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 [tex]dp[/tex]= 0.003 m
particle density = 1300 kg/m³
the bed void fraction [tex]\in =[/tex] 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 [tex]\rho[/tex] = 0.15 mol/dm ⁻³
viscosity of methane gas [tex]\mu[/tex] = 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 = [tex]0.1 5 *\dfrac{16}{0.1^3}[/tex]
Density = 2400
Density [tex]\rho_f[/tex] = 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
[tex]Re = \dfrac{dV \rho}{\mu}[/tex]
[tex]Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}[/tex]
[tex]Re=2276.317705[/tex]
For Re > 1000
[tex]\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}[/tex]
[tex]\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}[/tex]
[tex]\Delta P=8575.755212*2.5[/tex]
[tex]\Delta = 21439.38803 \ Pa[/tex]
To atm ; we have
[tex]\Delta P = \dfrac{21439.38803 }{101325}[/tex]
[tex]\Delta P =0.2115903087 \ atm[/tex]
ΔP ≅ 0.21159 atm
Thus; the pressure drop is 0.21159 atm
