91影视

In order to determine the volume rate of flow, firstly use the continuity equation and then apply the reduced form of Bernoulli鈥檚 equation.

The diameter of horizontal pipe at one end is \({d_1} = 6.0\;{\rm{cm}}\).

The diameter of horizontal pipe at other end is \({d_2} = 4.5\;{\rm{cm}}\).

The pressure of water at one end is \({P_1} = 33.5\;{\rm{kPa}}\).

The pressure of water at another end is \({P_2} = 22.6\;{\rm{kPa}}\).

The standard value for the density of water is \(\rho = 1000\;{\rm{kg/}}{{\rm{m}}^3}\).

The velocity relation between the ends of pipe by using discharge concept is calculated below:

\(\begin{array}{c}{A_1}{v_1} = {A_2}{v_2}\\\frac{\pi }{4}{\left( {{d_1}} \right)^2}{v_1} = \frac{\pi }{4}{\left( {{d_2}} \right)^2}{v_2}\\{v_2} = {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}{v_1}\end{array}\)

Here, \({v_1}\) and \({v_2}\) are the velocity of flow at two ends.

The velocity of flow at larger end is calculated below:

\(\begin{array}{l}{P_1} - {P_2} = \frac{1}{2}\rho \left( {{{\left( {{v_2}} \right)}^2} - {{\left( {{v_1}} \right)}^2}} \right)\\{P_1} - {P_2} = \frac{1}{2}\rho \left\{ {{{\left( {{{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)}^2}{v_1}} \right)}^2} - {{\left( {{v_1}} \right)}^2}} \right\}\\{P_1} - {P_2} = \frac{1}{2}\rho \left\{ {\left( {{{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)}^4}{{\left( {{v_1}} \right)}^2}} \right) - {{\left( {{v_1}} \right)}^2}} \right\}\\{P_1} - {P_2} = \frac{1}{2}\rho {\left( {{v_1}} \right)^2}\left( {{{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)}^4} - 1} \right)\end{array}\)

On further simplifying the above relation.

\(\left( {{v_1}} \right) = \sqrt {\frac{{2\left( {{P_1} - {P_2}} \right)}}{{\rho \left( {{{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)}^4} - 1} \right)}}} \)

Substitute the values in the above equation.

\(\begin{array}{c}{v_1} = \sqrt {\frac{{2\left( {33.5\;{\rm{kPa}} \times \frac{{{{10}^3}\;{\rm{Pa}}}}{{1\;{\rm{kPa}}}} - 22.6\;{\rm{kPa}} \times \frac{{{{10}^3}\;{\rm{Pa}}}}{{1\;{\rm{kPa}}}}} \right)}}{{\left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {{{\left( {\frac{{6\;{\rm{cm}}}}{{4.5\;{\rm{cm}}}}} \right)}^4} - 1} \right)}}} \\{v_1} = \sqrt {10.10\;{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}} \\{v_1} = 3.17\;{\rm{m/s}}\end{array}\)

The volume rate of flow is calculated below:

\(\begin{array}{c}Q = {A_1}{v_1}\\ = \left[ {\frac{\pi }{4}{{\left( {{d_1}} \right)}^2}} \right]{v_1}\end{array}\)

Substitute the values in the above equation.

\(\begin{array}{c}Q = \left[ {\frac{\pi }{4}{{\left( {6\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}} \right]\left( {3.17\;{\rm{m/s}}} \right)\\Q = 8.96 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}\end{array}\)

Hence, the volume rate of flow is \(8.96 \times {10^{ - 3}}\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}\).