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).
Discussion:
24 comments Page 2 of 3.
Kamran Islam said:
8 years ago
(10^-6* .001)/10^-6= .001= Re.
Cd = 24/Re =24/.001 = 24000 is the right answer.
Cd = 24/Re =24/.001 = 24000 is the right answer.
Komal said:
1 decade ago
But answer is 0.44.
Can anyone give correct solution with proper explanation.
Can anyone give correct solution with proper explanation.
Navi said:
5 years ago
Re = ρ * v * d/μ,
So, Re=1 and hence Stokes law regime and Cd=24/Re.
So, Re=1 and hence Stokes law regime and Cd=24/Re.
(1)
Vicky said:
4 years ago
Re = (0.001*10^-6)÷10^-6 = 0.001.
Re<0.2.
So 24/Re = 24/0.001=24000.
Re<0.2.
So 24/Re = 24/0.001=24000.
(1)
Daniel said:
1 decade ago
24000 is the correct answer.
Because in stokes regime - Cd =24/Re.
Because in stokes regime - Cd =24/Re.
Sanjeev said:
1 decade ago
How to calculate the drag coefficient when area is not given?
Ankit Garg said:
1 decade ago
Yes, Re=0.001. So answer is 24/0.001 = 24000.
Shyam said:
6 years ago
Can anyone explain the correct answer?
Salu said:
6 years ago
can anyone explain correct answer
Goutam said:
5 years ago
24000 is the correct answer.
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