Question

A venturi meter (venturi coefficient $C_v = 0.98$) is installed in a pipe of inner diameter 50 mm. Water (density 1000 kg m$^{-3}$) flows through the pipe. The pressure drop across the venturi meter is 50 kPa. If the venturi throat diameter is 20 mm, the estimated flow rate is \(\underline{\hspace{2cm}}\) $\times 10^{-3}$ m$^{3}$ s$^{-1}$ (rounded off to two decimal places).

Hint:

Venturi meters use the pressure drop caused by acceleration through the throat to calculate flow rate.

Answer 1

Correct Answer: 3.1 - 3.17

Pipe and throat areas:
$D = 0.05$ m, $d = 0.02$ m
$A_1 = \frac{\pi D^2}{4} = 1.963 \times 10^{-3}$ m$^2$
$A_2 = \frac{\pi d^2}{4} = 3.142 \times 10^{-4}$ m$^2$
Venturi equation:
\[ Q = C_v A_2 \sqrt{\frac{2\Delta P}{\rho(1-(A_2/A_1)^2)}} \] Compute area ratio:
$(A_2/A_1)^2 = (0.3142/1.963)^2 = 0.0256$
Now the flow rate:
\[ Q = 0.98 \times 3.142\times 10^{-4} \sqrt{\frac{2 \times 50{,}000}{1000(1 - 0.0256)}} \] Inside the square root:
\[ \frac{100000}{1000 \times 0.9744} = 102.65 \] \[ \sqrt{102.65} = 10.13 \] Thus:
\[ Q = 0.98 \times 3.142\times 10^{-4} \times 10.13 = 3.12 \times 10^{-3} \text{ m}^3\text{/s} \] Rounded: $3.12 \times 10^{-3}$ m$^3$ s$^{-1}$