Plug Flow Reactor Design Presentation...

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KUMASI

POL

YTECHNIC

SCHOOL OF ENGINEERING

DEPARTMENT OF CHEMICAL ENGINEERING

FINAL YEAR DESIGN REPORT ON PLUG FLOW REACTOR

COMPILED BY

BOA

TENG

MICHAEL

(CME05100016)

SUPER

VISOR:

DR.

FRANCIS A

TTIOGBE

JUNE, 2013

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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

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DESIGN PARAMETERS CALCULATIONS

Plug flow reactor design: Reactor conversion = 87.4%

Average temperature of reactor = 246.5o_C

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:

 =



(4)

But temperature of plug flow reactor used, T = 246.50_{C = 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: P_mV = nRT

P_m= C_ART

Methanol reaction rate: (

r

_m

) =

K

 : K__

=

 K__ (_: K__)

Putting –r _minto equation (1):

τ =

C

_AO



: K_K  d



_ 

(5)

Concentration of methanol initially(C_A):

C_A = CAO( 1  XA) (1+ ε_AX_A)

But from the reaction equation:

CH₃OH + 0.5O₂ CH₂O + H₂O

Extent of reaction, (Ԑ_A) = ⅀u _⅀ ;⅀ 

 

=; . . = 0.333

Also total molar flowrate components entering = 1281.5929kmol/hr And PV_o = nRT,

Volumetric flowrate of components, V_o= _

=. x . x . .6

= .mx h h x 6min = 455.25m3_/min

(6)

Mole flowrate of methanol: (F_methanol) = 6.kmol x h_{h x 6min}

= 6.27kmol/min

Thus initial concentration of methanol:

C

_AO

=

F

V_ =

6.    min .

= 0.0138kmol/m3

Thus residence time:

τ =

C

_AO



(: K)d

K__ _



But _

=

_{.  .}

=

0.0002

Initial rate constant, K ₁ = 1.1657, final rate constant, K ₂= 12474.5 conversion rate, X_AF= 0.874 Therefore: τ =

C

_AO



(.:.)d .6_ .  Since C_A= (; ) (:._) ,

(7)

The above relation then becomes

:

τ =

C

_AO



(:. ) _(; _) (.:6_)d_  . 

=

C

_AO



(:. ) _(; _) [.( :._ :._ ; _ ]d_ (:._) . 

Putting C

_AO

= 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))

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Residence time calculations continued:

= 0.8578 (0.0002[In

1  X

_A



_.

+

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 – X

_A

)



_.



X

_A



_.

]

+ 150.46)

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

= 0.8578 (0.0004 + 0.3983 + 150.46)

= 129.40

(9)

Thus residence time, τ = 129.40

=

.kmol x kg.h x m  m x kmol x kg

=

.6ℎ  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

(10)

Catalyst Dimensions:

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

Hence diameter of silver catalyst chosen = 3mm

Voidage = 55% - 65%

Chosen Voidage = 60% = 0.6

Density of silver catalyst used = 950kg/m

3

Volume of vessel, V

_V

= ?

Volume of reactor vessel

:

Voidage, (e)

=

V ; V V_

= 0.6

It implies that, V

_V

– 0.6V

_V

= 4

0.4V

_V

= 4

V

_V

=

 .

= 10m

3

(11)

Heat capacity of liquid mixture 150o_{C: [Cp}

mx= CP(CH3OH + H2O + N2 + O2+ CH2O)]

= 1920 + 4320 + 1039 + 910 + 847 = 9036J/kg. K

Mass flow rate of liquid: (ṁ_mix) = 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 =  ULM= 6. /. × 6. _{Provisional area = 104.5m}2 Number of tubes: N_t=__{ ()} =_{ ×. ×6.}. = 107tubes

(12)

Pitch diameter:

For square pitches,

Pitch used = 1.25 x OD

= 1.25 x 0.0508m = 0.0635m

Bundle diameter:

Bundle diameter, D

_B

= d

_o

x

(

N K_

)

 

But outer diameter, d

_o

= 2in = 0.0508m

Also for one pass and two tube passes, K

₁

= 0.215, n

₁

= 2.207

Therefore bundle diameter, D

_B

= 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

(13)

Pressure drop in the reactor:

Molar mass of liquid, M_L = M_W(O₂) + M_W(CH₃OH) + M_W(H₂O) + M_W(N₂) + M_W(CH₂O) = 0.2935(32) + 0.1483(32) + 0.00006(18) + 0.5579(28) + 0.0002(30) = 29.77kg/kmol Also, PV = nRT = m M

RT

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

(14)

Volume of reactor (V_r):

Volume of reactor, V_r= π x D



Let length of reactor, L = 3 x diameter of reactor = 3D V_r= π x D   Thus V_r=  x _ = 3.142 x D3 D = 

4.244

= 1.62m

Hence length of reactor, L = 3 x 1.62m = 4.86m Area of reactor (A_r): A_r= π x D  = .  (.6)  = 2.06m2

(15)

Superficial velocity of fluid (U_c): U_C = .kmol x .kg x h_{h x kmol x 6s} = 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:



= 0.8611U_C + 79.74(U_C)2

Putting U_C = 3.9m/s

P = 0.8611(3.9) + 79.74(3.9)2

= 3.3583 + 1212.88 = 1.22kPa

(16)

Tube Dimensions:

Assumptions:

Outer diameter of tube = 2in = 0.0508m

Length of tube chosen = 20ft = 6.09m

Therefore volume of tube, V

_t

=

π x d x L



=

. x (.) x . 

= 0.0435m

3

Log Mean Temperature Difference:

LMTD =

ℎ; ; (ℎ;) ln(−_{−})

=

; ; (6;) ln(−_{−})

=

6 .

= 96.7

o

C

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

Hence mean temperature difference (DTM) = f



_{LMTD = 1.0}



_{96.7 =}

(17)

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) = PD

f;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

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