Question

512 Hydraulically smooth pipe walls. water av -51/s 2m

Answer 1

flowrate qv = 5 L/s = 0.005 m³/s

velocity inside the pipe v = qv / Ac

Ac = cross-sectional area of the pipe.

d = 50 mm = 0.05 m

$v = \frac{q_{v}}{\frac{\pi }{4}d^{2}}$

$v = \frac{0.005}{\frac{\pi }{4}0.05^{2}} = 2.54648 \ m/s$

density of water = 1000 kg/m³

$\nu = \frac{\mu }{\rho }$

$\mu$ = 1000 x 1.3 x 10^-4 = 0.13 kg/m.s

$Re=\frac{\rho vd}{\mu } = \frac{1000 *2.54648*0.05}{0.13} = 979.415$

As Re < 2100, flow is laminar.

For laminar flow, the frictional head loss is given by

$h_{L} = \frac{32\mu vL}{\rho gd^{2}}$

$h_{L} = \frac{32*0.13*2.54648*10}{1000*9.81*0.05^{2}} = 4.3194 \ m$

Bernoulli's equation for point 1 and 2,

$\frac{P1}{\rho g}+\frac{v1^{2}}{2g}+z1 = \frac{P2}{\rho g}+\frac{v2^{2}}{2g}+z2 + h_{L}$

P2 = Po

v1 = v2 $\approx$ 0

z1 = 1.2 m

z2 = 0.8 m

$\frac{P_{1}}{1000*9.81}+0+1.2 = \frac{P_{0}}{1000*9.81}+0+0.8 + 4.3194$

.: P₁ - P₀ = 38449.314 Pa...............................................................Answer