The value of friction factor ‘f’ for smooth pipes for Reynolds number 106 is approximately equal to
A. 0.1
B. 0.01
C. 0.001
D. 0.0001
Answer: Option B
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A. 22.5 m/sec.
B. 33 m/sec.
C. 40 m/sec.
D. 90 m/sec.
A. the weight of the body
B. more than the weight of the body
C. less than the weight of the body
D. weight of the fluid displaced by the body
The difference of pressure between the inside and outside of a liquid drop is
A. $${\text{p}} = {\text{T}} \times {\text{r}}$$
B. $${\text{p}} = \frac{{\text{T}}}{{\text{r}}}$$
C. $${\text{p}} = \frac{{\text{T}}}{{2{\text{r}}}}$$
D. $${\text{p}} = \frac{{2{\text{T}}}}{{\text{r}}}$$
A. cannot be subjected to shear forces
B. always expands until it fills any container
C. has the same shear stress.at a point regardless of its motion
D. cannot remain at rest under action of any shear force
f = 0.0032 + [ 0.221 / (Re ^ 0.237) ]
For 10^5 < Re < 4* 10^7
Use this formula
f = 0.0032 + [ 0.221 / ( R ^ 0.237 ) ]
For 10^5 < Re < 4* 10^7
f = friction factor
Answer :- f = 0.0032 + [ 0.221 / (Re ^ 0.237) ]
For
Re between 10^5 to 4* 10^7
f = friction factor
F=0.316/(re)^0.25 is called balausious result for smooth pipe.
So, f= 0.316/(10^6)^0.25=0.01 is correct answer.
It is creast not sill
For smooth pipe f=0.316/(Re) ^0.25
f=0.316/1000000^0.25
f=0.01
f=0.046/Re^0.2
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