Chemical Engineering - Fluid Mechanics - Discussion

Discussion Forum : Fluid Mechanics - Section 1 (Q.No. 31)
31.
The drag co-efficient for a bacterium moving in water at 1 mm/s, will be of the following order of magnitude (assume size of the bacterium to be 1 micron and kinematic viscosity of water to be 10-6m2/s).
24000
24
0.24
0.44
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
24 comments Page 1 of 3.

Sanjeev said:   1 decade ago
How to calculate the drag coefficient when area is not given?

Paresh chaudhary said:   1 decade ago
Drag force cd = 24/Nre
Nre=dia*velocity/kinematic viscosity
= 24/.001
= 24000

Komal said:   1 decade ago
But answer is 0.44.

Can anyone give correct solution with proper explanation.

Daniel said:   1 decade ago
24000 is the correct answer.

Because in stokes regime - Cd =24/Re.

Dhanush P said:   1 decade ago
Cd=(24/Re), applies only when Re < 1(according to stokes law).

When Re>1000, newtons law applies, where Cd is almost constant. i.e 0.44.

Afeefa said:   1 decade ago
The Reynolds number is coming as (10^-3), and in stokes law regime cd=24/NRe which comes to be 24000.

Ashnu said:   1 decade ago
Correct answer is 24000. Because the range of motion of the particle is in stoke's law regime. Nre = 1x10-3 which is under stoke's law. Also Cd = b1/Nre^n where b1= 24 and n=1 for stoke's law.

Calculating all must result to 24000.

Vikash lamoria said:   1 decade ago
Answer is 24000.

Cd = 24/Re for Reynold no.<1 and Re = .001.
So answer is 24000.

Ankit Garg said:   1 decade ago
Yes, Re=0.001. So answer is 24/0.001 = 24000.

Lovakumar said:   1 decade ago
Cd = 24/Re for laminar.

Re = d*v/kine viscosity = 1000.

Since it is turbulent Cd = 0.44 which is constant for turbulent.


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