Question

Calculate the pressure drop of air flowing at 30 °C and 1 atm pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed has a 125 cm diameter and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 cp and the density is 0.001156 grams/cm

Answer 1

Step-by-step explanation:

The given data is as follows.

Mass flow rate of Air =
`60 kg/min * 1 min/60 sec`

= 1 kg/s

Density of Air (r) =
`0.001156 g/cm^(3) * (1000kg/m^(3)) / (g/cm^(3))`

= 1.156
`kg/m^(3)`

Viscosity of Air (m) =
`0.0182 cP * 10^(-3) kg/ms/ cP`

=
`0.0182 * 10^(-3) kg/(ms)`

Bed porosity (e) = 0.38

Diameter of bed (D)=
`125 cm * (1 m)/(100 cm)` = 1.25 m

Length of bed (L) =
`250 cm * (1 m)/(100 cm)` = 2.5 m

Diameter of particles (Dp) =
`1.25 cm * (1 m)/(100 cm)`

= 0.0125 m

Sphericity = 1

Volumetric flow rate =
`\frac{\text{mass flow rate}}{density}`

=
`(1 kg/s)/(1.156 kg/m^(3))`

= 0.865
`m^(3)/s`

Superficial velocity,
`V_(o)` =
`(0.865)/((3.14)/(4)) * D^(2)`

=
`(0.865)/((3.14)/(4)) times 1.25 m * 1.25 m`

= 0.705 m/s

`NR_(ePM)` =
`(D_(p)V_(o)r)/(m (1 - e))`

NRePM = \frac{0.0125 m \times 0.705 m/s \times 1.156 kg/m^{3}}{0.0182 times 10^{-3} kg/(ms) \times (1 - 0.38)[/tex]

= 903

As we known that 10 >
`NR_(ePM)` > 1000

Below 10 means laminar flow.

Higher than 1000 is turbulent flow.

As, Reynolds number is between 10 and 100, therefore it is in transition flow.

According to Ergun equation,

`(\Delta p \Phi_(s) D_(p) \varepsilon^(3))/(L_(p) V^(2)_(o) (1 - \varepsilon)) = (150(1 - \varepsilon))/(\Phi_(S) D_(P) V_(o) (\rho)/(\mu)) + 1.75`

`\Delta p * 1 * 0.0125 m * 0.383 / [2.5 m * 1.156 kg/m^(3) * 0.7052 m^(2)/s^(2) * (1 - 0.38)]`

=
`150 * (1 - 0.38) / [1 * 0.0125 m * 0.7052 m^(2)/s^(2) * 1.156 kg/m^(3) / 0.0182 * 10^(-3) kg/(ms)] + 1.75`

=
`\Delta p * 7.702 * 10^(-4)`

= 1.92

`\Delta p` =
`(1.92)/(7.702 * 10^(-4))`

`\Delta p = 2492.90 N/m^(2)`

Thus, we can conclude that the pressure drop of air under given conditions is
`2492.90 N/m^(2)`.