Answer to Question #101825 in Mechanics | Relativity for Emmanuel Tennyson

Question

A greenish pump has tap section pipe. The pipe is running full of water. The pipe diameter at the inlet and at the upper end are 1metre and 0.5metre respectively. The free water surface is 2metre above the inlet and centre upper end is 3metre above the top of the free water surface. The pressure at the top end of the pipe is 25cm of mercury and it is known that loss of head by friction between tap and butttom section is ¹tenth of the top section. Compute the discharge in line per second

Answer 1

The pressure of mercury at the top end of the pipe: $p_t$

The loss: $p_l=p_t/10$

Pressure of water at the upper end: $p_{w.top}$

Pressure of water at the bottom end: $p_{w.bottom}$

Bernoulli's law: pressure from the free water surface at the top end causes water to flow:

$$p_{w.top}+p_l+p_t=\frac{1}{2}\rho v^2+p_{w.bottom},\\ \space\\ v=\sqrt{\frac{2}{\rho_w}(p_{w.top}+p_l+p_t-p_{w.bottom})}=\\ \space\\ =\sqrt{2g(h_{w.top}+1.1\frac{\rho_m}{\rho_w}h_m-h_{w.bottom})}=\\ \space\\ =\sqrt{2\cdot9.8(5+1.1\frac{13600}{10^3}\cdot0.25-2)}=\\ \space\\ =11.5\text{ m/s}.$$

This speed the water would have if the pipe was of constant diameter. The volume of water at the top equals the volume at the bottom:

$$\frac{\pi D^2}{4}\cdot v\rho t=\frac{\pi d^2}{4}\cdot u\rho t,\\ \space\\ \bigg(\frac{D}{d}\bigg)^2=\frac{u}{v}.$$

$$v=u\bigg(\frac{d}{D}\bigg)^2=11.5\bigg(\frac{0.5}{1}\bigg)^2=2.88\text{ m/s}.$$