Question

What is the air speed v₁ at point 1?

What is the air speed v₂ at point 2?

What is the volume flow rate?

Answer 1

From the equation of continuity,

$$ \begin{array}{c} A_{1} v_{1}=A_{2} v_{2} \\ v_{2}=\frac{A_{1}}{A_{2}} v_{1} \end{array} $$

From the Bernoulli's principle,

$$ \begin{aligned} P_{1}+\frac{1}{2} \rho v_{1}^{2} &=P_{2}+\frac{1}{2} \rho v_{2}^{2} \\ \frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right) &=P_{2}-P_{1}=\rho_{\text {liquid }} g h \\ \frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right) &=\rho_{\text {liquid }} g h \\ \frac{1}{2} \rho\left(v_{1}^{2}-\left(\frac{A_{1} v_{1}}{A_{2}}\right)^{2}\right) &=\rho_{\text {liquid }} g h \\ v_{1}^{2}\left(\frac{1}{2} \rho\left(\frac{A_{2}^{2}-A_{1}^{2}}{A_{2}^{2}}\right)\right) &=\rho_{\text {liquid }} g h \\ v_{1} &=\sqrt{\frac{2 \rho_{\text {liquid }} g h}{\rho\left(A_{2}^{2}-A_{1}^{2}\right)}} \end{aligned} $$

The air speed \(v_{1}\) at point 1 is,

$$ \begin{aligned} v_{1} &=A_{2} \sqrt{\frac{2 \rho_{\text {liquid }} g h}{\rho\left(A_{2}^{2}-A_{1}^{2}\right)}} \\ &=\left(\pi r_{2}^{2}\right) \sqrt{\frac{2 \rho_{\text {liquid }} g h}{\rho\left(A_{2}^{2}-A_{1}^{2}\right)}} \\ &=\left(\pi r_{2}^{2}\right) \sqrt{\frac{2 \rho_{\text {liquid }} g h}{\rho\left(\pi r_{2}^{2}-\pi r_{1}^{2}\right)}} \\ &=(3.14)\left(0.5 \times 10^{-2} \mathrm{~m}\right)^{2} \sqrt{\frac{2\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)(9.8)(0.1 \mathrm{~m})}{\sqrt{\left(1.28 \mathrm{~kg} / \mathrm{m}^{3}\right) \pi\left(\left(0.5 \times 10^{-2}\right)^{2}-\left(1.0 \times 10^{-3}\right)^{2}\right)}}} \\ &=144 \mathrm{~m} / \mathrm{s} \end{aligned} $$

The air speed \(v_{2}\) at point 2 is,

\(\begin{aligned} v_{2} &=A_{1} \sqrt{\frac{2 \rho_{\text {lịuid }} g h}{\rho\left(A_{2}^{2}-A_{1}^{2}\right)}} \\ &=\left(\pi r_{1}^{2}\right) \sqrt{\frac{2 \rho_{\text {liguid }} g h}{\rho\left(A_{2}^{2}-A_{1}^{2}\right)}} \\ &=\left(\pi r_{1}^{2}\right) \sqrt{\frac{2 \rho_{\text {liquid }}^{2}}{\rho\left(\pi r_{2}^{2}-\pi r_{1}^{2}\right)}} \\ &=(3.14)\left(1.0 \times 10^{-3} \mathrm{~m}\right)^{2} \sqrt{\frac{2\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)(9.8)(0.1 \mathrm{~m})}{\left(1.28 \mathrm{~kg} / \mathrm{m}^{3}\right) \pi\left(\left(0.5 \times 10^{-2}\right)^{2}-\left(1.0 \times 10^{-3}\right)^{2}\right)}} \\ &=5.77 \mathrm{~m} / \mathrm{s} \end{aligned}\)

The volume flow rate is. \(\frac{d V}{d t}=A_{1} v_{1}\)

$$ \begin{array}{l} =\pi\left(1.0 \times 10^{-3} \mathrm{~m}^{-1}\right)^{2}(144) \\ =4.5 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s} \end{array} $$