Question

Your friend has designed a Packed Bed Reactor for his design project in 15 points which a gas phase reaction A+B+2C, which is limited by internal diffusion, is to be carried out. The intrinsic rate constant (k) was found to be 0.16x10^(-5) m3/m2.s. He chose spherical catalyst beads with 1cm in diameter and internal surface area = 500 m2/g. The effective diffusivity of the catalyst beads is 8x10^-6 m2/s and the bulk density is 0.0028 kg/m3. Feed at 240 mol/min, contains 2/5 fraction of A and the rest are inert gases. It enters the reactor at 400K and 5 atm. The rate law for this heterogeneous reaction can be taken as -TA' (moles/kg Cat. s) = kSaCAO(1-X)/(1+0.8X). He designed the reactor for 75% and it was set up for experimental verification. The pressure drop was found to be negligible, but conversion was found to be only 66%. Verify why he got less conversion and recalculate the amount of catalyst needed to achieve 75% conversion. (Hint: assume that the gas phase behaves like an ideal gas for calculating the Inlet volumetric flow rate)

advanced chemical reaction engineering (advanced CRE) subject .

need solution

Answer 1

m m² sec 16oogm Assumptions The gas phase behavey like an ideal gas, ② The presure drop across the neactor is negligible ss Properties intrinsic rate constant (K) = 0.16810 internal specific surface area of catalyst (sa) - soom² gm (sq) = soom² =sx 15 m² gm lig kg Total Feed rate to reactor (to) = =240 mol/min mole Fraction of A = 33 mole fraction of Inert goj=- Temprature (T) = 400k Pressure (P) - satm Analysis: - using ideal gas equation to find volumetric > flow rate, min, mol-ic р P ;=nRT いたらRT tsofuomole ) (8:30 (8131445 ) x (4001) (5x1013x10spa) V = 1.575 m min Now calculating the total concentration of feed, C = concentration of freed mixture = Fo quomol/min C= 152.381 mol 1.575 m² m/min concentration of A in Feed, = 152.381 mol ms concentration of inerts in feed, Cinect = cx. Molar Flow rate of A bed to the reactor, FAO = 240 mol .96 mol m3 x3 = 91.42 mo! m3 X alu min

0.75 خدا .0.75 015 Design calculations) dx where, w=weight of catalyst FAO EVA) (2A) = reaction rate based on mass of catalent (1+0.8%) dx KSACAOCI-X) 0.16x10x sx 18 x 669sx (1-x) (1+0.8x) (1+0.8x) dx (1+0.8x) dx 4826 48,76(1-x) 48.76 (1-x) O 0.75 0175 FAO 0.75 (1 +0.8x) = 1.895 1.895 48.76 1808 FAO This much = 0.0388 FAO w= 0.0388x96 This much weight of catalyst Wz 3.724 kg is required to acheive 75% Conversion as per design calculation since the rate is limited by internal diffusion, so the actual or observed rate would be different from Fral! Era actural = x(-A) where n= effectivens factor 0.66 FAO dla dx (1+0.8x) Era l'actual 4x48,76 (1) Now substituting all the values to bind (n). 010388 = I (1+0-876) dx n 48.76 (_-*) 010388 = x 1.4138 nxu876 0.66

this the actual reaction rate is smaller that the This is why the actual conversion will be lesser than To calculate the weight of catalyst required to achieve n= 114138 = 0.747 0.0388 x 48.76 n=0.747 the effectiveness factor is less than 1, Era used to do the design calculation the calculated one. Since of because Tsolo actual conversion, coins de 0.75 dx (1+0.8x) ایا 017478 48,76 C1-x) FAO n (ra)' 0.75 (1+0.8x) (1-x) FAO 0.74774876 w' FAO 0.747848.76 x 1.895 - ابيا - 96x1.895 0.747848.76 w = 4.994 kg - This is the actual amount of catalyst needed to achiewe 750% actual conversion.