Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 10 (Q.No. 28)
28.
What is the value of Fanning friction factor f ' for smooth pipe at NRe = 106 approximately ?
Discussion:
4 comments Page 1 of 1.
NITIN BIRAJDAR said:
8 years ago
F={0.0791}{Re^{0.25}}}} for turbulent flow.
(1)
Rana said:
5 years ago
Here, f = 0.0791/Re^0.25 = 0.003.
(1)
Weng said:
2 years ago
The answer is 2.9037 x 10^-3 or approximately 0.003.
Churchill equation can be used.
(1/f^0.5) = -4log[(0.27*ε/D)+(7/Re)^0.9].
For smooth pipes, ε = 0.
f = {-4log[(7/10^6)^0.9]}^-2.
f = 2.9037 x 10^-3 = 0.0029 = 0.003.
Churchill equation can be used.
(1/f^0.5) = -4log[(0.27*ε/D)+(7/Re)^0.9].
For smooth pipes, ε = 0.
f = {-4log[(7/10^6)^0.9]}^-2.
f = 2.9037 x 10^-3 = 0.0029 = 0.003.
Udangshree Boro said:
4 months ago
For turbulent flow when Re=10^6,
Darcy friction factor( fd) = 0.015.
we have relationship fd = 4 * fanning friction factor( f),
So, f = fd/4 = 0.015/4,
= 0.003.
Darcy friction factor( fd) = 0.015.
we have relationship fd = 4 * fanning friction factor( f),
So, f = fd/4 = 0.015/4,
= 0.003.
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