1. Pressure drop in Packed Bed Reactors Chemical Reaction Engineering I Aug 2011 - Dec 2011 Dept. Chem. Engg., IIT-Madras

2. Overview Notation PFR design equation (mass balance) Pressure drop equation for packed bed Modified for catalytic reactions Accounts for change in number of moles (due to reaction) Does not consider phase change Methodology, with examples

3. Basics Usual notation applies. Reaction is usually written as r A, in mol/vol/time For catalytic reactions, it is written as r A ’, in mol/g-catalyst/time Rate law is written as Where the rate constant k is in appropriate units Remember that C A is in mol/vol Packed bed: The PFR equation will be modified. Instead of dV, we will use dW

4. Design Equation For constant temperature and with ideal gas law

5. Pressure Drop Equation Ergun equation, written in terms of mass flow rate Superficial mass velocity G =  V sup Porosity  Diameter of particles is D p Need not memorize this  is a variable. It should be written in terms of X and P Similarly, z must be written in terms of W

6. Pressure drop equation Mass flow rate is a constant For ideal gas, constant temperature, no phase change

7. Pressure drop equation Relate ‘W’ to ‘z’ Weight of catalyst = Volume of catalyst * density of catalyst W = Az * (1-  ) *  c. dW = A (1-  ) *  c * dz For ideal gas, constant temperature, no phase change

8. Simplifications If there is no change in volume due to reaction (if  = 0) Pressure equation can be solved If parameter  is small, and X is also small, then we can neglect that in the pressure drop equation, but not in the design equation

9. Examples Pin = 5e5; %Pa; T = 400 ; %Kelvin Qin = 0.01 ; % m3/s M = 50 ; % molecular weight, g/gmol dia = 0.1; % diameter in m mu = 2e-5; % viscosity in Pa-s k = 0.001; % rate constant with correct units. lit/g-catalyst/s, for example n = 1; %order of reaction epsilon = -0.2; % fractional change in volume for 100% conversion, for the given feed phi = 0.3 ; % Void fraction, volume of void/total volume RhoC = 2000 * 1000; % catalyst density, in g/m^3 W_end = 1000; % in g Dp = 0.005; % diameter of particle, in m

10. Solution Rho in – 7.5174 kg/m3 Fain = 1.5035 gmol/s Area = 0.0079 m2 Va-in=1.2732 m/s G = 9.5715 kg/s Alpha = 5.155 E6, Pa2/g catalystt Beta = 1.6 e-9 units

11. Results

12. Solution If dia is changed from 10 cm to 5 cm