###
Pressure Drop Through Catalyst Tower

Home ->
ChE Learning Resources
->
Solved Problems
->
Fluid Mechanics->

A catalyst tower 50 ft high and 20 ft in diameter is packed with 1-in. diameter spheres. Gas enters the top of the bed at a temperature of 500^{o}F and leaves at the same temperature. The pressure at the bottom of the bed is 30 lb_{f}/in.^{2} abs. The bed porosity is 0.40. If the gas has average properties similar to propane and the time of contact (based on superficial velocity of gas) between the gas and the catalyst is 10 s, what is the inlet pressure?

Data:

Dia of tower (D) = 20 ft = 20 x 0.3038 m = 6.096 m

Height of tower (L) = 50 ft = 15.24 m

Dia of packing (D_{p}) = 1 inch = 2.54 cm = 0.0254 m

Temperature of gas (T) = 500^{o}F = (500 - 32) x 5/9 ^{o}C = 260^{o}C = (273 + 260) K = 533 K

Pressure at the bottom = 30 lb_{f}/in.^{2} abs = 30/14.7 atm(a) = 2.04 atm(a) = 206785.7 N/m^{2}(a)

Bed porosity (ε) = 0.4

Time of contact = 10 sec

Molecular weight of gas = 44 (molecular weight of propane (C_{3}H_{8}) )

Formulae:

Density of gas (ρ) = PM/(RT)

Ergun equation:

\(\displaystyle \frac{\Delta p \phi_sD_p\varepsilon^3}{L\rho V_o^2(1-\varepsilon)} = \frac{150\mu(1-\varepsilon)}{\phi_sD_pV_o\rho} + 1.75 \)

NRe_{PM }= D_{p}V_{o}ρ/((1 - ε)μ)

Calculations:

Density of the leaving gas,

ρ = 206785.7 x 44 / (8314 x 533) = 2.053 kg/m^{3}

V_{o} = Height / time = 15.24 / 10 = 1.524 m/sec

Taking the viscosity of gas as that of air (μ = 0.025 x 10^{-3} kg/(m.sec) )

NRe_{PM }= 0.0254 x 1.524 x 2.053 / ( (1 - 0.4) x 0.025 x 10^{-3} ) = 5298

For these NRe_{PM} Burke-Plummer equation can be used.(i.e. Turbulent part of the Ergun's equation)

Δp x 0.0254 x 0.4^{3} / ( 15.24 x 2.053 x 1.524^{2} x (1-0.4) ) = 1.75

Δp = 46937.5 N/m^{2}

Pressure at the inlet of the column = 206785.7 + 46937.5 = 253723.2 N/m^{2}(a) = **36.81 lb**_{f}/in^{2}(a)

[Index]

Last Modified on: 01-May-2024

Chemical Engineering Learning Resources - msubbu

e-mail: learn[AT]msubbu.academy

www.msubbu.in