Plug Flow Reactor Design Presentation

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KUMASI POLYTECHNIC

SCHOOL OF ENGINEERING

DEPARTMENT OF CHEMICAL ENGINEERING

FINAL YEAR DESIGN REPORT ON PLUG FLOW REACTOR

COMPILED BY

BOATENG MICHAEL

(CME05100016)

SUPERVISOR: DR. FRANCIS ATTIOGBE

JUNE, 2013



Design Item (Plug Flow Reactor) Specifications

Function To convert methanol by oxidation into

formaldehyde

Material of construction Carbon steel

Diameter of reactor 1.62m

Volume of reactor 10m3

Height of reactor 4.86m

Area of reactor 104.5m2

Thickness of reactor 9mm

Volume of silver catalyst 4m3

Weight of silver catalyst 3713.22kg

Pressure drop in the reactor 1.22kPa

Number of tubes 107

Length of tubes 6.09m

Volume of tubes 0.0435m3

Residence time achieved 8 minutes

Superficial velocity of fluid 3.9m/s

Pitch diameter 0.0635m

Log Mean Temperature Difference 96.7oC



DESIGN PARAMETERS CALCULATIONS

Plug flow reactor design:

Reactor conversion = 87.4%

Average temperature of reactor = 246.5oC

Average pressure of reactor = 202.65kPa

Mole of components = mole(CH3OH + CH2O + H2O + N2 + O2)

= 1281.5929kmol/hr

Reaction rate constant for main and side reactions = K1 and K2

Density of silver catalyst used = 950kg/m3

Calculating of the residence time in the reactor:

Basic performance equation for a Plug Flow Reactor is:

= ;

=

;

The reaction rate (-r), expression for methanol-formaldehyde system is:

=

:



But temperature of plug flow reactor used, T = 246.50C = 519.5K

Where Log10 K 1 = 10.79 -6

= 10.79

6

.= 0.0666

K 1 = antilog (0.0666) = 1.1657

Also Log10 K 2 = 11.43 -

= 11.43

.= 4.0960

K 2 =antilog (4.0960) = 12474.5

From equation the above relations:

PmV = nRT

Pm = C ART

Methanol reaction rate: (rm ) =K

: K=

K

( : K )

Putting –r m into equation (1):

τ = CAO

: K d

K



Concentration of methanol initially (CA):

CA =CAO( 1 XA)

(1+ εAXA)

But from the reaction equation:

CH3OH + 0.5O2 CH2O + H2O

Extent of reaction, (ԐA) =⅀u ;⅀

⅀

= ; ..

= 0.333

Also total molar flowrate components entering = 1281.5929kmol/hr

And PVo = nRT,

Volumetric flowrate of components, Vo =

=. x . x .

.6

=.m x h

h x 6min

= 455.25m3/min



Mole flowrate of methanol: (Fmethanol) =6.kmol x h

h x 6min

= 6.27kmol/min

Thus initial concentration of methanol: CAO =F

V=

6.

min .

= 0.0138kmol/m3

Thus residence time:

τ = CAO (

: K)d

K

But

=

. .= 0.0002

Initial rate constant, K 1 = 1.1657, final rate constant, K 2 = 12474.5 conversion rate, XAF =

0.874

Therefore: τ = CAO (.:.)d

.6

.

Since CA =(; )

(:.),



The above relation then becomes:

τ = CAO

(:. )

(; ) (.:6)d

.

= CAO (:. )

(; ) [.( :. :. ; ]d

(:.)

.

Putting CAO = 0.0138 into the above relation:

τ = 0.8578 [.( :. :. x . ; ]d

(; )

.

Solving integrally:

τ = [.( :. d(; )

. + 172.15 (; )

(; ).

+ .

= 0.8578 (0.0002[

(; )

.

+ 0.333

(; )

.

] + 172.15(0.874))



Residence time calculations continued:

= 0.8578 (0.0002[In 1 XA +

.

0.333

(;: )

(; )

.

] + 150.46)

= 0.8578(0.0002[-In(1 – 0.874) – In (1)]

+0.333[

(; )

. 0.333

(: )

(; )

. ] + 134.10)

= 0.8578 (0.0002(2.07) + 0.333[-In (1 – XA) .

XA ].

+ 150.46)

= 0.8578 [0.0004 + 0.333(2.07 – 0.874) + 150.46]

= 0.8578 (0.0004 + 0.3983 + 150.46)

= 129.40



Thus residence time, τ = 129.40

=.kmol x kg.h x m

m x kmol x kg

=.6ℎ 6

ℎ

= 8.2 minutes

Hence the residence time achieved is 8.2 minutes

Weight of silver catalyst (W):

Also, τ = W

F

It implies that: (W) = τ

=. 6.

.

= 3713.22

Volume of catalyst used: (VC) =Weight of catalyst

density of catalyst=

.kg

kg/m = 4m3



Catalyst Dimensions:

Diameter of catalyst from rules of thumbs ranges from 2mm - 5mm

Hence diameter of silver catalyst chosen = 3mmVoidage = 55% - 65%

Chosen Voidage = 60% = 0.6

Density of silver catalyst used = 950kg/m3

Volume of vessel, VV = ?

Volume of reactor vessel:

Voidage, (e) =V ; V

V= 0.6

It implies that, VV – 0.6VV = 4

0.4VV = 4

VV =

. = 10m3

Hence volume of vessel needed is 10m3



Heat capacity of liquid mixture 150oC: [Cpmx = CP (CH3OH + H2O + N2 + O2 + CH2O)]

= 1920 + 4320 + 1039 + 910 + 847

= 9036J/kg. K

Mass flow rate of liquid: (ṁmix) = ṁ (CH3OH + H2O + N2 + O2 + CH2O)

= Molar mass of mixture x Mole flow rate

= 140kg/kmol x 1281.5929kmol/hr

= 179423kg/hr

Heat supplied (Q):

Q =kg x 6J x K

h x kg.K = 86917883.89W

Provisional area (A):

Thus heat, Q = UALTM

A =

ULM=

6.

/. × 6.

Provisional area = 104.5m2

Number of tubes:

Nt =

()=

.

×. ×6.= 107tubes



Pitch diameter:

For square pitches,

Pitch used = 1.25 x OD

= 1.25 x 0.0508m = 0.0635m

Bundle diameter:

Bundle diameter, DB = do x (N

K)

But outer diameter, do = 2in = 0.0508m

Also for one pass and two tube passes, K 1 = 0.215, n1 = 2.207

Therefore bundle diameter, DB = 0.0508 x (

.)

.

= 0.8469m

Bundle diameter clearance with bundle diameter of 0.8469m is 0.17m from

bundle diameter clearance against bundle diameter charts



Pressure drop in the reactor:

Molar mass of liquid, ML = MW(O2) + MW(CH3OH) + MW(H2O) + MW(N2) + MW(CH2O)

= 0.2935(32) + 0.1483(32) + 0.00006(18) + 0.5579(28) + 0.0002(30)

= 29.77kg/kmol

Also, PV = nRT =m

MRT

PM =

, but

= ₱ (density)

= ₱RT

₱ =PM

=

.6kPa x .kg x kmol.K

kmol x . x K= 1.25kg/m3

Assumptions:

Viscosity of fluid, = 0.02cP = 0.00002N.s/m2

Length of tube, L = 6.09m

Diameter of tube, D = 2in = 0.0508m

Voidage, e = 0.6



Volume of reactor (Vr ):

Volume of reactor, Vr =π x D

Let length of reactor, L = 3 x diameter of reactor

= 3D

Vr =π x D

Thus Vr = x

= 3.142 x D3

D = 4.244

= 1.62m

Hence length of reactor, L = 3 x 1.62m

= 4.86m

Area of reactor (Ar ):

Ar =π x D

=

. (.6)

= 2.06m2



Superficial velocity of fluid (Uc):

UC =.kmol x .kg x h

h x kmol x 6s= 10.0771kg/s =

.kg x m

s x .kg x .6m

=. m/s

.

= 3.9m/s

Pressure drop in the reactor:

Using the Ergun equation:

=(;)( )

+.(;)(() ₱)

Putting values of the parameters in the above equation:

= 0.8611UC + 79.74(UC)2

Putting UC = 3.9m/s

P = 0.8611(3.9) + 79.74(3.9)2

= 3.3583 + 1212.88

= 1.22kPa

Hence pressure drop in the reactor, P =.kPa x .psi

.kPa= 0.18psi



Tube Dimensions:

Assumptions:

Outer diameter of tube = 2in = 0.0508mLength of tube chosen = 20ft = 6.09m

Therefore volume of tube, Vt =π x d x L

=

. x (.) x .

= 0.0435m

3

Log Mean Temperature Difference:

LMTD =ℎ; ; (ℎ;)

ln(−−

)=

; ; (6;)

ln(−−

)=

6

.= 96.7oC

For steam, it is assumed that temperature correction factor is (f) = 1.0;

Hence mean temperature difference (DTM) = f LMTD = 1.0 96.7 =96.7C



Thickness of plug flow reactor vessel:

Design pressure = 202.65kPa = 2.03bar = 0.203N/mm2

Taking 15% above operating pressure = (2.03) x 1.15

= 2.33bar

= 0.233N/mm2

Hence design pressure (P1) = 0.233 + 0.203

= 0.44N/mm2

For carbon steel, allowable stress (f) = 70N/mm2

Thus, cylindrical section allowance or Voidage (e) = PDf;P= . x .6 x

x ;.

= 5.10mm

Adding corrosion allowance of 30% for 15years:

Thickness of the vessel = 5.10 + (0.3 x 15)

= 9.6mm



THANK YOU

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Plug Flow Reactor Design Presentation