# Venturimeter | Measurement of Rate of Flow of Fluid

### Venturimeter

It is a device used to measure the rate of streamline flow of a fluid through a tube . A venturimeter works on Bernoulli’s principle.

It consists of a U-tube filled with mercury fitted to a straight tube, with varying cross-section, through which the rate of flow of the fluid is to be measured.

The area of cross-section in the broader section of the straight tube is A1 and that of the narrower section is A2 . The speed of flow of the fluid in the narrower section is v2 and that at the broader section is v1

Using the principle of continuity, we have

v1A1 = v2A2     ………(i)

$\large \frac{A_1}{A_2} = \frac{v_2}{v_1}$

The fluid is incompressible. Therefore,

v1 A1 ρ = v2 A2 ρ      ……..(ii)

Suppose that the pressure in the broader section is P1 and in the narrower section is P2

Therefore, P1 – P2 = hρ0g     ……..(iii)

where ρ0 is the density of mercury in the U-tube, and h represents the difference in the levels of mercury in the limbs of the U-tube.

Now, applying Bernoulli’s principle at the two positions, we have

$\displaystyle P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

$\displaystyle P_1 – P_2 = \frac{1}{2}\rho (v_2^2 – v_1^2 )$

$\displaystyle \rho_0 g h = \frac{1}{2}\rho (v_2^2 – v_1^2 )$

$\displaystyle \rho_0 g h = \frac{1}{2}\rho v_1^2(\frac{v_2^2}{v_1^2} – 1)$

$\displaystyle \rho_0 g h = \frac{1}{2}\rho v_1^2(\frac{A_1^2}{A_2^2} – 1)$ …(iv)

Solving equations (1) and (4) for v1, we have

$\displaystyle v_1 = A_2 \sqrt{\frac{2\rho_0 g h}{\rho (A_1^2 -A_2^2)}}$ …….(v)

Therefore the rate of flow of fluid is given by,

Q = A1 v1

$\displaystyle Q = A_1 A_2 \sqrt{\frac{2\rho_0 g h}{\rho (A_1^2 -A_2^2)}}$     ……(vi)

Solved Example : The ratio of the radius of the tube of a venturimeter is η (η >1). The ratio of the densities of the liquid in the manometer and the moving fluid is η1. If the difference in heights of the liquid column in the manometer is h, find the minimum speed of flow of the fluid.

Solution: The speed of flow is minimum when the cross-sectional area of the tube is maximum. The equation for minimum speed of flow of the fluid in a venturimeter is given as

$\displaystyle v_1 = A_2 \sqrt{\frac{2\rho_0 g h}{\rho (A_1^2 -A_2^2)}}$

$\displaystyle v_1 = \sqrt{\frac{2\rho_0 g h}{\rho ((A_1/A_2)^2 – 1)}}$

Putting ρo/ρ = η1 and ,

$\displaystyle \frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = (\frac{r_1}{r_2})^2 = \eta^2$

we obtain ,

$\displaystyle v_1 = \sqrt{\frac{2\eta_1 g h}{\eta^4 -1}}$

Exercise : In a venturimeter, the diameter of the pipe is 4 cm and that of constriction is 3 cm. The heights of the liquid of density 1.5 gm/cc. in the pressure tube is 20 cm at the pipe and 5 cm at the constrictions. What is the discharge rate of water flowing inside the pipe?