Venturi meter discharge and numerical

(a) Venturi meter

A Venturi meter consists of a converging cone, a short throat and a diverging cone. The pressure drop between inlet (section 1) and throat (section 2) is used to find the discharge.

   1                2                
   |   converging   |  throat   diverging
 ==\__              _____              __/==
        \__________/     \____________/
   p1 ---|          |--- p2

Apply Bernoulli's equation between 1 and 2 (horizontal, ideal fluid):

$$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} = \frac{p_2}{\rho g} + \frac{V_2^2}{2g}$$

Let $h = \dfrac{p_1 - p_2}{\rho g}$ (differential head). Continuity: $a_1V_1 = a_2V_2$, so $V_1 = V_2\,a_2/a_1$.

$$h = \frac{V_2^2}{2g}\left[1 - \left(\frac{a_2}{a_1}\right)^2\right] \quad \Rightarrow \quad V_2 = \frac{a_1}{\sqrt{a_1^2 - a_2^2}}\sqrt{2gh}$$

Theoretical discharge $Q_{th} = a_2V_2$. Including the coefficient of discharge $C_d$ (about 0.96 to 0.99) for losses:

$$Q = C_d\,\frac{a_1a_2}{\sqrt{a_1^2 - a_2^2}}\sqrt{2gh}$$

For a differential mercury manometer with reading $x$: $h = x\left(\dfrac{S_m}{S_f} - 1\right)$.

(b) Numerical

$$\begin{aligned} a_1 &= \frac{\pi}{4}(0.3)^2 = 0.07069\ \text{m}^2, \quad a_2 = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2 \\ h &= x\left(\frac{13.6}{1} - 1\right) = 0.2 \times 12.6 = 2.52\ \text{m of water} \end{aligned}$$ $$\begin{aligned} Q &= 0.98 \times \frac{0.07069 \times 0.01767}{\sqrt{0.07069^2 - 0.01767^2}} \times \sqrt{2 \times 9.81 \times 2.52} \\ &= 0.98 \times \frac{1.2492\times10^{-3}}{0.06844} \times 7.032 \\ &= 0.1258\ \text{m}^3/\text{s} \end{aligned}$$

Velocity at the inlet = $0.1258/0.07069 = 1.78\ \text{m/s}$, and at the throat = $7.12\ \text{m/s}$.

Answer: $Q = 0.126\ \text{m}^3/\text{s} = 126\ \text{L/s}$