Bernoulli’s Equation

Step 1: Realize that this problem is a continuity equation

Even before tackling the problem, think about it conceptually. A fluid is traveling through a pipe with a velocity ${\overrightarrow{v}}_1$. The pipe diameter changes from bigger to smaller, therefore the the cross sectional area goes from bigger to smaller as well. This change in area means that the velocity will increase. To keep the same volume flow rate through the pipe, the velocity will have to increase through the pipe with a smaller diameter.

Step 2: Use the continuity equation

$A_1{\overrightarrow{v}}_1=A_2{\overrightarrow{v}}_2$

where $A_1$ is the area of a circle in this case since we have a pipe. We are given the diameter so in order to solve for the area, we need the

Step 4: Solve for the initial velocity

We now have an overall expression for the initial velocity ( ${\overrightarrow{v}}_1$ ). Solving for ${\overrightarrow{v}}_2$ from the continuity equation gives us:

${\overrightarrow{v}}_2={\left(\frac{r_1}{r_2}\right)}^2\cdot {\overrightarrow{v}}_1$

Substituting ${\overrightarrow{v}}_2$ from this equation into the kinematics equation and solving for ${\overrightarrow{v}}_1$ will give:​
${\overrightarrow{v}}_1=\sqrt{\frac{-2{\overrightarrow{a}}_{x}L}{1-{\left(\frac{r_1}{r_2}\right)}^4}}$

We have all the given values, so let’s solve our problem.

radius. Since $A_{circle}=\pi r^2$ :

$\pi {\left(\frac{d_1}{2}\right)}^2{\overrightarrow{v}}_1=\pi {\left(\frac{d_2}{2}\right)}^2{\overrightarrow{v}}_2$

And simplifying it even further will give us:

$\pi {\left(r_1\right)}^2{\overrightarrow{v}}_1=\pi {\left(r_2\right)}^2{\overrightarrow{v}}_2$

Step 3: Use of kinematics

Since this fluid is accelerating over a given length in a given direction ( $x$ -direction) with an initial and final velocity, we can use kinematics to help us solve our problem. Remember your kinematics Equation 3.2.1.​
Where ${\overrightarrow{v}}_x$ would be the final velocity ${\overrightarrow{v}}_2$, ${\overrightarrow{v}}_0$ is the initial velocity ${\overrightarrow{v}}_1$ and $x-x_0$ is the total length that the fluid travels across, which is $L$.

${\left({\overrightarrow{v}}_2\right)}^2={\left({\overrightarrow{v}}_1\right)}^2+2\cdot {\overrightarrow{a}}_x\cdot L$​

v₁ = √[2(3 m/s²)(7 m)/((4 m/2 m)⁴ − 1)] = √(42/15) m/s.

v₁ = 1.673 m/s.

As a check, the outlet must be faster because its cross-sectional area is smaller.

Since d₁/d₂ = 2, the area ratio is A₁/A₂ = 4. Continuity gives v₂ = 4v₁, not 2v₁.

v₂ = 6.693 m/s. Check: v₂² − v₁² = 42.0 m²/s² = 2aL.

Step 1: Set up your Bernoulli Equation

It should make sense that we are using Bernoulli’s equation because this is a fluid with varying heights, pressures and velocities. Let’s use Bernoulli’s Equation:

$P_1+\frac{1}{2}\rho {\left({\overrightarrow{v}}_1\right)}^2+\rho g{H}_1=P_2+\frac{1}{2}\rho {\left({\overrightarrow{v}}_2\right)}^2+\rho g{H}_2$​
Here the left side of the equation represents the top of the tank, while the right hand side represents the bottom of the tank where the hole is.

Step 2: Realize that the pressures are equal

How is that? Well in the problem statement, it was specified that the water was exposed to the open air, meaning that the pressure at the top is atmospheric pressure. The water coming out from the hole is also exposed to atmospheric pressure, therefore the pressures at both

points are the same. To make our lives easier let’s make the claim:

$P_1=P_2=P$​
Meaning that we can simplify the Bernoulli equation to just:

$P+\frac{1}{2}\rho {\left({\overrightarrow{v}}_1\right)}^2+\rho g{H}_1=P+\frac{1}{2}\rho {\left({\overrightarrow{v}}_2\right)}^2+\rho g{H}_2$

Step 3: Eliminate variables

We can clean this equation up even more to make it simpler. Since the pressures are equal, those will cancel out. Also since we set our datum to be at where the hole is, the height at that point is $0$. So we can now write our equation as:

$\frac{1}{2}\rho {\left({\overrightarrow{v}}_1\right)}^2+\rho g{H}_1=\frac{1}{2}\rho {\left({\overrightarrow{v}}_2\right)}^2+\rho g\left(0\right)$

continuity to help you estimate certain values and make educated guesses.

Solving for ${\overrightarrow{v}}_2$ gives us:

${\overrightarrow{v}}_2=\frac{A_1}{A_2}{\overrightarrow{v}}_1$

The initial cross-sectional area of the vessel at

Which can be cleaned up to make it look like:

$\rho {\overrightarrow{v}}_1^2+2\rho g{H}_1=\rho {\overrightarrow{v}}_2^2$

And by canceling out the densities, we get:

${\overrightarrow{v}}_1^2+2g{H}_1={\overrightarrow{v}}_2^2$

Step 4: Solve for the final velocity ${\overrightarrow{v}}_2$

We can solve for ${\overrightarrow{v}}_2$ to get:

${\overrightarrow{v}}_2=\sqrt{{\overrightarrow{v}}_1^2+2g{H}_1}$

So this is our final equation.......SIKE. We can actually simplify this even further. How?

Step 5: Use concept of continuity to help simplify

What do we mean ? Use the equation of

the top of the tank is much larger than the hole that’s created at the bottom. Meaning that:

$A_1>A_2$

If the area at the top is much larger with respect to the area of the hole, the final velocity will be much larger than the initial velocity. So if $A_1>A_2$ holds to be true, then we can make another statement:

${\overrightarrow{v}}_2>{\overrightarrow{v}}_1$

Neglect the free-surface speed only if the tank surface area is much larger than the opening area, A₁ ≫ A₂. Then v₁/v₂ = A₂/A₁ ≪ 1, and the outlet speed is approximately √(2gH₁).

$\stackrel{\phantom{.}}{{\overrightarrow{v}}_2=\sqrt{2g{H}_1}}$

Section Breakdown

1. Equation of Continuity
2. Example 8.2.1-Water Flowing through a Pipe
3. Bernoulli’s Equation
4. Example 8.2.2 - Leaking Vessel