Question

An ideal liquid (water) flowing through a tube of non-uniform cross-sectional area, where area at A and B are \[40c{m^2}\]and \[20c{m^2}\]respectively. If the pressure difference between A and B is \[700\dfrac{N}{{{m^2}}}\], then volume flow rate is (density of water\[ = 1000\dfrac{{kg}}{{{m^3}}}\])
A. \[2720\dfrac{{c{m^3}}}{s}\]
B. \[2420\dfrac{{c{m^3}}}{s}\]
C. \[1810\dfrac{{c{m^3}}}{s}\]
D. \[3020\dfrac{{c{m^3}}}{s}\]

Answer 1

Hint: When the viscosity of the fluid is neglected then the energy of the fluid remains constant while flowing in the frictionless pipe. We use Bernoulli's equation to determine the flow of the fluid.

Formula used:
\[Q = Av\], here Q is the volume flow rate of the fluid in a cross-section area A and flow velocity v.
\[P + \dfrac{{\rho {v^2}}}{2} + \rho gh = cons\tan t\], here P is the pressure, \[\rho \]is the density of the fluid, v is the flow velocity, g is the acceleration of gravity and h is the pressure height.



Complete answer:
When the area of cross-sectional changes the mass of the fluid per unit time will remain same using conservation of mass. If the flow velocity of the fluid at A and B is \[{v_A}\]and \[{v_B}\]respectively then,
\[{A_A}{v_A} = {A_B}{v_B}\]
The pressure difference between A and B is given as \[700\dfrac{N}{{{m^2}}}\]
\[{P_A} - {P_B} = 700\dfrac{N}{{{m^2}}}\]
The pipe is at same horizontal level, i.e. \[{h_A} = {h_B}\]
Using Bernoulli’s equation,
\[{P_A} + \dfrac{{\rho v_A^2}}{2} + \rho g{h_A} = {P_B} + \dfrac{{\rho v_B^2}}{2} + \rho g{h_B}\]
On simplifying, we get
\[\dfrac{{\rho v_A^2}}{2} - \dfrac{{\rho v_B^2}}{2} = \left( {{P_B} - {P_A}} \right) + \left( {\rho g{h_B} - \rho g{h_A}} \right)\]
\[\dfrac{{\rho \left( {v_A^2 - v_B^2} \right)}}{2} = - \left( {{P_A} - {P_B}} \right) + \rho g\left( {{h_B} - {h_A}} \right)\]

\[\dfrac{{\rho v_A^2}}{2}\left( {1 - {{\left( {\dfrac{{{A_A}}}{{{A_B}}}} \right)}^2}} \right) = - \left( {{P_A} - {P_B}} \right)\]
\[v_A^2 = \dfrac{{ - 2\left( {{P_A} - {P_B}} \right)}}{{\rho \left( {1 - {{\left( {\dfrac{{{A_A}}}{{{A_B}}}} \right)}^2}} \right)}}\]
\[{v_A} = \sqrt {\dfrac{{ - 2\left( {{P_A} - {P_B}} \right)}}{{\rho \left( {1 - {{\left( {\dfrac{{{A_A}}}{{{A_B}}}} \right)}^2}} \right)}}} \]
Putting the values, we get
\[{v_A} = \sqrt {\dfrac{{ - 2 \times 700}}{{1000\left( {1 - {{\left( {\dfrac{{40}}{{20}}} \right)}^2}} \right)}}} \]
\[{v_A} = \sqrt {\dfrac{{ - 14}}{{10\left( {1 - 4} \right)}}} m/s\]
\[{v_A} = \sqrt {\dfrac{{ - 14}}{{ - 30}}} m/s\]
\[{v_A} = 0.683m/s\]
As the final answer is in C.G.S. unit, so we need to convert the obtained velocity in terms of \[\dfrac{{cm}}{s}\]
As 1 meter is equal to 100 centimetres, so the obtained flow velocity will be,
\[{v_A} = 100 \times 0.683\dfrac{{cm}}{s}\]
\[{v_A} = 68.3\dfrac{{cm}}{s}\]
The flow speed at the point A is 68.3 cm/s where the area of cross-section is 40\[c{m^2}\]
So, the volume flow rate of the fluid is,
\[Q = {A_A}{v_A}\]
\[Q = \left( {40c{m^2}} \right) \times \left( {68.3cm/s} \right)\]
\[Q = 2732.5\dfrac{{c{m^3}}}{s}\]
The approximate value of the volume flow rate closest to the available option is \[2720\dfrac{{c{m^3}}}{s}\]
Therefore, the correct option is (A).


Note: We must be careful about the nature of the surface of the pipe through which the fluid is flowing. If the surface of the pipe is rough then the energy of the fluid will be lost as work is done against the resistance offered by the pipe surface.