Question

The reading of a meter which reads pressure is fitted in a closed pipe is $\text{5}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{N}{{\text{m}}^{-2}}$ on the opening the value of the pipe, the reading of that meter reduces to $\text{5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{N}{{\text{m}}^{-2}}$. The speed of water flowing in the pipe is?

Answer 1

Hint: In order to solve this question we will apply Bernoulli's principle when the pipe was closed and after it was opened. Only the value of the initial and final pressure is given in the question. It is to be assumed that the atmospheric pressure remains constant. Bernoulli’s equation can be summarized as the total pressure is the sum of static pressure and dynamic pressure.

Formula used:
${{\text{P}}_{\text{i}}}\ \text{=}\ {{\text{P}}_{\text{f}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{V}}^{\text{2}}}$
Here ${{\text{P}}_{\text{i}}}$ is a Initial static pressure
${{\text{P}}_{\text{f}}}$ is a final pressure
$\text{P}$ is the density of water
$\text{V}$ is the velocity of water.

Complete step by step solution:
Using Bernoulli’s principle we get
${{\text{P}}_{\text{1}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{P}}_{\text{1}}}^{\text{2}}\text{+ }\!\!\rho\!\!\text{ g}{{\text{h}}_{\text{1}}}\ \text{=}\ {{\text{P}}_{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{V}}_{\text{2}}}^{\text{2}}\text{+ }\!\!\rho\!\!\text{ g}{{\text{h}}_{\text{2}\ }}\ \ \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\left( \text{1} \right)$
Initially the value is closed, so velocity of water I.e. ${{\text{V}}_{\text{1}}}\text{=0}$
$\text{ }\!\!\And\!\!\text{ }\ {{\text{h}}_{\text{1}\ }}\text{=}\ {{\text{h}}_{\text{2}\ }}\ \text{=0}$
Both are flowing at same reference point

Now, putting the value ${{\text{V}}_{\text{1}}}\text{=}\ \text{0}\ \text{ }\!\!\And\!\!\text{ }\ {{\text{h}}_{\text{1}}}\text{=}{{\text{h}}_{\text{2}\ }}\text{=0}$
We get,
${{\text{P}}_{\text{1}}}\ \text{=}\ {{\text{P}}_{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{V}}_{\text{2}}}^{\text{2}}$
Here ${{\text{P}}_{\text{1}}}$ is the initial pressure
${{\text{P}}_{\text{2}}}$ is the final pressure when the valve is open
So, further
$\Rightarrow {{\text{P}}_{\text{1}}}\text{=}{{\text{P}}_{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{V}}_{\text{2}}}^{\text{2}}$
${{\text{V}}_{\text{2}}}^{\text{2}}\ \text{=}\ \dfrac{\text{2}\left( {{\text{P}}_{\text{1}}}\text{-}{{\text{P}}_{\text{2}}} \right)}{\text{P}}$
Here $\text{ }\!\!\rho\!\!\text{ }\ \text{=}\ \text{1000kg/}{{\text{m}}^{\text{3}}}$
${{\text{P}}_{\text{1}}}\ \text{=}\ \text{5}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\text{N/}{{\text{M}}^{\text{2}}}$
${{\text{P}}_{\text{2}}}\ \text{=}\ \text{5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\text{N/}{{\text{M}}^{\text{2}}}$
${{\text{V}}_{\text{2}}}^{\text{2}}\ \text{=}\ \dfrac{\text{2}\left( \text{5}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}}\text{-5}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{15}}} \right)}{\text{1}{{\text{0}}^{\text{3}}}}$
$=\ \dfrac{2\left( 0.5\times {{10}^{15}} \right)}{{{10}^{3}}}$
${{\text{V}}_{\text{2}}}^{\text{2}}\text{=}\dfrac{\text{1}{{\text{0}}^{\text{15}}}}{\text{1}{{\text{0}}^{\text{3}}}}\ \text{=}\ \text{1}{{\text{0}}^{\text{15-3}}}\ \text{=}\ \text{1}{{\text{0}}^{\text{12}}}$
So,
${{\text{V}}_{\text{2}}}\text{=}\ \text{1}{{\text{0}}^{\text{6}}}\text{m/s}$

So, the speed of the water is $\text{1}{{\text{0}}^{\text{6}}}\text{m/s}$.

Note: 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.
Principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed.
Here is a way to express kinetic energy is to do work on it.
This is expressed by the work energy principle
$\text{ }\!\!\Delta\!\!\text{ }{{\text{W}}_{\text{external}}}\ \text{= }\!\!\Delta\!\!\text{ K}\ \text{=}\ \dfrac{\text{1}}{\text{2}}\text{m}{{\text{V}}_{\text{f}}}^{\text{2}}\text{-}\dfrac{\text{1}}{\text{2}}\text{m}{{\text{V}}_{\text{i}}}^{\text{2}}$
Bernoulli’s equation is usually used in isentropic fluids.
In order to solve we have to use Bernoulli’s equation which is given by
${{\text{P}}_{\text{1}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{P}}_{\text{1}}}^{\text{2}}\text{+ }\!\!\rho\!\!\text{ g}{{\text{h}}_{\text{1}}}\ \text{=}\ {{\text{P}}_{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{V}}_{\text{2}}}^{\text{2}}\text{+ }\!\!\rho\!\!\text{ g}{{\text{h}}_{\text{2}\ }}$
Here the value is closed initially so ${{\text{V}}_{\text{1}\ }}\text{=0 }\!\!\And\!\!\text{ }\ {{\text{h}}_{\text{1}\ }}\text{ }\!\!\And\!\!\text{ }{{\text{h}}_{\text{2}\ }}\text{=0,}$ So, by putting these values we can do this question.