Question

Water is flowing through a horizontal pipe in streamline flow. What is the velocity and pressure at the narrowest part of the pipe?

Answer 1

Concept used:

Principle of continuity:

1. According to the principle of continuity, if the fluid is in streamlined flow and is in-compressible then we can say that mass of fluid passing through different cross-sections are equal.
2. The equation for the principle of continuity is given as follows:

$A_1{V}_1=A_2{V}_2$

Where the areas of cross-sections are $A_1$ and $A_2$ respectively, and the flow speed is $V_1$ and $V_2$ as shown in the figure below.

Bernoulli's theorem:

1. Bernoulli’s principle states that the total mechanical energy of the moving fluid comprising the gravitational potential energy of elevation, the energy associated with the fluid pressure, and the kinetic energy of the fluid motion, remains constant.
2. Bernoulli’s equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container.
3. The formula for Bernoulli’s principle is given as follows:

$P+\frac{1}{2}\rho v^2+\rho gh=\mathrm{constant}$

Where $P$ is the pressure exerted by the fluid, $v$ is the velocity of the fluid, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the container.

Explanation:

1. In streamlined flow, the product of cross-section area and velocity remains unchanged. So in the narrowest part of the pipe, velocity will be maximum.
2. Since the pipe is horizontal, the potential energy $\left(\rho gh\right)$ remains constant.
3. As the velocity is maximum (by the equation of continuity) in the narrowest part, so the pressure will be minimum.

Hence, at the narrowest part of the pipe, velocity will be maximum and pressure will be minimum.