Question

A 20-cm in diameter pipeline with a relative roughness of 0.01 has a total length of 45 m. When water (viscosity is 1x10-7 m2/s)is pumped through it at a rate of 5 m3/min, the major head loss (m) is most nearly

A) 3 mB) 10 mC) 15 mD) 20 m

Answer 1

A) 3 m

Step-by-step explanation:

The major head loss is calculated by using the expression:
`f (L)/(D)( \frac {U^2}{2g})`

here:

U = Q/A

Using the continuity equation:

`U = (Q)/((\pi)/(4)D^2)`

`U = (5/60 \ m^3/s)/((\pi)/(4)(0.2)^2)`

U = 2.65 m/s

Reynolds no =
`(\rho U D)/(\mu)`

`= (998 \ kg/m^3 * 2.65 \ m/s * 0.2 m)/(8.93 * 10^(-4) \ Pa^-s)`

= 594,894

Thus, this implies that the flow is turbulent.

Using Moddy's diagram at 5.94 × 10⁻⁵ &;

the relative roughness of 0.01

f = 0.038

Thus, the major head loss =
`f (L)/(D)( \frac {U^2}{2g})`

`=0.038 * ((45\ m )/(0.2\ m ) )* ((2.65 \ m/s) ^2)/(2(9.81 \ m/s^2))`

= 3.06 m

`\simeq` 3 m