Question

Two horizontal pipes of different diameters are connected together by a valve. Their diameters are $ 3cm\,and\,9cm $ . Water flows at speed $ 6m.{s^{ - 1}} $ in the first pipe and the pressure in it is $ 2 \times {10^5}N.{m^{ - 2}} $ . Find the water pressure and speed in the second pipe.

Answer 1

Hint :In order to this question, to find the water pressure and speed of water flows in second pipe or Pipe2, we will first assume the cross-sectional area of both pipe and then we will apply Continuity equation to find the speed of water flows of second pipe. And then to find the water pressure of the second pipe, we will apply Bernauli’s Equation.

Complete Step By Step Answer:
Let the cross-sectional area of Pipe1 and Pipe2 are $ {A_1}\,and\,{A_2} $ respectively.
And the diameters of both pipe1 and pipe2 are given i.e.. $ 3cm\,and\,9cm $ respectively.
Therefore, Radius of pipe1 $ = \dfrac{3}{2}cm = \dfrac{3}{2} \times {10^{ - 2}}m $
And, Radius of Pipe $ = \dfrac{9}{2}cm = \dfrac{9}{2} \times {10^{ - 2}}m $
Let the speed of water flows in Pipe1 and Pipe2 are $ {v_1}\,and\,{v_2} $ respectively.
So, speed of water flows in Pipe1, $ {v_1} = 6m.{s^{ - 1}} $ (given)
Now, we have to find the speed of water flows in Pipe2, $ {v_2} = ? $
According to the continuity equation-
 $ \therefore {A_1}{v_1} = {A_2}{v_2} \\
   \Rightarrow \pi \times {(\dfrac{3}{2} \times {10^{ - 2}})^2} \times 6 = \pi \times {(\dfrac{9}{2} \times {10^{ - 2}})^2} \times {v_2} $
 $ \Rightarrow \dfrac{9}{4} \times 6 = \dfrac{{81}}{4} \times {v_2} \\
   \Rightarrow {v_2} = \dfrac{2}{3}m.{s^{ - 1}} = 0.66m.{s^{ - 1}} $
Hence, the speed of the water flows of Pipe1, $ {v_2} = 0.66m.{s^{ - 1}} $ .
Now,
Pressure in the Pipe1, $ {P_1} = 2 \times {10^5}N.{m^{ - 2}} $
And, we have to find the pressure in the Pipe2, $ {P_2} = ? $
So, applying Bernauli’s Equation to find the required pressure:-
 $ \therefore {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 the pipes are connected horizontally, so their heights are the same, i.e.. $ {h_1} = {h_2} $ and they cancel each other in the further calculation.
and as we know, the density for water, $ \rho = {10^3} $
 $ \Rightarrow 2 \times {10^5} + \dfrac{1}{2} \times {10^3} \times {6^2} = {P_2} + \dfrac{1}{2} \times {10^3} \times {(\dfrac{2}{3})^2} \\
   \Rightarrow 2 \times {10^5} + 18 \times {10^3} = {P_2} + 0.22 \times {10^3} \\
   \Rightarrow {P_2} = 2.18 \times {10^5}Pa \\ $
Hence, the pressure in the Pipe2 is, $ {P_2} = 2.18 \times {10^5}Pa $ .

Note :
The pressures over the entrance and output zones are constant if gravity is ignored. The Bernoulli equation and the one-dimensional continuity equation give, respectively, along a streamline on the centerline. Even when the flow is not one-dimensional, these two findings give an obvious guidance for studying fluid fluxes.