calculation 5

CHAPTER 5

DESIGNING OF CRUDE DISTILLATION UNIT

5.1 OVERALL PRODUCT DISTRIBUTION

streamEnd

Temperature( c)⁰Volume distribution Weight distribution

Volume % Range Weight % Rangeoff Gases 15 2.32 0-2.32 1.50 0-1.50

Naphtha 149 21.70 2.32-24.02 18.60 1.50-20.1

kerosene 232 17.08 24.02-41.1 16.10 20.1-36.2

diesel 342 21.06 41.1-62.16 21.25 36.2-57.45

over flash - 2 62.16-64.16 42.55 57.45-59.46

residue Greater then 342 37.85 62.16-100 2.09 57.45-100

5.2 MID-VOLUME PERCENT

5.2.1 Pre-flash

stream Mid volume percent

Off Gases 1.07

Light Naphtha 8.02

5.2.2 Atmospheric tower

stream Mid volume percent

Gases 2.23

Naphtha 18.87

kerosene 32.56

diesel 51.63

residue 81.08

5.3 OVERALL MATERIAL BALANCE

streamvolume %

volume bbl

weight %

Weightlb

Density lb/bbl

APIMolecular weight

Moles

feed 100.0035000.0

0100.00

10255000.0

0293.00 37.20 210.00 48833

off Gases 2.32 812.00 1.50 153825.00 189.44125.7

060.00 2564

Naphtha 21.70 7595.00 18.60 1907430.00 251.14 65.50 90.00 21194

kerosene 17.08 5978.00 16.10 1651055.00 276.19 47.70 140.00 11793

diesel 21.06 7371.00 21.25 2179187.50 295.64 35.90 215.00 10136

over flash 2.00 700.00 2.09 214329.50 306.19 30.00 270.00 794

residue 37.8513247.5

042.55 4363502.50 329.38 18.80 - -

Density and API is taken from APPENDIX C

Molecular weight is computed from graph 2: Molecular weight curve

5.3.1 PRODUCT DISTRIBUTION AROUND PRE-FLASH AND ATMOSPHERIC

TOWER

Pre-Flash:

streamvolume

%volume

bblweight

%Weight

lbfeed 100 35000 100 10255000

off Gases 2.14 750 1.4 143570

Naphtha 11.4 4000 9.6 984480

Bottom 86.5 30250 89 9126950

Atmospheric Tower:

streamVolume

%Volume

bblWeight

%weight lb/day

tower feed 86.50 30250.00 89.00 9126950.00

Gases 0.18 62.00 0.10 10257.00

Naphtha 10.30 3595.00 9.00 922950.00

kerosene 17.08 5978.00 16.10 1651055.00

diesel 21.06 7371.00 21.25 2179187.50

over flash 2.00 700.00 2.09 214329.50

residue 37.85 13247.50 42.55 4363502.50

5.3.2 DISTRIBUTION OF NAPTHA IN PREFLASH AND ATMOSPHERIC

TOWER

Naphtha Recover∈preflash=40007595

Naphtha recover in pre-flash = 52.7%

Naptharecover∈Atmosphric tower=35957595

Naphtha recover in atmospheric tower = 47.3%

5.3.3 MATERIAL BALANCE AROUND ATMOSPHERIC TOWER

streamvolume

%volume

bblweight

%weight lb/day

Densitylb/bbl

APIMolecular

weightMoles

tower feed 86.50 30250.00 89.00 9126950.00 301.71 32.50 240.00 37688

Gases 0.18 62.00 0.10 10257.00 193.54 - 65.00 185

Naphtha 10.30 3595.00 9.00 922950.00 256.73 61.50 110.00 8390

kerosene 17.08 5978.00 16.10 1651055.00 276.19 47.70 140.00 11793

diesel 21.06 7371.00 21.25 2179187.50 295.64 35.90 215.00 10136

over flash 2.00 700.00 2.09 214329.50 306.19 30.00 270.00 794

residue 37.85 13247.50 42.55 4363502.50 329.38 18.80 - -


Molecular weight is computed from graph 2: Molecular weight curve

5.3.4 MATERIAL BALANCE AROUND PRE-FLASH

streamvolume

%volume

bblweight

%Weightlb/day

Density lb/bbl

APIMolecular

weightMoles

feed 100 35000 100 10255000 293 37.2 210 48333

off Gases 2.14 750 1.4 143570 191.4 112 55 2610

Naphtha 11.4 4000 9.6 984480 246.12 69.5 80 12306

Bottom 86.5 30250 89 9126950 301.79 32.5 - -


Molecular weight is computed from APPENDIX B graph 2: Molecular weight curve

5.4 STEAM REQUIRED

In CDU steam are required in the following position:

Kerosene stripper steam.

HSD stripper steam.

Bottom stripping steam.

We have two types of steam available:

Saturated steam at 8 bar, 172 C.⁰

Super Heated Steam at *bar, 360 C.⁰

In the atmospheric tower we use saturated steam

5.4.1 AMOUNT OF STEAM REQUIRED FOR THE FOLLOWIMG STEAM

STREAM Lb/galLb/bbl of product

BPSD product

Lb/day of steamMoles of

steam per day

Kerosene 0.3 12.6 5978 75323 4185

Diesel 0.5 21 7371 154791 8599

Reduced

crude0.6 25.2 13248 333850 18547

Total 563964 31331

Lb/gal of steam is taken from APPENDEX B figure 7.41 gal = 42 bbl

Amount Steam required = 23498 Lb/hr.

Amount Steam required = 1305 mol/hr.

5.5 ENERGY BALANCE AROUND ATMOSPHERIC TOWER

5.5.1 Latent heat

STREAMLatent Heat

Btu/LbHeat [ Q=mλ ]

BtuKerosene 108 1651055*108= 178313940

Diesel 92 2179187.5*92= 200485250

Total 378799190

λ is taken from APPENDEX B figure 5.8

5.5.2 Sensible Heat

STREAMSpecific

HeatBtu/Lb. F⁰

Heat [ Q=mCp∆T ]Btu

Gases 0.58 10257*0.58*(662-248)= 2462910.84

Naphtha 0.56 922950*0.56*(662-248)= 213976728

kerosene 0.57 1651055.00*0.57*(662-398)= 248450756.4

diesel 0.61 2179187.50*0.61*(662-527)= 179455925.9

residue 0.74 4363502.50*0.74*(662-635)= 87162799.95

steam 0.5 563964*0.5*(342-248)= 26506308

Total 758015429

Cp is taken from APPENDEX B figure 5.1 and 5.2∆T=Tin – Tout

Total heat to be removed = total latent heat + total sensible heat

Total heat to be removed = 758015429+378799190

Total heat to be removed = 1136814619 btu/day

Total heat to be removed = 1136.81 MMbtu/day

5.6 QUANTITIES OF REFLUX

5.6.1 HOT REFLUX

Reflux=Totalheat ¿beremoved ¿latent heat

Lbofhot Reflux=1136814619Btu128 Btu/Lb

Lb of hot reflex = 8881364.21 Lb/day

moles ofhot Reflux=8881364.21110

Moles of hot reflex = 80740 mol/day

bbl ofhot Reflux=8881364.21256.73

Bbl of hot Reflux = 34594.18 BPSD

5.6.2 COLD REFLUX

Reflux=Totalheat ¿beremoved ¿latent heat+sensible heat

Lbofcold Reflux= 1136814619128+(248−113 )∗0.56

Lb of cold reflex = 5583568.85 lb/day

moles ofcold Reflux=5583568.85110

moles of cold reflex = 50760 mol/day

bbl ofcold Reflux=5583568.85256.73

bbl of cold reflex = 21748.79 BPSD

5.6.3 CIRCULATING REFLUX

Reflux=Totalheat ¿beremoved ¿sensible heat

Lbofcirculating Reflux= 1136814619(248−150 )∗0.58

Lb of circulating reflux = 20000257.2 lb/day

moles ofcirculating Reflux=20000257.2110

Moles of circulating reflux = 181820 mol/day

bbl ofcirculating Reflux=20000257.2256.73

bbl of circulating reflux = 77903.85 BPSD

5.7 TOP TEMPERATURE

We are estimating top temperature by using cold reflux

Stream Moles

Gases 185

Naphtha 8390

Reflux 59335

Total vapor 59335

steam 31331

Total 90666

∂ pressure= totalmoles of vaportotal moles

∂ pressure=5933590666

∗32.7

Partial pressure=21.4 Psia

The dew point of overhead is 246 F. by using Maxwell correlation ⁰ APPENDEX B graph

5.27 at 2400F and 21.4 Psia the top temperature is corrected to 260 0F

However the computed top temperature is usually about 3% too high. Resulting in actual

top temperature=260*0.97 = 252 F.⁰

5.8 SIDE DRAW TEPERATURE

5.8.1 KEROSENE DRAW

LATENT HEAT

STREAMLatent Heat

Btu/LbHeat [ Q=mλ ]

BtuKerosene 108 1651055*108= 178313940

Diesel 92 2179187.5*92= 200485250

Total 378799190


SENSIBLE HEAT

STREAMSpecific

HeatBtu/Lb. F⁰

Heat [ Q=mCp∆T ]Btu

Gases 0.62 10257*0.62*(662-398)= 1678865.76

Naphtha 0.59 922950*0.59*(662-398)= 143758692

kerosene 0.581651055.00*0.58*(662-

398)=252809541.6

diesel 0.612179187.50*0.61*(662-

527)=179455925.9

residue 0.744363502.50*0.74*(662-

635)=87162799.95

steam 0.5 563964*0.5*(342-398)= -13681948

Total 651183877.2

Cp is taken from APPENDEX B figure 5.1 and 5.2, ∆T=Tin – Tout

Total heat at kerosene draw = 651183877.2+378799190

Total heat at kerosene draw = 1030002863 Btu

Reflux=Totalheat ¿beremoved ¿latent heat∗Molecular weight

moles of internalreflux=1030002863140∗108

Moles of internal reflux = 68121 mol.

Moles of steam = 31339 mol.

Total moles = Moles of internal reflux+ Moles of steam

Total moles = 99460 mol.


∂ pressure=6812199460

∗38.7

Partial pressure = 26.4 Psia

The bubble point of kerosene is 350 F. by using Maxwell correlation ⁰ APPENDEX B graph

5.27 at 3500F and 26.4 Psia the top temperature is corrected to 398 0F

5.8.2 DIESEL DRAW

LATENT HEAT

STREAMLatent Heat

Btu/LbHeat [ Q = mλ ]

Btu

Diesel 92 2179187.5*92= 200485250


SENSIBLE HEAT:

STREAMSpecific

HeatBtu/Lb. F⁰

Heat [ Q = mCp∆T ]Btu

Gases 0.65 10257*0.65*(662-527)= 900051.75

Naphtha 0.62 922950*0.62*(662-527)= 77250915

kerosene 0.61 1651055.00*0.61*(662-527)= 135964379.3

diesel 0.61 2179187.50*0.61*(662-527)= 179455925.9

residue 0.74 4363502.50*0.74*(662-635)= 871837340

steam 0.5 563964*0.5*(342-527)= -30881125

Total 1234527487

Cp is taken from APPENDEX B figure 5.1 and 5.2∆T=Tin – Tout

Total heat at kerosene draw = 1234527487+200485250

Total heat at kerosene draw = 1435012553 Btu

Reflux=Totalheat ¿beremoved ¿latent heat∗Molecular weight

moles of internalreflux=1435012553215∗92

Moles of internal reflux = 72549 mol.

Molesof steam=33385018

Moles of steam=18547 mol.

Moles of gases=185 mol.

Moles of naphtha=8390 mol.

Total moles of vapor = 185+8390+72549

Total moles of vapor = 81124 mol.

Total moles = Total moles of vapor + Moles of steam

Total moles = 81124+18547



∂ pressure=8112499671

∗41

Partial pressure=33.4 psia

The bubble point of diesel is 450 F. by using Maxwell correlation ⁰ APPENDEX B graph 5.27

at 4500F and 33.4 Psia the top temperature is corrected to 525 0F

5.9 BOTTOM TEMPERATURE

∆ t=1.2 P+35DB

(T−Ta)+0.65SB

(T−Ts)

∆t = feed less then bottom temperature, F⁰

P = percent removed by stripping=4% (taken from APPENDEX B figure 7.4)

B = lb oil per hr=4363502.5 lb/day=181812.60 lb/hr

D = tower diameter, ft=10 ft (assumed)

S = lb steam per hr=333850 lb/day=13910.4 lb/hr

T = temperature of oil, F=662 F⁰ ⁰

Ta = temperature of air, F=85 F⁰ ⁰

Ts = temperature of incoming steam, F=342 F⁰ ⁰

∆ t=1.2(4)+3510

181812.60(662−85 )+0.65

13910.4181812.60

(662−342)

∆t = 4.8+1.11+15.91

∆t = 22 F⁰

∆t = feed temperature - bottom temperature

∆t = Tf - TB

TB = 662 - 22

TB = 640 F⁰

The actual bottom temperature is 635 F⁰

5.10 DIAMETER OF ATMOSPHERIC COLUMN

wa=K √ ρv ( ρL−ρv )

Where,

ρv = density of vapor, lb/cu ft

ρL = density of vapor, lb/cu ft

K is related to service and to tray spacing. For valve tray and tray spacing 22 inch

K = 1000 ft/hr

Density of liquid

Specific gravity at 248 F = 0.63⁰

Specific gravity= density of substancedensity of water at standard condition

ρL = 0.625*62.4

ρL = 39 lb/ft

DENSITY OF VAPOR

components moles Lb/day

Cold reflux 50760 5583568.85

Naphtha 8390 922950

Gases 185 10257

Steam 31331 563964

total 90666 7080739.85 lb/day

W = 295030.82 lb/hr

Volume of the top

V 2=

moles∗P1V 1

T 1

∗T 2

P2

V 2=

90666∗14.7∗379520

∗708

32.5

V2 = 21031792.0 ft3/day

ρ v=7080739.85 lb /day21031792.0 ft3/day

ρv=0.33 lb/ft3


295030.82a

=1000√0.33 (39−0.33)

a=82.6 ft2

Area=π d2

4

Diameter = 10 ft

Downcomer consist of 12% of total diameter so it becomes

Diameter = 1.12*10 ft

Diameter = 11 ft

Using diameter the corresponding linear velocity:

Volume = 21031792.0 ft3/day

Velocity=21031792.0 ft3/day82.6 ft3

Velocity=254622.179 ft /day24∗3600

Velocity = 2.94 ft/sec

5.11 EFV CURVE

Data is taken from TBP curve APPENDEX B graph 1

10% on TBP curve = 176 F⁰

70% on TBP curve = 698 F⁰

slope of TBP curve=698−17670−10

Slope of EFV curve = 5.8 (APPENDEX B figure 4.18)

50% on TBP = 580 (APPENDEX B graph 1)

50% on TBP-50% on EFV = 80 (APPENDEX B figure 4.19)

50% on EFV = 580-80

50% on EFV = 500

0% on EFV = 500-(50*5.8)

0% on EFV = 210

100% on EFV = 500+ (50*5.8)

100% on EFV = 790

By using 10%,50%,100% of EFV data we plot EFV curve APPENDEX B graph 4

From EFV curve APPENDEX B graph 4

Percent vaporize in the feed = 58.5%

Pressure at 34 plate = 28.7 psig

5.12 VAPORIZER TEMPERATURE

At 58.5% we have 550 F of temperature at equilibrium flash vaporization curve⁰

Stream Moles

Gases 185

Naphtha 8390

kerosene 11793

diesel 10136

over flash 794 .

Total vapor 31298

steam 18547

total 49845


∂ pressure=3129849845

∗43.4

Partial pressure = 27.2 psia

By using Maxwell correlation APPENDEX B graph 5.27 at 5500F and 27.2 Psia the vaporizer

temperature is corrected to 600 0F

5.13 FLASH ZONE ENERGY BALANCE

At the flash zone we have temperature of 663 F and 35 psig.corrected the temperature by⁰

Maxwell correlation APPENDEX B graph 5.27 so the corrected temperature we have 550 F ⁰

at 760 mm Hg.

Volume % of vapor=58.5 vol. % and Volume % of liquid=41.5 vol. %

V’ = 0.585*30250

V’ = 17696.2 BPSD

W’ = 0.415*30250

W’ = 12553.8 BPSD

weight % of vapor=54.54 wt. % and Volume % of liquid=45.46 wt. %

V ’= 0.5454*9126950

V’ = 4977838.53 Lb/day.

W’ = 0.4546*9126950

W’ = 4149111.47 Lb/day.

Density= weightvolume

Density of vapor= 4977838.5317696.2

Density of vapor = 281.3 Lb/bbl.

API of vapor = 44.5 (Density and API relationship is taken from APPENDIX C)

Density of Liquid=4149111.4712553.8

Density of liquid = 330.5 Lb/bbl.

API of liquid = 18.25 (Density and API relationship is taken from APPENDIX C)

QFZI = V’H + W’H (Watkins graph 1. Enthalpy of Petroleum Fractions. APPENDIX B)

QFZI = (490*4977838.53) + (360*4149111.47)

QFZI = 3932821009 Btu/day.

5.14 FURNACE OUTLET

T = 355 C = 671 F⁰ ⁰

P = 40 Psig = 54.7 psia.

Correcting the temperature at 760 mmHg by using Maxwell correlation APPENDEX B

graph 5.27 we have 540 F.⁰

Using EFV curve APPENDEX B graph 4

Vol. % vapor = 56%

Volume % of vapor = 56 vol. % and Volume % of liquid = 44 vol. %

V’ = 0.56*30250

V’ = 16940 BPSD

Mid volume % = 56-(56/2) = 28

API = 45 (APPENDEX B graph 3)

ρ = 280.35 Lb/bbl (Density and API relationship is taken from APPENDIX C)

V’ = 280.35 * 16940

V’ = 4749129 Lb/day

W’= 0.44*30250

W’= 13310 BPSD

Mid volume % = 100-(44/2) = 78


ρ = 326.68 Lb/bbl (Density and API relationship is taken from APPENDIX C)

V’ = 326.68 *13310

V’ = 4348110.8 Lb/day

QFo = V’H + W’H (Watkins graph 1. Enthalpy of Petroleum Fractions. APPENDIX B)

QFO = (495*4749129) + (365*4348110.8)

QFO = 3937879297 Btu/day

QFO

QFZI

=39378792973932821009

QFO

QFZI

=1.001

5.15 HEAT QUANTITIES AT FLASH ZONE

QL = Amount of liquid * Enthalpy of liquid at flash zone (APPENDIX B Watkins graph 1)

QL = 4363502.5 * 445

QL=1941758390btuday

QL=1941.75MMbtuday

Qsteam = amount of steam at flash zone * enthalpy of steam (APPENDIX B Watkins graph 1)

Qsteam = 333850 * 1190

Qsteam=397281500btuday

Qsteam=397.28MMbtuday

QFZI=3932.82MMbtuday

Qvapor=QFZI−QL

Qvapor = 3932.82 - 1941.75

1512.5 BPSD

5978 BPSD

7490.5 BPSD

75323 lb/day

Qvapor=1991.07MMbtuday

5.16 KEROSENE STRIPPER

Percent removed by stripping = 5% (APPENDIX B Watkins figure 7.4)

TBP cut range of kerosene = 149 C_232 C = 300 F_450 F⁰ ⁰ ⁰ ⁰

Draw tray temperature of kerosene = 398 F⁰

Stripping steam rate of kerosene = 12.6 lb/bbl

Product volume % (24.02-41.1) = 17.08

5.16.1 FEED

Feed volume% of crude= Product volume∗feedProduct

Feed volume% of crude=17.08∗7490.55958

Feed volume % of crude = 21.41 %

Mid Volume % = 41-(21.41/2)

Mid Volume % = 30.4 vol. %


ρ = 272.58 lb/day (Density and API relationship is taken from APPENDIX C)

Lb = 2041760.5 Lb/day

5.16.2 PRODUCT

Volume = 5978 BPSD

API = 47.7 (APPENDEX B graph 3)


Lb = 1651055 Lb/day

5.16.3 VAPOR

Volume = 1512.5 BPSD

Lb = 390705.5 Lb/day

ρ = 258.34 lb/day

API = 60 (Density and API relationship is taken from APPENDIX C)

5.16.4 STRIPPEDOUT FRACTION

Strippedout fraction=VaporFeed

Strippedout fraction=1512.57490.5

Stripped out fraction = SF = 20.1 vol.%

Steam free ∆T = 37 F⁰ (APPENDEX B Figure 2.20)

Feed temperature-stripping steam temperature = tf - ts = 398 - 342 = 56 F⁰

Temperature error = -1 F⁰ (APPENDEX B Figure 2.21)

Actual temperature = steam free ∆T-correction

Actual temperature = 37-(-1)

Actual temperature drop= 38 F⁰

Toptray exit vapor temperature=398−385

=3900 F

5.16.5 DENSITY OF LIQUID

Kerosene API = 47.7

60 F Density = 49.35 lb/ft⁰ 3

Using figure 5.14 from nelson


Specific gravity of kerosene = 0.62

ρL = 0.62 * 62.4

ρL = 38.68 lb/ft3

5.16.6 DENSITY OF VAPOR

Top tray Temperature = 388.5 F⁰

Top tray Pressure = 23.3 Psig = 38psia

Lb of vapor evaporated = 390705.5 lb/day

Molecular weight = 140

Moles = 279.1 mol.

Moles of steam = 75323 lb/day

Lb of steam = 4185 mol/day


Total lb = 466028.5 lb/day

W = 19417.85 lb/hr

By using general gas law

2359.5 BPSD

7371 BPSD

9730.5 BPSD

154791 lb/day

V 2=

moles∗P1V 1

T 1

∗T 2

P2

V 2=

6976∗14.7∗379520

∗850

38

V2 = 1668891 ft3/day

ρ v=lbof vapor

volumeof vapor

ρv = 0.27 lb/ft3


19417.85a

=1000√0 .27(38 .68−0 .27)

a = 6.03 ft2

Diameter = 2.7 ft

Down comer consist of 12% of total diameter so it becomes

Diameter = 2.7 * 1.12

Diameter = 3 ft

5.17 DIESEL STRIPPER

From the graph:

Percent removed by stripping = 7.8%

TBP cut range = 232 C_342 C = 450 F_648 F⁰ ⁰ ⁰ ⁰

Draw tray temperature of Diesel = 527 F⁰

Stripping steam rate of Diesel = 21 lb/bbl

Product volume % (41.1-62.16) = 21.06

5.17.1 FEED

Feed volume% of crude= Product volume∗feedProduct

Feed volume% of crude=21.06∗9730.57371

Feed volume % of crude = 27.8 %

Mid Volume % = 62.16-(27.8/2)

Mid Volume % = 48.25



Lb = 2857847.85 Lb/day

5.17.2 PRODUCT

API = 35.9 (APPENDEX B graph 3)


Lb = 2179185.5 Lb/day

Volume = 7371 BPSD

5.17.2 VAPOR

Volume = 2359.5 BPSD

Lb = 678662.35 Lb/day

ρ = 287.63 lb/day

API = 40.5 (Density and API relationship is taken from APPENDIX C)

5.17.3 STRIPPEDOUT FRACTION

Strippedout fraction=VaporFeed

Strippedout fraction=2359.59730.5

Stripped out fraction = SF = 24.2 vol.%

Steam free ∆T = 37 F⁰ (APPENDEX B Figure 2.20)

Feed temperature-stripping steam temperature = tf - ts = 527-342 = 185 F⁰

Temperature error = -6 F⁰ (APPENDEX B Figure 2.21)

Actual temperature = steam free ∆T-correction

Actual temperature = 37-(-6)

Actual temperature = 43 F⁰

Toptray exit vapor temperature=398−435

=5180F

5.17.4 DENSITY OF LIQUID

Diesel API = 35.9

60 F Density = 52.74 lb/ft⁰ 3



Specific gravity of Diesel = 0.63

ρL = 0.63*62.4

ρL = 39.3 lb/ft3


Temperature = 518 F⁰

Pressure = 25.5 Psig = 40.2psia

Lb of vapor evaporated = 678662.35 lb/day

Molecular weight = 215

Moles = 3156 mol.

Lb of steam = 154791 lb/day

Moles of steam = 8600 mol/day


Total lb = 833453.35 lb/day

W = 34727.24 lb/hr

V 2=

moles∗P1V 1

T 1

∗T 2

P2

V 2=

11756∗14.7∗379520

∗978

40.2

V2 = 3064260 ft3/day

ρ v=lbof vapor

volumeof vapor

ρv = 0.27 lb/ft3


34727.24a

=1000√0 .27 (39.3−0 .27)

a = 10.6 ft2

Diameter = 3.5 ft


Diameter = 3.5*1.12

Diameter = 4 ft

5.18 NUMBER OF TRAYS

Recommended practice table is taken from Petroleum Refinery Engineering by W.L.Nelson.

(APPENDIX B table 16.13)

For better separation and meet our product specification. We use the following number of trays.

Material separated trays

Naphtha to kerosene 16

Kerosene to diesel 6

Diesel to gasoline 3

Flash zone to 1st side draw 3

Steam stripping section 4

total 32

Kerosene side stripper = 5 trays

Diesel side stripper = 5 trays

Different trays have different efficiency range. We are using valve tray having 85%

efficiency

Actual no .of trays= 320.85

Actual no. of trays = 38 trays

Tray spacing = 22 inch

Tower height = 22*38 = 836 inch.

Tower height = 70 ft = 21 meter

In this column the no. of trays are somewhat higher than the theoretical limits especially

at the upper section. These are employing for better separation, and to work efficiently on

different crude. Also upper section contains TPA drawn from the tray 8 and return to

tray6.

5.19 WATER CONDENSER DUTY

Steam temperature = 248 F⁰

The vapor will first pass trough the air cooler and then water condenser. Set an approach

for 50 F for air cooler exchanger, so temperature at the water condenser will be 198 F. ⁰ ⁰

Hence, all the steam will condense.

Utility water available at 35 C and it can cool the mixture upto 40 C.⁰ ⁰

Therefore mixture in overhead condenser cools from 198 ⁰F to 105 ⁰F

Sensible heat required to cool water from 198 ⁰F to 105 ⁰F

Qw = mCp∆T

Qw = 563964 * 1 * (198 - 105) (Cp of water at 35 ⁰C is Btu/lb.⁰F)

Qw = 52448652 btu/day

Latent heat required for Naphtha

Q = mλ

Q = 922950*128

Q = 118137600 Btu/day

Sensible heat required for Naphtha to cool it from 198 ⁰F to 105 ⁰F

Q = mCp∆T

Q = 922950*0.55*(198-105)

Q = 47208892.5 btu/day

Total heat required ¿beremoved=Sensible heat of Naphtha+Latent heat of naphtha+Sensible heat required ¿ coolwater

Total heat required to be removed = 52448652 + 118137600 + 47208892.5

Total heat required to be removed = 217795144.5 Btu/day

5.20 COOLING WATER REQUIRED IN THE OVERHEAD WATER

CONDENSER

Water supply temperature = 95 ⁰F

Maximum water return temperature = 130 ⁰F

Allowable water, change in temperature = 35⁰F

∆T = (130-95) = 35⁰F

Amount of cooling water required:

Q = mCp∆T

217795144.5 = m * 1 * (130- 95) (Cp of water at 35 ⁰C is Btu/lb.⁰F)

m = 6222718.4 lb/day

m = 259280 lb/hr /2.204 lb/kg

m = 571453 kg/hr.

5.21 PRE-FLASH

5.21.1 ENERGY BALANCE

STREAM Specific HeatBtu/Lb. F⁰

Heat [ Q = mCp∆T ]Btu

Off Gases 0.5 143570*0.5*(374-230) = 10337040

Naphtha 0.47 984480*0.47*(374-230) = 66629606.4

Bottom 0.61 9126950*0.61*(374-347) = 150320866.5

steam 0.5 244800*0.5*(680-230) = 55080000

Total 282367513

Theoretical steam required is 2% of crude

Steam required = 0.02*35000

Steam required = 700 bbl/day

Superheated steam is available at 360 C, 8bar.⁰

Weight of steam required = 10200 lb/hr

Weight of steam required = 244800 lb/day

Moles of steam = 13600mol/day.

Cold Reflux:

Reflux=Totalheat ¿beremoved ¿latent heat+sensible heat

Lbof cold Reflux= 282367513125+(230−113 )∗0.58

Lb of Reflux = 1464104 lb/day

moles of cold Reflux=146410480

Moles of cold reflux = 18301 mol/day

bbl of cold Reflux=1464104

lbday

246.12lbbbl

bbl of cold reflux = 5948.7 bbl/day.

5.21.2 DIAMETER OF PRE-FLASH

Stream Moles Lb/day

Cold Reflux 18301 1464104

Light Naphtha 12306 984480

Off Gases 2610 143570

Steam 13600 244800

total 46817 2836954

W = 2836954/24 = 118206.4 lb/hr

5.21.2.3 DENSITY OF LIQUID

Diesel API = 69.5


Specific gravity at 230 F = 0.62⁰

ρL = 0.62 * 62.4

ρL = 38.6 lb/ft3


V 2=

moles∗P1V 1

T 1

∗T 2

P2

V 2=

46817∗14.7∗379520

∗690

32.7

V2 = 10584202.01 ft3/day

ρv = 0.2 lb/ft3


118206.4a

=1000 √0 .2(38.6−0 .2)

a = 42.6 ft2

Diameter = 7.2 ft


Diameter = 7.2*1.12

Diameter = 8 ft

Vapor velocity=10584202.01 ft3/day42.6 ft2

Vapor Velocity = 2484554 ft/day = 2.8 ft/sec

5.21.5 NUMBER OF TRAYS

Two-phase separator can also be used instead of pre-flash but for better separation,

reducing the load and duty in distillation column and optimum efficiency we employed a

pre-flash tower with 20 trays thus the height of the pre-flash tower will be:

Tray Spacing = 22 inch

Height = 20 * 22 = 440 inch

Height = 440/12 = 37 ft.

calculation 5

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