Project design subject distillation of methanol water mixture

Flash or equilibrium distillation Flash vaporization, or equilibrium distillation is a single stage operation where a liquid mixture is partilly vaporized, the vapor is allowed to come t

VIETNAM NATIONAL UNIVERSITY HCMC

HCMC UNIVERSITY OF TECHNOLOGY

FACULTY OF CHEMICAL ENGINEERING

PROJECT DESIGN SUBJECT: DISTILLATION OF METHANOL-WATER

CHAPTER 1: OVERVIEW

1.1 Introduction of distillation processes:

Distillation is a seperation method in which the components in a liquid mixture are seperated based on their relative volality

Flash or equilibrium distillation

Flash vaporization, or equilibrium distillation is a single stage operation where a liquid mixture is partilly vaporized, the vapor is allowed to come to equilibrium with the residual liquid, and the resulting vapor and liquid phases are separated and removed fromthe apparatus It may be batchwise or continuous

Differential/simple distillation

The simplest example of batch distillation is a single stage, differential distillation, starting with a still pot, initially full, heated at a constant rate In this process the vapour formed on boiling the liquid is removed at once from the system Since this vapour is richer in the more volatile component than the liquid, it follows that the liquid remaining becomes steadily weaker in this component, with the result that the composition of the product progressively alters Thus, whilst the vapour formed over a short period is in equilibrium with the liquid, the total vapour formed is not in equilibrium with the residualliquid At the end of the process the liquid which has not been vaporised is removed as the bottom product

Rectification/Fractionation distillation

The essential merit of rectification is that it enables a vapour to be obtained that is substantially richer in the more volatile component than is the liquid left in the still The fractionating column consists of a cylindrical structure divided into sections by a series ofperforated trays which permit the upward flow of vapour The liquid reflux flows across each tray, over a weir and down a downcomer to the tray below The vapour rising from

the top tray passes to a condenser and then through an accumulator or reflux drum and a reflux divider, where part is withdrawn as the overhead product D, and the remainder is returned to the top tray as reflux R.

Azeotropic distillation

The principle of azeotropic and of extraction distillation lies in the addition of a newsubstance to the mixture so as to increase the relative volatility of the two keycomponents, and thus make separation relatively easy

Steam distillation

Where a material to be distilled has a high boiling point, and particularly wheredecomposition might occur if direct distillation is employed, the process of steamdistillation may be used

1.2 Types of distillation column

Distillation may be carried out in plate columns or packed columns in which each plate constitutes a single stage where mass transfer is between a vapour and liquid in continuous countercurrent flow

* Types of trays:

cost of sieve trays (10,11)

Low 20% higher than

sieve trays (11)

than sieve trays

Moderate Moderate

Effects of corrosion High Low Low to moderate

solids

Low Low to moderate

1.3 Methanol-water mixture

Dynamic viscosity 510*10-6(poise) 1250*10-7(poise)

* Equilibrium data for Methanol-water mixture at 1atm (%mol)

y 0 26.8 41.8 57.9 66.5 72.9 77.9 82.5 87 91.5 95.8 100

T 100 92.3 87.7 81.7 78 75.3 73.1 71.2 69.3 67.6 66 64.5

* No azeotrope mixture

2.2 Process description

The mixture containing 20% molar methanol and 80% molar water is heated to obtain the saturated liquid mixture by the heat exchanger (2) Then, it is delivered to the fractionating column (3) at the feed stage At each stage, the plate receives the vapour from the below stage and the liquid from the above stage whose are well mixed For rectifying stages, a liquid recycle condenses the less volatile components from rising vapour, hence more volatile components are concentrated in vapour stream On the other hand, for rectifying stages, a vapour recycle vaporizes the more volatile components fromdownflow liquid, hence the more volatile component is removed

The vapour at the top of the column is collected and totally condensed by condenser (4) to recover the top product with the methanol concentration of 99% A portion of the top product liquid is sent back to the column as the reflux while the rest is transfer into tank (5) for storage.

A portion of liquid at the bottom of the column is totally vaporized by reboiler (6) that is used to return to the distillation column The rest is collected as bottom product from tank (7)

CHAPTER 3: MATERIAL AND

ENERGY BALANCE2.1 Material balance

Operating data:

Feed section

Inlet temperature (TF) 30oCMethanol concentration (xF) 20% molarState of materials Saturated liquid

Overhead product

section Methanol concentration (xD) 99% molar

Bottom product

section Methanol concentration (xB) 0.01% molar

Operating pressure Atmospheric pressure

* Mass balance:

Overall : F=B+ D;V =L+ D(1)

* Component balance:

M F =x F ∗M methanol+(1−x F)∗M methanol =0.2∗32.042+0.8∗18.015=20.8204 ( g mol)

M D =x D ∗M methanol+(1−x D)∗M methanol =0.99∗32.042+0.01∗18.015=31.9017( g mol)

Operating component balance line:

* For rectifying section:

CHAPTER 4: MAJOR EQUIPMENT

DESIGN4.1 Tower height:

4.1.1 Rectifying section height:

=807.8605 kg/m3

ρ y (recifying)=[y y(recifying)b ∗M m +(1− y y(recifying) )M w]∗273

22.4 T rectifying =[0.787∗32.042+(1−0.787)∗18.015]∗273

22.4∗346.175 =1.0229 kg/m3

P methanol sat =expexp(C 1+ C 2 T +C 3∗ln ln(T)+C 4∗T C 5)=expexp(81.768+ −6876346.175−8.7078∗ln ln(346.175)+7.1926∗10 −6 ∗346.175 2)=140523.7365(Pa)

P water sat =expexp(C 1+C 2 T +C 3∗ln ln(T)+C 4∗T C 5)(73.649+−7258.2346.175−7.3037∗ln ln(346.175)+4.1653∗10 −6 ∗346.175 2)=35488.9819(Pa)

log log(μ rectifying)=x rectifying ∗log ⁡(μ methanol)+(1−x rectifying)log ⁡(μ water)

4.1.3 Rectifying section diameter

log log(μ stripping)=x rectifying ∗log ⁡(μ¿¿methanol)+(1−x rectifying)log ⁡(μ¿¿water)=0.105∗0.255+(1−0.105)∗0.305 ¿¿

→μ stripping =0.2993 cP

P methanol sat =expexp(C 1+ C 2 T +C 3∗ln ln(T)+C 4∗T C 5)=expexp(81.768+ −6876362.965−8.7078∗ln ln(362.965)+7.1926∗10−6∗362.9652)=254018.4739(Pa)

P water sat =expexp(C 1+C 2

→Choose H tower =15(m)

Choose tray tower with downcomer:

Tray fabricated material Stainless steel

Tower diameter, DT (m) Rectifying section (m) 0.607

Stripping section (m) 0.604Area at the top of the downcomer AD or ADT 10% of total tray area, AT

4.1.5 Tray layout

Since downcomer area AD at the top account for 10% of the total tray area, we have:

Using irritation to solve the above equation, we have α=93.2061 °

* Outlet weir length:

4.2 Tower parameters summary:

Tray fabricated material Stainless steel

Tower diameter, DT (m) Rectifying section (m) 0.7

Stripping section (m) 0.7Area at the top of the downcomer AD or ADT 10% of total tray area, AT

Holes layout on tray Hexagonal

4.3 Hydraulic performance:

4.3.1 Dry pressure drop:

Resistance factor ξ=1.82 as fractional hole area Ahole = 0.07 to 0.10

For rectifying section:

Vapor velocity throughholes :ω ' rectifying=ω rectifying

8 % = 1.12250,08 =14.03125( m s)

ΔP dry(rectifying) =ξ¿¿

For stripping section:

Vapor velocity through perforations: ω ' stripping=ω stripping8% = 0.77070,08 =9.6338( m s )

ΔP dry(stripping) =ξ¿¿

4.3.2 Residual pressure drop:

As we choose hole diameter dhole = 3mm > 1mm, we have:

For rectifying section (T = 346.175K):

4.3.3 Pressure drop due to liquid on the tray:

For rectifying section:

L w =sin D rectifying sin α2=0.667∗sin sin 93,322 =0.4851(m)

Aswe have : Q rectifying

L w =sin D stripping sin α2=0.604∗sin sin 93,322 =0.4393(m)

Aswe have : Q stripping

4.3.4 Totoal pressure drop:

ΔP rectifying =ΔP dry(rectifying) +Δp R(rectifying) +ΔP t =183.2598+14.5099+187.5149=385.2846(N /m2 )

ΔP stripping = ΔP dry(stripping) + Δp R(stripping) + ΔP t =75.9889+13.4686+197.0918=286.5493(N /m2 )

CHAPTER 5: MECHANICAL

DESIGN5.1 Tower thickness

With D = 0.7 (m), we design the cylindrical body by electric arc welding, two-sided welding Parts of the column are joined by flanges (Table XIII.8, p 362) with weld joint efficiency: 𝜑ℎ = 0.95 (in case of closed wall)

5.1.1 Design pressure calculation

Operating temperature: toperating = tmax = 100oC

Operating pressure: P = 1atm = 104197.7856(N /m2)

Choose X18H10T as fabracating material:

Standard allowable stress: ¿

Correction factor: η=1 due to non-isolation shell

→ Allowable stress: [σ]=η.[σ ]¿=142( N

mm2 )

Design pressure calculation:

+ Hydrostatic pressure: p1=ρ x ∗g∗H tower=998∗9.81∗15=146855.7

+ Design pressure: P Design = p1+P=146855.7+101325=248180.7(N /m2)

5.1.3 Thickness addition coefficient:

Thickness additional coefficient:

C=C A +C B +C C +C0

Where

Ca: additional coefficient due to corrosion, time of use of equipment is 15 years,

corrosion rate of methanol is less than 0.05 mm/year → Ca = 0.75 (mm) = 0.00075 (m)Cb: additional coefficient due to abrasion (Cb = 0, we only calculate Cb when the velocity

≥ 20 m/s (liquid) and 100 m/s (vapor) or the medium contains a lot of solid particles.)Cc: addition coefficient due to thickness tolerances, Cc = 0.0005 (m) (Table XIII.9, p.362[2])

The thickness of cylindrical body: S = 4 (mm).

The outer diameter of the body: Do= 700 + 2 × 4 = 708 (mm)

5.2 Bottom and head thickness

Choose ellipsoidal head made by stainless steel X18H10T

The bottom thickness is calculating using equation XIII.47, p.385 [2]:

k: Dimensionless coefficient, k = 1

C: Additional coefficient as S - C = 4 - 1.25 = 2.75 𝑚𝑚 ≤

10mm) (XIII.17, p.386 relation) →ChooseC=0.00125+0.002=0.00325(m)

From (Table XII.4, p.310) with thickness from 4 – 25 mm, we can get the Tensile

strength limit 𝜎k= 550 106 (N/m2)

[σ k]

P Design k φ h=0.2481807550 ∗1∗0.95=2105.32084 ≥30

→P Design canbe neglected

And from (Table XIII.3, p.356), we get the nk=2.6, nc= 1.5, 𝜂 = 1:

Following the (XIII.1, XIII.2, p.355)

2∗175 +3.25=3.9062(mm)Choose S’ = 4 (mm)

Checking the wall stress using equation XIII.49, p.386 [2]:

5.3 Diameters of pipes – flanges connecting parts of equipment with pipes

Tubes are usually connected to the device by a removable or non-removable joint

In this device, we use removable joints For removable joints, a pipe is made with a shorttube and a flange or screw – thread to connect to the pipe Flanged type commonly usedwith tube diameter d > 10 mm Usually choose screw – thread with 𝑑 ≤ 10 𝑚𝑚,sometimes can choose 𝑑 ≤ 32 𝑚𝑚 Pipe is made by stainless steel X18H10T Also,flange is made by stainless steel X18H10T, structure is blind flanges We can determinediameter of pipes based on the Equation II.36, p.369 [1]:

Superficialvelocity DiamterOverhead vapor 64.65 1.15 0.99 0.259021 20.00 0.1612

5.4 Flanges and pipes

The flange is an important part used to connect parts of equipment as well as to

connect

other parts to the body Choose body joint flange, bottom and head made of steel

X18H10T The structure of the flange is a weld neck flange This type of flange is mainlyused for equipment working with low and medium pressure

5.4.1 Flanges connecting to equipment’s body

Choose Flange: type 1 Material: Stainless steel, where:

+ Dt: internal diameter, D: flange diameter, Db: length of two symmetric bolts

+ Do: external diameter, db: bolt diameter

+ Z: number of bolts, h: height of flange

Search for (Table XIII.27, p.421 [2]), at Py = 0.3 N/mm2 > PDesign

Typ

D (mm)

At (Table IX.5, p.170 [2]):

htray = 410 (mm), then distance between flanges = 1300 (mm)

Number of trays between 2 flanges: 3 trays

Number of actual flanges:

N t

N p+2=333 +2=13(flanges)

The tightness of the flanges is depended by the packed material The packed material

is softer than the flange material, when tighten the bolts, the packed layer is deformed and filled up to the ruggedness of the flange We choose a packed layer of asbestos, with

a thickness of 3 (mm)

5.4.2 Flanges connecting equipment parts with pipes

Choose Flanges: type 1 Material: Black metal, where:

+ Dy: standard diameter of pipes, D: flange diameter, Db: The length of two

symmetric bolts

+ h: Height of flange, db: bolt diameter

Search for (Table XIII.26, p.409 [2]):

(mm)

Based on the (Table XIII.32, p.434 [2]), we can get the pipe fitting values:

Based on (Table XII.7, p.313), with the material X18H10T: 𝜌𝑋18𝐻10𝑇 = 7900 (kg/m3)

5.5.1 Mass of flanges connected to equipment’s body

Number of flanges: n = 13 (flanges)

5.5.2 Mass of trays

As there is 20% of surface area for the downcomer and 8% for fractional hole area, so

the surface area of the tray is only 72% left

m tray =0.8∗n trays ∗D T2∗δ tray ∗ρ X 18 H 10 T=0.72∗33∗0.7 2∗0.003∗7900=216.7109(kg)

5.5.3 Mass of weirs and downcomer tracer:

Choose weir thickness: δweir = 0.003 (m)

Length of weir: Lweir =

m weirs =N trays ∗L weir∗(h weir +h downcomer)∗δ weir ∗ρ X 18 H 10T=33∗0.5086∗(0.025+0.35)∗0.003∗7900=149.166(kg)

5.5.4 Mass of tower shell:

Outer column diameter:

m tower shell = π4∗(D02−D T2)∗H Tower ∗ρ X18 H 10 T = π4∗(0.708 2 −0.7 2)∗15∗7900=1048.3369(kg)

5.5.5 Mass of head and bottom’s shells:

With DT = 700 (mm) and the thickness of bottom S = 4 (mm), we can search for the

weight of the bottom on (Table XIII.11, p.384): 19 (kg)

→m bottom+head =2∗19=38(kg)

5.5.6 Mass of liquid:

m liquid = π4∗(D T2)∗H Tower ∗ρ x = π4∗(0.7 2)∗15∗998=5761.1311(kg)

5.5.7 Total distillation column weight:

m T =m flange +m trays +m body shell +m liquid +m bottom+head +m weir =352.9537+216.7109+1048.3369+5761.1311+38+149.166=7566.2986 (kg)

In table (XIII.35, p.437, [2]), we have:

Hangers are made by CT3 steel Choose 4 hangers with allowable weight on each edge:

Gt = Gc = 18556.3474 (𝑁) In table (XIII.36, p.438, [2]), we have:

current

Counter-Inner fluid/

Tube side Feed stream Subcooled water Overhead product

Saturated watervapor

6.1.1 Heat flux in outer tube

h outer=1.28∗4√ λ water ∗k water3∗ρ water2

μ water ∗∆ T outer ∗d equivalent →q outer =h outer ∗∆ T outer

Where:

∆ T outer =T outer −T wall ,1 =120.61−T wall, 1

k water , ρ water ,∧μ water are searched at T = T outer +T wall ,1

6.1.2 Heat flux of through wall

Totalresistance :R total=δ wall

k wall +R inner +R outer= 0.00317.5 +0.0002+0.0002=1750()

Sinceq outer =q inner =q wall=T wall, 1 −T wall ,2

R total

→T wall , 2 =T wall ,1 −q outer ∗R total =75.6578℃

6.1.3 Heat flux in inner tube

∆ T inner =T wall ,2 −T inner=−55.85

λ F ,k F , ρ F ,∧μ F is …at T= T inner +T2 wall ,2=❑❑

6.1.5 Heat transfer area

Overall heat transfer coefficient

∆ T outer =T hot , ave −T wall , 1 =64.65−T wall ,1

k D , ρ D ,∧μ D is searchedat T= T hot , ave +T wall ,1

→q outer =h outer ∗∆ T outer=40108.35597

6.2.2 Heat flux of through wall

Totalresistance :R total=δ wall

k wall +R inner +R outer= 0.00317.5 +0.0002+0.0002=1750()

Sinceq outer =q inner =q wall=T wall, 1 −T wall ,2

R total

→T wall , 2 =T wall ,1 −q outer ∗R total =32.2938℃

6.2.3 Heat flux in inner tube

6.2.5 Heat transfer area

Overall heat transfer coefficient

Heat transfer area

U∗∆ T LMTD =15.1507083(m2 )

6.3 Overhead product cooler

6.3.1 Heat flux in outer tube

λ wall ,1 ,k wall ,1 , ρ wall ,1 ,∧μ wall , 1 is searched at T=T wall ,1

Assume: T wall ,1 =32.9984 ℃, we have:

6.3.2 Heat flux of through wall

Totalresistance :R total=δ wall

k wall +R inner +R outer= 0.003

17.5 +0.0002+0.0002=1750()