Question

For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger shell (just like a shell and tube heat exchanger). Consider one tube inside such a reactor that is 2.5 m long with an inside diameter of 0.025 m. The catalyst is alumina spheres with a diameter of 0.003 m. The particle density is 1300 kg/m3 and the bed void fraction is 0.38. Compute the pressure drop seen for a superficial mass flux of 4684 kg/m2hr. The feed is methane at a pressure of 5 bar and 400 K. At these conditions the density of the gas is 0.15 mol/dm-3 and the viscosity is 1.429 x 10-5 Pa s.

Answer 1

Answer:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction $\in =$  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  $0.1 5 *\dfrac{16}{0.1^3}$

Density =  2400

Density $\rho_f$ =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm