# Water is flowing through a pipe of diameter 200 mm with a velocity of 3 m/s. What is the head loss due to friction for a length of 5 m if the coeffici

1.1k views
in General
closed

Water is flowing through a pipe of diameter 200 mm with a velocity of 3 m/s. What is the head loss due to friction for a length of 5 m if the coefficient of friction is given by $f=0.02+\frac{0.09}{Re^{0.3}}$, where Re is Reynolds number?

(Take the kinematic viscosity of water as 0.01 stoke, g = 9.81 m/s2 and (6× 105)0.3 = 54.13)

1. 0.993 m of water
2. 0.783 m of water
3. 0.685 m of water
4. 0.552 m of water
5.

by (41.9k points)
selected

Correct Answer - Option 1 : 0.993 m of water

Concept:

Reynolds number is given by

$Re = \frac{{\rho VD}}{\mu }=\frac{VD}{ν}$

where, ρ = Density of fluid, V = velocity of fluid, D = Diameter of pipe, μ = Dynamic viscosity of the fluid and ν = kinematic viscosity of fluid.

1 Stokes = 10-4 m2/sec.

The darcy-Weisbach formula for head loss due to friction is,

$h_f = \dfrac{4flV^2}{2gD}$

where, f = Darcy's coefficient of friction, l = length of the pipe, v = velocity of the fluid in the pipe, D = diameter of the pipe.

Calculation:

Given:

D = 200 mm, V = 3 m/s, L = 5 m,

kinematic viscosity of water as 0.01 stoke,

g = 9.81 m/s2 and (6 × 105)0.3 = 54.13

$f=0.02+\frac{0.09}{Re^{o.3}}$

Reynolds number is given by

$Re=\frac{VD}{ν}$

$Re=\frac{3\times 0.2}{0.01\times10^{-4}}=6\times10^5$

Given $f=0.02+\frac{0.09}{Re^{0.3}}$

$f=0.02+\frac{0.09}{(6\times10^5)^{0.3}}=0.02+\frac{0.09}{54.13}=0.02166$

The darcy-Weisbach formula for head loss due to friction is,

$h_f = \dfrac{4flV^2}{2gD}$

$h_f = \dfrac{4\times 0.02166\times 5\times 3^2}{2\times 9.81\times 0.2}=0.993\;m\; of\;water$