condenser design

Shell Diameter:

Bundle diameter:

Db = d0 {Nt/K}1/n

Where,

d o = tube diameter,

Nt = no. of tubes=63

For single pass,

K=0.319 and n=2.142

From which,

Db = 19.05{63/0.319}1/2.142

Db = 224.624 or 225mm.

For clearance if we add 35 mm

Then shell diameter: 260 mm

Tube side Co-efficient:

Mean water temperature= {24+34}/2

= 29 0C

Tube cross-section area= [π/4]Di2

Where Di is inside diameter of tube = 15.7mm=0.0157m

Area = 0.0410 m2

Mean velocity = velocity of water/area of tube

= 2.99798/0.0410

= 73 Kg/m2.s

Linear velocity of water= mean velocity/density of water

= 73/995

= 0.0734 m/s

This velocity is too low

Therefore choose 4-passes

Tube cross-section area = {Nt/np} {π/4}Di2

Where np= no. of passes

From which tube cross section area= 0.003050 m2

And

water mean velocity= 2.9978/0.003050

= 982.94 kg/m2.s

And

Linear water velocity= 982.94/995

= 0.98788 m/s

hi=4200[1.35+0.02 t ]u0.8

d0.2

Where,

t = mean water temperature= 290C

u= linear water velocity = 0.98788 m/s

d= i.d. of tube = 15.7 mm

from which

hi = 4627.959 W/m2. 0C

with no. of passes changed,

new Shell diameter = Db = d0 {Nt/K}1/n

where, K= 0.175 and n=2.285

with these values,

new shell diameter =240 mm, for shell clearance of 20mm

new shell diameter = 260 mm.

Shell side Co-efficient:

hco=0.926K l ¿

Calculation of mean temperature of condensate film:

Let tw = tube wall temperature

Assume, hco = 1700 W/m2 0C,

At steady state,

Heat transfer rate through cold film= Over all heat transfer

hco *Ac *(tc-tw) = Uc * Ac * (tc-tavg)

Ac on both sides cancels and substituting the values of

hco = 1700 W/m2 0C and Uc =525 W/m2 0C

and tavg= (28.453+32)/2 , tc= 76.67 0C

we get tw = 62.6355

therefore, mean temperature of condensate film = (tc+tw)/2

= (76.67+62.6355)/2 = 69.65 0C

All the physical properties are calculated at this temperature

Kl = liquid thermal conductivity of mixture

For Butanol it is 0.18 W/m.K

And for MEK it is 0.15 W/m.K

Therefore,

Average thermal conductivity= KmekxWm+KbutanolxWbutanol

From which,

Kl = 0.16 W/m. K

ρl = density of liquid condensate = 806 Kg/m3

ρ = density of vapor = 2.97 kg/m3

µl = avg. viscosity of condensate liquid

Average viscosity of condensate liquid is calculated as follows:

1μ=W of mekμof mek

+W of Butanolμ of butanol

Where,

W is weight fraction of MEK= 0.897 and W of butanol = 0.103

µ of MEK and Butanol are 0.16 and 1.2 mNs/m2

Substituting the values µ = 0.17569 mNs/m2 = 0.17569x10-3 kg/m.s

τh= horizontal tube loading of condensate per unit length of tube (kg/m.s)

= W c

LN t

Where Wc = condensate flow Kg/s =640/3600

= 0.177 kg/s

L = length of tube= 1.83m and Nt= no. of tubes = 63

Substituting

τh= 0.0015 Kg/m.s

Nr = average no. of tubes in vertical tube row

= (2/3)Nr’

= (2/3) (Db/Pt)

Where, Db= diameter of bundle = 240 mm and Pt = pitch = 23.8125

Substituting the values we get

Nr = 6.719

Substituting the values in the equation

hco=0.926K l ¿

h co = 2342.5 W/m2. 0C.

for sub-cooling,

standard value from kern,

hosub= 283.77 W/m2 0C

Overall Heat transfer Co-Efficient:

Overall heat transfer co-efficient can be calculated from following formula:

U oc=1

1hoc

+ 1hod

+d0 ln

d0

d i2kw

+d0

d i

1

h i¿+d0

d i

1hid

¿

¿¿

Generally for organic vapors,

hod = 10000 W/m2 0C and

for cooling water hid = 4000 W/m2 0C

as SS-304L is selected:

Kw = 16.3 W/m0C

Substituting the values

U oc=1

12342.5

+1

10000+

0.01905 ln19.05

15.7482 x16.3

+19.05

15.748x

14627.959

+19.05

15.748x

14000

= 832 W/m2 0C

Hence required area for condensation:

Aco = Q cond

Uoc¿∗∆Tmc

¿ = 109075.654

832 X 45.725 = 2.86 m2

Required area for sub-cooling:

U osub=1

1hosub

+ 1hod

+d0 ln

d0

d i2kw

+d0

d i

1

h i¿+d0

d i

1hid

¿

¿¿

Substituting the values,

hosub= 283.77 and rest values as same

U osub= 232.70 W/m2 0C

Therefore,

Area required for subcooling=

Asub = Q¿

Uosub¿

∗∆Tmc¿ =

16443.88232.7 X 29.20 = 2.42 m2

Total area = 2.86+2.42

5.28 m2

Here, AassumedAcalculated

= 6.79/5.28 = 1.2859

i.e. 28.59% extra area is provided which is more than 10% than required in design.

Therefore we can go with the designed area of 6.79 m2.

Area calculated for condensation : 4.543 m2

Area calculated for Sub-cooling : 2.252 m2

Area provided for condensation in excess:

= 2.86+28% of 2.86

= 3.6608 m2

And for sub-cooling,

= 2.42X1.28

=3.0976 m2

Hence compared to total area of 6.79 m2

45.6% of total area should be provided for sub-cooling.

Assuming that tubes are uniformly distributed in the cross section of shell,

0.456= { y * D2i}/(π/4) D2

i

From which y = 0.456 X π

y = 0.358

for this value of `y’, h/Di = 0.386

hence,

38.6% of shell must be submerged in the pool of condensate to facilitate sub-cooling.

This can be achieved by providing U-loop seal in the drain line of condensate.

Therefore,

38.6% of 260 mm = 100 mm height.

Shell side Pressure drop:

Shell side flow area:

As= (Pt−do )B sD s

P tX x '

Where,

x’ = 1-(h/di) =1-0.386=0.618

pt = tube pitch = 23.8125

do = tube O.D.=19.05 mm

Bs = Baffle spacing = 0.260

Ds= Shell I.D. = 0.260


As = 0.008355 m2

Shell side velocity = Gs/ρv

ρv = 2.79 Kg/m3

Gs =m/As = (658/3600)/0.008355

= 21.8 Kg/m2.s

Therefore,

Shell side velocity = 21.8/2.79

= 7.84 m/s

Equivalent diameter for triangular pitch,

de=1.1d0

(Pt2−0.907do2 )

From which, de = 13.736 mm

Reynolds number Re = Gsdeμ

Substituting value µ = 0.01

Re= 9790

For 25% baffle cut segment and given Reynolds no. of 9790

Jf =0.08

∆ P=0.5∗8∗J f (Ds

de )( LB s )(ρv us

2

2 )( μμw )0.14

Substituting the values

∆ P = 3.987 KPa.

This is the calculated pressure drop on shell side

Tube side Pressure Drop:

Tube side pressure drop can be calculated for

∆ Pt=N p [8J h ( Ldsi )( μμw )−m

+2.5 ] ρu t2

2

Jh calculation: Re = (d*v*ρ)/µ

Where, ρ = 995 kg/m3, v = 0.9878 m/s, d=0.015748m

And µ= 0.8x10-3

Substituting, Re = 19337

Therefore, Jh= 4x10-3

L= 2000 mm, di= 15.748mm, ut = 0.9878 m/s, Np = 4

Substituting in pressure drop formula,

∆ Pt= 12.680 KPa

FINAL HEAT EXCHANGER DATA:

HEAT EXCHANGER SPECIFICATION SHEET

EQUIPMENT NAME AND TYPE: MEK CONDENSER AND SHELL AND TUBE (FIXED-TUBE SHEET)

SERVICE: CONDENSING AND SUB-COOLING OF MEK AND BUTANOL VAPORS GENERATED FROM REACTOR IN PRESENCE OF HYDROGEN AS A NON-CONDENSIBLES

FLUID PROPERTIES DATA: SHELLSIDE TUBESIDE

FLUID STATE VAPOR LIQUID

TEMPERATURE IN (OC) 157.50C 240C

TEMPERATURE OUT (OC) 320C 340C

DENSITY (KG/M3) 2.97 995

VISCOSITY (Kg/m.s) 0.17569x10-3 0.001

SENSIBLE HEAT (J/KG/OK) 1.984 4.1868

LATENT HEAT (KJ/KG) 432.78 - PROCESS DATA: SHELLSIDE TUBESIDE

HEAT DUTY (KW) 125.519 -

FLOW RATE (KG/HR) 658 10792.72

PRESSURE DROP (KPA) 3.987 12.680

TUBE VELOCITY (M/S)

CONSTRUCTION DATA:

HEAT TRANSFER AREA (M2) 6.79

NUMBER OF TUBES/SHELL 63

TUBE OUTSIDE DIAMETER (MM) 19.05

TUBE BWG NUMBER 16

TUBE LENGTH (METER) 1.83

TUBE PITCH AND ORIENTATION TRIANGULAR

INSIDE SHELL DIAMETER (MM) 260

SHELL MATERIAL CARBON STEELl

TUBE MATERIAL STAINLESS STEELl

BAFFLE SPACING (MM) 260 mm

MECHANICAL DESIGN:

Shell thickness:

Shell material is chosen as Carbon steel. And operating pressure is 1.064X105 N/m2

Therefore design pressure is 10% more than operating pressure and that comes to 1.11X105 N/m2

Tsh =PdD

2 fj+P

Here, Pd= 0.11, D = 260, f = 95 N/m2 ( for carbon steel) and

J= 0.85 (joint efficiency)


Tsh = 0.170 mm

Taking minimum thickness as 6 mm and corrosion allowance of 2 mm total thickness comes as 8 mm

Taking standard thickness,

Shell thickness: 10 mm.

Head : (Torrispherical head)

T h=P¿Rc

¿

2 fJW

Where, = W=14⌊3+√

RcR ic

⌋

Rc= crown radius = 260 mm

Ric = 6% of crown radius=15.6 mm

With these values W = 1.77

And substituting the values of W and P* of 0.11 N/m2

F=95 and J = 1

We get thickness of head = Th = 0.25 mm

So, same thickness of shell i.e. 10 mm of head is selected.

Thickness of baffles is selected as a 6 mm standard.

BAFFLES SPACING:

Segmental baffles are used.

Baffle spacing = = 260/5 =52 mm

Thickness of baffles = 6 mm

Tie rods and spaces

Diameter of tie rod = 10 mm

Number of tie rods = 6

Design of Flanges:

Shell thickness = go = 10 mm

Flange material – IS: 2004 – 1962 class 2

Gasket material – asbestos composition

Bolting steel = 5% Cr Mo steel

Allowable stress of flange material – 100 MN / m2

Allowable stress of bolting material, Sg – 138 MN/m2

Outside diameter = B = 260 + (2 x 10)

= 280 mm

Gasket width

do / di = [(y- pm)/ (y- p{m+1})] 0.5

m – gasket factor – 2.75

y – min design seating stress – 25.5 MN / m2

let’s assume Gasket thickness = 1.6 mm

Thus,

do / di = 1.002

Let di of the gasket equal 290 mm [10 mm greater than shell

dia]

do = 0.290 x 1.002.= 0.290.58m=290.58mm

Mean gasket width = (0.290.58 – 0.290) /2

= 0.29 x 10-4

Taking gasket width of 12 mm,

do = 0.291 m

Basic gasket seating width, bo = 5mm

Diameter of location of gasket load reaction is,

G = di + N

= 0.290 + 0.012

= 0.302m

Estimation of bolt loads

Load due to design pressure:

H = π G2 P / 4 = (3.14 x 0.302 x 0.11) / 4

= 0.0260 MN

Load to keep joint tight under operation:

Hp = πG(2b)mp

=3.14 x0.302 x 2 x 5 x 10-3 x 2.75 x 0.11

=0.00286 MN

Total load = 0.0260+0.00286

=0.0288 MN

Load to seat gasket under bolting up condition

Wg = πGby

= 3.14x 0.302 x 0.005 x 25.5

= 0.1209 MN

Controlling load = 0.1589 MN

Minimum bolting area = Am = Wg/Sg

= 0.1209/138

= 8.76 x 10-4 m2

Lets take M 18 x 2

Root area of bolt = 1.54 x 10-4 m2

Therefore no. bolts required

= 8.76 x 10-4/1.54 x 10-4 =5.689

8 no. of bolts are selected.

R = 0.027m

g1 = go/0.707 = 1.415 g0 for weld leg

go = 10mm

Bolt circle diameter,

C = I.D +2(g1+R)

C = 0.260 + 2 (1.415 x 0.010 + 0.027)

= 0.3423m

Calculation of outside diameter of flange:

A = C + Bolt diameter + 0.02 (minimum)

= 0.3423+0.018+.02

= 0.3641m

Check gasket width:

= (Ag * Sb)/πGN

= 8x1.51x10-4 x 138/3.14 x 0.302 x 0.012

= 14.6496

Which is less than 2y (2 x25.5)

Therefore condition is satisfied.

Flange moment computation:

a) For operating condition:

Wo = W1 + W2 + W3

W 1=π B2

4P

W 1=3.14¿¿

= 0.00676

W2 = H-W1

Where H =(π/4)G2= 0.785 x 0.3022 =0.07161

W2= 0.07163 – 0.00676

= 0.0648 MN

W3 = Wo-H = Hp (gasket load)

= 0.00286 MN

Total flange moment, Mo = W1a1 + W2a2 + W3a3

a1 = (C-B) / 2 = (0.3423-0.280)/2 = 0.03115

a3 = (C-B) / 2 = (0.3423-0.302)/ 2 = 0.02015

a3 = (a1+a2) /2 = 0.02565

M0 = 2.838 x 10 -4 MJ

b) For bolting up condition (no internal pressure):

Mg = W. a3

W = (Am +Ab)/(2). Sg

Am = area of bolt

= 8 x 1.56 x 10 -4

= 1.248 x 10 -3 m 2

Am = Minimum bolt area. =1.38 x 10 -3 m 2

Sg = 138N/mm 2

Therefore, W = 0.1813 MN

a3 = 0.02015

Mg= 0.1813 x 0.02015

Mg = 3.65 x 10-3 MJ

MG is greater than M0. Therefore Mg is controlling.

Flange thickness :

t 2=M CY Y

B ST

K = A/B

Where, A = flange outside diameter = 0.3641 m

B = flange inside diameter = 0.280

K = (0.3641/0.280)

= 1.300

From figure given in 7.6

At K value of 1.300, Y = 10

Assume, Cf =1 and M = 3.65 x 10-3 MJ

St = Allowable stress =100MN / m 2

t2 =( 3.65 x 10-3 x 10) / (0.280 x 100)

= 0.0361m

t = 0.074m

Nozzle for condenser:

Total 5 no. of nozzles must be provided for condenser. Two nozzles on

tube side each cooling water supply in and out. 4 No. nozzles are provided

On shell side

Heat exchanger Piping:

Shell side fluid:

Vapor inlet pipe sizing:

Vapor flow rate: 658 Kg/hr.=0.1827 kg/s

ρg = 2.75 Kg/m3

volumetric flow rate: 658/2.75 = 243.7 m3hr = 0.06769 m3/s.

Optimum pipe size can be calculated from equation:

For stainless steel,

dopt=293G0.53ρ−0.37

Where,

Dopt= optimum pipe size

G = Flow rate, Kg/s=0.1827 kg/s

Ρ= density of vapor =2.75 kg/m3


Dopt= 81.85 mm

Standard of 80 NB (SCH.10) can be used.

Uncondensed vapor:

Flow rate =112.34 kg/hr = 0.0312 kg/s

And for same density of 2.75 kg/m3,

From the above formula,

The pipe size of 25 NB (SCH.10) can be fixed.

Condensate drain line:

Flow rate: 545.66 kg/hr=0.15157 Kg/s

Density of condensate : 807 kg/m3

Substituting the values in the same equation,

The pipe size of 15 NB (SCH.10) is fixed.

Cooling water lines:

Flow rate: 2.99 kg/s

Density: 985 kg/m3

Substituting the values in same equation:

50 NB (SCH.10) line size is fixed The same size of return line can be used.

As the final process-design stage is Complete, it becomes possible to make

accurate cost estimation because detailed equipment specification and definite

plant facility information are available. Direct price quotation based on detailed

specification can then be obtained from various manufacturers. However

design project should proceed to the final stages before costs are considered

and cost estimate should be made through out all the early stages of the design

when complete specifications are not available. Evaluation of costs in the

preliminary design is said pre-design cost estimation. Such estimation should be

capable of providing a basis for company management to decide if further capital

should be invested in the project. Evaluation of costs in the preliminary design

phase is some time called guesses estimations. A plant design obviously must

present a process that is capable of operating under condition which will yield a

profit.

As the final process-design stage is Complete, it becomes possible to make

accurate cost estimation because detailed equipment specification and definite

plant facility information are available. Direct price quotation based on detailed

specification can then be obtained from various manufacturers. However

design project should proceed to the final stages before costs are considered

and cost estimate should be made through out all the early stages of the design

when complete specifications are not available. Evaluation of costs in the

preliminary design is said pre-design cost estimation. Such estimation should be

capable of providing a basis for company management to decide if further capital

should be invested in the project. Evaluation of costs in the preliminary design

phase is some time called guesses estimations. A plant design obviously must

present a process that is capable of operating under condition which will yield a

profit.

condenser design

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