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.

TANMAY SHAH said:   2 years ago
Nre = d * v * ρ / μ

d = 1 micron = 10^-6 meter
v = 1 mm/s = 10^-3 m/s
ρ = 10^3 kg/m3
μ = 10^-6 × 1000 = 10^-3 kg/m.s
[ because; dynamic μ = kinematic μ × ρ ]

Nre = 10^-6 × 10^-3 × 10^3 /10^-3.
Nre = 10^-3 = 0.001.

Now for Nre < 1 it is called laminar flow region.

In the laminar flow region, called the Stokes law region the drag coefficient,

Cd= 24 µ/d ρ ν = 24 / Nre
Cd = 24/0.001 = 24000.

So, option-A is the right answer.
(10)

Athraa said:   3 years ago
Here, Re =1000.

Amar said:   4 years ago
24000 is the correct answer.
(2)

Vicky said:   4 years ago
Re = (0.001*10^-6)÷10^-6 = 0.001.

Re<0.2.
So 24/Re = 24/0.001=24000.
(1)

Raju said:   4 years ago
24000 is the correct answer.
(1)

Navi said:   5 years ago
Re = ρ * v * d/μ,
So, Re=1 and hence Stokes law regime and Cd=24/Re.
(1)

Goutam said:   5 years ago
24000 is the 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.

Salu said:   6 years ago
can anyone explain 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.


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