Equipments DesignEthanol plant

Prof. M. FahimEng. Yousef Ismael

Done By:

Mona Al-Khalaf

Agenda

• Distillation Columns Design( T-104) .

• Compressors Design ( K-101)&( K-102) .

• Heat Exchangers Design( E-105) .

To produce main product Ethanol

Distillation Columns

Distillation Column T-(104) design

Distillation Columns:

• Is a separation unit based on the difference between a liquid mixture and the vapor formed from it.

Objective:

To separate of ethanol from

propanol.

Number of traysShortcut method:

• It's used To determine the minimum number of trays• the method is also called the Fenske equation as follows:

Nm =log (Xlk/XHk)d * (XHk/XLk)b /(logαLK)

Average relative volatility of the light key with respect to the heavy key,

where αLK=(αlD*αlW)^.5αi=(Pi/PHK)Pi=Partial pressure for i key.PHK= Partial pressure for heavy Key.

Distillation Column Design.

• Assumptions:

1. Column Efficiency=70%.2. Tray spacing=0.55.3. Flooding Percentage at

maximum flow rate=85%.4. Percent of downcomer area

of total area=12%.5. Hole area( 10% of Active

area). 6. Weir height=50mm.7. Hole diameter =5 mm.8. Plate Thickness=5mm.9. Turn down Percentage

(70%)10. Material of column is carbon

steel.

Good Design:

1. actual minimum vapor velocity should be greater than Uh( vapor velocity).

2. Back-up in downcommer (hb) less than tray spacing to accept tray spacing.

3. Residence time exceeds 3 secs.

For column diameter

1-calculate the liquid –vapors flow factor for top and bottom

FLV= LW / Vw * (ρv / ρl) ½

Where:-

LW = liquid mole flow rate in kmol /h

Vw = vapors mole flow rate in kmol / h

ρv = density of the vapors in kg / m³

ρl = density of the liquid in kg / m³

Main design procedures:

2-from fig 11.27 get constant for the top and the bottom top k1 and bottom k1

3-calculate the correction factor for top and the bottom

K = (σ / 0.02) ^0.2 * K1

Where:-

σ = liquid surface tension in N/m

4-calculate the flooding velocity for top and bottom

Uf = K *( (ρl –ρV) / ρ v)½

Where:-

Uf = flooding vapour velocity in m/sK= constant obtain from fig 11.27

ρl = density of liquid in kg / m³ρv = density of vapour in kg / m³

5-Assume the flooding percentage is 85% at max flow rate for the top and the bottom

UV = 0.85 * Uf

6-calculate the net area for the top and the bottom

An = Vmax / UVWhere:

An = net area in m²

V = Volumetric flow rate in m³ / s

UV = vapour velocity in m/s

vMwtrateflowmolarvaporV /max

7-Assume as first - trail take down comer as 12% of total cross sectional area

for the top and the bottom

Ad = An / 0.88

Where:

Ad = area of the down comer in m²

An = net area in m²

8-calculate the diameter for the top and the bottom

D = ((4 /3.14) * Ad) ½

Where:

D = Diameter in m

Ad = area of the down comer in m²

Double pass plate is used (from figure 11.28)

9-calculate the liquid flow pattern

Max liquid volumetric flow rate = Lm *MW / ρL * 3600

10-calculate the areas Ac = (3.14 / 4)*D²

Where:

Ac = total cross sectional area in m²

Ad = 0.12 * Ac

Where:

Ad = area of the down comer in m²

An = Ac –A d

Where:

An = net area in m²

Aa = Ac – 2Ad

Where:

Aa = active area in m²

Ah = 0.1 * Aa

Where:

Ah = hole area in m²

11- Use fig 11.31 to get LW / Dc

12-Assume weir length = 50 mm

Hole diameter = 5 mm

Plate thickness = 5 mm

13-Check weeping how max = 750 * (Lwd max / (LW*ρl)) ^2/3how min = 750 * (Lwd min / (LW*ρl)) ^2/3At min rate = hw + how

Where:-how=Weir liquid crest

14-calculate the weep point

Where:Uh = min vapor velocity through the hole in m/sDh = hole diameter in mK2 = constant from fig 11.30

Uh = k2- 0.9 *(25.4-dh)/ρv½

15-calculate the actual vapour velocity

Calculate the actual vapour velocity = min vapour rate / Ah

16-Calculate the pressure drop

UH = Vv / Ah

Where:

Vv = volumetric flow rate in m³ / s

Ah = net area in m²

HD = 51 * (Uh/ C0)² * ρ V / ρL

Where:Hd = dry plate pressure dropUh = min vapour velocity in m/s Co is the orifice coefficient from figure (11.34)

Hr = 12.5E3 / ρL

Where:

Hr = residual head

Ht = HD + HW + HOW + HR

Where:

Ht = total pressure drop in mm

17-down comer liquid backup

Hap = HW – 10

Aap = LW * hap *0.001

Where:

Aap = area under apron

Hdc = 166 * LW / (ρ l * Aap)

Backup down comer

Hb = hdc + ht + how max + hw

from figure 11.29 get ψ

18-Calculate the residence time

TR = (Ad * hb * ρ l) / lwd

19-Calculate the flooding percentage

Flooding percentage = UV / uf * 100

20-Calculate the area of the hole

A = (3.14 / 4 ) * (dh * 0.001 )²

21-Calculate number of hole

Number of hole = A h / A

22-Calculate the thickness ,

Where:

t: thickness of the separator in (in)

P: operating pressure in Pisa

ri: radius of the separator in (in)

S: is the stress value of carbon steel = 13700 Pisa

Ej: joint efficiency (Ej=0.85 for spot examined welding)

C0: corrosion allowance = 0.125

Pr

0.6i

oj

t CSE P

23- calculate the cost

• From fig. we get the Cost of one tray at year 1990

• And we can get the cost of one tray by use the fig or by use the law which is depend on the original cost(1990)

• The cost of total trays=cost of one tray*actual stage number

• Cost of distillation=cost of vessel+cost of total trays+cost of condenser+cost of reboiler.

Compressors Design:( K-101)&( K-102)

Design two compressors.

Compressors

• Compressor :• Is a gas compressor is a mechanical device that increases the pressure of a

gas and naturally increases its temperature by reducing its volume.

• Objective

• To increase the pressure of feed before it's inters to cooler E(105)

• To increase the pressure of feed before it's inters to cooler E(101)

Design Procedures and Equations:

(1) Calculate the compression factor using the following equation:

Where P1,2 is the pressure of inlet and outlet respectively (psia),And T1,2 is the temperature of the inlet and outlet respectively (R).

(2) Calculate the work done in Btu/lbmol by:

Where R is the ratio of the specific heat capacities (Cp/Cv).

(3) Calculate the horse power, Hp using the following equation:

Where M is the molar flow rate in lbmol/s.

11 1

2 2

n

nP T

P T

*Hp W M

(4) Calculate the efficiency of the compressor using the following equation:

Where

and Mw is the molecular weight of the gas in the stream and Cp is the specific heat capacity (Btu/lb.F).

(5) Depending on the horse power we can decided which type of compressor we going to use and calculate its cost.

(6) Finally calculate the cost of the compressor from www.matche.com by using horse power value.

(7) Calculate the inflation rate cost from year 2007 to year 2009 from draw graph between (years and Nelson-Farrar refinery construction index), so we get linear equation (y=46.6785714x-91600.995).

1

1

nnEp

KK

1.986p

p

MwCK

MwC

We used this fig to know the type of compressor

Cooler ( E-105) (Heat Exchangers Design)

Design of first cooler

• a heat exchanger is a device made for efficient heat transfer from one fluid to another across a solid surface.

• Objective

• To cooled the feed stream and prepare it to inter the compressor(k-102)

Shell & Tube Heat Exchanger Design.• Assumptions:

1. For cooling the fluid, cooled water has been used.2. Assuming value for overall heat transfer coefficient based on table 12.1 which

=360, which should be closed to the calculated value.3. The type of heat exchanger is shell and tube, while the material of construction is

carbon steel.4. Assume the outer=40, and the tube length=5.3.5. Assume the inlet and outlet temperature and pressure for the water in the tube

side.6. Choose 25% baffle cut.7. Because the inlet stream flow rate was very high, so we divided it in to three

stream and the cost was multiply to three.

• a good design :

1. The assumed overall heat coefficient has to be close to the calculated overall heat transfer coefficient.,

2. The pressure drop in the tube side should be smaller value.3. The pressure drop in the shell side should be smaller value.

Shell & Tube Heat Exchanger Design.

Heat Load Log mean Temperature

Where;∆Tlm = log mean temperature differace

T1 is temperature of inlet hot stream. (oC)

T2 is the temperature of outlet hot stream. (oC)

. t1 is the temperature of inlet cold stream. (oC)

.t2 is the temperature of outlet cold stream. (oC).

T2)-(T1phothotphot cMTcMQ

lmtm

lm

TFT

tT

ttS

tt

TTR

tTtT

tTtTT

11

12

12

21

12

21

1221

;

ln

Where, Ft = the temperature correction factorUsing one shell pass and two or more even tube passes

Shell & Tube Heat Exchanger

Heat Transfer Area

mTU

QA

DensityPassArea

FlowRateuvelocity

areatoncrossPasstubespassArea

dareaSectioncross

sesAssumedPas

tubesPassTubes

ubeareaOfOneT

totalAreatubes

LdubeAreaOfOneT

t

i

o

*/

sec//

25.0

#/

#

**25.0

2

2

Number of tubes

10.12.Re

.,; 11

1

1

1

FigadingClearanceDD

PassesNofnKK

NdD

bs

nt

ob

Shell and Bundle diameter

Where; Nt is the number of tubes=Provisional area/Area of one tube. .

K1, n1 are constants .do=Outside diameter (mm) Db is the bundle diameter (mm)

Ds is the shell diameter. (mm)

where:   

A = provisional area m2= 

Q = heat load (kW)  

U = overall heat transfer coefficient (W/m2 oC)

Assuming U from table12.1=360(W/m2 oC)

DLA

Shell & tube heat exchanger Design

i

fi

ih

wh

pit

d

kNuh

d

LfjjNu

k

cdu

)(;PrRe

Pr;Re

14.033.0

Tube Side Heat Transfer Coefficient

Where is the density of fluid (kg/m3).

is the thermal conductivity (W/m.C).is specific heat (kJ/kg.k).

Re is the Reynolds number.Pr is the Prandtl number.Nu is the Nusselt number.

is the convective heat transfer coefficient (W/m2.C).

k

pc

e

fs

hw

h

pes

oto

e

ss

t

Bsots

ot

d

kNuh

cutbufflefjjNu

k

cdu

dpd

d

A

FlowRateu

p

lDdpA

dp

_Re,;PrRe

Pr;Re

917.01.1

25.1

14.033.0

22

Shell side heat Transfer Coefficient

Where ;.pt is the tube pitch (mm).

.lB is the baffle spacing (mm).As is the cross flow area (m2)

us is the velocity (m/s).de is the equivalent diameter for triangular arrangement

(mm).jh is the heat transfer factor

hs is the convective heat transfer coefficient (W/m2.C).

Shell & Tube Heat Exchanger Design .

ii

o

w

i

oo

oo hd

d

k

ddd

hU

1

2

ln11

Overall Heat Transfer coefficient

Tube side pressure Drop

25.28

2t

m

wifpt

u

d

LjNP

28

214.0

s

wBe

sfs

u

l

L

d

DjP

Shell Side Pressure Drop

cj

j

j

CPSE

t

Dr

6.0

Pr2

Thickness

Where; D is the shell diameter in m Rj is internal radius in (in) .

P is the operating pressure in psi S is the working stress (psi) .

E is the joint efficiency

• We calculate the cost of heat exchanger at year 2007 based on Heat transfer area from (www.matche.com) and we can calculate the cost of heat exchanger at year 2009 by adding the inflation rate value (y=46.6785714x-91600.995, where (X) is the year) to the cost of heat exchanger at year 2007.

• The cost of three heat exchanger= (cost of one heat exchanger) * (3)

Cost of heat exchanger

Thank you for listening