Solveeit Logo

Question

Question: In the figure shown the velocity and pressure of the liquid at the cross section (2) are given by (i...

In the figure shown the velocity and pressure of the liquid at the cross section (2) are given by (if P0{{\text{P}}_{\text{0}}} is the atmospheric pressure):
(A) 2hg,P0+ρhg2\left( {\text{A}} \right){\text{ }}\sqrt {{\text{2hg}}} ,{{\text{P}}_0} + \dfrac{{\rho {\text{hg}}}}{2}
(B) hg,P0+ρhg2\left( {\text{B}} \right){\text{ }}\sqrt {{\text{hg}}} ,{{\text{P}}_0} + \dfrac{{\rho {\text{hg}}}}{2}
(C) hg2,P0+3ρhg4\left( {\text{C}} \right){\text{ }}\sqrt {\dfrac{{{\text{hg}}}}{2}} ,{{\text{P}}_0} + \dfrac{{3\rho {\text{hg}}}}{4}
(D) hg2,P0+3ρhg4\left( {\text{D}} \right){\text{ }}\sqrt {\dfrac{{{\text{hg}}}}{2}} ,{{\text{P}}_0} + \dfrac{{3\rho {\text{hg}}}}{4}

Explanation

Solution

Viscosity of fluid: It is the property of the fluid by virtue under 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.
Streamline flow: It is the specific flow of fluid such that each particle of the liquid passing a given point moves along the same path and has the same velocity as its predecessor.
Equation of continuity: It defines that during a streamline flow of the non-viscous and incompressible fluid through a pipe of varying cross-section, the product of the area of cross-section of the pipe and the normal fluid velocity inside the pipe (av) remains constant throughout the flow.
Bernoulli’s principle: It states that the sum of pressure energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline.

Formula used:
Equation of continuity, a1v1=a2v2{{\text{a}}_1}{{\text{v}}_1} = {{\text{a}}_2}{{\text{v}}_2}, here a1, a2={a_1},{\text{ }}{a_2} = area of cross-section of the pipe, v1, v2={v_1},{\text{ }}{v_2} = velocity of the fluid
Bernoulli’s equation, P + 12ρv2+ρgh = constant{\text{P + }}\dfrac{1}{2}\rho {{\text{v}}^2} + \rho {\text{gh = constant}}
Here  P=\;P = fluid pressure,  ρ=\;\rho = density of the fluid, v=v = velocity of the fluid,
g=g = Acceleration due to gravity,  h=\;h = height of the fluid in the pipe

Complete step by step answer:
It is given that P0={P_0} = atmospheric pressure, A=A = area of the cross section of region1, A/2=A/2 = area of the cross section of region22, A/4=A/4 = area of the cross section of region3
Therefore, we use the above Bernoulli’s equations between the top and point 33,P0+ρgh = P0+12ρv32{{\text{P}}_0} + \rho {\text{gh = }}{{\text{P}}_0} + \dfrac{1}{2}\rho {{\text{v}}_3}^2,
Here at the top of the pipe the velocity (v1) = 0\left( {{v_1}} \right){\text{ }} = {\text{ }}0 and the height at point 33, h=0h = 0,
We get the value of v3=2gh{{\text{v}}_3} = \sqrt {2{\text{gh}}}
By using the equation of continuity between point 11 and point 22, we get,
A2v2=A4v3\dfrac{{\text{A}}}{2}{{\text{v}}_2} = \dfrac{{\text{A}}}{4}{{\text{v}}_3}
v2=v32\Rightarrow {{\text{v}}_2} = \dfrac{{{{\text{v}}_3}}}{2}
Substituting the value of v3{v_3} in the equation we get,
v2=2gh2{{\text{v}}_2} = \dfrac{{\sqrt {2{\text{gh}}} }}{2}
gh2\Rightarrow \sqrt {\dfrac{{{\text{gh}}}}{2}}
We use the above Bernoulli’s equations between the top and point 22,
P0+ρgh = P2+12ρv22{{\text{P}}_0} + \rho {\text{gh = }}{{\text{P}}_2} + \dfrac{1}{2}\rho {{\text{v}}_2}^2,
Here at the top of the pipe the velocity (v1) = 0\left( {{v_1}} \right){\text{ }} = {\text{ }}0 and the height at point22, h=0h = 0, we get the value of   P2\;{P_2},
P2=P0+ρgh - 12ρv22{{\text{P}}_2} = {{\text{P}}_0} + \rho {\text{gh - }}\dfrac{1}{2}\rho {{\text{v}}_2}^2,
Now we will substitute the value of v2{v_2} in the equation
P2=P0+ρgh - 12ρ(gh2)\Rightarrow {{\text{P}}_2} = {{\text{P}}_0} + \rho {\text{gh - }}\dfrac{1}{2}\rho \left( {\dfrac{{{\text{gh}}}}{2}} \right)
On multiplying the terms we get,
P2=P0+ρghρgh4\Rightarrow {{\text{P}}_2} = {{\text{P}}_0} + \rho {\text{gh}} - \dfrac{{\rho {\text{gh}}}}{4}
Taking LCM on the second and third term we get,
P2=P0+34ρgh\Rightarrow {{\text{P}}_2} = {{\text{P}}_0} + \dfrac{3}{4}\rho {\text{gh}}
Hence velocity gh2\sqrt {\dfrac{{{\text{gh}}}}{2}} and pressure of the liquid P2=P0+34ρgh{{\text{P}}_2} = {{\text{P}}_0} + \dfrac{3}{4}\rho {\text{gh}}

\therefore The correct option is (C)\left( {\text{C}} \right).

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 converted into heat energy due to the work done against the internal energy or friction or the viscous forces.
The angular momentum of the fluid is not taken into consideration in Bernoulli’s equation.