Area ratio using Bernoulli's equation

Question

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.

Answer 1

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!

Answer 2

Adding to alephzero,

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.

Answer 3

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

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