Civil Engineering - Hydraulics - Discussion

Discussion Forum : Hydraulics - Section 1 (Q.No. 40)
40.
In a circular pipe of length L and diameter d, a viscous liquid is flowing with a velocity v. The loss in head, is
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Harsha said:   9 years ago
Friction factor f = 64/re re-reynolds no.

Man singh said:   8 years ago
Friction factor for reynold no f = 16/re.

Anomi said:   7 years ago
hf=f.l.v^2/2gd,
f=64/Re.

Naveen Venkat said:   7 years ago
F = 64/Re - friction factor.

Co-efficient of friction f = 16/Re,
Therefore F = 4f =4 * 16/Re.

Nilraj said:   7 years ago
Thanks @Naveen.

Bk prajapati said:   4 years ago
I think B is the right answer.

Hemanth said:   3 years ago
hf = FLV^2/2gd.
Coefficient of friction f = 16/Re,
F=4f=4*16/Re.

Substitute all in the formula of hf.

So, A is the correct answer.
(2)

Priyan said:   6 months ago
The correct answer to the question can be determined using the Darcy-Weisbach equation for head loss in a pipe:

hf = frac{4 f L v^2}{2 g d}.

For laminar flow, the Darcy friction factor () is given by:
f = frac{16}{Re}.

where Reynolds number () is:
Re = frac{ρ v d}{mu}.

Substituting into the equation:

hf = frac{16}{Re} times frac{4L}{d} times frac{v^2}{2g}.
hf = frac{16}{Re} times frac{4L}{d} times frac{v^2}{2g}.

Comparing this with the given options, option (A) is the correct answer:
frac{16}{Re}times frac{4L}{d} times frac{v^2}{2g}.

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