Process heat transferDouble pipe heat exchanger

Group members:

Sannan salabat butt (2007-CHEM-19)

Harris mehmood khan (2007-CHEM-99)

DiscussionDouble pipe heat exchangerInternal partsDiagrams Flow arrangementsCalculations 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 gaskets

Nitrile 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

731 TTTTT inc

inh

1062 TTTTT outc

outh

COUNTER CURRENT FLOW

1062 TTTTT inc

outh

731 TTTTT outc

inh

T1T2

T4 T5

T3

T7 T8 T9

T10

T6

Counter - Current Flow

T1 T2T4 T5

T6T3

T7T8 T9

T10

Parallel Flow

Log Mean Temperature evaluation

T1

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)

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 and

pressure 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 areas Higher 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 fluid As 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 side Benzene 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/secReynolds 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)=0.2086mhydraulic 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/sReynolds number, Re = dvp/u = 14395 Prandtl number,Pr = Cpu/k = 4.51Calculation 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(D2

2-D12)/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=(0.615x7.25/0.067)0.33=4.05

tc =304F c=0.585Btu/lbF (graph) K=0.073Btu/hrft2(F/ft)

(graph) Pr=(cµ/

k)0.33=(0.585x2.01/0.073)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:

Heating Media IN

Heating Media OUT

ProductOUT

ProductIN

0 LLENGTH OF PIPE, ft

PR

ES

SU

RE

T

HR

OU

GH

PIP

E,

psi

Product

Heating Media

Heating Media IN

Heating Media OUT

ProductOUT

ProductIN

0 LLENGTH OF PIPE, ft

PR

ES

SU

RE

T

HR

OU

GH

PIP

E,

psi

atmospheric atmospheric

Co-Current Counter-Current

FIGURE 1: Double Pipe Co vs. Counter-Current

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:

HOLD TUBES

DOUBLE PIPE HEAT EXCHANGERCOUNTER CURRENT

PRODUCT TANK

TO FILLER

DRAIN

Water Steam

DRAIN

CONTROL VALVE

FIGURE 5: Final Process Setup

Inner outer diameter description:

OD of pipe

ID of pipe

OD of

tube

ID of tube

Double pipe description:

Graphical interpretation:

Pressure drop description:

Heating Media IN

Heating Media OUT

ProductOUT

ProductIN

0 LLENGTH OF PIPE, ft

PR

ES

SU

RE

T

HR

OU

GH

PIP

E,

psi

Product

Heating Media

Heating Media IN

Heating Media OUT

ProductOUT

ProductIN

0 LLENGTH OF PIPE, ft

PR

ES

SU

RE

T

HR

OU

GH

PIP

E,

psi

atmospheric atmospheric

Co-Current Counter-Current

FIGURE 1: Double Pipe Co vs. Counter-Current

Industrial setup flow sheet of double pipe heat exchanger: Process description:

HOLD TUBES

DOUBLE PIPE HEAT EXCHANGERCOUNTER CURRENT

PRODUCT TANK

TO FILLER

DRAIN

Water Steam

DRAIN

CONTROL VALVE

FIGURE 5: Final Process Setup

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_managingreboilerops.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

Any question….