1

FIRST SESSIONAL ASSIGNMENT

Design a 1-2horizental condenser for the condensation of 60,000lb/hr of pure n

propanol coming from the top of distillation column operating at 15 psig, at which it

boils at 244 oF. Water at 85

Submitted to:

Dr. ING. NAVEED RAMZAN

Prepared by:

ASIF NADEEM TABISH 2005

ZEESHAN TALIB 2005

USMAN FAROOQ 2005

QAIM ALI

MUHAMMAD TARIQ 2005

FIRST SESSIONAL ASSIGNMENT

2horizental condenser for the condensation of 60,000lb/hr of pure n

propanol coming from the top of distillation column operating at 15 psig, at which it

F. Water at 85 oF will be used as cooling medium.

Dr. ING. NAVEED RAMZAN

ASIF NADEEM TABISH 2005-Chem-03

ZEESHAN TALIB 2005-Chem-07

USMAN FAROOQ 2005-Chem-53

QAIM ALI 2005-Chem-87

MUHAMMAD TARIQ 2005-Chem-89

2horizental condenser for the condensation of 60,000lb/hr of pure n-

propanol coming from the top of distillation column operating at 15 psig, at which it

03

07

53

87

89

2

WORKING PRINCIPLE OF EQUIPMENT

Heat Transfer from a hot stream to cold stream can be accomplished by following heat

transfer mechanism

1. Conduction

2. Convection

3. Radiation

In exchangers heat transfer occurs by a combination of 2 or 3 of these heat transfer

mechanisms.

Steady State Heat Transfer

When the rate of heat transfer remains constant and is unaffected by time,

then the flow of heat is steady state.

Unsteady State Heat Transfer

An unsteady state exists when the rate of heat transfer at any point varies

with time.

Most industrial processes in which heat transfer is involved are assumed to

operate under steady state conditions even though in some cases unsteady state

conditions are observed e.g.

• During start up

• Cool down

• Surge conditions

Unsteady State Conditions are observed in:

• Batch Process

• Cooling & heating of material .e.g. polymer & glass

• Certain types of regeneration

Steady State Heat Transfer Considerations:

Most cases of heat transfer in exchangers involve the flow of heat from one

fluid through a retaining wall to another fluid. The heat that is transferred flows from

warmer fluid to colder fluid through several thermal resistances in series.

Total resistance to heat transfer

Rtot = (Th _ Tc) /q Where,

Th = Temperature of hot fluid

3

Tc = Temperature of cold fluid

Rtot is comprised of

• Resistance due to convective heat transfer 1 / hh

• Resistance due to fouling on warm side Rhf

• Resistance to heat transfer through wall x/ kw

• Resistance due to fouling on colder side Rcf

• Resistance due to convective heat transfer in cold fluid 1/ hc

so

Rtot = 1/U = 1/ hh + Rhf + x/ kw + Rcf + 1/ hc

Consequences of Heat Transfer

Heat transfer to a fluid or from a fluid may lead to

• Vaporization (The change of Phase from liquid to vapor)

• Condensation (The change from vapor to liquid)

• Super heating of vapor ( Change in sensible heat of vapor)

• Sub-cooling of condensed liquid ( Change in sensible heat o liquid)

4

CONDENSATION

The process of heat transfer accompanied by a phase change from vapor

to liquid at constant pressure is called condensation.

• Since vapor liquid heat transfer usually occurs at constant pressure, the

condensation of a single compound normally occurs isothermally.

• If a mixture of vapors instead of a single compound is condensed then

condensation may not take place isothermally, there may be sub-

cooling of condensed liquid.

But in our case we have pure n-propanol, so there is only phase change at

isothermal conditions.

CONDENSATION MECHANISMS

1) Drop-wise Condensation

When a saturated pure vapor comes in contact with a cold surface such as

tubes, it condenses and may form liquid droplets on the surface of tube, such a type is

drop wise condensation.

2) Film-wise Condensation

When a distinct film of condensed vapors appears & coats the tube and

additional vapor is required to condense onto the liquid film, then it is film wise

condensation

COMPARISON OF TWO CONDENSATION MECHANISMS

Due to the resistance of condensate film to the heat passing through it the

heat transfer coefficient for drop wise condensation are 4 to 8 times greater than film

wise condensation.

But drop wise condensation occurs rarely because special conditions are

required for it to occur. It occurs by the presence of dirt on the surface or by the use of

a contaminant that adheres to the surface. Steam is the only pure vapor known to

condense in a drop wise manner. Drop wise condensation also occurs when several

materials condense simultaneously as mixture and where the condensate mixture is

not miscible as in the case of hydrocarbons and steam but we have pure n-propanol so

mechanism is of film wise.

Film wise condensation may be considered as self diffusion process. The

saturation pressure of the vapor in the vapor body is greater than the saturation

pressure of the cold condensate in contact with the cold surface. This pressure

5

difference provides potential for driving vapors out the vapor body at a great rate.

Here the controlling resistance is the film of condensate on cold tube wall. It is the

slowness with which the heat of condensation passes through this film that

determines the condensing coefficient.

NUSSELT THEORY FOR CONDENSATION ON SURFACE

The following assumptions are involved for condensation on surfaces as given by

Nusselt.

• The heat delivered by the vapor is latent heat only.

• The drainage of the condensate film from the surface is by laminar flow

only and the heat is transferred through the film by conduction.

• The thickness of the film at any point is a function of the mean velocity

of flow and of the amount of condensate passing at that point.

• The velocity of the individual layers of the film is a function of the

relation between frictional shearing force and weight of the film.

• The quantity of the condensate is proportional to the quantity of heat

transferred which is in turn related to the thickness of the film and of

the temperature difference between the vapor and the surface.

• The condensate film is so thin that temperature through it is linear.

• The surface is assumed to be relatively smooth and clean.

• The temperature of the solid surface is assumed to be constant.

6

MAIN PARTS OF HORIZONTAL CONDENSER

The principal components of horizontal condenser are:

1. Shell

2. Tubes

3. Tube sheets

4. Tube Bundles

5. Baffles

6. Shell Cover

7. Inlet & Outlet Nozzles

8. Channel Or Bonnet 9. Channel Cover

10. Pass Partition

11. Flanges

12. Expansion Joint

13. Tie Rods

14. Connections

15. Support Saddles

7

EXPLANATION:

Usually, for the case of condensers processes streams are taken in the shell

side and utilities are inside the tubes to ensure the tube outside condensation. But, it

also depends upon the requirements and nature of the fluids. They are commonly

termed as shell side and tube side fluids respectively. Shell side fluids are non

corrosive in nature and their flow rate is kept low.

Using a small tube diameter makes the exchanger both economical and

compact. However, it is more likely for the heat exchanger to foul up faster and the

small size makes mechanical cleaning of the fouling difficult. To prevail over the

fouling and cleaning problems, larger tube diameters can be used. Thus to determine

the tube diameter, the available space, cost and the fouling nature of the fluids must

be considered. Typically, exchangers are usually cheaper when they have a smaller

shell diameter and a long tube length.

Tube sheets are usually fixed tube sheets but in some cases floating tube

sheets are also used. Tube bundles are set of tubes and can be made up of several

types of tubes plain, longitudinally finned etc.

Baffles are used to increase the heat transfer by creating the turbulence, and

to reduce the virbational motion of the tube bundles. Three types of baffles mostly

used are:

1. Segmented baffles

2. Disc and doughnut baffles

3. Orifice baffles

Specially, for the case of condensers longitudinal baffles are preferred.

Cover is provided to the outer surface of the shell to prevent the early damage

of the equipment due to change in atmospheric conditions. Shell cover also serves as

insulator to prevent the heat loss to the atmosphere up to some extent.

Inlet and Outlet nozzles are used for the entrance and exit of the tube side and

shell side fluids. It is commonly termed as Tube inlet, Tube outlet, Shell inlet, Shell

outlet.

Channel or bonnet is simply throws the fluid to tubes through tube sheets.

Usually, it is connected to both ends of the exchangers (rare & front end).They are

stationary. Channel cover is provided for the protection of fluids inside the channel

due to change in environmental constraints.

8

Heat Exchangers are categorized on the basis of passes involved in tube side

and shell side. It depends upon heat requirement and their use for special purposes

that what type of exchanger should be installed. 1-2 pass heat exchanger indicates

that there is one shell pass and two tube passes.

Flanges are used to connect the tube sheet to channel and with shells. ie. shell

flange rare, shell flange stationary, shell cover flange. All these are used for connecting

by using nuts and bolts. Using of flanges is preferable than welding because of easy

cleaning and they are also easily removable.

Due to change in atmospheric conditions like change in temperature causes

the expansion of the parts of the exchangers. In order to accommodate this expansion,

we use expansion joints.

The basic purpose of tie rods is to provide the support for tubes. They are

preferably used when we are dealing with the longer horizontal exchangers.

Different kinds of connections are applied to the exchangers to maintain the

process conditions inside the exchangers like

• Vent Connection

• Drain Connection

• Instrument connection

Vent connection is usually applied to remove the non condensible gases and to

maintain the required pressure. It is applied on the top of the exchangers. Drain

connection is applied at the bottom of the exchanger to remove the settled material in

the shell side fluid. Drain connection is applied at the rare end of the exchangers. The

purpose of instrument connection is to maintain the process conditions by installing

any type of instrument like thermocouples and pressure gauges.

Saddles are also used for the support purpose of the exchangers. They also

provide us the security of the equipment like corrosion of different parts of the

equipment.

9

DESIGN PROCEDURE:

CALCULATION OF HEAT LOAD:

Heat balance (Q) = m Cp ∆T (Btu/hr)

CALCULATION OF LMTD:

T1 T2

t2 t1 LMTD = (∆ T1 - ∆ T2) / ln (∆ T1/∆ T2)

= (T1-t2) – (T2 –t1) / ln {(T1-t2)/ (T2 –t1)}

Average temperature of cooling water ta = (t1 + t2) / 2

ASSUMTION BASED CALCULATIONS:

Assume Design overall Coefficient (UD) = 75 – 150 (Btu/ ft2.hr.0F)

Heat Transfer Area A = Q / UD ∆TLMTD ( ft2)

Tube Lay out & size i.e. length, OD, BWG, pitch and no of passes

Number of tubes Nt = A / �D0 L

Comparing the Number of tubes with standard value

Selection of Shell layout and size i.e. ID, Baffle spacing and no. of passes

Correction of UD and flow area on the basis of standard number of tubes

TUBE SIDE CALCULATIONS:

Calculation of Flow area at = Ntat/144n ( ft2)

Calculation of Mass Velocity tt amG /&= ( lb/hr ft2)

Velocity of water in tubes ρ3600

tGv = ( ft/s)

Determination of viscosity of fluid inside tube at ta = μ (lb/ft hr)

Diameter of tube = Dt (ft)

CONDENSER

10

Reynolds Number calculation µ

tte

GDR =

Determination of tube inside heat transfer coefficient = hi (Btu/hr.ft2.oF)

Calculation of hio OD

IDhh iio = (Btu/hr.ft2.oF)

Determination of fanning friction factor ƒ

Calculation of tube side Pressure Drop

t

tt sDx

LfGP

ϕ10

2

1022.5=∆

Calculation of Return Loss

′

=∆g

v

s

nPr 2

4 2

Total pressure drop along tube side ∆PT = ∆Pt + ∆Pr psi

SHELL SIDE CALCULATIONS

Calculation of Flow area as = (Ds.PD.LB) / (144PT) (ft2)

Calculation of Mass Velocity s

s a

mG

&= (lb/hr ft2)

Equivalent Dia. of Shell side ( )

o

otte d

dppD

ππ 4/86.0*4 2−= (ft)

Calculation of viscosity of fluid inside shell = μ (lb/ft hr)

Reynolds Number calculation µ

see

GDR =

Determination of fanning friction factor for above Reynolds Number ƒ

Determination of specific gravity of n-propanol

Calculation of shell side Pressure Drop ( )

se

sssd sDx

NDfGP

ϕ10

2

1022.5

1+=∆ (psi)

11

CONDENSATIOM ZONE CALCULATIONS:

Calculation of Condensation loading 3/2tLN

WG =′′ (lb/ft2hr)

Assume ho (Btu/hr.ft2.oF)

Calculation of Temp of tube wall ( )avoio

oaw tT

hh

htt −

++= (oF)

Calculation of Ave temp of film tf = ( Tv + tw)/2 (oF)

Where,

Tv =Ave. Temp. of condensation (oF)

ta =Ave temp of cold fluid (oF)

Determination of

Thermal conductivity (k),

Viscosity ( μ) and

Specific_gravity at tf

Verifying the value of ho from graph 12.9

Calculation of Clean overall coefficient for Condensation zone

oio

oioc hh

hhU

+= Btu/hr.ft2.oF

Calculation of Dirt factor RD = (UC – UD) / UCUD hr.ft2.oF/Btu

12

DESIGN CALCULATIONS OF CONDENSER:

Vapors T1= 244oF condensate T2= 244oF

Hot Water t2= Cooling water t1 = 85 oF

Inlet Temperature of n-propanol T1= 244oF

Outlet Temperature of n-propanol T2 = 244oF

Mass flow rate of n-propanol m1 = 60,000 lb/hr

LATENT HEAT OF n-Propanol

λ = ∑ [Ci(1-Tr)i/3

]|i=(1-6) …………….…………. (1

Where latent heat of n-propanol is expressed in kj/kg and temperature is expressed in

Kelvin

Critical Temperature of propanol Tc = 536.78 oC

Reduced Temperature Tr = 0.7282

For n-propanol

C1 C2 C3 C4 C5 C6

2222.5 -8368.5 14567 -2870.3 -9655.7 4966.0

λ = [{2222.5*(1-0.7282)1/3} – {8368.4*(1-0.7282)2/3} + {14567*(1-0.7282)3/3} –

{2870.3*(1-0.7282)4/3} – {9655.7*(1-0.7282)5/3} + {4966.0*(1-0.7282)6/3}]

= 1439.65 - 3511.36 + 3959.30 – 505.35 – 1101.20 + 366.86

= 648.0 kj/kg = 278.55 Btu/lb

13

HEAT LOAD:

Heat load for condensing the vapors: Q1 = m1 λ

= 60,000 * 278.55 Btu/hr

= 16,713,000 Btu/hr

HEAT BALANCE FOR WATER:

Assumption:

Cooling Water flow rate = 500,000 lb/hr

Q2 = m2Cp(Tout – Tin)

16,713,000 = 50,000 * 1 * (Tout - 85)

Tout = 118.42 oF

So

Outlet Temperature of water = 118.42oF

CALCULATION OF LMTD:

As LMTD = (T2 – t1) – (T1 – t2) / ln {(T2 – t1) / (T1 – t2)}

= (224 - 85) – (224 – 118.42) / ln{(224 - 85)/ (224 – 118.42)}

= 141.63 oF

Average temperature of cooling water = (85+118.42)/2

= 101.71 oF

ASSUMED CALCULATIONS:

Assume Design overall Coefficient (Ud) = 110 Btu/hr.ft2.oF

Heat Transfer Area: wd

t

tU

QA

∆=

.

= 16713000/(110*141.63)

= 1072.72ft2

14

Tube Lay out & size:

Length = 12 ft

OD, BWG, pitch = 1in, 12 BWG, 1-1/4Triangular pitch

Passes = 2

Out side surface area per linear ft a’ = 0.2618

No. of tubes = Nt = A/ a”L

= 1072.77 / (0.2618*12)

= 341

Nearest standard value for 1” OD tube on 1-1/4” triangular pitch and shell

dia 27’’ = 334 tubes ……………………(2

Area of tubes = (Nt * a’ * L)

= (334 * 0.2618 * 12)

= 1049.29 ft2

Corrected value of UD = 16,713,000 / (141.63*1049.29)

= 112.46 Btu/ft2.hr.oF

Close to out assumed value (UD = 110) so our assumption is right.

Shell side:

ID = 27in

Baffle spacing = 36in

Baffles are segmental with 25% cut and side to side flow mechanism

Passes ss = 1

TUBE SIDE CALCULATIONS:

Flow area per tube: at’ = 0.479 in2

at = Ntat/144n

= (334*0.479)/144*2

= 0.555ft2

15

Mass Velocity:

tt amG /&=

= (500,000/0.555)

= 900078.76 lb/hr ft2

Velocity of water in tubes ρ3600

tGv =

= (900078.76/58.79*3600)

= 4.25 ft/s

Tube inside coefficient hi = 1070 Btu/hr.ft2.oF ………………………. (3

Viscosity at average temperature 101.71oF

μ = 1.74 lb/ft hr

Tube inside diameter Dt = 0.0652 ft ………………………… (4

Reynolds No µ

tte

GDR =

= (0.0652 * 900078.76)/1.74

= 33727.09

Tube outside coefficient OD

IDhh iio =

= (1070 * 0.0652) / 0.0833

= 837.50 Btu/hr.ft2.oF

Friction Factor for given Reynolds No

= 0.00021 ……………………………… (5

16

Pressure Drop along Tube side

ΔPt = fG2Ln / 5.22-E10*Dsφ …….. ……………………..(6

= (0.00021*900078.762*12*2) / (5.22-E10*0.0652*1.0*1.0)

=1.20psi

Return Loss ΔPr = (4n*V2)/(2sg) …………………..……..(7

= (4*2*4.252)/(2*32.15)

= 2.247psi

Total Pressure drop in tube side ΔPT = (ΔPt + ΔPr)

= 1.20 + 2.247

= 3.447psi

SHELL SIDE CALCULATIONS:

Baffle spacing LB = 36 in

Number of Baffles = 4 (with side to side flow mechanism)

Shell Diameter Ds = 27 in

Number of shell passes = 1

Tube Dimensions

OD, BWG, pitch = 1in, 12 BWG, 1.1/4Triangular pitch.

Tube clearance PD = 1.25 - 0.25

= 0.25 in

Shell inside flow area as = (Ds.PD.LB) / (144PT)

= (27*0.25*36) / (1.25*144)

= 1.35 ft2

Mass Velocity s

s a

mG

&=

= 60,000 / 1.35 = 4.44x104 lb/hr ft2

17

Equivalent Dia. of Shell side

( )

o

otte d

dppD

ππ 4/86.0*4 2−= ……………………………….. (8

= 4(1.25*0.86*1.25 – π *12*0.25) / ( π*12)

= 0.059 ft

Viscosity of n-propanol μ = 0.0242 lb/hr.ft

Reynolds No

µ

see

GDR =

= (0.059*4.44x104)/0.0242

= 108247.93

For above Reynolds Number ƒ = 0.0014 ft2/in2

Density of n-propanol vapors ρ = PM/RT

= (29.7*60.10*703.67) / (14.7*359/492)

= 0.236 lb/ft2

Specific gravity of n-propanol s = 0.00377

Diameter of shell Ds = 27 / 12

= 2.25 ft

Pressure drop on shell side ( )

se

sssd sDx

NDfGP

ϕ10

2

1022.5

1+=∆

Assuming φs =1

= 0.0014*(4.44*104)2*2.25*5)

2*5.22*1010*0.059*0.00377

= 1.34 psi

18

CONDENSATION ZONE:

Condensation loading 3/2tLN

WG =′′

= 60,000/12*3342/3

= 103.86 lb/ft2hr

Let ho = 150 Btu/hr.ft2.oF

hio = 837.50 Btu/hr.ft2.oF

Temp of tube wall ( )avoio

oaw tT

hh

htt −

++=

…………………………. (9

= 101.75 + {175/ (837.50 + 175)}*(244 – 101.75)

= 126.33oF

Ave temp of film = tf = ( Tv + tw)/2

= (244 + 126.24)/2

= 185.12oF

Thermal conductivity of propanol condensate can be calculated by using equation for

reduced temperature between 0.25 and 0.80 and below 3.5 MPA ……………………….. (10

And critical temperature of n-propanol is 536.78K

So kL = 1.811*10-4*13.38*60.101.001*18.43/14.82

kL = 0.102 Btu/ft hr oF

19

μ = 0.62 lb/ft hr

s = 0.80 (sp. gravity)

By using these values from graph 12.9 we get

ho = 170 Btu/hr.ft2.oF

Close to assumed (ho= 175 Btu/hr.ft2.oF ) so our assumption is right.

Now Clean overall coefficient for Condensation zone is:

oio

oioc hh

hhU

+=

= (837.50*175) / (837.50 + 175

= 144.75 Btu/hr.ft2.oF

Dirt factor:

RD = (UC – UD) / UCUD

= (144.75-112.46)/144.75*112.46

= 0.002 hr.ft2.oF/Btu

REFERENCES:

1) Eq. 31 Korean J. chem. Engg 17(1) 93-100

2) Table-9 Process Heat Transfer by D.Q.Kern

3) Fig-25 Process Heat Transfer by D.Q.Kern

4) Table-10 Process Heat Transfer by D.Q.Kern

5) Fig-26 Process Heat Transfer by D.Q.Kern

6) Eq.7.45 Process Heat Transfer by D.Q.Kern

7) Eq.7.46 Process Heat Transfer by D.Q.Kern

8) Eq.14-34 Plant Design and Economics by Peter Timmerhaus

9) Eq.5-33 Process Heat Transfer by D.Q.Kern

10) Eq.2-135 Perry’s Chemical Engineers Handbook 7th edition

20

MECHENICAL DESIGN:

STREAMS INVOLVED:

1) Process fluid :

N-propane at 224oC

2) Service fluid :

Water at 85 co

NATURE OF FLUID:

1) N-propane

2) Water

MATERIAL OF CONSTRUCTION:

Shell side: Mild steel

Tube side: stainless steel 430

CHARACTERISTICS OF STAINLESS STEEL 430:

Composition:

Cr = 14.00-18.00%

Ni = not present

C = 0.12 %

Major characteristics:

Good corrosion resistance properties against oxidizing media.

Applications:

Chemical processing towers, condensers, furnaces etc.

*(Ref: plant design by timerhaus 5th edition)

TYPE OF CONDENSER SELECTED:

Bundle dia; Db = do (Nt/K1) 1/n1

Where:

do = tube outside dia = 1”

21

Db = Bundle dia =?

Nt = No. of tubes = 334

K1 & n1 are constants

K1 = 0.249 & n1 = 2.207

So:

Bundle dia calculated = 0.65m

&

Shell inside dia = 27” = 685.8mm

Shell inside dia – bundle dia = 35.8mm

From graph:

22

STANDERD AEP TYPE CONDENSER:

A = Chanel & removable cover

E = One shell pass

P = outside packed floating head

TEMA standards important parameters for AEP type:

Definition Generally for petroleum and related processing

application.

Corrosion allowance 1/8 inch

Tube diameters ¾, 1, 1 ¼ ,1 ½ & 2 inch OD

Tube pitch 1.25 inch

Minimum shell diameter 8 inch tabulated

Longitudinal baffle

thickness

¼ inch minimum

Minimum tie rod diameter 3/8 inch

Floating head cover 1.3 times tube flow area

Significant feature:

One tube sheet “floats “in shell or with shell, tube bundle may or may not

be removable from shell, but back cover can be removed to expose tube ends.

23

Area of application:

High temperature differentials, above about 200°F.extremes; dirty

fluids requiring cleaning of inside as well as outside of shell, horizontal or vertical .

Limitations:

Internal gaskets offer danger of leaking. Corrosiveness of fluids on shell side

floating parts. Usually confined to horizontal units.

Maintenance:

Provision for differential expansion floating head

Removable bundle yes

Replacement bundle possible yes

Individual tubes replaceable yes

Tube interiors cleanable yes, mechanically or chemically

Tube exteriors with triangular pitch

cleanable

chemically only

Tube exteriors with square pitch cleanable yes, mechanically or chemically

Number of tube passes normally no limitations

Internal gaskets eliminated yes

*(ref : Rules of Thumb for Chemical Engineers (3rd Ed) by Carl R. Branan)