Question

A horizontal pipe of non-uniform cross section allows water to flow through it with a velocity of $1\,m{\text{ }}{s^{ - 1}}$ when $50\,kPa$ pressure is exerted at a point. If the velocity of the flow has to be $2\,m{\text{ }}{{\text{s}}^{ - 1}}$ at some other point, what will be the pressure at this point?
A. $50\,kPa$
B. $100\,kPa$
C. $48.5\,kPa$
D. $24.25\,kPa$

Answer 1

Hint:In this question, we will simply apply Bernoulli's principle and put all the given data in question to get the required result. In fluid dynamics, Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy.

Formula used:
${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{h_2}$
Where, ${P_1}$ is the pressure at elevation 1, ${P_2}$ is the pressure at elevation 2, ${v_1}$ is the velocity at elevation 1, ${V_2}$ is the velocity at elevation 2, ${h_1}$ is the height of elevation 1, ${h_2}$ is the height of elevation 2, $\rho $ is the fluid density and $g$ is the acceleration due to gravity.

Complete step by step answer:
Viscosity of fluid:It is the property of the fluid by virtue of which an internal force of friction comes into action when the fluid is in motion and which opposes the relative motion between different layers of the fluid.

It is given that, ${v_1}$ is $1m{\text{ }}{s^{ - 1}}$, ${P_1} = 50 \times {10^3}Pa$ and ${v_2} = 2m{\text{ }}{{\text{s}}^{ - 1}}$. Now, let’s apply the above Bernoulli's equation to calculate the ${P_2}$. As we know that,
${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{h_2}$
As per the question they both are at same height,
Therefore, $\rho gh$ will get cancel and the new formula will be,
${P_1} + \dfrac{1}{2}\rho v_1^2 = {P_2} + \dfrac{1}{2}\rho v_2^2$

Now, substituting the values in above equation,
${P_2} = {P_1} + \dfrac{1}{2}\rho (v_1^2 - v_2^2) \\
\Rightarrow {P_2} = 50 \times {10^3} + \dfrac{1}{2} \times {10^3}(1 - 4) \\
\Rightarrow {P_2} = (50 - 1.5) \times {10^3} \\
\therefore {P_2} = 48.5\,kPa $
So, the pressure at this point is ${P_2} = 48.5\,kPa$ .

Hence, the correct option is C.

Note:Bernoulli’s equation is derived from the assumption that there is no loss of energy due to friction among fluid particles. Some of the kinetic energy is converted into heat energy due to the work done against the internal energy or friction or the viscous forces.Angular momentum of the fluid is not taken into consideration in Bernoulli’s equation.