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

Explanation:

The given data  is as follows.

Mass flow rate of Air =

                                   = 1 kg/s

Density of Air (r) =

                             = 1.156

Viscosity of Air (m) =

                                 =

Bed porosity (e) = 0.38

Diameter of bed (D)= = 1.25 m

Length of bed (L) = = 2.5 m

Diameter of particles (Dp) =

                                           = 0.0125 m

Sphericity = 1

Volumetric flow rate =

                                =

                                = 0.865

Superficial velocity, =

                                  =

                                  = 0.705 m/s

=

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 > > 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,

        =

          =

         = 1.92

      =

Thus, we can conclude that the pressure drop of air under given conditions is .

answer

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