Question

In: Physics

Air flows through the tube shown in the figure. Assume that air is an ideal fluid.

Air flows through the tube shown in the figure. Assume that air is an ideal fluid.

What is the air speed v1 at point 1?

What is the air speed v2 at point 2?

What is the volume flow rate?

Solutions

Expert Solution

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,

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

The air speed \(v_{2}\) at point 2 is \(v_{2}=A_{1} \sqrt{\frac{2 \rho_{l i q u i} g h}{\rho\left(A_{2}^{2}-A_{1}^{2}\right)}}\)

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

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


The air speed \(v_{1}\) at point 1 is 144 m/s.

The air speed \(v_{2}\) at point 1 is 5.77 m/s.

The volume flow rate is \(4.5 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\).

 

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