# Area ratio using Bernoulli's equation

I am using Bernoulli's equation in order to calculate a relative area between two points in a pipe. Problem is, based on the solution I have obtained, I end up with imaginary numbers and not sure what to make of this practically. Here is my math:

$$\begin{gather} P_1 + \dfrac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2 \\ v_1 A_1 = v_2 A_2 \\ \dfrac{A_1}{A_2} = \sqrt{\dfrac{v_2^2}{v_2^2+\dfrac{2}{\rho}\left(P_2-P_1\right)}} \end{gather}$$

I know my velocity at location 2, density, and pressures. I have no change in height so I got rid of third element in Bernoulli's equation. I am trying to obtain an area ratio between both points.

It is very possible in my model that I will have a low P2 and high P1 leading to a negative denominator and an imaginary solution. I feel like I am missing something practical or mathematical here that is simple but my mind has drawn a blank.

As a side note, I am assuming non-compressible airflow. Cannot assume laminar flow.

Unless you are putting "randomly chosen numbers" into your equation, or you have measured the numbers but your assumptions about Bernoulli and compressibility don't apply to your real-world flow situation, you shouldn't have a problem.

Look at it this way: for given values of $P_2$ and $v_2$, physically $P_1$ must always be less than some maximum value - namely, the value when $v_1 = 0$.

That follows directly from your first equation (your statement of Bernoulli's principle).

That maximum value of $P_1$ that is physically possible will make the denominator of your square-root equal to zero, or in other words, $A_1$ will be "infinite" - which of course is correct, if $v_1$ is zero!

• Ahhh, so simple! You are right. I have yet to actually collect my v2 data so I've been doing mock calculations and running into my problem. So will see how the data actually shakes out, practically, you are totally correct.
– A M
Aug 13 '18 at 1:30

Imagine you are getting a negative denominator,

$v_{2}^2 + \frac{2}{\rho}(P_2 - P_1) < 0$,

$\Rightarrow \frac{1}{2}{\rho}v_{2}^2 +(P_2 - P_1) < 0$ ,

$\Rightarrow \frac{1}{2}{\rho}v_{2}^2 +P_2 < P_1$ ,

$\Rightarrow P_{02} < P_1$ , This violates the Bernoulli's principle. So you will never get an imaginary velocity value.

As alephzero clearly mentioned, $P_1 \rightarrow P_0$ when $v_1 \rightarrow 0$.

Note: $P_0$ is the total pressure and it will be constant along the streamline on wihch you apply the Bernoulli's equation.

As a side note, I am assuming non-compressible airflow. Cannot assume laminar flow.

The assumptions that $\dot{V}_1 = v_1 A_1 = v_2 A_2 = \dot{V}_2$ and that $\rho =$ constant are only valid for incompressible flow. Air is compressible, not incompressible. Use the ideal gas law with $\rho = M_{air}(p/RT)$ and $v = \dot{V}/A = \dot{n}RT/pA$. This gives

$$\dot{n}_1 = \dot{n}_2 = \dot{n}$$ $$\frac{RT_1}{M_{air}} + \frac{1}{2}\left(\frac{\dot{n}RT_1}{p_1 A_1}\right)^2 = \frac{RT_2}{M_{air}} + \frac{1}{2}\left(\frac{\dot{n}RT_2}{p_2 A_2}\right)^2$$

The proper analysis for a compressible (ideal) gas must consider pressure and temperature not pressure and velocity. Corrections for non-ideality may be applied depending on how drastically the gas is compressed and/or how accurate the answer really must be.

The ratio of areas as a function of velocities for ideal gas flow is expressed as

$$\frac{A_1}{A_2} = \frac{\dot{V}_1 v_2}{\dot{V}_2 v_1} = \frac{T_1 v_2 p_2}{T_2 v_1 p_1}$$