Plug Flow Reactor Design Presentation...
Short Description
Design equations and specifications of a typical tubular reactor...
Description
KUMASI POL POLYTECHNIC YTECHNIC SCHOOL OF ENGINEERING DEPARTMENT OF CHEMICAL ENGINEERING FINAL YEAR DESIGN REPORT ON PLUG FLOW REACTOR COMPILED BY BOATENG MICHAEL BOATENG (CME05100016) SUPERVISOR: SUPER VISOR: DR. FRANCIS ATTIOGBE 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 = 246.5 oC Average pressure of reactor = 202.65kPa Mole of components = mole(CH 3OH + CH2O + H2O + N2 + O2) = 1281.5929kmol/hr Reaction rate constant for main and side reactions = K 1 and K2 Density of silver catalyst used = 950kg/m 3 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 = 246.5 0C = 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 ) = Putting –r m into equation (1): : K d τ = CAO K
K : K
=
K ( : K )
Concentration of methanol initially (CA):
CA =
CAO ( 1 XA ) (1+ εA XA )
But from the reaction equation: CH3OH + 0.5O2
CH2O + H2O ⅀u ;⅀ ⅀ ; . = .
Extent of reaction, (ԐA) =
= 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 6. = V min .
= 0.0138kmol/m3 Thus residence time: ( : K )d τ = CAO K
But
=
. .
= 0.0002
Initial rate constant, K 1 = 1.1657, final rate constant, K 2 = 12474.5 conversion rate, X AF = 0.874 . (.:.)d .6
Therefore: τ = CAO Since CA =
(; ) , (:.)
The above relation then becomes : . (:. ) (.:6 )d (; )
τ = CAO
. (:. ) [.( :. :. ; ]d (; ) (:.)
= CAO
Putting CAO = 0.0138 into the above relation: τ=
. [.( :. :. x . ; ]d (; )
0.8578
Solving integrally: . [.( :. d . (; ) . + 172.15 + τ = (; ) (; ) . . = 0.8578 (0.0002[ + 0.333 ] + 172.15(0.874)) (; ) (; )
Residence time calculations continued: = 0.8578 (0.0002[In
.
1 XA
. (;: ) ] + 150.46) (; )
+ 0.333
= 0.8578(0.0002[-In(1 – 0.874) – In (1)] . . (: ) 0.333 ] + 134.10) (; ) (; )
+0.333[
.
= 0.8578 (0.0002(2.07) + 0.333[-In (1 – XA)
= 0.8578 [0.0004 + 0.333(2.07 – 0.874) + 150.46] = 0.8578 (0.0004 + 0.3983 + 150.46) = 129.40
.
XA
] + 150.46)
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. .
Volume of catalyst used: (VC) =
= 3713.22
Weight of catalyst density of catalyst
=
.kg = 4m3 kg/m
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/m3 Volume of vessel, V V = ? Volume of reactor vessel:
Voidage, (e) =
V ; V = 0.6 V
It implies that, VV – 0.6VV = 4 0.4VV = 4 VV =
= 10m3 .
Hence volume of vessel needed is 10m 3
Heat capacity of liquid mixture 150 oC: [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 = 86917883.89W h x kg.K
Provisional area (A):
Thus heat, Q = UALTM A=
6. = ULM / . × 6.
Provisional
area = 104.5m2
Number of tubes:
Nt =
. = = 107tubes () ×. ×6.
Pitch diameter:
For square pitches, Pitch used = 1.25 x OD = 1.25 x 0.0508m = 0.0635m Bundle diameter:
N Bundle diameter, DB = do x ( ) 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, M L = 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 m RT M , but = ₱ (density)
Also, PV = nRT = PM =
= ₱RT ₱=
PM .6kPa x .kg x kmol.K = = 1.25kg/m3 kmol x . x K
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 ): π x D Volume of reactor, Vr =
Let length of reactor, L = 3 x diameter of reactor = 3D π x D Vr =
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 ): π x D . (.6) Ar = = = 2.06m2
Superficial velocity of fluid (U c):
.kmol x .kg x h .kg x m UC = = 10.0771kg/s = h x kmol x 6s 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 = 0.18psi .kPa
Tube Dimensions: Assumptions: Outer diameter of tube = 2in = 0.0508m Length of tube chosen = 20ft = 6.09m π x d x L . x (.) x . Therefore volume of tube, V t = =
= 0.0435m3
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/mm 2 Thus, cylindrical section allowance or Voidage (e) =
P D . x .6 x = f;P 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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