###
Sphericity of Cylinder

Home ->
Solved Problems
->
Mechanical Operations ->

Calculate the sphericity of a cylinder of dia 1 cm and height 3 cm.

Calculations:

Volume of particle = p r_{c}^{2} h = p x 0.5^{2} x 3 = 2.356 cm^{3}

Radius of sphere of volume 2.356 cm^{3}:

4 p r_{s}^{3 }/ 3 = 2.356

r_{s} = 0.8255 cm

Surface area of sphere of same volume as the particle = 4 p r_{s}^{2} = 4 x p x 0.8255^{2} = 8.563 cm^{2}

Surface area of particle = 2 p r_{c} (h + r_{c}) = 2 x p x 0.5 x (3 + 0.5) = 10.996 cm^{2}

Sphericity (f
_{s}) = 8.563/10.996 = 0.779

Sphericity could also be found from the formula,

Sphericity (f
_{s}) = 6 V_{p} / (D_{p}S_{p})

Where V_{p} = volume of particle

D_{p} = Equivalent diameter of particle. (Equivalent diameter is defined as the diameter of a sphere of equal volume)

S_{p} = surface area of particle

V_{p} = p r_{c}^{2} h = 2.356 cm^{3}

D_{p} = 2 r_{s} = 2 x 0.8255 = 1.651 cm

S_{p} = 2 p r_{c} (h + r_{c}) = 10.996 cm^{2}

f
_{s} = 6 x 2.356 / (1.651 x 10.996) = 0.779

HOME

Last Modified on: 11-Sep-2014

Chemical Engineering Learning Resources - msubbu

e-mail: msubbu.in[AT]gmail.com

Web: http://www.msubbu.in