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 .

Kerosene is lighter than water and so it floats on the water and continues to burn. Carbon (IV) oxide blankets the flame cutting off the supply of oxygen and extinguishing the fire.

Explanation:

As water is heavier than kerosene, it slips down permitting the kerosene to rise to the surface and continue to burn. Besides, the existing temperature is so high that the water poured on the fire would evaporate before it can extinguish the fire. Thus the kerosene fire cannot be extinguished with water.