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

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 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