Question 6.14: A venturimeter having throat diameter of 150 mm is set in a ......
A venturimeter having throat diameter of 150 mm is set in a vertical pipe of 300 mm diameter to measure the discharge of an oil of specific gravity 0.85 which is flowing through the pipe in upward direction. The difference in elevations of the throat section and entrance section of the venturimeter is 4 cm. The differential U-tube mercury manometer shows a gauge deflection of 25 cm. If the coefficient of discharge of the venturimeter is 0.95, calculate the discharge of oil flowing through the pipe.
Given data:
Diameter at inlet D_1=300 \mathrm{~mm}=0.3 \mathrm{~m}
Diameter at throat D_2=150 \mathrm{~mm}=0.15 \mathrm{~m}
A_2=\frac{\pi}{4} D_2^2=\frac{\pi}{4} \times(0.15)^2=0.01767 \mathrm{~m}^2
Cross-sectional area of venturimeter at throat is
A_1=\frac{\pi}{4} D_1^2=\frac{\pi}{4} \times(0.3)^2=0.07068 \mathrm{~m}^2Differential manometer reading Δh = 25 cm = 0.25 m
Coefficient of discharge C_d=0.95
The discharge is given by Eq. (6.23) as
Q=\frac{C_d A_1 A_2}{\sqrt{A_1^2-A_2^2}} \sqrt{2 g\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{0.95 \times 0.07068 \times 0.01767 \times \sqrt{2 \times 9.81 \times\left(\frac{13.6}{0.85}-1\right) \times 0.25}}{\sqrt{(0.07068)^2-(0.01767)^2}}
= 0.1487 m³/s