Answer:
Check the explanation
Explanation:
A diagrams showing the process in p-v and T-s and the Model engineering <u><em>(which is the quest in constructing miniature working representations of proportionally-scaled in full-sized machines. It is a subdivision of metalworking with a sharp importance on artisanry, in contrast to mass production.)</em></u> can be seen in the attached images below.
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 = 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