Question

: A pipe is running full of water. At point A it tapers from \[200{\rm{ c}}{{\rm{m}}^2}\] cross-sectional area to \[100{\rm{ c}}{{\rm{m}}^2}\] at other point B. If the pressure difference between A and B is \[15{\rm{ cm}}\] of water column, the rate of volume flow of water through the pipe in litres/sec \[\left( {g = 10{\rm{ m}}{{\rm{s}}^{ - 2}}} \right)\]
(1) 20
(2) 30
(3) 15
(4) 5

Answer 1

Hint: Hint: From Bernoulli’s equation, we know that the summation of pressure energy, kinetic energy, and gravitational energy of water is constant. Continuity tells us that the mass flow rate of water is constant. We will be using Bernoulli’s equation and the continuity equation to calculate the flow of water.

Complete step by step answer:
Given:
The cross-sectional area of pipe at point A is \[{A_1} = 200{\rm{ c}}{{\rm{m}}^2}\].
The cross-sectional area of pipe at point P is \[{A_2} = 100{\rm{ c}}{{\rm{m}}^2}\].
The pressure difference between point A and point B is \[{P_1} - {P_2} = 15{\rm{ cm}}\]of the water column.
The acceleration due to gravity is \[g = 10{\rm{ m}}{{\rm{s}}^{ - 2}}\].
We have to calculate the flow rate of water through the pipe.
Let us write the continuity equation for water between point A and point B.
\[
m = \rho {A_1}{V_1}\\
= \rho {A_2}{V_2}
\]……(1)
Here \[{V_1}\] is the velocity at point A and \[{V_2}\] is the velocity at point B.
From the above expression, we can write:
\[{A_1}{V_1} = {A_2}{V_2}\]
On substituting \[200{\rm{ c}}{{\rm{m}}^2}\] for \[{A_1}\] and \[100{\rm{ c}}{{\rm{m}}^2}\] for \[{A_2}\] in the above expression, we can write:
\[
\left( {200{\rm{ c}}{{\rm{m}}^2}} \right){V_1} = \left( {100{\rm{ c}}{{\rm{m}}^2}} \right){V_2}\\
{V_2} = 2{V_1}
\]
We know that one millimeter of the water column is \[9.80665{\rm{ Pa}}\] so we can convert the pressure into Pascal as below:
\[
{P_1} - {P_2} = 9.80665 \times 15{\rm{ Pa}}\\
= 1470.99{\rm{ Pa}}
\]
We can apply the concept of Bernoulli’s equation between point A and point B.
\[{P_1} + \dfrac{1}{2}\rho V_1^2 + g{z_1} = {P_2} + \dfrac{1}{2}\rho V_2^2 + g{z_2}\]……(2)
Here \[{P_1}\] is the pressure at point A, \[{P_2}\] is the pressure at point B, \[{z_1}\] is the head at point A and \[{z_2}\] is the head at point B.
We can consider pipe at a horizontal pipe, so the head at point A is equal to point B.
\[{z_1} = {z_2}\]
On substituting \[{z_2}\] for \[{z_1}\] in equation (2), we can write:
\[
{P_1} + \dfrac{1}{2}\rho V_1^2 + g{z_2} = {P_2} + \dfrac{1}{2}\rho V_2^2 + g{z_2}\\
{P_1} + \dfrac{1}{2}\rho V_1^2 = {P_2} + \dfrac{1}{2}\rho V_2^2\\
{P_1} - {P_2} = \dfrac{1}{2}\rho \left( {V_2^2 - V_1^2} \right)
\]
On substituting \[2{V_1}\] for \[{V_2}\], \[1470.99{\rm{ Pa}}\] for \[\left( {{P_1} - {P_2}}
\right)\] and \[1000{\rm{ }}{{{\rm{kg}}} {\left/
{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
} {{{\rm{m}}^3}}}\] for \[\rho \] in the above expression, we get:
\[
1470.99{\rm{ Pa }} \times \left( {\dfrac{{{{\rm{N}} {\left/
{\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}}}{{{\rm{Pa}}}}} \right) \times \left( {\dfrac{{{\rm{kg}}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}}}{{\rm{N}}}} \right) = \dfrac{1}{2}\left( {1000{\rm{ }}{{{\rm{kg}}} {\left/
{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
} {{{\rm{m}}^3}}}} \right)\left[ {\left( {2V} \right)_1^2 - V_1^2} \right]\\
3V_1^2 = 2.942\\
{V_1} = 0.99{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\\
\cong 1{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}
\]
Let us write the expression for the flow rate of water at point A.
\[Q = {A_1}{V_1}\]
On substituting \[1{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\] for \[{V_1}\], \[1000{\rm{ }}{{{\rm{kg}}} {\left/
{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
} {{{\rm{m}}^3}}}\] for \[\rho \] , and \[200{\rm{ c}}{{\rm{m}}^2}\] for \[{A_2}\] in equation (1), we get:
\[
Q = \left( {200{\rm{ c}}{{\rm{m}}^2} \times \dfrac{{{{\rm{m}}^2}}}{{{{10}^4}{\rm{
c}}{{\rm{m}}^2}}}} \right)\left( {1{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}} \right)\\
= 0.02{\rm{ }}{{{{\rm{m}}^3}} {\left/
{\vphantom {{{{\rm{m}}^3}} {\rm{s}}}} \right.
} {\rm{s}}} \times \left( {\dfrac{{1000\,{\rm{litre}}}}{{{{\rm{m}}^3}}}} \right)\\
= 20{{{\rm{ litre}}} {\left/
{\vphantom {{{\rm{ litre}}} {\rm{s}}}} \right.
} {\rm{s}}}
\]
Therefore, the rate of the volume flow of water through the pipe is \[20{{{\rm{ litre}}} {\left/{\vphantom {{{\rm{ litre}}} {\rm{s}}}} \right.} {\rm{s}}}\], and option (1) is correct.

Note: We can note that the final answer for the flow rate is asked in liter/sec, so do not calculate the mass flow rate (kg/sec). Alternatively, we can calculate the flow rate of water by cross-sectional area and velocity at point B.