries with solvents, sugar, acetic acid, ammoniacal
liquor, and so on.

170
Comparison of STHE and PHE
STHE:
5 m long, ¾ inch OD tubes placed on 1 inch triangular
pitch, 150 tubes
Shell ID = 15¼ inch
Total surface area = 45 m2
Volume = 0.6 m3
Compactness = 76 m2/m3

171
PHE:
Plate area = 0.5 m2
Number of plates = 90
Total surface area = 45 m2
Plate spacing = 5 mm
Volume = 0.225 m3
Compactness = 200 m2/m3

172
Close Temperature Approach
Hot stream: 5 kg/s, CP = 1000 J/(kg K), cooling from 71
to 31 0C (Q = 200000 W).
Cold Stream: 5 kg/s, CP = 4000 J/(kg K), heating from
30 to 40 0C.
U = 500 J/(s m2 K)
Y = R = MCCPC/MHCPH = 4
X = P = (TC,OUT-TC,IN)/(TH,IN-TC,IN) = 0.244
LMTD = (31 – 1)/ln(31/1) = 8.74 K
A = 200000/(500 × 8.74) = 45 m2

173
LMTD correction factor F for a 1,2-STHE

174
LMTD correction factor F for a 2,4-STHE

175
Spiral heat exchangers
Two fluids flow through the channels formed between
spiral plates.
Fluid velocities may be as high as 2.1 m/s and overall
heat transfer coefficients (U) ≈ 2800 W/(m2 K) can be
obtained.
Inner heat transfer coefficient is almost DOUBLE that
for a straight tube.
Cost is comparable or even less than that of STHE,
particularly when they are fabricated from alloy steels.

176
High surface area per unit volume of shell and the high
inside heat transfer coefficient.

179
Spiral Flow – Spiral Flow HE

180
Cross Flow – Spiral Flow HE

181
Combination Cross Flow and Spiral Flow – Spiral Flow
HE

182
General arrangement of a Plate-and-Shell HE
Cooling medium flows
between the plate
pairs
“Closed” model has
an outer (pressure
vessel) shell which is
welded together to
enclose the plate pack
Plate pairs are welded
together from two
individual plates
“Open” model has removable end cap / plate pack for cleaning

183
TEFLON® (PTFE: polytetrafluoroethylene) STHE

184
Thermal Insulation
Heat Losses through Lagging
Heat loss (gain) from vessels and utility piping:
radiation, conduction, convection.
Radiation: f(T1
4–T2
4) = increases rapidly with (T1–T2).
Conduction: Air is a very poor heat conductor ⇒ heat
loss by conduction (by air) is very small.
Convection: Convection currents form very easily ⇒
heat loss from an unlagged surface is considerable.

185
Lagging of hot and cold surfaces is necessary for:
Conservation of energy,
To achieve acceptable working conditions.
For example: surface temperature of furnaces is
reduced by using a series of “poor-heat-conducting”
insulating bricks.
Main requirements of a good lagging material:
Low thermal conductivity,
It should suppress convection currents.

186
The materials that are frequently used are:
Cork: very good insulator but becomes charred at
moderate temperatures and is used mainly in
refrigerating plants.
85 per cent magnesia: widely used for lagging steam
pipes: applied either as a hot plastic material
(cannot be reused) or in preformed sections (can be
reused).
Glass wool: cannot be reused.
Vermiculite: a reusable inorganic material.

187
The rate of heat loss per unit area is given by:
temperature difference
thermal resistance
∑
∑
For example: in case of heat loss to atmosphere from a
lagged (insulated) steam pipe, thermal resistance is:
Due to steam condensate film inside pipe
(conVEctive),
Due to dirt / scaling inside the pipe (conDUctive),
Due to pipe wall (conDUctive),
Due to Lagging / insulation (conDUctive),
Due to air film outside the lagging (conVEctive).

188
For a lagged (insulated) pipe:
( )( )
( )( )
2 L T overallQ
i r r
po lago
ln ln
r r
pi po1 1
r h k k r h
pi i p lag lago o
Q 2 T overalli
L r
po
ln l
r
lag
n
r r
pi p
o
k r h
lag lago o
HEAT LOSS RATE per
o1 1
r h k
pi, i
unit p
p
π Δ
⎡ ⎤⎛ ⎞ ⎛ ⎞
⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥+ + +
⎢ ⎥
⎢ ⎥⎣ ⎦
π Δ
⎡ ⎤⎛ ⎞ ⎛ ⎞
⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
=
⇒ =
=
⎢ ⎥+ + +
⎢ ⎥
⎢ ⎥⎣ ⎦
ipe length

189
( )
where, heat loass per unit pipe length,
overall temperature difference,
inside radius of pipe,
outside rad
Q
1 1i J s m
L
T Koverall
r m
pi
r m
po
r m
l
ius of pipe,
outside radius of lagging,
convective
a
h
i
go
=
=
=
=
Δ
=
− −
=
film heat transfer coefficient inside pipe,
convective film heat transfer coefficient outside lagging,
thermal conductivity of pipe mat
1 2 1J s m K
h 1 2 1o J s m K
1 1 1k erial, J s
thermal
m K
p
k
lag
co
⎧
⎨ − − −
⎩
− − −
−=
=
− −
⎧
= ⎨
⎩
nductivity of lagging 1 1 1J smateri Ka ml, − − −

190
A steam pipe, 150 mm i.d. and 168 mm o.d., is carrying
steam at 444 K and is lagged with 50 mm of 85%
magnesia. What is the heat loss to air at 294 K?
Given:
Temperature on the outside of lagging ≈ 314 K
hi = steam-side convective film heat transfer
coefficient = 8500 W m–2 K–1
ho = convective film heat transfer coefficient outside
lagging = 10 W m–2 K–1
kp = thermal conductivity of pipe material
= 45 W m–1 K–1
klag = thermal conductivity of lagging material
= 0.073 W m–1 K–1

191
( )( )
( )T ove
Q 2 T overalli
L r r
po lago
ln ln
r r
pi po1 1
r h k k r h
pi, i p lag lago o
444 294 150 K
150 3mm 75 mm 75 10 m
2
168 3mm 84 mm 84 1
rall
r
pi
r
po
r r
lago po
0 m
2
50 84 50 mm 134 mm 134 10
π Δ
=
⎡ ⎤⎛ ⎞ ⎛ ⎞
⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥+ + +
⎢ ⎥
⎢ ⎥⎣ ⎦
= − =
−= = = ×
−= = = ×
−= + = + = ×
Δ
= 3 m

192
( )
( ) ( )
Q 2 150i
L 84 134
ln ln
1 175 84
3 345 0.07375 10 8500 134 10 10
Q 942.478i
3 3 1L 1.569 10 2.518 10 6.398 7.463 10
Q 942.478i 131.85
L 7.14
s
8
1 1J m
π
=
⎡ ⎤⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎢ ⎥
⎝ ⎠ ⎝ ⎠⎢ ⎥+ + +
− −⎢ ⎥× ×⎣ ⎦
⇒ =
− − −⎡ ⎤× + × + + ×⎣
= −
⎦
= −⇒

193
The temperature on the outside of the lagging may now
be cross-checked as follows:
( )
( )
( )
( )
( )
( )
( ) ( )
( )
lagging's thermal resistance
total thermal resistance
T 6.398 2lagging
0.8950
T 7.148 2overall
T 0.8950 Tlagging overall
0.8950 444 294 134.25
T lagging
T ov
K
Temperature outside lagging
neglect n
er
i
all
= ⇒
Δ π
= = ⇒
Δ π
Δ = × Δ
= × − =
⇒
Δ
Δ
( )g T-drop in pipe wall
444 134.25 309.75 K
⎫
⎬
⎭
= − =

194
In the absence of lagging, under otherwise the same
conditions, the heat loss per unit pipe length will be:
( )( )
( )
( ) ( )
Q 2 T overalli
L r
po
ln
r
pi1 1
r h k r h
pi, i p po o
Q 2 150i
L 84
ln
1 175
3 34575 10 8500 84 10 10
π Δ
=
⎡ ⎤⎛ ⎞
⎢ ⎥⎜ ⎟
⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥+ +
⎢ ⎥
⎢ ⎥⎣ ⎦
π
⇒ =
⎡ ⎤⎛ ⎞
⎜ ⎟⎢ ⎥
⎝ ⎠⎢ ⎥+ +
− −⎢ ⎥× ×⎣ ⎦

195
Q 942.478i
3 3L 1.569 10 2.518 10 1.190
Q 942.478i
L 1.195
Q
i 789.0
L
Therefore, in this case, losses are 6 times higher
than those compar
1 1J s m
ed to the "lagged" case.
⇒ =
− −⎡ ⎤× + × +⎣ ⎦
⇒ =
≈
− −⇒ =

196
Economic Thickness of Lagging
Thickness of lagging ↑, loss of heat ↓ ⇒ saving in
operating costs.
However, cost of lagging ↑ with thickness ⇒ there will
be an optimum thickness of lagging when further
increase does not save sufficient heat to justify the cost.

197
Typical Recommendations (loose):
373-423 K, 150 mm dia. pipe: 25 mm thickness of
85% magnesia lagging
373-423 K, > 230 mm dia. pipe: 50 mm thickness of
85% magnesia lagging
470-520 K, < 75 mm dia. pipe: 38 mm thickness of
85% magnesia lagging.
470-520 K, 75 mm < dpipe ≤ 230 mm : 50 mm
thickness of 85% magnesia lagging.

198
Critical Thickness of Lagging
Lagging thickness ↑, resistance to heat transfer by
thermal conduction ↑, AND the outside area (circular
pipes) from which heat is lost to surroundings also ↑ ⇒
possibility of increased heat loss.
The above argument can be explained if we consider
the lagging to work as a large circular fin having very
low thermal conductivity.

199
TS TL TA
rp xlag
rp = radius of pipe
xlag = lagging thickness
TS = pipe temperature
TA = surrounding temperature
TL = outside temperature of lagging

200
Consider heat loss from a pipe (TS ) to surroundings
(TA).
Heat flows through lagging of thickness xlag across
which the temperature falls from a TS (at its radius rp),
to TL (at rp+xlag).
Heat loss rate Q from a pipe length L is given by (by
considering heat loss from outside of lagging),
( ) ( )
( )
... conVEctive
and,
Q h 2 L r x T T
o p lag L A
T T
S LQ k 2 L
lag
... conDUcti
x
r e
LM
v
lag
⎡ ⎤= π + −
⎣ ⎦
−
= π⎡ ⎤
⎣ ⎦

201
Equating Q given in equations above:
( ) ( )
( )
Q h 2 L r x T T
o p lag L A
x T Tlag S LQ k 2 L
lag r x x
p lag lag
l
an
n
r
p
d,
⎡ ⎤= π + −
⎣ ⎦
⎡ ⎤ −
⎢ ⎥= π
+⎛ ⎞⎢ ⎥
⎜ ⎟⎢ ⎥
⎜ ⎟⎢ ⎥
⎝⎣
⇒
⎠ ⎦
( ) ( ) ( )
k
lag
h r x T T T T
o p lag L A S Lr x
p lag
ln
r
p
⎡ ⎤+ − = −
+⎛ ⎞⎣ ⎦
⎜ ⎟
⎜ ⎟
⎝ ⎠

202
( )
( ) ( ) ( )
( )
( )
T T
S L a say
T T
L A
T T aT aT
S L L A
aT T
A ST
L 1 a
T
r xh p lago r x ln
p lagk r
lag pi
Q h 2 L r
herefore
x T
o p lag A
T T
S AQ h 2 L r
,
aT T
A S
x
o p lag
1 a
1 a
−
⇒
−
⇒ − −
+
⇒
+
+
+⎛ ⎞
⎜ ⎟= + =
⎜ ⎟
⎝ ⎠
=
=
⎛ ⎞
⎡ ⎤= π + −⎜ ⎟
⎝ ⎠⎣ ⎦
−⎛ ⎞
⎡ ⎤= π + ⎜ ⎟
+⎝⎦
+
⇒
⎠⎣

203
( )( )
( )
( )( )
( )
Also,
Theref
h 2 L r x T T
o p lag S A
Q
1 a
r xh p lagoa r x ln
p lagk r
lag p
h 2 L r x T T
o p lag S A
Q
r xh p lago1 r x ln
p
ore
lagk r
l p
,
ag
π + −
=
+
+⎛ ⎞
⎜ ⎟= +
⎜ ⎟
⎝ ⎠
π + −
=
⎧ ⎫+⎛ ⎞
⎪ ⎪⎜ ⎟+ +⎨ ⎬
⎜ ⎟⎪ ⎪⎝⎩
⇒
⇒
⎠⎭

204
( )( )
( )
( )
( )
( )
Differentia
h 2 L r x T T
o p lag
Q
h 2 L T T
o
S A
Q
r xh p lago1 r x ln
p lagk r
la
S A
r x
p l
ting
ag
r xh p lago1 r x ln
p lagk r
l
w.r.t. ,
g pi
x
l
a
g
g p
a
⎡ ⎤π −
⎣ ⎦
+
=
π + −
=
⎧ ⎫+⎛ ⎞
⎪ ⎪⎜ ⎟+ +⎨ ⎬
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⎧ ⎫+⎛ ⎞
⎪ ⎪⎜ ⎟+ +⎨ ⎬
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⇒

205
( )
( )
( )
( ) ( )
( )
r xh p lago1 r x ln
p lagk r
lag p
rh 1po
1
h 2 L T T
o S A
r xh p lago ln
k r
la
dQ
dx
g p
r x
p lag
r xh p lago1 r x ln
p
r x
p lag k r x r
lag p lag p
lagk r
lag p
lag
+⎛ ⎞
⎜ ⎟+
⇒
⎡ ⎤π −
⎣ ⎦
⎧ ⎫
⎪ ⎪
⎪ ⎪
⎪ ⎪
+⎪ ⎪⎛ ⎞
⎪ ⎪⎜ ⎟⎨ ⎬
⎜ ⎟⎪ ⎪⎝ ⎠− +⎪ ⎪
⎪ ⎪
+
⎜ ⎟
⎝
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣
+⎪ ⎪
⎪ ⎪⎩ ⎭=
⎧ ⎫+⎛ ⎞
⎪ ⎜ ⎟
⎠
+ +⎨ ⎬
⎜ ⎟⎪ ⎝
+
⎩
+
⎦
⎠
2
⎪
⎪⎭

206
Maximum value of Q (= Qmax) occurs at dQ/dxlag = 0.
( )
gives
... any addition of laggi
kh lag01 r x 0 x r
p lag lag pk h
lag 0
ng heat
k k h r
lag lag 0 p
x 0 r 1 1
lag ph h r k
0 0 p lag
DECREASES
k k h r
lag lag 0 p
x 0 r 1 1
l
loss
Also, gives
ag ph h r k
0 0 p lag
− + = = −
≤ ≤ ≤ ≥
> >
⇒ ⇒
⇒ ⇒ ⇒
⇒ ⇒> <

207
Thin layers of lagging heat loss
...
It is necessary to exceed MUCH BEYOND
critical thickness for reducing heat loss
... give
increases
h r
0 p
1
k
lag
Critical lagging thickness
k
lag
x r
lag ph
lag
MAXs I
⎧
⎪
⎪
⎨
⎪
= −
⎪⎩
=
<
heatMUM loss

208
( ) ( )
( )
( )
h 2 L T T r x
o S A p lag
Q
max r xh p lago1 r x ln
p lagk r
lag p
k
lag
h 2 L T T
o S A h
oQ
max k k hh lag lag oo1 ln
k h r
lag o p
⎡ ⎤π − +
⎣ ⎦
=
⎧ ⎫+⎛ ⎞
⎪ ⎪⎜ ⎟+ +⎨ ⎬
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⎡ ⎤π −
⎣ ⎦
=
⎧ ⎫⎛ ⎞
⎪ ⎪⎜ ⎟+⎨ ⎬
⎜ ⎟⎪ ⎠⎩
⇒
⎝
⇒
⎪⎭

209
( )k 2 L T T
lag S A
Q
max k
lag
1 ln
h r
o p
⎡ ⎤π −
⎣ ⎦=
⎧ ⎫⎛ ⎞
⎪ ⎪⎜ ⎟+⎨ ⎬
⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⇒
For an unlagged pipe: TL = TS, and xlag = 0. The rate of
heat loss Q0 for an unlagged pipe is given as:
( ) ( )
( ) ( )
Q h 2 L r x T T
0 o p lag S A
Q h 2 L r T T
0 o p S A
⎡ ⎤= π + −
⎣ ⎦
⎡ ⎤
⎣ ⎦
⇒ = π −

210
k
lag
h rQ o pmax
kQ
lag0
1 ln
h r
o p
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎝ ⎠=
⎛ ⎞
⎜ ⎟+
⎜
⇒
⎟
⎝ ⎠
The ratio Q/Q0 is plotted as a function of xlag (see next
slide)

211
( )Q Q
0 max
Q
Q
0
⎛ ⎞
⎜
↑
⎟⎜ ⎟
⎝ ⎠
lagging thickness xlag →
h r
0 p
1
k
lag
>
h r
0 p
1
k
lag
=
h r
0 p
1
k
lag
<
Critical thickness of lagging

applications of the principles of heat transfer to design of heat exchangers

Recomendados

Recomendados

Mais conteúdo relacionado

Mais procurados

Mais procurados (20)

Semelhante a applications of the principles of heat transfer to design of heat exchangers

Semelhante a applications of the principles of heat transfer to design of heat exchangers (20)

Último

Último (20)

applications of the principles of heat transfer to design of heat exchangers