applications of the principles of heat transfer to design of heat exchangers
DESCRIPTION
This file contain a very good description for the processes design of heat ex changer. the file courtesy is Prof. Anand Patwardhan ICT Mumbai (Deemed University)TRANSCRIPT
1
Momentum and Heat TransferApplications of the principles of heat transfer to design of
heat exchangers
Dr. Anand V. PatwardhanProfessor of Chemical EngineeringInstitute of Chemical Technology
Nathalal M. Parikh RoadMatunga (East), Mumbai-400019
2
Temperature driving force (log-mean ΔT)Consider a Double Pipe Heat Exchanger (DPHE):Cold stream OUT
T = Tc2
Cold stream INT = Tc1
Hot stream INT = Th1
Hot stream OUTT = 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 Hh
m H
h
Hc 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 Th h
Q m
P
c T Tc c Pc c
h h2 1
c
h
1 2Qc
= −
== −
4
Differential steady state heat balance for hot stream:
( ) ( ) ( )lU 2m c dT Th P R dh To o ch h= π −
RO = outer radius of the inner pipe
UO = overall heat transfer coefficient based on RO
Rearranging the above differential balance:
( )( ) ( )
( )( ) ( )
lFor
lFor
cold
hot stream
stream:
: ...
...
TcU 2 R dd
U 2 R ddT
T o
o oh 1T m
o
ch h
c 2T m cc
P
Th 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 1hT m c
d T 1c
h h PhU 2 R do oT m cc c Pc
−− ⎛ ⎞= π⎜ ⎟⎜ ⎟−
⎝ ⎠
Assuming UO independent of length l, and integrating between length zero (∆T1=Th1–Tc2) to L (∆T2=Th2–Tc1) :
( )T 1h2T m c
T 1c1ln U 2 R Lo oT m cc2 c Pch1 h Ph
− ⎛ ⎞= π⎜ ⎟⎜ ⎟−⎝ ⎠
−
6
( ) ( )( )
( ) ( )
( )( )
QhQ m c T T m ch h P
QcQ m c T T m cc
h h2 h1 h Ph
c Pc c1 c2 c Pc
Now,
T Tc1 c2
T T1 c
T Th2 h1
T T1 h2 h1m ch Ph T T m ch2 h1 h Ph
1 c2m c Qc Pc c
QcQc
= − ⇒ =
= − ⇒
−
−= =
−⇒ ⇒
=−
−⇒ =
7
( )( ) ( )
( )( ) ( )
( )( ) ( )
T TT h2 h1h2Th1
T Th2 h1
T TT c1 c2c1ln U 2 R Lo oT Q Qc2 c c
T Tc1 c2U 2 R Lo o Qc
T TT Th2 h1Th2Th1
c1 c2Q U 2 R Lc o o Tc1lnTc2
⎛ ⎞−−⎜ ⎟= π⎜ ⎟−⎝ ⎠
⎛ ⎞− − −⎜
−−
⎟= π⎜ ⎟⎝ ⎠
⎧ ⎫− − −⎪ ⎪= π ⎪ ⎪−⎨ ⎬⎪ ⎪−⎪ ⎪⎩ ⎭
⇒
⇒
8
( )( ) ( )
( )( ) ( )
( ) ( )
( ) ( )
log-mean
Similarly, log-mean
T T2 1T2lnT1
T T1 2T1ln
U 2 R Lo o
U 2 R Lo o
Q U 2 R Lc o o
Q 2 Lc Ui
T
T
TRi
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
ΔT1ΔT2
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
Exchangersupport Shell-side
inlet
Shell-sideoutlet
Baffles Tube-sideoutlet
Tube-sideinletFixed
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)
16
Shell-side flow
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-sideinlet
Shell-sideoutlet
Tube-sideoutlet
Tube-sideinlet
Fixedtube sheet
Floatinghead
Splitring
Tubesupport
Baffles
Exchangersupport
Tube-passpartition
Floatingtube sheet
STHE with floating tube sheet and head
20A typical tube bundle
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-sideinlet
Shell-sideoutlet
Tube-sideoutlet
Tube-sideinlet
Tubesheet
Saddlesupport
Tube supportsand baffles
Tube-passpartition
GasketsShellvent 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 onannual operating cost of condenser
Water velocity →
Cos
t →
Total overall cost
Depreciation
Operating cost
27
Classification of STHEBasis 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 endstationaryhead type
Rear endhead 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 shellwith
longitudinalbaffle
Double split flow Kettle typereboiler
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-passshell withlongitudinalbaffle
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 →
Cle
aran
ce =
(she
ll ID
–tu
be b
undl
e di
amet
er) →
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 Pd dt G fdG
P
fdt
gasket diameter, m
2design pressure, MN m2allowable working stress, MN m
tube sh
floating head tube
eet thickness, m
Thickness of is usually
calcul
ated as:
sheet
d 2 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
PTXC
C = clearancePTX = pitchPTY = pitch
Tube layout and pitch
47
PTY
PTX C
C = clearancePTX = pitchPTY = pitch
Staggered layout, Rectangular pitch
48
Equilateral triangular pitch
Φ = 300
CV = 0.5 PTCH = 0.866 PT
CV
CH
PTΦ
Y
49
In-line layout, Square pitch
Φ = 900
CV = PTCH = PT
PTΦ
PT
CH
CV
50
Staggered layout, Square pitch
CV
CH
PT
Φ
P T
Φ = 450
CV = 0.707 PTCH = 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:
bdbNumber of tubes N at do
⎛ ⎞= = ⎜ ⎟
⎜ ⎟⎝ ⎠
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 bas N d dt 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
DrillingsShell
Tubes
Baffles
58
Disk and doughnut baffle
Shell DoughnutDisk
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 T1 2Y
2 1
−=
θ − θ
( ) ( )X T 2 1 1 1= θ − θ − θ →
Cor
rect
ion
fact
or F
→
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 →
Cor
rect
ion
fact
or F
→
0.5
0.6
0.7
0.8
0.9
1.0
1.00.90.1 0.3 0.5 0.7
( )( )T T1 2Y
2 1
−=
θ − θ
T1
T2
θ1
θ2
Correction for LMTD for 2,4-STHE
62
Correction for LMTD for 3,6-STHE
63
Correction for LMTD for 4,8-STHE
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 TtubeOUT tubeIN shellIN shellOUTX ; Y
TshellIN tubeIN tubeOUT tubeIN
T T2 1 1 2X ; YT1 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 Th,I
T Tc,OUT c,INTc,IN
McP coldT Tc,
Th,I
N h,OUTY R in some books
N
MOUT c,IN cP hot
=−
−=
−=
−
68
Temperature profiles in 1,2-STHET1
T2θ1
θ2
T1
T2θ1
θ2
T1
T2θ1
θ2
T1
T2θ1
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 passSTHE, 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
T2θ1
θ2
T2
T1
θ1
θ2
(c)
71
Shell-side (longitudinal) baffles: difficult to fit andserious 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 unitsin series; initial cost is higher.
72
3 × 1,2-STHE in seriesEffectively: 3,6-STHE in series
θ1
θ2 T1
T1
73
For example, a STHE is to operate as following:T 455 K, T 372 K1 2
283 K, 388 K1 2388 2832 1X 0.61
T 455 2831 1T T 455 3721 2Y 0.79
388 2832 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, θ2should 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 0.922 573 473 373 473 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 aminimum 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 Transfer0.8 0.4Nu 0.023 Re Pr
0.40.8 cd vh d pi avei i 0.023
... Dittus-Boelter equation
0.7 Pr 1604... Re 10
L 1i
k k0
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 lnL4 6... 0.3 Pr 300; 4 10
r
Re 10 ; 10di
⎡ ⎤= −⎣ ⎦
< < × < < >
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
ulkh , Ti T T wallbulk 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:
0.14hi,corrected bulkh , T Ti 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 Dereplaces diameter d.
4 cross-sectional areaDe 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 OilDensity, ρ (kg/m3) 0.955 974 854Kinematic 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-Boelter 3 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 vi ave i ave d vi aveRe
ρ= = =
μ μ ρ ν
( )Re di
Re vave di
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 khi di
=
86
From hi, compute the convective heat transfer flux q:
( ) ( ) ( )q h T T h 372 328 h 44i 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 di
ρΔ=
Computation summary is presented on the next slide:
87
Fluid Nu hiJ/(s m2K)
qJ/(s m2)
vavem/s
ΔP/L(N/m2)/m
Air 199 239 10529 84 2456Water 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 transferand
pressure loss calculationsfor
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 0.14shell eq eq p0.36k k wall
0.55 1 3h d
vshell
mshd c 0.14shell eq eq p0 e.36k 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
hshelldeqkmsh
l mass fel
low rate of f
l
=
=
==
=luid on shell-side
shell cross-flow area total mass flow rate on shell-side
SS 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 temperawa l el turμ
94
PTC
C = clearancePT = 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
dsh
n a
ellPT
shels l diamat
C
LB
=
=
=
=
, and
Flow area associated with each tube betwe
dsh
en
ellN
baffl
e
e
T P
L
r
T
B
sC
=
=
=
96
PT
C
Flowarea
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 shellSS
dshell C LBT B
The
shell diam
equivalent diameter i
t
s
a er
PT
= = ×
= × =
4
deq
2 2 2 24 P d 4 P dT o4 flow area T o4 4dwetted perimeter do o
defined as,
where d outside diameter of tubeso
4
44
⎡ ⎤π⎛ ⎞ π⎡ ⎤− −⎜ ⎟⎢ ⎥ ⎢ ⎥× ×⎝ ⎠⎣ ⎦ ⎣ ⎦= = =π π⎛ ⎞
⎜ ⎟⎝ ⎠
=
98
Flowarea
PT
C
TRIANGULAR pitch tube layout
99
( )( )
( )
( )
22 dP 3 oT4 8triangular pitch deq d 2o
shell-side pressure loss22 N 1 f d mshell shellPshell 0.14 de
For tube layout:
Kern 1950 correlation for :
where, shell inside diameq wall
dshellf
ter
friction factor;
π−
=π
+Δ =
μ
=
=
ρ μ
[ ]
Number of
= number of across the
Does not account
N bafflesBN
for "leak
tube bund
ages" bet
+
ween baffle spaces
1 fluid pas lesesB
=
⇒
100
Shell-side Reynolds number Re →
Shel
l-sid
e fr
ictio
n fa
ctor
f→
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 24 0.
2 24
031
P dT o4deq do
0.75 0.0254
4 02513 m0.0254
π⎡ ⎤−
π⎡ ⎤−⎢ ⎥
⎢ ⎥⎣
⎦
=
⎣π
⎦ =π
=
Equivalent shell diameter deq:
( )( )0.5398 0.00635 0.12700.031
dshellS C LS BPT20.0137 m
75
1
= =
=
Cross-flow area at shell diameter (shell centre-line) SS:
106
total mass flow ratemshell at diameter of shellM kgT
shell cross-flow area 5.51880.01371
402.5 2S s mS
=
= ==
Mass flux (mass velocity) mshell:
v mave shed deq eqRe 252240.025l 1shel
.ll
3 402 5×= =
μ=
μ μ
ρ=
Reynolds number Re:
( )( )3 62.470 10 40cpPr 1 100.1324
7.481k
−= ==
×μ ×
Prandtl number Pr:
107
Shell-side heat transfer coefficient hshell:
( ) ( )
( ) ( ) ( )
h dshell eq 0.55 1 3Nu 0.36 Re Prshell shell shellk
0.55 1 3d m c0.36 k eq shel
0.36 0.13
l phshell d keq
W977.9
24 0.55 1 325224 7.4810.
4
025
2m
1
K
3
= ≈
μ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟≈ ⎜ ⎟μ ⎝ ⎠⎝
=
=
⎠
108
Calculation of pressure drop (loss) ΔPshell:
Calculate number of baffles in the shell (N):LS
L TB1 1
1 1
shell lengthNBB
4.8768 4.8768 35.950
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 mB shell shell B shel
22 3
l shellPshell 0.14 d d e
6 1 0.063 0.5
qeq wal
398 402
l
222.5730 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 ct 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:
QEQmax
=
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
Tem
pera
ture
→
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
Tem
pera
ture
→
Distance →
T2,2
T1
118
Cooler-Evaporator(non-isothermal hot fluid; isothermal cold fluid)
T2
T1,1
T1,2
T2 T2
M1
M2
Tem
pera
ture
→
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
Tem
pera
ture
→
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
M2T
empe
ratu
re →
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 1Q M c T Tmax P ,1 1 ,12= −
Over heat balance gives:
( ) ( )Q M c T T M c T TP ,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 TP ,1 ,2 ,1 ,2EM c T T T TP ,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 TP ,2 ,1EM c T TP
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,ORGΔTORG = 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 kgM 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
( ) ( )
( )
QEQmax
Actual heat lo
Actual heat
adMC T maxP m
loadMaximum he
in
400001333.
at load
3 350 290
0.5
=
=
=Δ
−
=
⇒
=
126
Heat Transfer Units
Number of heat transfer units (NHTU) is defined by:
( )( ) ( ) ( )
U ANHTUMcP min
Mc M c M cP 1 P1 2 P2minwhere 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 T1 2= −
T2,1
T1,1
T1,2
T2,1 T2,2
M1
M2
Tem
pera
ture
→
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 2T T 1 2
T T1 2
dQ M c dT M c dT2 P2 2 1 P1 1dQ dQdT dT2 1M c M c2
difference
Also,
andP2 1 P1
dT dT d T T1 2 1 2
= − = θ
θ
= = −
−= =
− =
= −
⇒ −
=
⇒
130
( )
1 1dQ U dA d U dA M c M c1 P1 2 P2
d 1 1U dAM c M c1 P1 2 P2
1 12 U AM c M c1 1 P1 2 P2
T T 1 11,2 2,2 U AT T M c M c1,1 2,
1 1d dQM c M c1 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 c1 P1 2 P2 1 P1 1 P1minT T 1 11,2 2,2 U AT T M c M c1,1 2,1 1 P1 2 P2
U A U ANHTUMc M cP 1 P1min
T T M cU A1,2 2,2 1 P11T T M c M c1,1 2,1 1 P1 2 P2
M c1 P1NHTU 1M c
If
ln
ln
T T1,2 2,2T T1,1 2,1 2 P
exp2
< =
− ⎡ ⎤= − +⎢ ⎥− ⎣ ⎦
= =
− ⎡ ⎤−= +⎢ ⎥
− ⎢
⇒
⇒
⇒⎥⎣ ⎦
⎡ ⎤= − +⎢ ⎥
⎢⎣
−
− ⎦⇒
⎥
⎧ ⎫⎪ ⎪⎨ ⎬⎪ ⎪⎩ ⎭
132
( )( )
( )
( )
( )
M c T TT T 2 P2 2,2 2,11,1 1,2 E ET T M c T T1,1 2,1 1 P1 1,1 2,1
T T E T T 1,1 1,2 1,1 2,1M c1 P1T T E T T2,2 2,1 1,1 2,1M c2 P2
M c1 P1T T E 11,2 2
Since and
and
Adding:
T T T T1,1 2,1 1,1 2,1,2 M c2 PDividing both sides by
2T1 1 ,
−−= =
− −
⇒ − = −
− = −
⎡ ⎤− + = +⎢ ⎥
⎢ ⎥⎣− −
⎦
( )T2,1− ⇒
133
M c1 P11 E 1M c2 P2
M c M c1 P1 1 P11 exp NHTU 1 E 1M
T T1,2 2,2T T1,1 2,1
M c1 P11 exp NHTU 1M c2 P2E
M c1 P11M c2 P
c M c2 P2
2
2 P2
⎡ ⎤⇒ − = +⎢ ⎥
⎢ ⎥⎣ ⎦
⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⇒ − − + =
−
−
⎧ ⎫⎡ ⎤⎪ ⎪− − +⎢ ⎥⎨ ⎬⎢ ⎥⎪
+⎢ ⎥ ⎢ ⎥⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦
⎪⎣ ⎦⎩ ⎭=⎡ ⎤+⎢ ⎥
⎢⎣
⎦⎩ ⎭
⎦
⎣
⎥
134
( ){ }[ ]
{ }[ ]
( )
{ }[ ]
[ ]
1 exp NHTU 1 1E1 1
0.5 1 exp 2
Special case: M c M c1 P1 2 P2
Special case: NHTU very hig
NHTU
1 expE2
1 0 0.52
h
=
⇒
→
− − +=
+
= − −
− −∞=
−= =
∞
⇒
135
T1,2
T1,1
T1,2
T2,1 T2,2
M1
M2
Tem
pera
ture
→
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 c1 P11 exp NHTU 1M c2 P2E
M c M c1 P1 1 P11 exp NHTU 1M c M c2 P2 2 P2
⎧ ⎫⎡ ⎤⎪ ⎪− − −⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭=
⎧ ⎫⎡ ⎤⎪ ⎪− − −⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
137
[ ]{ }1 exp NHTU 1 1EM c M c1 P1 1 P11 exp NHTU 1M c M c2 P2 2 P
Special case: M c M c1
INDETERM
P1
INAT
2 P2
21 1 0
1E
1 0
− − −=
⎧ ⎫⎡ ⎤⎪ ⎪− − −⎢ ⎥⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭
−= =
−
=
=
Therefore, expanding the exponential term gives:NHTUE
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 →
Eff
ectiv
enes
s E→
0.0
0.2
0.4
0.6
0.8
1.0
0 1 2 3 4 5
M c M c 01 P1 2 P2 =0.250.500.75
1.0
140
1,2-STHE
NHTU →
Eff
ectiv
enes
s E→
0.0
0.2
0.4
0.6
0.8
1.0
0 1 2 3 4 5
M c M c 01 P1 2 P2 =
0.250.500.751.0
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, 360or 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 loadMaximum hea
M c 167201 P1 0.5M c 334402 P2
QEQmaxM c 340 320 M c 340 3201 P1 1 P1
MC T 320 M c T 320P hot,1
t load
1 P1 hot,1min
= =
= =
− −= =
− −
⇒
143
( )
( ) ( )Therefore,
for T : Ehot,1
for T K :
340 320 20ET 320 T 32
Ehot,1
for T K : Ehot
0hot,1 hot,1
380 K 0.3333
370 0.4
36,1
for T K : Ehot
0 0.5
350 0.66,1 67
⇒
= =
= =
= =
= =
−= =
− −
144
( )P MIN
From graph for 1,2-STHE, is found as: T : NHTU 0.45hot,1 T K : NHTU 0.6hot,1 T K : NHTU 0.9hot,1 T K : NHTU 1.7hot,1
The area is calculat
NHTUfor 380 K
for 370
for 360
for 350
A NHTU(Mc )ed as:
for Th t,1
U
o
= =
= =
= =
= =
=
= : A
for T K : Ahot,1
for T K : Aho
2380 K 5.02 m
2370 6.69 m
2360 10t,1
for T K : Ahot
.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 ExchangerPlate
149
Because of shape of the plates, the developed area of surface is appreciably greater than the projected area.
150
Plate Heat ExchangerPlate and Gasket
151
Plate Heat ExchangerPlate and Gasket
Plate of suitable metal
Gasket of suitable polymer
152
Two-pass plate and frame flow arrangement
153
154
Welded PHE
155
Welded PHE
156
Plate Heat Exchanger (PHE)
157
Plate Heat Exchanger (PHE)
158
Plate Heat Exchanger (PHE)
159
Plate Heat Exchanger (PHE)
160
Air Cooled Heat Exchanger
161
Plate Heat Exchanger (PHE)
162
Plate Heat Exchanger (PHE)
163
Plate Heat Exchanger (PHE)
164
Plate Heat Exchanger (PHE)Schematic
165
Plate Heat Exchanger (PHE)Schematic
166
Plate Heat Exchanger (PHE)Schematic
167
Plate Heat Exchanger (PHE)Functioning
168
Plate Heat Exchanger (PHE)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, ammoniacalliquor, 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 ApproachHot 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.
177
Spiral HE - manufacture
178
Spiral Heat Exchanger
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 InsulationHeat Losses through Lagging
Heat loss (gain) from vessels and utility piping: radiation, conduction, convection.
Radiation: f(T14–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 overallQi r rpo lagoln lnr rpi po1 1
r h k k r hpi i p lag lago oQ 2 T overalliL rpoln l
rlagnr rpi p
o
k r hlag lago oHEAT LOSS RATE per
o1 1r h kpi, i
unit pp
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 mL
T Koverallr mpir mpo
r ml
ius of pipe,
outside radius of lagging,
convectivea
hi
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 K1 1 1k erial, J s
thermal
m Kp
klag 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 Khi = 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 overalliL r rpo lagoln ln
r rpi po1 1r h k k r hpi, i p lag lago o
444 294 150 K150 3 mm 75 mm 75 10 m
2168 3 mm 84 mm 84 1
rall
rpi
rpo
r rlago po
0 m2
50 84 50 mm 134 mm 134 10
π Δ=⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎢ ⎥+ + +⎢ ⎥⎢ ⎥⎣ ⎦
= − =
−= = = ×
−= = = ×
−= + = + = ×
Δ
= 3 m
192
( )
( ) ( )
Q 2 150iL 84 134ln ln
1 175 843 345 0.07375 10 8500 134 10 10
Q 942.478i3 3 1L 1.569 10 2.518 10 6.398 7.463 10
Q 942.478i 131.85 L 7.14
s8
1 1J m
π=⎡ ⎤⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎢ ⎥+ + +− −⎢ ⎥× ×⎣ ⎦
⇒ =− − −⎡ ⎤× + × + + ×⎣
= −
⎦
= −⇒
193
The temperature on the outside of the lagging may now be cross-checked as follows:( )( )
( )( )
( )( )
( ) ( )
( )
lagging's thermal resis tancetotal thermal resistance
T 6.398 2lagging 0.8950T 7.148 2overallT 0.8950 Tlagging overall
0.8950 444 294 134.25
T laggingT ov
KTemperature 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 overalliL rpoln
rpi1 1r h k r hpi, i p po o
Q 2 150iL 84ln
1 1753 34575 10 8500 84 10 10
π Δ=⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥+ +⎢ ⎥⎢ ⎥⎣ ⎦
π⇒ =
⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥+ +
− −⎢ ⎥× ×⎣ ⎦
195
Q 942.478i3 3L 1.569 10 2.518 10 1.190
Q 942.478iL 1.195
Qi 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 pipexlag = lagging thicknessTS = pipe temperatureTA = surrounding temperatureTL = 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 TS LQ k 2 L lag ... conDUctix
r eLM vlag
⎡ ⎤= π + −⎣ ⎦
−= π⎡ ⎤⎣ ⎦
201
Equating Q given in equations above:
( ) ( )( )
Q h 2 L r x T To p lag L A
x T Tlag S LQ k 2 Llag r x xp lag lagl
an
nrp
d,
⎡ ⎤= π + −⎣ ⎦
⎡ ⎤ −⎢ ⎥= π
+⎛ ⎞⎢ ⎥⎜ ⎟⎢ ⎥⎜ ⎟⎢ ⎥⎝⎣
⇒
⎠ ⎦
( ) ( ) ( )klagh r x T T T To p lag L A S Lr xp lagln
rp
⎡ ⎤+ − = −+⎛ ⎞⎣ ⎦
⎜ ⎟⎜ ⎟⎝ ⎠
202
( )( ) ( ) ( )
( )
( )
T TS L a sayT TL A
T T aT aTS L L AaT TA STL 1 a
T
r xh p lago r x lnp lagk rlag pi
Q h 2 L r
herefore
x To p lag A
T TS AQ h 2 L r
,aT TA S
xo p lag
1 a
1 a
−⇒
−
⇒ − −
+⇒
+
+
+⎛ ⎞⎜ ⎟= + =⎜ ⎟⎝ ⎠
=
=
⎛ ⎞⎡ ⎤= π + −⎜ ⎟
⎝ ⎠⎣ ⎦−⎛ ⎞
⎡ ⎤= π + ⎜ ⎟+⎝⎦
+
⇒⎠⎣
203
( ) ( )
( )
( ) ( )
( )
Also,
Theref
h 2 L r x T To p lag S AQ
1 ar xh p lagoa r x lnp lagk rlag p
h 2 L r x T To p lag S AQ
r xh p lago1 r x lnp
ore
lagk rl p
,
ag
π + −=
++⎛ ⎞
⎜ ⎟= +⎜ ⎟⎝ ⎠
π + −=⎧ ⎫+⎛ ⎞⎪ ⎪⎜ ⎟+ +⎨ ⎬⎜ ⎟⎪ ⎪⎝⎩
⇒
⇒
⎠⎭
204
( ) ( )
( )
( )( )
( )Differentia
h 2 L r x T To p lag
Qh 2 L T To
S AQ
r xh p lago1 r x lnp lagk rla
S Ar xp l
ting
agr xh p lago1 r x lnp lagk rl
w.r.t. ,
g pi
xl
a
g
g p
a
⎡ ⎤π −⎣ ⎦
+=
π + −=⎧ ⎫+⎛ ⎞⎪ ⎪⎜ ⎟+ +⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⎧ ⎫+⎛ ⎞⎪ ⎪⎜ ⎟+ +⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⇒
205
( )
( )
( )( ) ( )
( )
r xh p lago1 r x lnp lagk rlag p
rh 1po
1h 2 L T To S A
r xh p lago lnk rla
dQdx
g pr xp lag
r xh p lago1 r x lnp
r xp lag k r x rlag p lag p
lagk rlag 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 rp lag lag pk hlag 0
ng heat
k k h rlag lag 0 px 0 r 1 1lag ph h r k0 0 p lag
DECREASES
k k h rlag lag 0 px 0 r 1 1l
loss
Also, givesag ph h r k0 0 p lag
− + = = −
≤ ≤ ≤ ≥
> >
⇒ ⇒
⇒ ⇒ ⇒
⇒ ⇒> <
207
Thin layers of lagging heat loss
... It is necessary to exceed MUCH BEYONDcritical thickness for reducing heat loss
... give
increases h r0 p 1 klag
Critical lagging thickness
klagx rlag phlagMAXs I
⎧⎪⎪⎨⎪
= −
⎪⎩
=
<
heat MUM loss
208
( ) ( )
( )
( )
h 2 L T T r xo S A p lagQmax r xh p lago1 r x lnp lagk rlag p
klagh 2 L T To S A hoQmax k k hh lag lag oo1 lnk h rlag o p
⎡ ⎤π − +⎣ ⎦=
⎧ ⎫+⎛ ⎞⎪ ⎪⎜ ⎟+ +⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
⎡ ⎤π −⎣ ⎦
=⎧ ⎫⎛ ⎞⎪ ⎪⎜ ⎟+⎨ ⎬⎜ ⎟⎪ ⎠⎩
⇒
⎝
⇒
⎪⎭
209
( )k 2 L T Tlag S AQmax klag1 lnh ro 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 T0 o p lag S A
Q h 2 L r T T0 o p S A
⎡ ⎤= π + −⎣ ⎦
⎡ ⎤⎣ ⎦
⇒ = π −
210
klagh rQ o pmax
kQ lag0 1 lnh ro p
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠=
⎛ ⎞⎜ ⎟+⎜
⇒
⎟⎝ ⎠
The ratio Q/Q0 is plotted as a function of xlag (see next slide)
211
( )Q Q0 max
QQ0
⎛ ⎞⎜
↑
⎟⎜ ⎟⎝ ⎠
lagging thickness xlag →
h r0 p 1klag
>h r0 p 1klag
=
h r0 p 1klag
<
Critical thickness of lagging