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.
Laxmi said:
9 years ago
In the Stokes Regime.
Cd = 24/Re.
But here since flow is turbulent, Re= 1000 So, Cd is almost constant and is 0.44.
Cd = 24/Re.
But here since flow is turbulent, Re= 1000 So, Cd is almost constant and is 0.44.
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.
BaT said:
6 years ago
24000 is the correct answer.
Shyam said:
6 years ago
Can anyone explain the correct answer?
Makda said:
6 years ago
The correct answer is 24000 because Re = 1000(please make sure dimensional similarly ;i.e 1micron = 10-6 and 1mm = 0.001m.
Salu said:
6 years ago
can anyone explain correct answer
Amit said:
5 years ago
Nre = D*V/ν = 10^-6*10^-3/10^6=10^-3=0.001. SO IT IS UNDER STOKES LAW.
SO CD = 24/Nre = 24/0.001=24000.
SO CD = 24/Nre = 24/0.001=24000.
Goutam said:
5 years ago
24000 is the correct answer.
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)
Raju said:
4 years ago
24000 is the correct answer.
(1)
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