Question 5.10: Water flows at the rate of 0.015 m³/s through a 100 mm diame...
Water flows at the rate of 0.015 m³/s through a 100 mm diameter orifice used in a 200 mm pipe. What is the difference in pressure head between the upstream section and the vena contracta section? (Take coefficient of contraction C_{c} = 0.60 and C_{v} = 1.0).
We know from Eq. (5.67)
\begin{array}{l}Q=C_{c} A_{0} C_{v} \sqrt{\frac{2 g\left(\rho_{m} / \rho-1\right)}{\left(1-C_{c}^{2} A_{0}^{2} / A_{1}^{2}\right)} \Delta h}\\=C_{v} C_{c} A_{0} \sqrt{\frac{2 g}{\left(1-C_{c}^{2} A_{0}^{2} / A_{1}^{2}\right.}} \sqrt{\left(\rho_{m} / \rho-1\right) \Delta h}\\=C \sqrt{\left(\rho_{m} / \rho-1\right) \Delta h}\end{array} (5.67)
Q=C \sqrt{\frac{\Delta p}{\rho g}}
where, C=C_{v} C_{c} A_{0} \sqrt{\frac{2 g}{\left(1-C_{c}^{2} A_{0}^{2} / A_{1}^{2}\right.}}
For the present problem,
C=1.0 \times 0.60 \times \frac{\pi}{4}(0.1)^{2} \sqrt{\frac{2 \times 9.81}{\left[\left(1-(0.60)^{2}(1 / 2)^{4}\right)\right]}}
= 0.0211
Hence, 0.015=0.0211 \sqrt{\Delta p / \rho g}
or \Delta p / \rho g=0.505 m of water