__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 A_{1} and that of the narrower section is A_{2} . The speed of flow of the fluid in the narrower section is v_{2} and that at the broader section is v_{1}

Using the principle of continuity, we have

v_{1}A_{1} = v_{2}A_{2} ………(i)

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

The fluid is incompressible. Therefore,

v_{1} A_{1} ρ = v_{2} A_{2} ρ ……..(ii)

Suppose that the pressure in the broader section is P_{1} and in the narrower section is P_{2}

Therefore, P_{1} – P_{2} = hρ_{0}g ……..(iii)

where ρ_{0} 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 v_{1}, 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 = A_{1} v_{1}

$ \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 η_{1}. 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 ρ_{o}/ρ = η_{1} 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?