Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 7 (Q.No. 49)
49.
A gas (density = 1.5 kg/m3 , viscosity = 2x 10-5 kg/m.s) flowing through a packed bed (particle size = 0.5 cm, porosity = 0.5) at a superficial velocity of 2 m/s causes a pressure drop of 8400 Pa/m. The pressure drop for another gas, with density of 1.5kg/m3and viscosity of 3 x 10-5kg/m.s flowing at 3 m/s will be
Discussion:
4 comments Page 1 of 1.
Broskie said:
12 months ago
@All.
I agree C is the correct answer.
First, calculate the Reynold's number:
Case I:
NRe of porous medium = Dp*v*ρ/(1-ε)*μ.
= 0.005 m * 2 m/s * 1.5 kg/m3/(1-0.5) * 2x10-5 kg/m.s.
= 1500.
Since NRe > 1000, use the Burke-Plummer equation.
-ΔP/L = 1.75ρv^2/Dp * (1-ε)/ε^3.
Solving for -ΔP/L, we will get 8400 Pa/m, which verifies the pressure drop stated in the problem.
Case II:
NRe = 0.005 m * 3 m/s * 1.5 kg/m3 / (1-0.5) * 3x10-5 kg/m.s.
NRe = 1500.
Again use the Burke-Plummer equation since NRe > 1000.
Solve for -ΔP/L, you will get 18900 Pa/m.
I agree C is the correct answer.
First, calculate the Reynold's number:
Case I:
NRe of porous medium = Dp*v*ρ/(1-ε)*μ.
= 0.005 m * 2 m/s * 1.5 kg/m3/(1-0.5) * 2x10-5 kg/m.s.
= 1500.
Since NRe > 1000, use the Burke-Plummer equation.
-ΔP/L = 1.75ρv^2/Dp * (1-ε)/ε^3.
Solving for -ΔP/L, we will get 8400 Pa/m, which verifies the pressure drop stated in the problem.
Case II:
NRe = 0.005 m * 3 m/s * 1.5 kg/m3 / (1-0.5) * 3x10-5 kg/m.s.
NRe = 1500.
Again use the Burke-Plummer equation since NRe > 1000.
Solve for -ΔP/L, you will get 18900 Pa/m.
Darshan said:
5 years ago
Option C is correct.
Siddhant ju said:
5 years ago
Option C is the correct answer. Burke Plummer equation should be used because NRe is 1500.
Rana said:
5 years ago
Please explain the answer.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers