Question

The difference of pressure between two points along a horizontal pipe, through which water is flowing is 5.6 cm of mercury. If due to non-uniform cross section, the speed of flow at a point of greater cross section is $0.6m/s$ . Calculate the speed of flow at other points. The density of mercury is $13600kg/{m^3}$ .

Answer 1

Hint : In a pipe the pressure varies with the change in cross section. As the cross section increases the pressure will decrease and as the cross section decreases the pressure will increase. In this way the cross section and pressure are inversely proportional to each other.
Whatever happens to the cross section the flow rate of liquid across the pipe will remain constant
i.e. ${A_1}{V_1} = {A_2}{V_2}$
A: Area of cross section
V:Velocity of cross section
From Bernoulli’s equation we can write
${P_1} + \dfrac{1}{2}\rho {v_1}^2 = {P_2} + \dfrac{1}{2}\rho {v_2}^2$
P: Pressure
v: velocity
$\rho$ :Density of fluid.

Complete Step By Step Answer:
We have given the data as the pressure difference between points of different cross section is 5.6cm of mercury (density= $13600kg/{m^3}$ )
$= 5.6 \times {10^{ - 2}} \times 13600 \times 9.8 \\\ \Rightarrow 7463.5Pa$
This can also be written as
$0.07463\,atm$
At greater cross section $v = 0.6m/s$
From Bernoulli’s equation we can write
${P_1} + \dfrac{1}{2}\rho {v_1}^2 = {P_2} + \dfrac{1}{2}\rho {v_2}^2$
From greater cross section to lower cross section pressure decreases $\Rightarrow {P_2} > {P_1}$
In similar manner the velocity at smaller section will be greater $\Rightarrow {v_2} > {v_1}$

\Rightarrow ({P_1} - {P_2}) - \dfrac{1}{2}\rho {v_1}^2 = \dfrac{1}{2}\rho {v_2}^2 $$ If we substitute the values we have then we get LHS $$= 7463.5 - \dfrac{1}{2}(1000 \times 0.36) \\\ \Rightarrow 7463.5 - (500 \times 0.36) $$ Solving both sides $ 7283.5 = \dfrac{1}{2}(1000) \times {({v_2})^2} \\\ rearranging \\\ \Rightarrow {({v_2})^2} = \dfrac{{7283.5 \times 2}}{{1000}} $ Solving and taking the square root we get, $ {v_2}^2 = 7.2835 \times 2 \\\ \Rightarrow {v_2}^2 = 14.567 $ Taking square root on both sides $ {v_2} = 3.82m/s $ . **Note :** There are different energy losses like head loss, friction loss due to the surface of pipe, pressure loss are observed in flows through pipes. In any process as we know the energy will always remain constant it can never be created or destroyed, So, the problems can be solved by using energy balance at any point inside the pipe.