Minimum throat diameter of venturimeter

Given: $d_1 = 0.15$ m, $Q = 240$ m³/h $= 0.06667$ m³/s, $p_1/\gamma = 18$ m (gauge), minimum throat pressure head $p_2/\gamma = -7$ m (gauge). Friction loss $= 10\%$ of the pressure head difference.

Smaller throat gives higher velocity and lower pressure, so the minimum throat diameter corresponds to $p_2/\gamma = -7$ m.

$$\begin{aligned} \Delta\left(\frac{p}{\gamma}\right) &= 18 - (-7) = 25\ \text{m}\\ h_L &= 0.1\times 25 = 2.5\ \text{m} \end{aligned}$$

Inlet velocity:

$$A_1 = \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2,\qquad V_1 = \frac{0.06667}{0.017671} = 3.7726\ \text{m/s},\qquad \frac{V_1^2}{2g} = 0.7254\ \text{m}$$

Bernoulli (horizontal pipe):

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_L$$ $$\begin{aligned} \frac{V_2^2}{2g} &= 18 + 0.7254 - (-7) - 2.5 = 23.2254\ \text{m}\\ V_2 &= \sqrt{2\times 9.81\times 23.2254} = 21.347\ \text{m/s} \end{aligned}$$

Throat area and diameter:

$$A_2 = \frac{Q}{V_2} = \frac{0.06667}{21.347} = 3.123\times 10^{-3}\ \text{m}^2$$ $$d_2 = \sqrt{\frac{4A_2}{\pi}} = 0.0631\ \text{m}$$

Answer: minimum throat diameter $\approx 63$ mm (0.0631 m). The throat velocity is 21.35 m/s and the loss is 2.5 m.