Venturimeter with reservoir suction tube

The figure values are only partly readable, so the following reading is assumed: the pipe diameters are $D_1 = 15$ cm and $D_2 = 9$ cm (throat); the piezometer at the inlet stands 1 cm above the pipe axis ($p_1/\gamma = 0.01$ m), and the piezometer difference between the inlet and the throat is 5 cm of water. The venturimeter is horizontal and the throat is connected by a tube to a reservoir open to the atmosphere.

(i) Velocity in the smaller diameter section (throat)

Bernoulli between 1 and 2 (horizontal, no loss) and continuity $A_1V_1 = A_2V_2$:

$$\frac{p_1 - p_2}{\gamma} = \frac{V_2^2 - V_1^2}{2g} = \frac{V_2^2}{2g}\left[1 - \left(\frac{D_2}{D_1}\right)^4\right]$$ $$\begin{aligned} \left(\frac{9}{15}\right)^4 &= 0.1296\\ V_2 &= \sqrt{\frac{2g\,(0.05)}{1 - 0.1296}} = \sqrt{\frac{0.981}{0.8704}} = 1.062\ \text{m/s} \end{aligned}$$

Flow $Q = A_2V_2 = \dfrac{\pi}{4}(0.09)^2(1.062) = 6.76\times 10^{-3}$ m³/s.

(ii) Maximum height h lifted from the reservoir

Pressure head at the throat:

$$\frac{p_2}{\gamma} = \frac{p_1}{\gamma} - 0.05 = 0.01 - 0.05 = -0.04\ \text{m (gauge)}$$

The fluid in the lifting tube is at rest, so the throat pressure just balances a column of fluid of height $h$ above the reservoir surface (open to atmosphere): $p_2 = -\gamma h$.

$$h_{max} = \left|\frac{p_2}{\gamma}\right| = 0.04\ \text{m} = 4\ \text{cm}$$

Fluid rises up the tube until it reaches the height at which the throat suction is balanced.

   inlet 1          throat 2
 ----------\      /-----------
            \____/
                |   <- tube, h = 4 cm lift
 ~~~~~~~~~~~~~~~|~~~ reservoir (open to air)

Answer: (i) $V_2 \approx 1.06$ m/s (so $Q \approx 6.8$ L/s); (ii) $h_{max} = 0.04$ m. (Method: $V_2 = \sqrt{2g\Delta h/(1-(D_2/D_1)^4)}$ and $h = -p_2/\gamma$.)