Question

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss for the water supply over a distance of 200 m? Hint: Use an iterative approach for the empirical equation.

Answer 1

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

    $= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor  

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where,$\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m