PED 2 Report Group 8

8/19/2019 PED 2 Report Group 8

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PROCESS EQUIPMENT

DESIGN –

II

ASSIGNMENT-I

SUBMITTED TO:

PROF. JAYANTA CHAKRABORTY

SUBMITTED BY:GROUP 8

RAHUL KUMAR 13CH10034

RAHUL YADAV 13CH10035

RAJA BARNWAL 13CH10036

RAKESH CHHIPA 13CH10037


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TABLE OF

CONTENTS

CONTENT PAGE

PROBLEM STATEMENT 03

BASIC THEORY 04

PROCESS DEISGN 05

MECHANICAL DESIGN 14

REFERENCES 21


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PROBLEM STATEMENT

A sieve plate column is to be designed to separate 8000 kg/hr of feedhaving 40 mole % Methanol (A) & 55 mole % Water (B) into an

overhead product containing 96 mole % A and a bottom product

containing mole 98 mole % B. The feed enters as an equilibrium

mixture of 30% liquid & 70% vapor. A reflux equal to 1.5 times the

minimum is to be used. Also an external reboiler is necessary to

remove the bottom product from the reboiler. The condensate

C & the

reflux enters the column at this temperature. Assume the Murphree

P ffiiy 70%. Gi αAB=3.91

(a) Complete the process design calculation.

(b) Mechanical design of the column.

(c) Enclose a drawing showing the details of the column.


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BASIC THEORY

Optimization stems from the need for improvement. Typical problems in

chemical engineering arise in process design, process control, model

development, process identification, and real-time optimization.

Distillation, being the most common industrial process for separation of

chemical components, is the focal unit of petroleum refining. While distillation

can be economically and easily scaled to different production levels, it is highly

energy intensive, consuming up to 80-90% of the total energy of a typical

petrochemical process. This makes the need for optimization essential to

maximize profitability of the entire process.

The performance of a distillation column is determined by many factors, such

as distillation feeds, internal liquid and fluid flow conditions, state of trays and

even ambient conditions which change over time, which makes the response

of real-time optimization to these changes a key contributor to successful

operation. In this particular process, detailed analysis of the operation and

design of the distillation column is performed involving the optimisation of

reflux ratio subject to various practical constraints and also the different

problems and cost analysis regarding the setting up of a distillation column inreal time.

In this design problem our task is to first find optimum value of reflux ratio at

which the total cost (operating + fixed cost) for setting up the distillation

facility for required separation of methanol is minimum. There are two

opposite factors contributing to the total cost. As the reflux ratio is increased

the number of trays required for a given degree of separation reduces which in

turn reduces the column height. However because of higher liquid and vapour

flow rates there is a increase in column diameter. Also the operating cost

increases because of higher heat duty of both the re-boiler and the condenser.

The optimum value of reflux ratio is determined by determining the minima in

a plot of Total cost Vs reflux ratio. The total cost calculated in our case is

actually annualised total cost which is calculated by dividing the total cost by

im “if” f qim f kig into account the interest

rates on various cost items.


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PROCESS DESIGN

In this problem, we need to separate methanol from a mixture of water and

methanol by using a bubble cap distillation column.

Given Data

Mol. Wt. (Methanol)(MWm) 32.04 gm/mol

Operating Pr. 14.7 Psig

Temperature (T) 148.5 F

Liquid Density (rhoL) 47.1645 lb/ft3

Vapour Density (rhoV) 0.08 lb/ft3

Liq. Surface Tension(sigma) 19.3 dyne/cm

Liquid Rate(L) 14875 lb/hr

Vapour Rate(V) 25581 lb/hr

Liquid rate (expressed in gpm) = 39.32 gpm

Vapour rate (expressed in cfs) = V*1/rhoV*3600

= (25581*1)/(.08*3600)

= 88.82292 cfs


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Cost optimization

F = 8000/(0.45*32+0.55*18) = 329.2181 kmol/hr

D = =(0.45-0.02)/(0.96-0.02)*F = 150.5998 kmol/hr

B = F-D = 178.6183 kmol/hr

Roptimum = Rmin * (R/Rmin)optimum

Refer to Excel sheet for detailed calculations. The optimization has been done

in MATLAB. The following is the result)

Roptimum = 1.3984

Theoretical number of trays have been derived from McCabe Thiele method.For calculation of column diameter, qc, qr, and cost of column, condenser,

reboiler, reboiler, cooling water and total annual cost, refer attached excel

sheet. Also refer [4]

Theoretical number of trays = 20

McCabe Thiele (Equilibrium curve and Equilibrium line)


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232000

233000

234000

235000

236000

237000

238000

239000

1 1.02 1.04 1.06 1.08 1.1 1.12

T o t a l A n n u a l C o s t

R/Rmin

Cost Optimization

R/Rmin |optimum = 1.0393


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Calculation of tower diameter:

F(lv) = (L /V)*sqrt(rhoV/rhoL)

= 0.023948418

Assume tray spacing = 24 inch

Cab= 0.4 [1](Figure 14.4)

Cab(corrected) = 0.4*(sigma/20)0.2

= 0.4*(19.3/20)0.2

= 0.397159956 (No measurable change)

Vapour velocity term (Un) = 0.4/(sqrt(rhoV/(rhoL-rhoV)))= 0.4/((sqrt(0.08/(47.1645-0.08)))

= 9.704071311 fps

Take flooding = 75%

Net tray area (An) = (1/0.75)*(88.82292/9.704071311)

= 12.20933562 ft2

Take 12% for downcomerColumn cross section area (At) = An/0.88

= 13.87424503 ft2

Dt = sqrt(4*At/pie)

= 4.203003381 ft

Next available standard diameter, Dn = 6 ft

From [1] Table 14.5, we get 24 inch tray spacing is suitable for a 6 ft diameter

tower

Area (new), An = pie* Dn2/4

= 28.27433388 ft2

From [1] Table 14.2, we get REVERSE FLOW

We take,

Riser area 0.1 * At

Down flow area 0.12 * At


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Max. active area 0.88 * At

Side weir length 0.62 * Dn

Max liq. flow path 0.3 * Dn

Dynamics slot submergence of 1.5 inch (Assumed)

Estimated tray pressure drop (ht)

= 0.53*(rhoV/rhoL)*(V/(0.1*An))+0.8*1.5+1

= 0.53*(0.08/47.1645)*(88.82292/(0.1*28.27433388))+0.8*1.5+1

= 2.228241208 inch of fluid

Tray Layout

Tower diameter, Dn = 6 ft

Tray spacing = 24 inch

Flow Type = Reverse Flow

Layout = Triangular

Bubble Cap Calculations

Bubble cap diameter = 4 inch

Area of 1 cap = pie*42/4

= 12.56637061 in2

Total bubble area, ABt = 0.7*An

= 19.79203372 ft2

Total number of bubble caps = ABt*144/area of 1 cap

= 226.8

Actual number of bubble caps = 227

Service is non-corrosive. We use CARBON STEEL – A 53B


10/21


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Summary (areas)

Area Formulae Per tray (ft2) % of An

Riser Area 4.8*227/144 7.566666667 26.76160895

Slot Area 8.12*227/144 12.80027778 45.27172181Downcomer Area 0.12*An 3.392920066 12

Active Area 0.88*An 24.88141382 88

Area under apron Area under baffle 0.5 1.768388257

Tower area An 28.27433388 100

Net area 0.88*An 24.88141382 88

Flooding

Vapour velocity based on net area, U = V/An

= 88.82292 cfs/24.88141382 ft2

= 3.569850062 ft/s

% flooding = U/Un = 36.787 % (Under limits)

Entrainment

Phi = 0.055 [1] (Figure 14.5)

It is well within limits

Shape factor = Rs = 0.167/0.333 = 0.501501502

For Rs = 0.50, we have Cs = 0.74 [1] (Figure 14.6)

Q max = Cs*Slot area*sqrt(slot height*(rhoL-rhoV)/rhoV)

= 229.7973954 cfs

Vapour load = 100*V/Q max = 38.65 % of slot capacity

Slot opening = 55% [1] (Figure)

Slot height = 1*0.55 = 0.55 inch


12/21

Liquid crest over weir

Liquid load = 39.32 gpm

HOW = 0.48*(L/lw)2/3

= 0.51 inch

From Figure 14.7 [1],

L/lw2.5

= 2.2522

Weir correction = Fs = 1.04

HOW (corrected) = Fs*HOW = 0.53 inch

Liquid gradient

Arithmetic average flow width = 2.5 ft (approximated)Liquid load per unit width = L/2.5 = 15.728 gpm/ft

CD = 0.75 [1] (Figure 14.9 and 14.10)

Hl = 2.75+0.53+1 = 5 (approximated)

Del = 0.5 inch (using figure 14.9 and 14.10)

Vapour pressure Drop

hcd = 0.53*(0.08/47.1645)*(D15/(0.1*B30)) = 0.03 inch of liquid

Mean dynamic slot seal, hda = 0.5+0.53+0.5/2 = 1.28 inch of liquid

Fva = 1.62 [1](Equation 14.40)

hal = 0.6*hda = 0.98 inch of liquid [1](figure 14.15 and equation 14.37)

Total tray pressure drop, ht = 0.03+0.51+0.98 = 1.52 inch of liquid

Vapour distribution Ratio

hc = 0.03+0.51 = 0.54 inch of liquid

Rcd = 0.5/0.54 = 0.925


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Downcomer dynamics

had = 0.2 inch of liquid [1](using Equation 14.41 and 14.42)

hdc = 6.2 inch of liquid [1](using Equation 14.43)

hfd = 6.2/.4 =15.5 inch of liquid

tdc = 3.8 seconds [1](using Equation 14.47)

TRAY AND BUBBLE CAP DESIGN (all dimensions in mm)


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MECHANICAL DESIGN

Shell ID (Di) 1828.8 mm

No. of trays 20

Tray spacing (t) 609.6 mm

Hole Diameter (dh) 101.6 mm

Plate Thicknes (tp) 101.6 mm

Weir height (hw) 69.85 mm

Material for tray Carbon Steel A 53B

A 53BShell material Carbon Steel

Allowable stress for shell mat.(Pa) 1.03E+08 PaDensity of shell mat 7850 kg/m3

skirt height 25.4 Mm

Operating pr (p) 112405 Pa

Design pr (pD) 123645.5 Pa

Operating temp 64.7 to 100 *C

Design temp 110 *C

Top disengaging space 609.6 mm

Bottom separator space 1000 mmInsulation material taken Asbestos

Insulation thickness considered 50 mm

Density of insulation 270 kg/m3

Formulae source: [2] and [6]

Shell thickness

J = 0.8

Minimum shell thickness, ts = p*Dt/(2*f*J-p)

= 1.367819 mm

After adding corrosion allowance, ts = 8mm

Do = Di+2*ts

= 1844.8 mm

th = pD*Di/(2*J*Pa-0.2*pD)

= 1.367001 mm


15/21

Adding corrosion allowance, th = 8 mm

Column height = (t + 19*t + 20*tp + 1000)/1000 = 15.224 m

(Refer to Excel sheet along with this file for the calculations)

Stresses developed in column

Formulae

Source [2]

Axial Stress (f as) 8910204 Pa Eq. 9.3.1

Circumferential Stress (f cs) 17820408 Pa Eq. 9.3.1

Compressive stresses caused by dead loads

Wshell 54066.25 N

Mshell 6620.357 kg

f(dead wt shell) 29434.89 Pa Eq. 9.3.2

M(insulation) 1191.137 kg

f(dead wt insulation) 955285 Pa Eq. 9.3.3

Active Area 2.31 m2

Liquid depth on trays 83.2977 mm

Mass of liquid 5002.86 Kg

f(dead wt liquid) 4012265 Pa Eq. 9.3.4

Stress induced by column attachments

f(dead wt attachments) 5309488 Pa Eq. 9.3.5

f(dead total) 10306473 Pa Eq. 9.3.6

f(dead total)


16/21

Below the guy ring

(Formulae used, [2] Eq. 9.3.8-9.4.4)

M(w,H,guy)(max) 26723.1 N

f(wind,H,below guy) 5269799 Pa

f(com guy) 641969.5 Pa

Analysis of stress

For upwind side

Total stress 82920516 Pa

For downwind side

Total stress 92472689 Pa

Torispherical Head has a range of 0.1 to 1.5 MPa. It is, therefore,

suitable to use for our column design.

Design of head

Do 1844.8 mm

ro 110.688 mm 0.06*Do

ho 312.394 mm Eq. 4.2.22

Do2/4*Ro 461.2 mm

(Do*ro/2)^0.5 319.5287 mm

hE, the minimum of above 3 312.394 mm

hE/Do 0.169338 mm

As the diameter of the vessel is not very large, head can be fabricated

from single plate, therefore J = 1

t/DoC 0.000544

C 3.679807 Table 4.1A

t/Do 0.002

t 3.689856 mm Eq. 4.2.20Corrosion allowance 2 mm


17/21

Uncorroded plate thickness 6 mm

Torispherical Head Volume 0.623133 m3

Skirt Support Design

Mshell 6620.357 Kg

Wshell 64879.5 N

Desity of Carbon Steel 7850 Kg/m3

WD 47937.64 N

Wmin 112817.1 NWi 11673.14 N

Wl 391901.7 N

Wmax 516392 N

Period of vibration at min dead wt

Tmin 0.178766 s


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Tmax 0.382461 s


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Design of skirt bearing plate

Stress_max(compressive) 0.099138 MN/m2

Thickness of bearing plate, tbp 5.363147 mm

According to IS code [3] 7 mm

As thickness is


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Area available from shell reinforcrment

As 0.003650572 m2

Area available from nozzle

reinforcrment, An 2H1(tn-tr'-c)

tr' 0.004160723 m

H1 0.05396295 m

An 0.000953987 m2

Reinforcement area available from shell and

nozzle is

As+An 0.004604559 m2

Area remained to be

compansated

A-(As+An) 0.00227287 m2

Ar >= 0.00227287 m2

tp 0.010538229

Ring Pad dimension

Inner Diameter, do 0.25 m

Outer diameter 0.448 m

Thickness 0.025 m


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REFERENCES

[1] Smi, B.D., “Dig f Eqiibim Sg P”, MGw-Hill Book Co.

[2] Introduction to Chemical Equipment Design by B.C Bhattacharya, 2012

[3] IS : 2825-1969, the Indian Standard codes

[4] Plant Design and Economics for Chemical Engineers Fourth Edition

by Max S. Peters & Klaus D. Timmerhaus

[5] R.H.Py, D.W.G; Py’ Cmi Egi’ Hbk7 Eii

[6] R. K. Si, C Ri’ Chemical Engineering: Chemical

Engineering Design (vol. 6), Butterworth-Heinemann, 3rd ed. 1999

PED 2 Report Group 8

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