Bernoulli's equation:

$\overline{){{\mathbf{P}}}_{{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho}}{{{\mathbf{v}}}_{{\mathbf{1}}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{h}}}_{{\mathbf{1}}}{\mathbf{\rho}}{\mathbf{g}}{\mathbf{=}}{{\mathbf{P}}}_{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho}}{{\mathbf{v}}_{\mathbf{2}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{h}}}_{{\mathbf{2}}}{\mathbf{\rho}}{\mathbf{g}}}$ where **ρ** is the density of the moving fluid, **h** is the difference in height between the two points.

Continuity equation:

$\overline{){\mathbf{Q}}{\mathbf{=}}{{\mathbf{A}}}_{{\mathbf{1}}}{{\mathbf{v}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{A}}}_{{\mathbf{2}}}{{\mathbf{v}}}_{{\mathbf{2}}}{\mathbf{=}}{{\mathbf{A}}}_{{\mathbf{n}}}{{\mathbf{v}}}_{{\mathbf{n}}}}$

Pressure in liquids:

$\overline{){\mathit{P}}{\mathbf{=}}{\mathit{\rho}}{\mathit{g}}{\mathit{h}}}$

We'll use Bernoulli's equation to solve for the velocity.

The pressure is greater in the wider cross section. P_{2} is greater than P_{1}.

The elevation at points 1 and 2 is equal, h_{1} = h_{2}. Thus, we can drop the terms h_{1}ρ_{air}g and h_{2}ρ_{air}g.

Using Bernoulli's equation:

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

What is the air speed v_{1} at point 1?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Bernoulli's Equation concept. If you need more Bernoulli's Equation practice, you can also practice Bernoulli's Equation practice problems.