Question 6.15: A vertical venturimeter has an area ratio of 5. It has a thr......
A vertical venturimeter has an area ratio of 5. It has a throat diameter of 1 cm. When oil of specific gravity 0.85 flows through it the mercury in the differential gauge indicates a difference in height of 20 cm. Find the discharge through the venturimeter. Take coefficient of discharge of the venturimeter as 0.98.
Given data:
Diameter at throat D_2=1 \mathrm{~cm}=0.01 \mathrm{~m}
Cross-sectional area of venturimeter at throat is
A_2=\frac{\pi}{4} \times(0.01)^2=7.854 \times 10^{-5} \mathrm{~m}^2
Cross-sectional area of pipe is
A_1=5 A_2=3.927 \times 10^{-4} \mathrm{~m}^2\left(\text { area ratio } \frac{A_1}{A_2}=5\right)
Differential manometer reading Δh = 20 cm = 0.2 m
Coefficient of discharge C_d=0.98
The discharge through the venturimeter is given by Eq. (6.23) as
Q={\frac{C_{d}A_{1}A_{2}}{\sqrt{A_{1}^{2}-A_{2}^{2}}}}{\sqrt{2g\left({\frac{\rho_{m}}{\rho_{w}}}-1\right)\Delta h}} (6.23)
Q=\frac{C_{d}A_{1}A_{2}\sqrt{2g\left(\frac{\rho_{m}}{\rho_{w}}-1\right)\Delta h}}{\sqrt{A_{1}^{2}-A_{2}^{2}}}=\frac{C_{d}A_{2}\sqrt{2g\left(\frac{\rho_{m}}{\rho_{w}}-1\right)\Delta h}}{\sqrt{1-\left(\frac{A_{2}}{A_{1}}\right)^{2}}}
={\frac{0.98\times7.854\times10^{-5}\times{\sqrt{2\times9.81\times{\left({\frac{13.6}{0.85}}-1\right)\times0.2}}}}{\sqrt{1-\left({\frac{1}{5}}\right)^{2}}}}
= 0.00059 m³ /s