Answered

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 :

Explanation:

The given data  is as follows.

Mass flow rate of Air = [tex]60 kg/min \times 1 min/60 sec[/tex]

                                   = 1 kg/s

Density of Air (r) = [tex]0.001156 g/cm^{3} \times (1000kg/m^{3}) / (g/cm^{3})[/tex]

                             = 1.156 [tex]kg/m^{3}[/tex]

Viscosity of Air (m) = [tex]0.0182 cP \times 10^{-3} kg/ms/ cP[/tex]

                                 = [tex]0.0182 \times 10^{-3} kg/(ms)[/tex]

Bed porosity (e) = 0.38

Diameter of bed (D)= [tex]125 cm \times \frac{1 m}{100 cm}[/tex] = 1.25 m

Length of bed (L) = [tex]250 cm \times \frac{1 m}{100 cm}[/tex] = 2.5 m

Diameter of particles (Dp) = [tex]1.25 cm \times \frac{1 m}{100 cm}[/tex]

                                           = 0.0125 m

Sphericity = 1

Volumetric flow rate = [tex]\frac{\text{mass flow rate}}{density}[/tex]

                                = [tex]\frac{1 kg/s}{1.156 kg/m^{3}}[/tex]

                                = 0.865 [tex]m^{3}/s[/tex]

Superficial velocity, [tex]V_{o}[/tex] = [tex]\frac{0.865}{\frac{3.14}{4}} \times D^{2}[/tex]

                                  = [tex]\frac{0.865}{\frac{3.14}{4}} times 1.25 m \times 1.25 m[/tex]

                                  = 0.705 m/s

[tex]NR_{ePM}[/tex] = [tex]\frac{D_{p}V_{o}r}{m (1 - e)}[/tex]

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 > [tex]NR_{ePM}[/tex] > 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,

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

[tex]\Delta p \times 1 \times 0.0125 m \times 0.383 / [2.5 m \times 1.156 kg/m^{3} \times 0.7052 m^{2}/s^{2} \times (1 - 0.38)][/tex]

        = [tex]150 \times (1 - 0.38) / [1 \times 0.0125 m \times 0.7052 m^{2}/s^{2} \times 1.156 kg/m^{3} / 0.0182 \times 10^{-3} kg/(ms)] + 1.75[/tex]

          = [tex]\Delta p \times 7.702 \times 10^{-4}[/tex]

         = 1.92

      [tex]\Delta p[/tex] = [tex]\frac{1.92}{7.702 \times 10^{-4}}[/tex]

      [tex]\Delta p = 2492.90 N/m^{2}[/tex]

Thus, we can conclude that the pressure drop of air under given conditions is [tex]2492.90 N/m^{2}[/tex].