Heat transfer Section 01

EDS 2004/EXC 1-1

Training Services

Shell & TubeHeat Exchangers

Presentation on shell and tube exchangers.

EDS 2004/EXC 1-2

Heat Exchangers - Course Content

Introduction to Heat Transfer MechanismsConduction, Convection, and RadiationOverall Heat Transfer CoefficientLog Mean Temperature DifferenceExample Problem - FCC Flue Gas CoolerHydraulicsExample Problems - Tube side Pressure Drop

Section 1 - Introduction to Heat Transfer

Table of Contents of First Section - Introduction and Basics of Heat Transfer

This may be a review of basic concepts but it is useful for establishing nomenclature and symbols.

EDS 2004/EXC 1-3

ConductionConvectionRadiation

Heat Transfer Basics

Three Forms of Heat Transfer

The three forms of heat transfer encountered in industrial situations are generally these.

Radiation heat transfer is of little consequence in most refinery heat exchangers and is generally ignored.

EDS 2004/EXC 1-4

Conduction Heat Transfer

Heat transfer within a material.Fourier’s equation governs.

Q = A k dTdx

Conduction heat transfer is the transfer of heat within a material. The material can be solid, liquid or vapor.

Q is the amount of heat transferred.

A is the area across with the heat is transferred.

k is the thermal conductivity of the material. It is a unique value for each material. It is normally a function of temperature. In this equation, the thermal conductivity at the average temperature should be used.

T is the temperature. dT is the differential temperature.

x is the distance that the heat is transferred. dX is the differential distance.

EDS 2004/EXC 1-5

Convection Heat Transfer

Heat Transfer between solids and liquids/vapors

Q = A h ∆T

where h is a function of physical properties and geometry

Convection heat transfer occurs between solids and either liquids and vapors in contact with the solid.

Q is the amount of heat transferred.

A is the area across with the heat is transferred.

h is the heat transfer coefficient. It is generally empirically found for fluids depending on their physical properties and the geometry of the flowing setup. Therefore, the h value is found using different equations for flow inside a tube and outside a tube, against a flat plate, using non-round shapes etc.

T is the temperature. ∆T is the difference in temperature.

EDS 2004/EXC 1-6

Conduction Heat Transfer

Thermal Conductivity of Various Materials

EXC-Roo-01

This chart shows the thermal conductivity of various materials.

Note that the vapor materials occurs at the bottom of the chart, liquids occur higher up, and solids near the top have the highest values of thermal conductivity.

The higher the value the easier the material transfers heat.

Most materials transfer heat better at higher temperatures, but not all.

EDS 2004/EXC 1-7

Example - Conduction

T = 100°C

BoilingWater

T = 30°C

Hand

100°C

30°C

dT/dx = - 35000 °C/m

T

x

2 mm

k (steel) = 46 kcal/(hr m°C)

dTdx

(100-30) °C0.002 m= = 35 000 °C/m

EXC-Roo-02

A simple example everyone can relate to - touching a pot of boiling water with your finger.

The inside of the the metal pot has water at its boiling point. The heat is conducted through the metal of the pot to your finger on the outside.

Knowing the thermal conductivity of the pot, the thickness of the metal, and the temperature of the water and your finger -

Continued

EDS 2004/EXC 1-8

Example - Conduction(continued)

Touch boiling kettle with a finger, A ≈ 1 cm x 1 cm = 0.0001 m2

⇒ Q = 0.0001 m2 x 1.61x106 kcal/hr/m2 = 161 kcal/hr = 189 W= 368 BTU/hr

=mC3500

Cm hrkcal46

AQ o

o

The amount of heat transferred to your finger can be calculated.

Using the various sets of units of measure.

It will be noted that the values shown here are actually very low compared to the heat transfer rates in refinery type equipment.

EDS 2004/EXC 1-9

Resistances in SeriesL 1 L 2 L 3

T 1 T 2

k 1 k 2 k 3

T hot T cold

T

x

Tube wall FoulingFouling

Tube metal has higher conductivity than foulingEXC-Roo-03

A slightly more complicated problem involves conduction through multiple layers in series.

In this case, three layers are shown. This could represent the case of a tube with fouling on both sides.

The amount the temperature drops in each layer is a function of the relative values of thermal conductivity. The higher the value the less the temperature drop as there is less resistance to heat transfer.

EDS 2004/EXC 1-10

Resistances in Series (continued)

Heat flux is calculated as follows:

( )cThTeffk

3k3L

2k2L

1k1L

cThTAQ

−=

++

−=

1

3k3L

2k2L

1k1L

effk−

++=so

The last example can be substituted into the conduction equation. By making up a factor called k eff (effective thermal conductivity), the series conduction problem looks just like the single layer problem.

In other words, the effective resistance is the sum of all the resistances. L1/k1 is the “resistance” of the layer 1.

Once you have Q, any intermediate temperatures can be calculated:

Qk

LA

Th T1

1

1

= − get T1, etc.⇒

EDS 2004/EXC 1-11

Convection Heat TransferTubeside heat transfer coefficientUse Dittus-Boelter Equation:

h kD

. C

k D v

ii

p/

i.

=

0 0231 3 0 8µ ρ

µ

collect variables:

( )h . kD

C v i

/

i. p

/ . .= −0 0232 3

0 21 3 0 467 0 8µ ρ

h DD

hioi

oi=

increase this hi ....k increasesDi DECREASESCp increasesµ DECREASESρ increasesv increases

k = Thermal conductivityCp = Specific heatµ = Viscosityρ = Densityv = Velocity

Di = Inside diameter

Correct for Outside Surface Area:

Convection heat transfer is the second type to be discussed.

In convection the heat transfer is found empirically for each type of geometry. This equation covers single phase heat transfer in a circular tube. The heat transfer coefficient is a function of the physical properties of the fluid.

The table on the left shows the effect of changing each of the variables in the equation. Note that the two factors that hurt heat transfer are increasing diameter and increasing viscosity. Heat transfer is worse in larger tubes and with more viscous fluids.

The last equation corrects this coefficient, which is calculated based on inside diameter to the outside diameter of the tubes. All of the heat transfer equations refer to the area across which the heat is transferred. To make that a consistent area is used, all calculations are based on the area of the outside diameter of the tubes. Therefore, this inside coefficient must be corrected.

EDS 2004/EXC 1-12

Convection Heat Transfer(continued)

Shellside CoefficientEquation from Kern:

h . kD

D v C ko

e

e.

p/

w

.

=

0 36

0 55 1 3 0 14ρ

µ

µ µµ

Collect Variables:

( )h . k

D v C o

/

e. .

.p

/

w

.

=

0 36

2 3

0 45 0 22

0 55 1 30 14

µρ

µµ

The last term (viscosity correction) is usually small, and we will ignore it.

increase this ho ....k increasesDe DECREASESCp increasesµ DECREASESρ increasesv increases

This equation is the convection equation for fluid flowing on the outside of a circular tube. The form of the equation is similar. An additional term that corrects for the viscosity of the fluid at the wall temperature is present.

Again, increasing the diameter and viscosity hurts the heat transfer rate.

The diameter here is an equivalent diameter that will be covered later.

EDS 2004/EXC 1-13

Combined Conductionand Convection Coefficients

L

k tube

h 1 h2

T

X

Tube wall

hot cold

EXC-R00-04

Combining the conduction and convection problems.

Consider a flat surface with fluids on each side. The heat is transferred from the hot fluid to the surface by convection. Conduction occurs in the surface. A second convection rate is on the other side of the surface.

This looks similar to the series conduction problem.

EDS 2004/EXC 1-14

Combined Conduction and Convection(continued)

( )QA

T TLk

Lk

Lk

k T Th ceff h c=

−

+ +

= −1

1

2

2

3

3

substitute as shownh1 h2 U for keff (convention)

( )QA

T T

hL

k h

U T Th c

tube

h c=−

+ +

= −1 1

1 2

There may be a few solid layers present (tube, 2 fouling layers).

If you look at the equation for the series conduction problem and substitute the convection heat transfer terms, our effective conductivity term becomes an effective heat transfer term for the overall problem. This is call U or the Overall Heat Transfer Coefficient.

EDS 2004/EXC 1-15


The overall resistance is the sum of the individual resistances.

( )ch

ofo

fo

tubeif

if

i

ch TTU

h1

kx

kL

k

xh1

TTAQ

−=

++++

−=

or

Q = U A ∆T

The U value term can be expanded to include the fouling layers on both sides of the surface.

The U value becomes an easy way to relate the amount of heat being transferred to the area and the temperature difference.

EDS 2004/EXC 1-16

Tube wall is curved; outside area is different from inside area: tubewall resistance is calculated as:

1

otube

i

oeo

i

o

i h1

krrlogr

rr

h1U

−

+

•

+

=

or if you include fouling resistances1

foi

ofi

otube

i

oeo

i

o

ir

rrr

h1

krrlogr

rr

h1U

−

+

++

•

+

=


Round Geometry

Up to this point, the surface we showed were flat. Most tubular heat exchangers use round geometries. A slight correction needs to be made due to the curvature of the tubes versus a flat surface.

Therefore, the bottom equation describes the U value for most tubular type heat exchangers.

This equation shows that BEFORE combining resistances they must be based on the same area. The equation shown is based on the OUTER tube area, which is the convention in the heat transfer industry.

EDS 2004/EXC 1-17

When calculating duty, what ∆T should be used here?

t out = 130

t in = 80

T in = 150

T out = 90

Temperature Difference

mean log

inout

outine

inoutoutineff T

tTtTlog

)t(T)t(TT ∆=

−−

−−−=∆

Q=UA∆T is the basic equation for heat transfer in exchangers.

We have just talked about the U value.

The A or area is that of the outside of the tubes.

We still need to discuss the ∆T term.

In a typical exchanger, there are two fluids entering and leaving. Each of these four points has a temperature. Which temperature difference do we use?

For the exchanger shown in the last picture, one where the two fluids flow in precisely opposite direction (counter current flow), the temperature difference is found using this formula. This is referred to as the log mean temperature difference of LMTD.

EDS 2004/EXC 1-18

Temperature Difference(continued)

In practice, pure countercurrent flow situations often do not exist.For example: Most exchangers have more than one tube-pass.

coldincoldout

hotin

hotout

a

b c

d

EXC-Roo-06

However, few if any exchangers are really using counter current flow. This diagram shows a more typical exchanger. The tube side fluid travels the length of the exchanger twice. One trip in the bottom half and the return in the top half. At the same time, the shell side fluid travels once along the length of the exchanger, but many times back and forth across it.

Again, which temperature difference is used here?

EDS 2004/EXC 1-19

Temperature Difference(continued)

∆T in the top half of the bundle > ∆T bottom half.What is the EFFECTIVE TEMPERATURE

difference in the exchanger?What is ∆T at a, b and c and d?

This is a difficult question. The answer varies depending on all 4 inlet/outlet temperatures.

We know that ∆T effective is LESS than ∆Tlm.

⇒

The temperature differences are different at each end of the bundle and the difference between the temperature of the two fluids is different depending where along its journey it is.

The bottom line is that the temperature difference is less than that in a true counter current flow case.

We will return to this problem later.

EDS 2004/EXC 1-20

Example Calculation

FCC Flue Gas Cooler

Flue gas on the tubeside.Boiler feed water is on the shellside.

Flue Gas: 51500 kg/hr, in at 740°C, density = 0.81 kg/m3,Cp = 0.259 kcal/kg/°C, µ = 2.2 x 10-5 kg/s-m,k = 0.029 kcal/hr/m/°C.

Example calculation problem.

The flue gases from an FCC regenerator are often cooled to protect pollution control equipment or merely to conserve energy. The cooling is typically done in a waste heat steam generator. In this case, the flue gas will flow on the inside of the tubes while boiler feed water and steam is on the shell side.

The flue gas conditions and properties are listed here.

EDS 2004/EXC 1-21

Example Calculation(continued)

a) Calculate hio if there are 350 tubes, 1.75" OD (44.45 mm), 0.2" (5.1 mm) wall thickness.

b) Given the BFW coefficient as

ho =8 600kcal/hr/m2/°C, bulk water temperature = 240°C and using tube metal thermal conductivityof = 37 kcal/hr/m/°C, calculate tubewall temperatures.

c) What happens if BFW supply is lost ?

Part A - Calculated the inside heat transfer coefficient for the flue gas stream using the tube data.

Part B - Given the shell side heat transfer coefficient and conditions, calculate the tube wall temperatures.

Part C - Predict what will happen if the supply of boiler feed water is lost.

EDS 2004/EXC 1-22

Solution to Problem - FCC Flue Gas CoolerPart a)1) Calculate velocity:

Volumetric flow ms

ms

=•

=51500

0 81 360017 66

3 3

..

( )tube flow area m=π − •

=3500 04445 2 0 0051

40 323

22. .

.

2) Calculate Reynolds number

Re ( . . ) . ..

=µ

=

− • • •

=−

D vx

i ρ 0 04445 2 0 0051 54 6 0 812 2 10

68 8515

3) Calculate Pr

Pr. .

..=

µ

=

• •

=

−Ck

xp 0 259 2 2 10 36000 029

0 7065

velocity m s= =17 660 323

54 6..

.

Part A -

First calculate the flue gas velocity.

Then calculate the Reynolds Number.

Then find the Prandtl Number.

EDS 2004/EXC 1-23

Solution to Problem - FCC Flue Gas Cooler (continued)

4) Calculate inside heat transfer coefficient

( ) ( )

h kD

Ck

D v

h

h kcal hr m C

ii

p i

i

i

=µ

µ

=−

= °

0 023

0 0290 04445 2 0 0051

0 023 0 706 68851

128 6

1 3 0 8

1 3 0 8

2

.

.. ( . )

. .

. / / /

/ .

/ .

ρ

5) Correct to tube outside diameter

hio = 0.7705 hi = 99.1 kcal/hr/m2/°C

Now calculate the inside heat transfer coefficient and correct it for the outside area of the tubes.

EDS 2004/EXC 1-24

Solution to Problem - FCC Flue Gas Cooler(continued)

Part b)1) Calculate tubewall resistance

resis cer r

rk

Chr mkcal

o eo

i

tubetan

log

. log .. . .

=•

=•

− •

=

°0 04445

20 04445

0 04445 2 0 005137

0 000152

Part B -

Calculate the resistance in the tube wall itself. Remember the tube is round.

EDS 2004/EXC 1-25


2) Calculate U

Uh

r rr

k hr r

rr

U kcalhr m C

io

o eo

i

tube ofi

o

ifo= +

•

+ +

+

= + + + +

=°

−

−

1 1

199 1

0 00015 18600

0 0 96 55

1

1

2

log

.. .

QA

kcalhr moutside

= − =96 55 740 240 48 276 2. ( )⇒

Using the inside heat transfer coefficient, the given outside coefficient, the calculated tube wall resistance, and ignoring any fouling resistances, the Overall Heat Transfer Coefficient (U) is now calculated.

EDS 2004/EXC 1-26


3) Calculate hot side tubewall temperature:

( )

( ) ( )

( )

QA

U T

QA

kcal h T T

T C

T C

io w hot w hot

w hot

w hot

=

= = − = − =

= − = ° =

= = °

∆

48 276 740 99 1 740

740 487 1

252 9

hr m2 , ,

,

,

.

.

.

Knowing the value of Q/A and U, the temperature differences in each layer can be calculated. Using the inside heat transfer coefficient, and the inside bulk fluid temperature, the temperature at the outside of the tubewall can be found.

EDS 2004/EXC 1-27


4) Calculate cold side tubewall temperature:

( )

( )

QA

kcalhr m resis ce

. T

. T kcalhr m

. C hr mkcal

. C

T . . C . C

w,cold

w,cold

w,cold

= = • − =

− = •°

= °

= − ° = °

48 276 1 252 9

252 9 48 276 0 00015 7 3

252 9 7 3 245 6

2

2

2

tan

Summary:hio = 99.1 kcal/hr/m2/°Cho = 8 600 kcal/hr/m2/°C Tubewall temperature: 253°C to 246°C

Using the fact that the Q/A term is a constant, the temperature drop in the tube itself can be found.

In this case, note that the heat transfer coefficient is very high on the outside of the tubes compared to that on the inside. Therefore, most of the temperature drop is in the flue gas itself. The temperature of the tube is much closer to that of the water than the temperature of the flue gas.

EDS 2004/EXC 1-28

Solution to Problem - FCC Flue Gas Cooler (continued)

Part c)

If BFW is lost, tube will soon reach 740°C since ho approaches 0CS tubes cannot tolerate 740°C, and will crack Time to failure is in a few minutesTube failure allows steam into flue gas line

Part C -

It is obvious that if the water on the outside of the tubes is lost, the temperature of the tubes will approach the temperature of the flue gas. This will result in a tube failure.

EDS 2004/EXC 1-29

Hydraulics

Tubeside ∆P is easy to calculate.Calculations similar to pipe flow calculations.

19.0i

t

i

2t

t

vD046.0f where

DnLvf2P

−

µρ

≈

ρ=∆

L = Length of tubesn = Number of passesft = Friction factor

Changing topics:

Pressure drop on the tube side of an exchanger is similar to pipe flow pressure drop. The flow is generally split among many tubes in parallel and may have several directional changes.

The equation for fluid flow in pipes is adequate.

The friction factor is a function of NRe and tube diameter and is found in the literature.

At high NRe, f is constant for a fix tube size.

Tube ID, in. f 0.2 0.0360.25 0.0340.3 0.0320.4 0.0290.5 0.0270.75 0.0241.0 0.0231.5 0.021

EDS 2004/EXC 1-30

Friction Factor Chart

EXC-Roo-07

The friction factor needs to be found using a chart like this or the simplifying equation on the last page.

EDS 2004/EXC 1-31

Hydraulics(continued)

Shellside ∆P Need to know what area to use. Use flow area between tubes at center of shell:

shell

tube-tubeclearance=p-D

bundle-shellclearance

# tubes = D s/p

t

( )A Dp

p D Bfss

t s= −EXC-Roo-08

Ds = Shell diameterp = pitch

Bs = Baffle spacing

Pressure drop on the shell side is a much harder problem. As the flow progresses, it is flowing radially and longitudinally in a cylinder. The number of tubes it is flowing past and the diameter of the cylinder is constantly changing.

This procedure provides a simplified solution to the problem based on using the geometry at the center of the bundle. It assumes a uniform baffle spacing along the axis of the exchanger.

To get more accurate calculations, use a computer simulation such as HTRI or HTFS etc.

EDS 2004/EXC 1-32

Hydraulics(continued)

For small bundles or accurate calculation, allow for bundle-shell clearance.

∆ = +

=

=µ

−

P f D N vD

v WA

f D v

ss s b

e

s

fs

se

( )

..

12

1 79

2

0 19

ρ

ρ

Dp D

De

t

t

=−π

π

44

22

Nb = Number of bafflesWs = Volumetric flow rateDe = Equivalent diameter

Additional parameters for the calculation are given here. Note this is where the equivalent diameter mentioned for the shell side heat transfer coefficient is defined.

EDS 2004/EXC 1-33

Hydraulics in General

∆varies with flow rate, length, and diameter scales:

∆ ∝ ⋅ ⋅ ⋅

⇒∆∆

=

P f geometry fluid flow v LengthScaleDiamScale

PP

vv

LL

DD

( , , ) 2

2

1

2

1

22

1

1

2

ρ

Length Scale (L) is length measured in direction of flow (tube length, or shell diameter).Diameter Scale is a length measured perpendicular to flow (tube ID or equivalent diameter).This general form is valid for both shell and tubesideFor highly turbulent flow, f does not change much with increase of flow rateExceptions: crude and atmospheric residue, asphalt etc. Then variation of f with flow rate must be included.

In general, it is possible to compare the effect on pressure drop by modification to the exchanger quite simply.

The pressure drop ratio is the ratio of the velocity squared, the difference in flowing length, and the difference in flowing diameter.

This is true for either side of an exchanger. It is slightly less precise on the shell side due to the more complicated flow paths.

EDS 2004/EXC 1-34

Quick Estimations

Example 1If existing exchanger has tube ∆P = 70 kPa, and revamp needs extra 10% flow:

∆Prevamp ≈ 70 kPa (1.1)2 = 70 (1.21) = 85 kPa

Example 2A heat exchanger design has 2 tube passes, ∆Ptube = 15 kPa (too low).

Someone proposes changing to 4 tube passesvelocity doubles, number of passes doubles:

∴

∆ = • • = • =P kPa15 2 2 15 8 1202 (too high).

Example 1 - Increase the flow by 10%

This increases the velocity by 10% and has no effect on length or diameter so the result is simply the square of the velocity or 21%.

Example 2 - Going from 2 to 4 tube passes

This doubles the velocity as half the tubes are available for each pass. It also doubles the length as the flow goes along the axis of the exchanger 4 times instead of twice. Therefore, the pressure drop goes up by 2 squared due to velocity times another 2 for length for a total of 8 times.

EDS 2004/EXC 1-35

Quick Estimations(continued)

Example 3Retube existing exchanger from 1" OD 16 BWG (0.065" wall) tubes to 1" OD 14 BWG (0.083" wall) tubes (same flow rate)

Old Tube ID is = 1" - 2 (0.065") = 0.870"New Tube ID is = 1" - 2 (0.083") = 0.834”

Therefore, this retubing will increase the pressure drop by 24%.

∆∆

=

=

=

=

PP

vv

14

16

14

16

2

2 2 5

0 8700 834

0 8700 834

0 8700 834

0 8700 834

1 24

.

.

.

...

.

..

Example 3 - Retubing the exchanger with slightly thicker tubes

The velocity increase by the square of the change in inside diameter. Therefore, the pressure drop increases by the ratio of the diameters to the fourth due to the velocity and once more due to the change in diameter. Therefore, the ratio of pressure drops is the 5th power of the diameter ratio. So the 4% change in diameter results in a 24% change in pressure drop.

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Heat transfer Section 01