Respuesta :
The pressure drop of air in the bed is 14.5 kPa.
Explanation:
To calculate Re:
[tex]R e=\frac{1}{1-\varepsilon} \frac{\rho q d_{p}}{\mu}[/tex]
From the tables air property
[tex]\mu_{394 k}=2.27 \times 10^{-5}[/tex]
Ideal gas law is used to calculate the density:
ρ = [tex]\frac{2.2}{2.83 \times 10^{-3} \times 394.3}[/tex]
ρ = 1.97 Kg / [tex]m^{3}[/tex]
ρ = [tex]\frac{P}{RT}[/tex]
R = [tex]\frac{R_{c} }{M}[/tex] = 8.2 × [tex]10^{-5}[/tex] / 28.97×[tex]10^{-3}[/tex]
R = 2.83 × [tex]10^{-3}[/tex] [tex]m^{3}[/tex] atm / K Kg
q is expressed in the unit m/s
[tex]q=\frac{2.45}{1.97}[/tex]
q = 1.24 m/s
Re = [tex]\frac{1}{1-0.4} \frac{1.97 \times 1.24 \times 0.0127}{2.27 \times 10^{-5}}[/tex]
Re = 2278
The Ergun equation is used when Re > 10,
[tex]\frac{\Delta P}{L}=\frac{180 \mu}{d_{p}^{2}} \frac{(1-\varepsilon)^{2}}{\varepsilon^{3}} q+\frac{7}{4} \frac{\rho}{d_{p}} \frac{(1-\varepsilon)}{\varepsilon^{3}} q^{2}[/tex]
[tex]\frac{\Delta P}{L}=\frac{180 \times 2.27 \times 10^{-5}}{0.0127^{2}} \frac{(1-0.4)^{2}}{0.4^{3}} 1.24[/tex] [tex]+\frac{7}{4} \frac{1.97}{0.0127} \frac{(1-0.4)}{0.4^{3}} 1.24^{2}[/tex]
= 4089.748 Pa/m
ΔP = 4089.748 × 3.66
ΔP = 14.5 kPa
