Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 13 (Q.No. 26)
26.
Reynolds number for flow of water at room temperature through 2 cm dia pipe at an average velocity of 5 cm/sec is around
Discussion:
5 comments Page 1 of 1.
Rajkumar Pegu said:
3 years ago
Given:
Diameter= 2cm
Velocity=5cm/sec
Since,
Density of water= 1g/cc
Viscosity of water= .01poise.
Reynolds number= Density * diameterr * Velocity/ dynamic viscosity.
So, Re = 1*2*5/.01.
= 10/10-^2,
= 10*10^2,
= 1000.
Diameter= 2cm
Velocity=5cm/sec
Since,
Density of water= 1g/cc
Viscosity of water= .01poise.
Reynolds number= Density * diameterr * Velocity/ dynamic viscosity.
So, Re = 1*2*5/.01.
= 10/10-^2,
= 10*10^2,
= 1000.
(5)
Sudhanshu said:
3 years ago
Nre = density.v.d/viscosity.
= 1 * .02*.05/10^(4),
= 1000.
Take unit in cm and gram.
= 1 * .02*.05/10^(4),
= 1000.
Take unit in cm and gram.
(2)
Rahul said:
3 years ago
The viscosity of water is 0.01 poise = 0.001 kg/(m^2s).
So, the answer is 1000.
So, the answer is 1000.
Viji said:
5 years ago
@Satya.
Ans is 100, the viscosity of water is .001.
Ans is 100, the viscosity of water is .001.
Satya said:
6 years ago
Nre = density.v.d/kinamatic viscosity.
= 1000 * .02*.05/10^(-4),
= 1000.
= 1000 * .02*.05/10^(-4),
= 1000.
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