Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 7 (Q.No. 40)
40.
A bed consists of particles of density 2000 kg/m3. If the height of the bed is 1.5 metres and its porosity 0.6, the pressure drop required to fluidise the bed by air is
Discussion:
4 comments Page 1 of 1.
Pushkar sharma said:
4 years ago
The Right answer is option B.
Akhlaque Ahmad said:
4 years ago
Pressure drop = height of bed*(1-porosity)*(density of solids - density of fluid)* acceleration due to gravity.
Density of air = 1.225 kg/m^3.
P = 1.5 m *(1-0.6)*(2000-1.225) kg/m^3 * 9.8 m/s^2.
P = 11.75 KPa.
Density of air = 1.225 kg/m^3.
P = 1.5 m *(1-0.6)*(2000-1.225) kg/m^3 * 9.8 m/s^2.
P = 11.75 KPa.
(1)
Sharad yadav said:
6 years ago
Thanks @Venkat.
Venkat said:
7 years ago
Pressure drop = height of bed*(1-porosity)*(density of solids - density of fluid)* accleration due to gravity.
Density of air = 1.225 kg/m^3.
P = 1.5 m *(1-0.6)*(2000-1.225) kg/m^3 * 9.8 m/s^2.
P = 11.75 KPa.
Density of air = 1.225 kg/m^3.
P = 1.5 m *(1-0.6)*(2000-1.225) kg/m^3 * 9.8 m/s^2.
P = 11.75 KPa.
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