Question

An ideal liquid is flowing in 2 pipes AC and BD. One is inclined and the second is horizontal. Both the pipes are connected by 2 vertical tubes. Assuming streamlines flow everywhere, if velocity of liquid at A, B and C are $2m/s$, $4m/s$ and $4m/s$ respectively, what will be the velocity at D.

Answer 1

Hint
Here we can use Bernoulli's theorem to find the equation of the flow of water at the points A, B, C and D. Then we can write the equations as point A and C as equal and from there find the pressure difference between point A and C. Using that in the equations of point B and D we can find the answer.
In this solution we will be using the following equation,
$\Rightarrow P + \dfrac{1}{2}\rho {v^2} + h\rho g = {\text{constant}}$
where $P$ is the pressure,
$\rho$ is the density of the liquid
$v$ is the velocity of the liquid
$h$ is the height and $g$ is the acceleration due to gravity.

Complete step by step answer
Let us consider the height between the points A and B is ${h_1}$ and that between the points C and D is ${h_2}$. According to Bernoulli's theorem, for steady, irrotational motion of a fluid, the addition of the pressure head, kinetic energy head and the gravitational head is constant.
Therefore we can write, $P + \dfrac{1}{2}\rho {v^2} + h\rho g = {\text{constant}}$
We consider the pressure at the point A, B, C and D ${P_A}$, ${P_B}$, ${P_C}$ and ${P_D}$ respectively and the velocities as, ${v_A}$, ${v_B}$, ${v_C}$ and ${v_D}$ .
Therefore, according to the figure, we can write the equation for the points A and C as,
$\Rightarrow {P_A} + \dfrac{1}{2}\rho {v_A}^2 + {h_1}\rho g = {\text{constant}}$ for the point A.
and ${P_C} + \dfrac{1}{2}\rho {v_C}^2 + {h_2}\rho g = {\text{constant}}$ for the point C.
We can equate them and hence get,
$\Rightarrow {P_A} + \dfrac{1}{2}\rho {v_A}^2 + {h_1}\rho g = {P_C} + \dfrac{1}{2}\rho {v_C}^2 + {h_2}\rho g$
We take the terms containing ${h_1}$ and ${h_2}$ to the RHS and get,
$\Rightarrow {P_A} + \dfrac{1}{2}\rho {v_A}^2 = {P_C} + \dfrac{1}{2}\rho {v_C}^2 + \left( {{h_2} - {h_1}} \right)\rho g$
We can take the pressure terms and bring them to the LHS and take the rest terms to RHS and get
$\Rightarrow {P_A} - {P_C} = \dfrac{1}{2}\rho \left( {{v_C}^2 - {v_A}^2} \right) + \left( {{h_2} - {h_1}} \right)\rho g$
Similarly the Bernoulli’s equation for the points B and D are
$\Rightarrow {P_B} + \dfrac{1}{2}\rho {v_B}^2 = {\text{constant}}$, as there is no height.
and ${P_D} + \dfrac{1}{2}\rho {v_D}^2 = {\text{constant}}$
Equating these two we get,
$\Rightarrow {P_B} + \dfrac{1}{2}\rho {v_B}^2 = {P_D} + \dfrac{1}{2}\rho {v_D}^2$
Now we can write the pressure at point B is equal to, ${P_B} = {P_A} + {h_1}\rho g$
and the pressure at the point D is equal to ${P_D} = {P_C} + {h_2}\rho g$
So substituting these in the equation we get,
$\Rightarrow {P_A} + {h_1}\rho g + \dfrac{1}{2}\rho {v_B}^2 = {P_C} + {h_2}\rho g + \dfrac{1}{2}\rho {v_D}^2$
Now we bring the pressure terms to the LHS and take the other terms to the RHS and get,
$\Rightarrow {P_A} - {P_C} = \left( {{h_2} - {h_1}} \right)\rho g + \dfrac{1}{2}\rho \left( {{v_D}^2 - {v_B}^2} \right)$
Now in place of ${P_A} - {P_C}$ we can substitute ${P_A} - {P_C} = \dfrac{1}{2}\rho \left( {{v_C}^2 - {v_A}^2} \right) + \left( {{h_2} - {h_1}} \right)\rho g$
Therefore, we get
$\Rightarrow \dfrac{1}{2}\rho \left( {{v_C}^2 - {v_A}^2} \right) + \left( {{h_2} - {h_1}} \right)\rho g = \left( {{h_2} - {h_1}} \right)\rho g + \dfrac{1}{2}\rho \left( {{v_D}^2 - {v_B}^2} \right)$
The term $\left( {{h_2} - {h_1}} \right)\rho g$ gets cancelled from both sides of the equation. Therefore we have,
$\Rightarrow \dfrac{1}{2}\rho \left( {{v_C}^2 - {v_A}^2} \right) = \dfrac{1}{2}\rho \left( {{v_D}^2 - {v_B}^2} \right)$
Further cancelling the $\dfrac{1}{2}\rho$ from both sides we get,
$\Rightarrow {v_C}^2 - {v_A}^2 = {v_D}^2 - {v_B}^2$
Now keeping the velocity of water at the point D on one side and taking all the other terms to the other side we get,
$\Rightarrow {v_D}^2 = {v_C}^2 - {v_A}^2 + {v_B}^2$
In the question we are given ${v_A} = 2m/s$, ${v_B} = 4m/s$ and ${v_C} = 4m/s$
Therefore, substituting the values we get,
$\Rightarrow {v_D}^2 = {\left( 4 \right)^2} - {\left( 2 \right)^2} + {\left( 4 \right)^2}$
On doing the squares and adding, the value we get is,
$\Rightarrow {v_D}^2 = 16 - 4 + 16 = 28$
Therefore, taking square root on both sides,
$\Rightarrow {v_D} = \sqrt {28} m/s$
Hence the velocity of water at the point D is $\sqrt {28} m/s$ .

Note
Bernoulli's theorem in fluid dynamics is a relation between the pressure, velocity and the elevation of the moving fluid where we consider the fluid to be incompressible, having negligible viscosity and in a steady irrotational flow. It can be derived from the law of conservation of energy, which states that in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points.