Tagged: Bernoulli, compressibility, fluiddynamics, incompressibility, massflow
 This topic has 0 replies, 1 voice, and was last updated 3 years ago by freixas.

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September 21, 2020 at 12:39 pm #258freixasKeymaster
Please read The Compressibility of Air before proceeding.
Mass Flow
The phrase “1 liter of air per second” describes what is called a volume flow rate. Because we treat air as incompressible, 1 liter of air will have a constant fluid density and thus a constant mass. 1 liter of air weighs about 1.2929 grams. Therefore, a flow of 1 liter per second is the same as a flow of 1.2929 grams per second. The later is called the mass flow rate (often shorted to mass flow). I use these terms interchangeably.
The Continuity Equation
Since air flowing through a melodica is incompressible, if I push 1 liter of air per second into a melodica, exactly 1 liter of air per second will come out (plus or minus a few molecules).
The size of the exit opening is irrelevant. Unless we compress the air, there is no other option but for the amount flowing in to match the amount flowing out.
Since this is true for any tube as a whole, it is equally true for any subset of the tube. That is, we could cut the tube at any point to establish a new exit and the rule would be true. The mass flow in would equal the mass flow out. In general terms, the mass flow though any crosssection of a tube is a constant. This is one of the most basic principles of fluid dynamics. From this, we can derive the continuity equation.
Let’s start with a tube with different crosssections:
The red blocks represent two equal volumes of air. The labels A_{1} and A_{2} point to two crosssectional areas within the tube.
For any unit of time t, the same volume of air will flow past A_{1} as A_{2}.$$\begin{align*} V_1 &= V_2\\
A_1 v_1 t &= A_2 v_2 t\\
A_1 v_1 &= A_2 v_2
\end{align*}$$where V stands for volume, A for area, v for velocity and t for time.
This is the continuity equation. It lets us calculate the velocity of the air flowing by one point relative to another. If we know A_{1}, v_{1} and A_{2}, we can calculate v_{2} as $$v_2 = \frac {A_1}{A_2} v_1$$For example, if A_{1} is three times larger than A_{2}, then v_{2} will be three times faster than v_{1}. The air will flow faster in the narrower section than in the wider one.
What this means for a melodica is that:
 Air flows at different velocities throughout the system, and
 The velocity of the air through any crosssection can be computed from the velocity of the air flowing into the melodica. The shape of anything in between doesn’t matter.
But…
What I should have said is that the continuity equation tells us the average velocity of the air through any crosssection. But it doesn’t tell us the exact velocity at any point through that crosssection.
Let’s look at this diagram:
The two red sections again represent equal volumes of air and we could use the continuity equation to tell us the average velocity of the air flowing through A_{2} if we know the velocity flowing through A_{1}. But the air molecules will not instantly fill the wider crosssection. Some molecules may travel as fast as at A_{1}, but molecules elsewhere will move much slower.
To exaggerate, consider this:
The gray sections represent air whose velocity is 0. The air moving in the white area between the gray sections would then move just as fast as the air at A_{1}, but the average velocity through crosssection A_{2} would be slower than through A_{1}. It’s unlikely this precise situation would happen—some molecules will spread out into the wider chamber—but it doesn’t violate the continuity equation.
The Bernoulli Equation
For the final part of this background post, I will briefly cover Bernoulli’s equation. In an open tube with a steady flow, the following values is a constant at every point in the system: $$P + \frac {1}{2}\rho v^2 + \rho g h$$
There are a lot of letter in this one, so let me list each one:
 P is pressure
 ρ is the fluid density (which is itself a constant)
 v is the velocity
 g is the gravitational constant
 h is the height
Let’s simplify this by assuming we have a horizontal tube so that we can factor out the effects of gravity. And let me rewrite it for two points along the tube. $$P_1 + \frac {1}{2}\rho v_1^2 = P_2 + \frac {1}{2}\rho v_2^2$$
The bottom line is this: If the velocity increases, the pressure drops. If the velocity drops, the pressure increases. In a melodica, air in the air chamber has the slowest velocity and the highest pressure.
This might seem odd, so think of it this way: the P in the equation is also called “static pressure”—think of it as the pressure that is exerted equally in all directions from molecules moving randomly. The $\frac {1}{2}\rho v^2$ term is called “dynamic pressure”—it come from the motion of molecules in a specific direction in the tube. As molecules gain more directed motion, they lose random motion and vice versa, but the total energy in the molecules remains the same.
Everything in this post applies to an ideal fluid, one that is completely incompressible and has no viscosity (friction), but we can apply these formulas when the the fluid we have behaves “close enough” to an ideal fluid that the differences aren’t significant.
 This topic was modified 3 years ago by freixas.

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