CBE343Feb 15, 2012

Topics

P5-11BFlow through a packed bedPressure drop in a packed bed

P5-11B (modified)The dehydration of butanol on alumina is carried out over a silica alumina catalyst at 680 K.

The rate law is rBu=kPBu/(1+KBuPBu)2

k= 0.054 mol/gcat.hr.atmKBu=0.32 atm-1Pure butanol enters a thin-tubed packed-bed reactor at a molar flow rate of 50 kmol/hr and a pressure of 10 atm.

What PFR catalyst weight is necessary to achieve 80%, 85% and 90% conversionin absence of pressure drop. Plot X, y, and reaction rate, -rA as a function of catalyst weight. (b) Repeat (a) when there is pressure drop with pressure drop parameter a=0.0006 kg-1. Do you observe a maximum in the rate of reaction and if so, why? What catalystweight is necessary to achieve 60% and 70% conversion. Compare this weight with that for no pressure drop to achieve the same conversion.

Pressure drop in Packed Bed Reactors (PBR)The only parameter that varies with pressure on the RHS is

Page 171 for definitions:

Steady state conditions meansNow we combine

For tubular packed-bed reactors we need to look at the catalyst weightWeight of catalystVolume of SolidsDensity of solid catalystDifferentiate W with respect to z, then replace dz with dWForm of the equation we use when working with multiple reactions or when there is pressure drop in a membrane reactors

For single reactions in PBRs we want to put the Ergun equation in terms of conversion, X. Remember thatSubstitue in:Form we are going to use the mostTwo coupled ODE that we can solve. We are going to use Polymath the most.

The conversion will be less when you have pressure drop versus no pressure drop.

d(x)/d(w)=-rBu/FBuOFBuO=50PBuO=10PBu=PBuO*(1-x)/(1+x)k=0.054K=0.32rBu=-k*PBu/(1+K*PBu)^2w(0)=0w(f)=2500x(0)=0(a)

From table: w = 1054 kgCalculated values of DEQ variables

VariableInitial valueMinimal valueMaximal valueFinal value1FBuO50.50.50.50.2k0.0540.0540.0540.0543K0.320.320.320.324PBu10.0.000736910.0.00073695PBuO10.10.10.10.6rBu-0.0306122-0.0421873-3.977E-05-3.977E-057w002500.2500.8x000.99985260.9998526

From table: w = 1054 kg

d(x)/d(w)=-rBu/FBuOd(y)/d(w)=-alpha*(1+x)/2/yFBuO=50f=(1+x)/yPBuO=10PBu=PBuO*(1-x)/(1+x)*yk=0.054K=0.32rBu=-k*PBu/(1+K*PBu)^2alpha=0.0006y(0)=1w(0)=0w(f)=1000x(0)=0(b)

(b)Calculated values of DEQ variables

VariableInitial valueMinimal valueMaximal valueFinal value1 alpha 0.0006 0.0006 0.0006 0.0006 2 f 1. 1. 4.117227 4.117227 3 FBuO 50. 50. 50. 50. 4 K 0.32 0.32 0.32 0.32 5 k 0.054 0.054 0.054 0.054 6 PBu 10. 0.6417032 10. 0.6417032 7 PBuO 10. 10. 10. 10. 8 rBu -0.0306122 -0.0421852 -0.0238509 -0.0238509 9 w 0 0 1000. 1000. 10 x 0 0 0.7357962 0.7357962 11 y 1. 0.4215935 1. 0.4215935

(b)

(b)Rate maximum at catalyst weight of = 600 kg

(b)Catalyst weight for 70% conversion (with pressure drop) = 932 kg

Catalyst weight for 70% conversion (no pressure drop) = 915 kg