Process heat transfer

Double pipe heat exchanger

Group members:

Sannan salabat butt (2007-CHEM-19)

Harris mehmood khan (2007-CHEM-99)

Discussion

�Double pipe heat exchanger

�Internal parts

�Diagrams

�Flow arrangements

�Calculations for L.M.T.D

� Advantages

� Limitations

� Comparison with conventional shall

and tube heat exchanger

� Design types

� Cost estimation

� Numerical problems

HEAT EXCHANGER:

Heat exchanger is a device in which two fluid streams , one hot & another cold are brought into ‘’ thermal contact

‘’ in order to effect transfer of heat from the hot fluid

stream to the cold.

DOUBLE PIPE HEAT EXCHANGER:

A typical double pipe heat exchanger basically consists

of a tube or pipe fixed concentrically inside a larger pipe

or tube.

OR

Heat exchanger which are used when the flow rates of

the fluids and the heat duty are small (less than 500 kW)

Construction of double pipe

� Hair pin: union of two legs

hairpin construction is preferred because it requires less space

� Packing & gland: The packing and gland provides sealing to the annulus and support the inner pipe.

� Return bend: The opposite ends are joined by a U-bend through welded joints.

� Support lugs: Support lugs may be fitted at these ends to hold the inner pipe position.

� Flange: The outer pipes are joined by flanges at the return ends in order that the assembly may be opened or dismantled for cleaning and maintenance.

� Union joint: For joining the inner tube with U-bend.

Contd….

� Nozzles: small sections of pipes welded to the shell or to the

channel which acts as the inlet or outlet of the fluids are called nozzles.

� Gaskets: Gaskets are placed between the two flanges to make the joint leak-free.

� Different types of gasketsNitrile rubber. Used up to 110 oC for mineral oils, dilute

mineral acids, and aliphatic hydrocarbons.

EPDM.

(ethylene-propylene-diene monomer)

Used up to 160 oC for mineral acids, or bases,

aqeuous solutions or steam

Viton.

( copolymer of vinylidine flouride and

hexafluoro-propylene)

Used up to 100 oC for hydrocarbons and

chlorinated hydrocarbons

Double Pipe Heat Exchangers

fluid flow passages & configuration

Basically there are two flow arrangements of double pipe heat exchanger:

� Co-current� Counter current

configuration

� Series & parallel arrangement

Co-current counter current

� Counter current

� max. heat transfer within minimum area due to more L.M.T.D

� Co-current

� Used for viscous fluids & gives lesser value of L.M.T.D

� Co & counter current gives same value of L.M.T.D if one of the fluid stream is isothermal (e.g steam)

� Series-parallel arrangement

This configuration is used when value of pressure exceeds its limits (500psig shell side and 500 psig tube side) .pressure drop problem can be solved by:

� Reversing the location of streams

� By-passing one of the fluid streams

� Dividing of stream at higher pressure drop( series-parallel arrag.)

CO CURRENT FLOW

∆

∆

∆−∆=∆

1

2

12

lnT

T

TTTLn

731TTTTT

in

c

in

h−=−=∆

1062TTTTT

out

c

out

h−=−=∆

COUNTER CURRENT FLOW

1062TTTTT

in

c

out

h−=−=∆

731TTTTT

out

c

in

h−=−=∆

T1T2

T4 T5

T3

T7 T8 T9

T10

T6

Counter - Current Flow

T1 T2T4 T5

T6T3

T7

T8 T9

T10

Parallel Flow

Log Mean Temperature evaluation

T 1

A

1 2

T2

T3

T6

T4 T6

T7

T8

T9

T10

Wall∆T1

∆ T2

∆ A

A

1 2

ADVANTAGES….

� Compactness

� Very high heat transfer coefficients on both sides of the exchanger

� Close approach temperatures in counter-current flow

� Ease of maintenance.

� Heat transfer area can be added or subtracted with out complete dismantling the equipment.

� High pressure ranges

(30 MPa shell side , 140 MPa tube side)

� High temperatures range

(600 C)

� Approach temperature: for counter flow arrangement the

approach is the number of degrees temperature between

the hot fluid inlet and the cold fluid outlet (T1-t2) or hot

fluid outlet & cold fluid inlet (T2-t1)whichever is small.

similar term is

� Range of temperature: The actual temperature rise or

fall for the hot fluid(T1-T2) & the cold fluid (t2-t1)

CONTD…..

� Ease of inspection on both sides

� Ease of cleaning

� Low cost

� No Local over heating and possibility of stagnant

zones is also reduced

� Fouling tendency is less

� low pressure loss

� Used for small applications

LIMITATIONS

� It is not as cost effective as most shell and tube

exchangers

� It requires special gaskets

� Limited volumetric capacity

� Fouling?

Contd..

Fouling :formation of a scale or a deposit on a heat transfer surface is called fouling

Types of fouling:

� Precipitation fouling ( due to dissolved salts of Ca & Mg )

� Particulate fouling( due to suspended particles )

� Corrosion fouling

� Chemical reaction fouling (due to deposits formed by chemical reactions)

� Bio fouling ( due to the attachment of bio chemical species )

� Solidification fouling ( due to sub cooling of fluids )

Comparison with shell & tube heat exchanger

shell & tube heat exchangers are:

� designed to withstand the greatest temperature andpressure condition

� Ideal for large scale applications

� Commonly used in petrochemical industry where

dangerous substances are present (protective

shell)

� Consists of very bulky or heavy construction, baffles are used to increase mixing

� Subject to water hammer and corrosion

� High pressure loses

Design types

In case of any design equipment , the design of a heat exchanger may be divided into two parts.

Process design Mechanical design

(Thermal design)

� Estimation of heat transfer area. Material of construction

� Determination of tube diameter. Thickness of tubes

� Number & length of tubes. Flanges, gaskets, support design

� Tube layout ( series or parallel )

� Shell & tube side pressure drops.(hydraulic design)

.

Designtypes

Mechanical design

Double pipe Heat exchangers

can be made with various materials:

� Carbon steel

� Alloy steels

� Copper alloys

� Exotic materials (tantalum)

Cost of heat exchanger

� Some of the major factors which influence the cost of heat

exchanger are :

� Heat transfer area

� Tube diameter and thickness

� Tube length

� Pressure of fluids

� Materials of construction

� Special design features ( finned surface,U-bends,removeable

bundles e.t.c )

DESIGN STEPS WITH SOLVED

EXAMPLE

1)Thermal design.

2) Hydraulic design.

ASSUMPTIONS

� The heat exchanger operates under steady state

conditions.

� No phase change occurs: both fluids are single phase and are unmixed.

� Heat losses are negligible

� The temperature in the fluid streams is uniform over

the flow cross section.

� There is no thermal energy source or sink in the heat

exchanger.

� The fluids have constant specific heats.

� The fouling resistance is negligible.

In thermal design we tabulate physical

properties of:

hot stream(Benzene)

cold stream(Water)

� Benzene(hot stream)� entering temp.= 75°C

� Leaving temp.=50°C

� average temp=62.5°C

� Sp.heat=1.88 kJ/kg °C

� Viscosity=0.37cP

� density = 860 kg./m3

� thermal conductivity = 0.154 W/m K.

� Flow rate = 1000 Kg/hr

� outer pipe spec.� i.d. = 41 mm

� o.d. = 48 mm.

� LMTD = ?

� Uo = ?

� Water(cold stream)� entering temp.= 30°C

� Leaving temp.=40°C

� average temp=35°C

� Sp.heat=4.187 kJ/kg °C

� Viscosity=0.8cP

� density = 1000 kg./m3

� thermal conductivity = 0.623 W/m K.

� Flow rate = ?

� Inner tube spec.� i.d=21mm

� O.d=25.4mm

� Wall thickness=2.2mm

� thermal conductivity of wall=74.5 W/m K.

Selection of tube & pipe fluid & flow

passage type� Flow rates

Cannot be considered because water side flow rate is not given

� Flow areasHigher mass flow rate stream is passed through greater flow area which cannot be considered because we don't know which stream is of higher flow rate

� Tube side fluidAs we know that water causes a lot of fouling and corrosion hence we will take water in the tube side in this way it would cause lesser damage to the heat exchanger.

� Pipe side/annulus sideBenzene will be taken on annulus side

� Flow arrangement

Counter current flow is selected because it reduces the required surface area

General design equation & steps

Q =Uo A (∆T)

� Step 1: Calculate (∆T) LMTD

� Step 2: Calculate heat duty Q

� Step 3: Calculate overall heat transfer co-efficient on the

basis of outer diameter of tube

� Putting all the three values will give us the required heat transmission area of double pipe.

� Such a problem in which we have to calculate size of

heat exchanger is called sizing problem

Calculation of LMTD (step 1)

benzene 75 C 50 C

water 40 C 30 C

∆t1=75-40=35°C ∆t2=50-30=20°C

L.M.T.D= (∆t1- ∆t2) / Ln (∆t1/ ∆t2)

LMTD =(35 – 20)/Ln(35/20)

= 26.8°C

Heat duty calculations(step 2)

� SOLUTION

(a) 1000 kg of benzene is cooled from 75°C to 50°C per hour.

Therefore,

Heat duty (Q) = m Cp (T2-T1)

= (1000 kg,/h)(1.88 kJ/kg °C)(75 – 50)°C

= 47,000 kJ/h

Heat given by the hot stream = Heat taken by the cold stream

Water is heated from 30°C to 40°C Therefore,

Water flow rate = Q / Cp x (t2-t1)

= 47000/(4187)(10)

=1122 kg/h

overall heat transfer co-efficient(step 3)

� Calculate convective heat transfer coefficient

for tube side (hi).

� Calculate convective heat transfer coefficient

for shell side (ho).

� Outside surface area of tube (Ao)

� Inside surface area of tube (Ai )

� Mean surface area (Am)

� 1/Uo=1/ho +(Ao/Am)x(ro-ri/kw)+Ao/Ai(1/hi)

Calculating hi( tube side water )

Velocity = volumetric flow rate / flow area

=0.9 m/sec

Reynolds number, Re = dvp/u

= (21 x 10-3)(0.9)(1000)/8 x 10-4

=23,625Prandtl number, Pr = Cpu/k

=(4.187)(1000)(8 x 10-4)/0.623

= 5.37

Use of Dittus-Boelter equation to calculate hi,

Nu = hidi/k = 0.023(Re)0.8(Pr)0.3

= (0.023)(23,625)0.8 (5.37)0.3 =120

Thus,hi=120x(k/di)=35660W/m2°C

Calculating ho( annulus side benzene )for annulus calculation we calculate hydraulic diameter

Flow area annulus = inner cross-section of the pipe - outer cross-section of the tube

= Pi/4(iD2) - Pi/4(OD1)=8.13x10-4 m2

wetted perimeter= Pi(iD2+OD1)=0.2086m

hydraulic diameter of annulus dh=4 x ( flow area/wetted perimeter)

=0.0156m

Contd…

Benzene mass flow rate = 1000 kg/h

Benzene volumetric flow rate = (1000)/(860) = 1.163 m3/hr

Velocity = volumetric flow rate / flow area = 0.397 m/s

Reynolds number, Re = dvp/u

= 14395

Prandtl number,Pr = Cpu/k

= 4.51

Calculation of ho from the Dittus-Boelter equation

Nu = hodi/k = 0.023(Re)0.8(Pr)0.3

=(0.023)(14395)0.8(4.51)0.4 = 89.12

ho = (89.12 x k/dh) = 879.8W/m2C

Contd…

outside area of tube = A0 = ∏ OD L = ∏(0.0254)(L)

inside area of tube = Ai = ∏ ID L = ∏ (0.021)(L)

Am = (OD-ID) / Ln (OD/ID)

= (0.0254 - 0.021)(∏L)/ Ln (0.0254/0.021)

= 0.023 (∏L)

A0/Am = 1.098

A0/Ai = 1.21

1/Uo=1/ho +(Ao/Am)x(ro-ri/kw)+Ao/Ai(1/hi)

Uo = 662.3W/m2K

Length of double pipe

Now calculate the required area from

Q = UoAo∆Tm

where,

Q = 1122 kg/h

Uo = 662.3W/m2K

∆Tm= 26.8 C

Ao = Q / Uo∆Tm= 0.74m2

Tube length necessary, L = Ao / ∏ OD1 L

= 0.74 / ∏ (0.0254)

= 9.3 m

Hydraulic design

� In hydraulic design involves calculations of

pressure drop on:

� The pipe side (annulus side)

� The tube side

Contd…

� ∆P = f G2 L / 2 g p Di Φ Where,

� F = friction factor

� G = mass velocity of the fluid

� L = length of the tube

� G =9.8m/s2

� p = density of tube fluid

� Di = inside diameter of tube

� Φ = dimensionless viscosity ratio

� ∆P =pressure drop

� ∆P( tube side ) = 1.476 x 10-4 kgf/m2

� ∆P( pipe /annulus side ) = 2.50 x 10-4 kgf/m2

Calculation on software

Auto-cad design (2D & 3D)

DESIGN PROBLEM :

Double Pipe Heat Exchanger

• Double pipe lube oil crude oil exchanger:6900lb/hr of 26 API lube oil must be cooled from 450 to 350F by 72500lb/hr of 34 API mid continent crude oil. The crude oil will be heated from 300 to 310F.

• A fouling factor of 0.003 should be provided for each stream, and the allowable pressure drop on each stream will be 10psi.

CONTINUED…

• A number of 20-ft hairpins of 3 by 2inch IPS are

available. How many must be used, and how shall they

be arranged? The viscosity of crude oil may be obtained

from graph. For the lube oil, viscosities are 1.4cp at

500F, 3.0 at 400F and 7.7 at 300F. These are enough to introduce an error if (u/uw)0.14=1 is assumed.

GIVEN DATA:

• Lube Oil:

• Mass flow

rate=wL=6900lb/hr

• 26 API

• Entering temp.=450F

• Leaving temp.=350F

• Viscosity =3.0cp at 400F

• Crude Oil:

• Mass flow

rate=wc=72500lb/hr

• 34 API

• Entering temp.=300F

• Leaving temp.=310F

• Viscosity = use graph

(1)HEAT DUTY CALCULATION :

� For lube oil: Q=Wcp(T1-T2)

=6900x0.62(450-350) cp(graph) =427000Btu/hr

. For crude oil:

Q=wcp(t2-t1)

=72500x0.585(310-300) cp(graph)

=427000Btu/hr

(2)a LMTD Calculation:

• LMTD = (∆ t)a- (∆ t)b/ln (∆ t)a/ (∆ t)b

(∆ t) = 87.5 F

It will be impossible to put the 72,500lb/hr into single pipe or

annulus, since the flow area of each is too small. Assume

it will be employed in two parallel streams.

(2)bTemperature difference (∆ t):

Hot fluid Temp. Cold fluid Diff.

450 F Higher temp. 310 F 140 F (∆ t)a

350 F Lower temp. 300 F 50 F (∆ t)b

_ _ _ 90 F

(∆ t)a - (∆ t)b

Concept of caloric temperature:

� In our problem we are given with petroleum fractions so we won’t use arithematic temperatures for evaluating physical properties. As in case of petroleum fractions, there viscosities show sharp variations with temperature and also overall heat transfer coefficient doesn’t remain constant. That is why we will use average caloric temperature for evaluating physical properties like viscosity, specific heat etc

(3)Caloric temperatures:

� (∆ t)c/ (∆ t)h =50/140

= 0.357

Kc factor =0.43

caloric temp. fraction (Fc) =0.395 (graph)

� Tc=350x0.395(450-350)=389.5 F

� tc =300x0.395(310-300)=304 F

Basic objective:

� In order to calculate clean overall heat

transfer coefficient Uc , we require two things.

� ho ( from annulus) lube oil

� hio (from inner pipe) crude oil

� Since Uc=hio xho/hio +ho

Concept of outer and inner diameter:

� We will always take inner diameter of inner pipe while

calculating the flow area in tube.

� In case of annulus inner diameter of outer pipe and outer

diameter of inner pipe (equivalent diameter) is considered.. table

Flow area calculations:

� Hot fluid (annulus)

� D2 =3.068/12 =0.256ft

� D1 =2.38/12 =0.199ft

� aa = 3.14(D22-D1

2)/4

� =0.0206ft2

� Equivalent dia. De=(D22-

D12)/D1

� = 0.13ft

� Cold fluid (inner pipe)

� D =2.067/12=0.172ft

� ap =3.14D2/4

� =0.0233ft2

� Since two parallel streams have been

assumed so half will flow

in each pipe.

Mass velocity calculations:

� Ga=W/aa

� =6900/0.0206

� =335000lb/hrft2

� At Tc=389.5F µ=3.0cp

� =3x2.42=7.25lb/hrft

� Rea=DeGa/µ=0.13x335000/7.25=6000

� If only two hairpins in series are required,L/D will be 2x40/0.13=614

� Use L/D=600

� jH=20.5

� Ga=w/ap

�

=72500/(2x0.0233)=1560000lb/hrft2

� At tc=304F, µ=0.83cp

� Rep=DGp/µ

�

=0.172x1560000/2.01=133500

� jH=320

Calculation of hio and ho :

� Tc=389.50F ,

� C=0.615Btu/lbF (graph)

� K=0.067Btu/hrft2(F/ft)

(graph)

� Pr=(cµ/k)0.33=(0.615x7.2

5/0.067)0.33=4.05

� tc =304F

� c=0.585Btu/lbF (graph)

� K=0.073Btu/hrft2(F/ft)

(graph)

� Pr=(cµ/k)0.33=(0.585x2.0

1/0.073)0.33=2.52

Continued…

� ho= jHxk/De(cµ/k)0.33xΦa

� ho/Φa =20.5x0.067x4.05/0.13 =42.7btu/hrft2F

� tw=tc+ (ho/Φa)/(hio/Φp)+(ho/Φa)x(Tc-tc)

� hi= jHxk/D(cµ/k)0.33xΦp

� hi/Φp =� 320x0.073x2.52/0.172=

34btu/hrft2F� (hio/Φp)=(hio/Φp)x(ID/

OD)

� =342x2.067/2.38=297

Continued….

� tw=304+42.7/(297+42.7)x(

389.5-304)

� =314F

� µw=6.6x2.42=16lb/fthr

� Φa=(µ/µw)0.14=0.9

� ho= ho/Φa xΦe

� =38.4

� As tw is calculated

� µw=0.77x2.42=1.86

� Φp=(µ/µw)0.14=1.0

� ho= hw/Φp xΦw

� =297x1.0=297

Clean overall & design overall co-

efficient….

• Uc=(hioxho)/(hio

ho)=297x38.4/(297+38.4)

=34.0btu/hrft2F

• 1/Ud=1/Uc+Rd

• Rd=0.003x2=0.006hrft2F/Btu

• Ud=28.2

38.4 h (outside) 297

Uc 34 ---

Ud 28.2 ---

Surface area….

� A=Q/(Udx∆t)=173ft2

� External surface per unit ft=0.622ft

� Required length=173/0.622=278lin ft

� This is equivalent to more than six 20-feet hairpins or 240 lin ft. since two parallel streams are employed, use eight hairpins or 320 lin ft. The hairpin should have the annuli connected in series and the tubes in two parallel banks of four exchangers. the corrected Ud will be =24.5.the corrected dirt factor will Rd =1/Ud-1/Uc=0.0114

Pressure drop calculations :

� De = D2 – D1

� = 0.058 ft

� Rea=( De x Ga ) /u

� =2680

� f = 0.0035+0.264/26800.42

� s =0.775 , p=62.5x0.775

= 48.4

� For Rep =133500

� f = 0.0035 +

0.0264/1335000.042

� = 0.005375

� s = 0.076 ,p = 62.5x0.76

=47.5

Continued…

� ∆Fa = 4f Ga2La / 2Gp2De

� =16.07 ft

� V=Ga / 3600 x p

� = 1.9 fps

� ∆ Fl =8(v2 /2G)

� =0.45 ft

� ∆Pa= (16.7 + 0.45) x 48.4 /144

� =5.8 Psi

� Allowable pressure drop =10Psi

� ∆Fp = 4f Ga2La / 2Gp2De

� =25.7 ft

� ∆Pp = 25.7 x47.5/144

� = 8.5 Psi

� Allowable pressure drop =10Psi

Graphical interpretation:

� Pressure drop description:

Designed double pipe heat exchanger:

� Configurated heat exchanger:

Double pipe heat exchange software

calculations:

Software calculations :

Continued…

Continued….

Industrial setup flow sheet of double pipe

heat exchanger:

� Process description:

Inner outer diameter description:

OD of pipe

ID of pipe

OD of tube

ID of tube

� Double pipe description:

Cross view of double pipe:

� Cross sectional view :

rori

Graphical interpretation:

� Pressure drop description:

Industrial setup flow sheet of double pipe

heat exchanger:

� Process description:

� Donald .Q. Kern (1950) ,heat transfer & applications( 2nd Design problem )

� Binay K.Datta,heat transfer principles and applications( 1st Design problem )

� Max S. Peters, Klaus D.Timmerhaus,Ronald E.West ,plant design and economics for chemical engineers (fifth edition)

� Yunus A.Cengel,Heat & Mass transfer,a practical approach (third edition)

� Y.V.C Rao , heat transfer principles � Incropera,F.P.,Dewitt D.P., Fundamentals of Heat and

Mass Transfer, 5th ed.,John Wiley & Sons Inc., NY,2000� Kakaç S. Heat exchangers selection, rating & thermal

design CRC Press, Fla, 1998

Books references

Internet references

� http://chentserver.uwaterloo.ca/courses/Che025Lab/perry/Chap11.pdf

� http://en.wikipedia.org/wiki/Heat_exchanger#Flow_arrangement� http://www.advantageengineering.com/fyi/110/advantageFYI110

.php� http://www.buildingdesign.co.uk/mech/guntner/dry-air-

coolers.htm� http://www.engineeringpage.com/heat_exchangers/tema.html� http://www.martechsystems.com/downloads/tech_managingreb

oilerops.pdf� http://www.me.wustl.edu/ME/labs/thermal/me372b5.htm� http://www.pacificconsultant.net/compact_heat_exchanger.htm� http://www.rwholland.com/hairpin.htm� http://www.taftan.com/thermodynamics/EXCHANGE.HTM� http://www.thomasnet.com/about/exchangers-heat-shell-tube-

26641001.html

Thermal conductivity from graph:

Specific heat calculation graph:

JH factor calculation using graph:

Caloric temperature graph:

Any question?.