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Unformatted text preview: " REACTOR DESIGN " Volume Calculations
REACTIONS OCCURING 1 CO + CH3OH (G) (L) 2 CH3OH + CH3COOH (L) (L) 3 CO (G) + H20 (L) CO2 (G) CH3COOH ( MAIN REACTION ) (L) CH3COOCH3 + H20 + H2 (G) 4 4CO (G) + 3H2O (L) STEP 1: First of all we will calculate gas hold up ( ЄG) According to Hassan and Robinson ( 1997 ) it is given as ЄG = 0.25(QVG*NR2/σ)0.45 As Gas mass flowrate ( MG ) = ρG QVG It is given that NR σ Putting the values in equation A we get ЄG STEP 2 = Assume a supeficial gas velocity ( Vsg) Vsg STEP 3 Calculate area A = QVG/Vsg A STEP 4 Calculate diameter As A =( π / 4 )^ D2 So D2 D STEP 5 Volume determination As ЄG + ЄL ЄL where ЄL Also ЄL = VL/VT Also we have residence time TR Now VL = Where Vo TR * Vo = Mass flowrate of methanol ( ML) = = Density ( ρ ) = Vo = putting in equation B we get VL = Now from equation B VT ACCORDING TO BOOK " Chemical Process Engineering Design WE HAVE TO GIVE HEADSPACE IN REACTOR VOLUME . BY F If V < 1.9 m^3 than 15 % allowance is must If V > 1.9 m^3 than 10 % allowance is must As our reactor volume is greater than 1.9 m3 so we have to give 10 Therefore finally VR = Step 6: Verification of superficial velocity by optimum HR/DR ratio ( reactor HR HR HR/DR = VR/A = = RESULTS VR HR DR VSG Internal Dimension Calculations
Diameter of the impeller blade = d2 = Distance from the bottom of the reactor bottom of impeller = h2 = Height of the impeller blade = h3 = Width of the impeller blade = L3 = Width of the baffles = J = Power absorbed by the impeller First we will calculate the power absorbed when only liquid stream is present i.e Reynolds # is given as = Re = (ρL*NR*(d2^2))/µL) where ρL =
COMPONENTS CH3COOH CH3COOCH3 H2O CH3I
TOTAL ρL = Similarly µL=
COMPONENTS CH3COOH CH3COOCH3 H2O CH3I TOTAL µL= ρL NR µL σ g KL Re = density of liquid rotational speed of impeller viscosity of liquid phase surface tension of liquid phase acceleration due to gravity thermal conductivity 1882355.6 Against this reynold # the power # Np is given in fig 4.20 of book " CHEMICAL Np( against Re from Fig is) As Np is PL/(ρL*NR^3*d2^5) so PL/(ρL*NR^3*d2^5) this implies PL = = 14723.52 14.72 2.8 3.8 = = Aeration Number= NQG=QVG/(NR*d2^3) Fot this we need QVG which is mass flowrate divided by density of gas wh QVG = 0.03 NQG = 0.042 Now from fig 4.21 page # 276 book "chemical reactors " by Trembuze this PLG/PL = PLG = 0.58 8.466 11.4 2 Liquid Holdup:
Denoted by Є11 or ЄL is given as = Gas Holdup:
Denoted by Є1 or ЄG is given as = Sparger Calculations:
First we will have to calculate bubble diameter dB According to Jiang et al (1995) it is given as below dB^2 = ((VSG*UL/σ)^0.04) ((ρL*σ^3)/g*UL^4)^0.12) ((ρL/ρg)^0.22) σ/g*ρL dB^2 dB = = According to Trembuze et al when bubble diameter comes thi therefore no need to determine the number of holes. Reactor Jacket Calculations Jacket Specifications: Jacket height Spacing between jacket and reactor Pitch Tentering Tleaving Mass flow rate (m) Jacket type Heat Transfer Coefficent Calculation: 1. Heat transfer coefficient at the outside wall of reactor:The baffle forms a continous spiral section of 75 * 200 mm Number of spirals( N ) = height of jacket/pitch Outside diameter of the reactor (Do) Length of channel( L ) = DNπ Cross Sectional Area of channel ( A ) Hydraulic mean diameter ( de ) Mean Temperature Physical properties of dowtherm Q at mean temperature are as foll Density ( ρ ) Specific heat ( Cp ) Viscosity ( u ) Thermal conductivity ( Kf ) Velocity through the channel ( v ) Reynolds number ( Re ) Prandtl number ( Pr ) Using the following equation to calculate heat transter coefficient a Nu = C * ( Re )^0.8 * ( Pr )^0.33 As dowtherm Q is non viscous so C Nu As Nu = ho * de / Kf So ho 2. Heat transfer coefficient at inside wall of reactor Rotational speed of the impeller ( NR ) Physical properties of reactor contents density of liquid ( ρL ) viscosity of liquid ( uL ) Cp Thermal conductivity ( KL) Reynold number ( Re ) Prandtl number ( Pr ) For turbine type agitator the formula for nusselt number is Nu = 1.10Re^0.62*Pr^0.33( u/uw)^0.14 neglecting the viscosity factor at wall we calculate as follows Nu As Nu= hi D/KL hi OVERALL HEAT TRANSFER COEFFICIENT 1/U Thickness of reactor wall ( Xw) Thermal conductivity ( K ) 1/U U Pressure Drop Calculations: From equation 12.18 given in coulson and richardson volume 6 we have ΔP = 8Jf( L / de ) ρv^2/2 from fig 12.24 we have for reynolds number Re Jf Substituting te values in equation 12.18 we get ΔP ΔP ACTOR DESIGN " GAS LIQUID CSTR 3COOH ( MAIN REACTION ) CH3COOCH3 + H20 (L) (L) O2 ) + H2 (G) ( SIDE REACTION 3CO2 (G) + CH4 (G) s hold up ( ЄG) inson ( 1997 ) it is given as eq ( A ) 2250 0.63 = = 19.9 0.031 = = A we get 0.40 ocity ( Vsg) 150 2.5 0.07 rpm rps N/m 0.01 m/s 3 m2 4.00 2.00 m2 m = = = 1 0.60 Liquid holdup eq ( B ) 60 min 3600 sec eq (C) Volumetric flowrate of entering liquid 2470 0.69 787 0.00087 3 5.2 emical Process Engineering Design and Economics by H. Silla (2003) " ACE IN REACTOR VOLUME . BY FOLLOWING RULE ance is must ance is must er than 1.9 m3 so we have to give 10 % allowance 6 m3 ity by optimum HR/DR ratio ( reactor height to reactor diameter) 2m 1 6 m3 2m 2m 0.01 m/s HEIGHT OF THE LIQU A' Height of liquid column Diameter = D1' = optimum H1'/D1' ratio 0.67 m tom of impeller = h2 = 0.13 m 0.67 m 0.17 m 0.17 m when only liquid stream is present i.e PL Average Density
WT FRACTIONS (Xi) = 1/(Σxi/ρi)
DENSITY(ρi)(kg/m^3) Xi/ρi 0.59 0.240 0.018 0.158 1.000 1102.75 kg/m^3 1043 927.26 995.01 2264.97 5230.24 0.00056 0.00026 0.00002 0.00007 0.00091 Average Viscosity =1/(Σxi/µi)
Xi/µi AMOUNT/HR WT FRACTIONS (Xi) Viscosity(µi)kg/m.s 0.59 0.240 0.018 0.0011 0.0004 0.0009 524.66 660.12 19.59 2786.34 1142.89 83.49 0.158 1.000 0.00065 kg/m.s 0.0005 0.0028 332.95 1537.32 750.25 4762.97 1102.8 Kg/m^3 150.0 rpm 0.0007 Kg/m.s 0.07 N/m 9.8 m/s 0.16 W/m.C 2.500 en in fig 4.20 of book " CHEMICAL REACTORS" by Tremboze and Euzen 6.5 6.5 J Kw Kw/ m3 hp/m3 wrate divided by density of gas which has been calculated below as Vo m^3/s mical reactors " by Trembuze this corresponds to Kw hp hp/m3 3.14 m^3 2.07 m^3 te bubble diameter dB ) it is given as below 8.8*((VSG*UL/σ)^0.04)*((ρL*σ^3)/g*UL^4)^0.12)*((ρL/ρg)^0.22)*σ/g*ρL 1.45 0.04 2.42 6.5E06 8.7E06 0.003 m 0.3 cm 3 mm l when bubble diameter comes this small than sintered metal is used. ine the number of holes. 1.36 m 75 mm 200 mm 200 deg C 190 deg C 390159.43 Kg/hr Spiral baffle jacket 0.08 m 0.2 m lculation: t the outside wall of reactor:piral section of 75 * 200 mm = = = = = 5 2.01 m 31.6 m 75*200*10^6 m 0.02 m 4*( cross sectional area)/(wetted perimeter) 4*(0.075*0.200)/(2*(0.075+0.200) 0.109 m = (200 + 190)/2 deg C 195 deg C m Q at mean temperature are as follows: = = = = = = = 833.1 Kg/m^3 2.19 Kj/Kg.K 0.32 m Pa s 0 Pa s 0.1 W/m.K m / (ρ*A) 8.7 m/s ρ* v *de/u 2440250.4 Cp* u /Kf 7.27 o calculate heat transter coefficient at outside wall of reactor ( from Coulson and Richa = = 0.02 5702.04 = 5090.97 W/m2.C t inside wall of reactor = contents = = = = = = rmula for nusselt number is /uw)^0.14 1102.75 Kg/m^3 0.000650 Kg/m.s 2.54 Kj/Kg.K 0.16 W/m.C ρNd2^2/u 1882355.55 10.640 150.0 rpm 2.500 rps at wall we calculate as follows 18646.3 1448.30 W/m.C COEFFICIENT = = = = = (1/hi) + ( 1/ho) + ( Xw / K ) 0.009 m 22 W/m.C 0.0013 760 W/m2.C son and richardson volume 6 we have s number = = 12.18 we get = = 2440250.37 0 17412 Pa 0.2 atm 2.5 psi IDE REACTIONS ) T OF THE LIQUID COLUMN 1.05 m^2 of liquid column = H1'=V1/A' ter = D1' = m H1'/D1' ratio #REF! m 1.158 m #REF! AMOUNT/s WT FRACTIONSTHERMAL COND 0.774 0.317 0.023 0.59 0.24 0.02 0.16 0.15 0.61 0.208 1.323 0.16 1 0.09 1.01 rps rotations per second .22)*σ/g*ρL lson and Richardson vol 6 ) " MECHANICAL DESIGN 1. Design Pressure:
of the liquid in reactor. Pdesign Poperating Phydrostatic Pdesign = = = = = = Poperating + 0.075Poperating + It is taken as the pressure at which the relief device is se 2735775 Pa 27 atm 19720.2755 Pa 0.19 atm 2962157.42 Pa 29.23 atm 2. Design Temperature: It is taken as maximum working temperature of material of construction of reactor. Tdesign = 1852 deg C 3. Material Selection: Selected material : Zirconium 99.5 percent (Zircadyne 702) 4. Design Stress:
From data it is given as ( at reactor conditions ) f = = 369.3 N/mm^2 369300000 N/m^2 5. Wall thickness:
For cylindrical vessels the wall thickness required can be obtained from e where e = wall thickness putting the values in the above equation we get e = = 0.01 m 7.44 mm = PiDi/(2*fPi) Giving 2mm corrosion allowance we get efinal = 9.4 mm = 0.0094 m Which is very close to standards given in coulson and richardson vol 6 p Outside diameter = 2.01 6. Head thickness: Reactor has ellipsodial head.For ellipsodial head the formula for minimum e = PiDi/(2Jf0.2Pi) where e Pi Di J f Putting in 2 we get e = 0.0074 = = = = = minimum wall thickness Internal Pressure Internal Diameter Joint factor Dedign stress Giving a corrosion allowance of 2 mm the final thickness efinal = 9.4 Observe the similarity in ' A ' and ' B ' . This shows th 7.Weight Of Vessel:
Wv where = CvπρmDmg(Hv+0.8Dm)t Wv Cv Hv g ρm Dm π = = = = = = = = Wieght of vessel A factor to account for the weight 1.08 Length of cylindrical section Acceleration due to gravity Density of vessel material Mean diameter of vessel ( Di + t ) pi putting values in 4 Wv = 15081.4 8.Longitudinal Stress:
σL σL = = = PiDi/4t 145311354.8 145.3 9.Circumferential Stress:
σh σh = = = PiDi/2t 290622709.6 290.6 10.Direct Stress: It is due to the weight of vessel and its contents. The stress will be tensil points above the plane of the vessel support. As we have the weight of vessel: Wv = 15081 The wt of contents ( that means the liquid present in it) can be determine with acceleration due to gravity. WL = 33918 So total wt is giben as follows. W σw = = = 48999 W/π(Di+t)t 824553 0.82 11. Principal Stresses: For determining the principal stresses we have following equations σ1 σ2 σ3 where ĩ σz σz = = = = = = (1/2)*(σh+σz+sqrt((σhσz)^2+4ĩ^2))= longitud (1/2)*(σh+σzsqrt((σhσz)^2+4ĩ^2))=circumfe 0.5(P)=radial eq 6 Torsional shear stess is very sma total longitudinal stress = σL + ### 146.14 From eq 4 we have σ1 = 290622709.6 290.6 From eq 5 we have σ2 = From eq 6 we have σ3 = 1481078.71 1.48 ### 146.14 The value is in accordance with what is given in 12. Allowable stress intensity:
σ1σ2 σ1σ3 σ2σ3 calculating we get σ1σ2 σ1σ3 σ2σ3 = = = = = 144486801.7 144.5 ### 289.1 144654829.2 144.7 The maximum stress intensity taken at any point for the d The maximum stress intensity 290.345 N/mm^2 is les 13. Vessel support :
I have selected " SKIRT SUPPORT " for my reactor HANICAL DESIGN OF REACTOR " hich the relief device is set. It is 510 percent above normal pressure of reactor plus th 075Poperating + Phydrostatic rial of construction of reactor. As material of construction is Zirconium so ent (Zircadyne 702) ired can be obtained from the following formula. A son and richardson vol 6 page # 811 m d the formula for minimum thickness required is as follows. eq 2 m wall thickness l Pressure l Diameter = = = = = ? 2735775 Pa 2.00 m 1 369.3 N/mm^2 = ### m 7.41 mm of 2 mm the final thickness becomes mm = t = 0.0094 ' and ' B ' . This shows that the choice of ellipsodial head is right. Dmg(Hv+0.8Dm)t eq 4 of vessel r to account for the weight of manways, internal supports etc. In our case it can be take of cylindrical section ration due to gravity of vessel material iameter of vessel ( Di + t ) 2.00 m 9.81 m/s^2 6640 Kg/m^3 2.01 m 3.14 N eq 3 N/m^2 N/mm^2 N/m^2 N/mm^2 ts. The stress will be tensile( +ive ) for the points below the plane of vessel. And compr N ent in it) can be determined by multiplyin the density of liquid content with the reactor v N N N/m^2 N/mm^2 σ1 e following equations hσz)^2+4ĩ^2))= longitudinal eq 4 hσz)^2+4ĩ^2))=circumferential eq 5 σ2 σ3 al shear stess is very small and is not considered gitudinal stress = σL + σw N/m^2 N/mm^2 σ1 N/m^2 N/mm^2 N/m^2 N/mm^2 N/m^2 N/mm^2 ance with what is given in Coulson and Richardson vol 6. That radial stress is negligib taken at any point for the design purpose is the numerical greater value of the followin N/m^2 N/mm^2 N/m^2 N/mm^2 N/m^2 N/mm^2 A B C ( GREATEST ) ity 290.345 N/mm^2 is less than design stress which is 369.3 N/mm^2 so Zirconiu σ1σ3 < f PORT " for my reactor sure of reactor plus the hydrostatic pressure eq 1 N/m^2 m B ur case it can be taken as of vessel. And compressive(ve) for tent with the reactor volume and σ3 σ2 adial stress is negligibly small. r value of the following. N/mm^2 so Zirconium is best choice ...
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