Question 6.20: An inclined venturimeter is used to measure the flow of wate......

Given data:

Diameter at inlet      D_1=15 \mathrm{~cm}=0.15 \mathrm{~m}

Cross-sectional area of pipe is

A_1=\frac{\pi}{4} D_1^2=\frac{\pi}{4} \times(0.15)^2=0.01767 \mathrm{~m}^2

Volume flow rate is given as              Q=30 \mathrm{lit} / \mathrm{s}=30 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s}

Coefficient of discharge                      C_d=0.98

Differential manometer reading      Δh = 30 cm = 0.30 m

Let the diameter at throat and cross-sectional area at throat be D_2 \text{and} A_2, respectively.
The discharge through the venturimeter is given by Eq. (6.23) as

Q=\frac{C_d A_1 A_2 \sqrt{2 g\left(\frac{\rho_m}{\rho_w}-1\right) \cdot \Delta h}}{\sqrt{A_1^2-A_2^2}}

Substituting the values, we have

0.026=\frac{0.98 \times 0.01767 \times A_2 \times \sqrt{2 \times 9.81 \times\left(\frac{13.6}{1}-1\right) \times 0.3}}{\sqrt{(0.01767)^2-A_2^2}}

or                \sqrt{(0.01767)^2-A_2^2}=5.7357 A_2

or                (0.01767)^2-A_2^2=32.8983 A_2^2

or                33.8983 A_2^2=0.000312

or                A_2^2=\frac{0.000312}{33.8983}=9.204 \times 10^{-6}

or                A_2=\sqrt{9.204 \times 10^{-6}}=3.0338 \times 10^{-3} \mathrm{~m}^2

or                  \frac{\pi}{4} D_2^2=3.0338 \times 10^{-3}

or                  D_2^2=\frac{4 \times 3.0338 \times 10^{-3}}{\pi}

or                    D_2=\sqrt{\frac{4 \times 3.0338 \times 10^{-3}}{\pi}}=0.062 \mathrm{~m}=6.2 \mathrm{~cm}