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.

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 = 0.003 m

particle density  = 1300 kg/m³

the bed void fraction  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 = 0.15 mol/dm ⁻³

viscosity of methane gas = 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 =  

Density =  2400

Density =  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

For Re > 1000

To atm ; we have

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

how is this a question? in order for me to answer it i have to understand the question

explanation: