Question

Flow and Pressure Drop of Gases in Packed Bed. Air at 394.3 K flows through a packed bed of cylinders having a diameter of 0.0127 m and length the same as the diameter. The bed void fraction is 0.40 and the length of the packed bed is 3.66 m. The air enters the bed at 2.20 atm abs at the rate of 2.45 kg/m2 · s based on the empty cross section of the bed. Calculate the pressure drop of air in the bed.

Ans: Δp = 0.1547 × 105 Pa

Answer 1

The pressure drop of air in the bed is 14.5 kPa.

Step-by-step explanation:

To calculate Re:

`R e=(1)/(1-\varepsilon) (\rho q d_(p))/(\mu)`

From the tables air property

`\mu_(394 k)=2.27 * 10^(-5)`

Ideal gas law is used to calculate the density:

ρ =
`(2.2)/(2.83 * 10^(-3) * 394.3)`

ρ = 1.97 Kg /
`m^(3)`

ρ =
`(P)/(RT)`

R =
`(R_(c) )/(M)` = 8.2 ×
`10^(-5)` / 28.97×
`10^(-3)`

R = 2.83 ×
`10^(-3)`
`m^(3)` atm / K Kg

q is expressed in the unit m/s

`q=(2.45)/(1.97)`

q = 1.24 m/s

Re =
`(1)/(1-0.4) (1.97 * 1.24 * 0.0127)/(2.27 * 10^(-5))`

Re = 2278

The Ergun equation is used when Re > 10,

`(\Delta P)/(L)=(180 \mu)/(d_(p)^(2)) ((1-\varepsilon)^(2))/(\varepsilon^(3)) q+(7)/(4) (\rho)/(d_(p)) ((1-\varepsilon))/(\varepsilon^(3)) q^(2)`

`(\Delta P)/(L)=(180 * 2.27 * 10^(-5))/(0.0127^(2)) ((1-0.4)^(2))/(0.4^(3)) 1.24`
`+(7)/(4) (1.97)/(0.0127) ((1-0.4))/(0.4^(3)) 1.24^(2)`

= 4089.748 Pa/m

ΔP = 4089.748 × 3.66

ΔP = 14.5 kPa