Can I make the water boil simply by spinning it in a glass? Correct me if I'm wrong, I think the pressure in a fluid reduces when the speed increases(The airplane rises because the air above the airfoil moves faster than the air below it).

Next, looking at the air above the surface of the water which is in a spinning glass, I'm wondering if this relative motion decreases the pressure in the air just above the water surface... If yes, does the boiling point of the water reduce due to the loweing of this air pressure over the surface ? (A layer of air just above the water surface spins with the water, so there will be a relative motion between this layer of air and the air above it.) • With brisk enough stirring, one gets cavitation. It's not exactly boiling, but does create a bit of a vapor bubble. Mar 31 '17 at 9:25
• xkcd already did this one Mar 31 '17 at 9:44
• The wing in your example does not rise because air moves faster than the air below it. That image is extremely misleading — a wing only provides lift when it diverts air downward, and that image shows no change in the motion of the air. Now, in a real wing, the air on top will flow faster than the air underneath, but isn't what causes the wing to generate lift. Apr 1 '17 at 3:05

No, mostly

You mostly can't boil water by spinning the glass. "Mostly" because some weird stuff is possible under extreme conditions like in a rotary evaporator; in such cases, whether or not there's "boiling" starts to become an issue of semantics.

That explanation of aerodynamic lift is a common misconception

First, to correct a misconception:

Correct me if I'm wrong, I think the pressure in a fluid reduces when the speed increases(The airplane rises because the air above the airfoil moves faster than the air below it).

This statement is a common misconception supported even by some authoritative sources. However the logic behind it presumes that equilibrium behavior applies in dynamic scenarios which doesn't hold; more in the comments below.

"What really allows airplanes to fly?" has some excellent discussion on airfoils.

Spinning water in a glass can decrease pressure

If you spin a glass of water fast enough, you can get a vortex going in the center. This vortex is sorta like a tornado, with lower pressure in the center and higher pressure at the boundaries, against the glass.

In principle, if you do this fast enough, you could lower the inner pressure down to the static boiling point. Checking a phase diagram for water, it looks like water can boil at room temperature if we drop the pressure down to a just few percent of normal atmospheric pressure.

Would additional evaporation be due to "boiling"?

In a classical sense, boiling is when a material in the liquid state reaches the point where it'll start turning into its vapor state with appreciable stability throughout its volume. Non-boiling liquids can still turn into vapor, but they usually do so at their boundaries, in which case we call it "evaporation" instead of "boiling".

The distinction between rapid evaporation and boiling kinda breaks down under extreme conditions since the classical sense in which we defined those terms no longer applies, but I think that most people would find the term "boiling" to be misleading in this case.

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
Apr 1 '17 at 2:26

No, you cannot make water boil simply by spinning it in a glass. It's impossible. The pressure inside the water will not decrease, but increase. Even at the water surface.

The situation

Assume there is a glass cylinder, which is closed at the bottom and initially open at the top. It is partially filled with water and spinning at a very high velocity (a very, very, very high velocity if you like). The water is dragged along with the glass cylinder and at some point spins with the same velocity as the glass, generating a vortex.

At some velocity (depending on the length of the glass and the amount of water), the water will start to spill out on top of the glass. To prevent this, we place either a ring on top, leaving a hole in the center, or we close the glass completely. This does not make a big difference.

There are two phases involved: Water and air. Let's first look at ...

The water.

The pressure distribution inside the water can be calculated using e.g. the idea of potential energy. There effectively are two pressure gradients which are superposed:

1. the usual hydrostatic pressure gradient due to Earth's gravity from top to bottom (roughly 1 bar per 10 m),

2. a pressure gradient due to the rotation from the center line outwards.

The following formula can be derived for the pressure $p$ inside the water: $$p = p_0 - \rho g (z-z_0) + \tfrac 12 \rho \omega^2 r^2$$ ($z$ vertical coordinate pointing upward, $r$ radial coordinate pointing outward, $\rho$ fluid density, $g$ gravity of Earth, $\omega$ angular velocity, $p_0=p(z=z_0,r=0)$).

Note that the assumptions that result in this pressure distribution are only a spinning tube in Earth's gravity field, filled with incompressible fluid. No assumption has been made yet on the position of the boundary between water and air.

This means that the pressure is increasing with increasing $r$. At the boundary between water and air, the pressure is lowest, while at the inner wall of the cylindrical part of the glass, the pressure is highest. Due to Earth's gravity, the highest pressure is reached at the bottom of the glass.

Important statement: The pressure inside the water is increasing when moving from the inside (surface) to the outside (glass).

Now let's look at ...

The air.

For the air that spins inside the cylinder, the same principle holds true as for the water. Since air is compressible, the above equation cannot be used directly for air, but a similar statement can be made:

In the air, too, the pressure increases when we move from the inside (rotation axis) to the outside (water surface).

This by the way also holds true for tornados: The outside pressure is greater than the inside pressure. For tornados, the outside pressure is set by the surrounding atmospheric pressure.

One big difference is of course the density: $\rho_{\mathrm{water}} \approx 1000 \mathrm{kg/m^3}$, $\rho_{\mathrm{air}} \approx 1.2 \mathrm{kg/m^3}$. This results in a much smaller pressure gradient inside the air when compared to the pressure gradient in the water. But for very, very high velocities, the pressure gradient in air can become non-negligible.

Now we need ...

The Euler–Cauchy stress principle.

This is one of the fundamental assumption of continuum mechanics. It states that the stress vector is continuous. For the water and the air in our glass this means that the pressure at the water-air boundary is equal inside the water and the air. Put in other words: The pressure between water and air is continuous and does not jump.

Now we finally need the ...

Reference pressure.

This is of course der springende Punkt, as we say in German ("the jumping point" - the important point).

If the tube is partially open, the pressure will equalize at the opening. Since the tube is spinning really, really fast, there is a pressure gradient inside the air as well. This will result in air flowing into the tube at the rotation axis, and air flowing out of the tube at the outer rim of the opening. So we get a secondary flow here. This secondary flow leads to a decreased pressure through the Bernoulli equation (see below), but the pressure at the opening cannot drop below atmospheric pressure, because otherwise the secondary flow would stop.

For a stationary rotation, this secondary flow is stationary. Because the air pressure near the rotation axis must be smaller than the air pressure at the water surface, air must flow in near the rotation axis and exit the tube near the outer rim of the opening (assuming we do not consume or generate matter inside the tube).

Therefore, the pressure at the outer part of the opening must be above atmospheric pressure.

Since the pressure gradient from rotation axis to water surface extends throughout the whole length of the tube, the secondary flow also extends through the whole length of the tube. In order to keep going, the pressure at the water surface at the bottom of the tube must therefore be slightly above the pressure at the outer rim of the opening.

Therefore, the pressure at the entire water surface must be above atmospheric pressure.

The closed tube is easier. There is no secondary flow, just a pressure increase towards the water. Since the amount of air inside the tube has not changed, and the pressure at the water surface is above the pressure at the tube center, the pressure at the water surface must be above atmospheric pressure. The pressure at the tube center is below atmospheric pressure.

In total, all you can get is a pressure increase at the water-air boundary as @Peter.A.Schneider pointed out in his comment. This is independent on how fast the glass spins.

For an open glass, the secondary flow will by the way result in water being evaporated, which will lead to a cooling of the water. So instead of boiling it is more likely that the water will freeze.

Bernoulli equation

Finally, the original posts states:

I think the pressure in a fluid reduces when the speed increases

A pressure decrease in a moving fluid is relative to the fluid at rest. You are referring to the momentum conservation equation for an inviscid fluids, aka Bernoulli equation: $$\tfrac12 (v_2^2-v_1^2) + \frac 1 \rho (p_2-p_1)=0$$ (here without the gravitation term). This is valid along a streamline only. It is not a general principle that the pressure decreases in fast moving fluids.

In the spinning glass, the streamlines are circles concentric to the spin axis. Along these streamlines, neither velocity nor pressure changes. In short, using Bernoulli's equation on your spinning glass does not tell you anything on the pressure in the water.

• I think that people are looking at this question with too many simplifying assumptions. For example, consider an extreme case in which a cup is continuously spun until it creates a vortex in the local environment and generates a literal tornado around it. Then pressure drop, right? Not that someone's likely to do that, just "impossible"'s going too far. That said, I agree that it'd take a lot of contriving to get there. =P Anyway, there're a few more routes toward getting pressure drops; it's definitely possible.
– Nat
Mar 31 '17 at 14:13
• @Nat, I extended my answer. Yes, the tornado can be generated. No, this can never lead to a pressure decrease at the surface of the water. It can lead to a pressure decrease in the center of the tornado, but there is no water there to boil. Mar 31 '17 at 18:32
• @Nat No, the pressure would not drop. Why would it? In weather, the updraft creates a local pressure drop. There is no updraft because air is spinning circularly in a glass, I think. Mar 31 '17 at 19:34
• Actually, if you have a concave wall and stir really fast, wouldn't the water heat up from compression? I know it's not very compressible, but it's not completely incompressible either. Mar 31 '17 at 19:35
• Hm... looking at the phase diagram one can see that stirring fast enough always creates ice, which I think is the opposite of what the OP wanted ;-). Mar 31 '17 at 19:44

You are trying to use the Bernoulli Principle were it is not applicable! If we temporarily ignore the radial force created by the spinning, we would have the water spinning at some angular velocity $w$. Because of drag, the air above it would also spin. However, the fastest angular velocity that the air could attain, would be the same as the water. For the Bernoulli Principle to apply, the air would have to spin faster than the water, which can't happen - under the given conditions!

If a liquid is starting to boil I'd usually apply that to the whole liquid not just a part of it.

If you just make a part of it evaporate due to extremely low pressure in one part that does not make the whole liquid boil.

• irrelevant and nonstandard Mar 31 '17 at 12:02
• It does make sense to me. Why is it not relevant @CarlWitthoft Mar 31 '17 at 14:06

Here is a link saturated vapour pressure

There is plot in this link and I attached here. So if we want the water to boil, we need bring either pressure or temperature to the curve. I think the above discussion is assuming the temperature is room temperature e.g. 20-25 degrees C. So the focus is how to bring the pressure down by fast stirring. This is also the reasoning in the question. I think making your stirring spoon having a shape of airfoil can reduce local pressure down to the vapor pressure if you spin fast. You will get local evaporation (cold though).

It could be thought in another way, fast stirring can increase the temperature to 100 degrees and thus boil the water. The stirring does not to be necessary in one direction. The goal is to let the work dissipated to heat by viscosity.

• If you use an airfoil to stir the water, you can (at least in theory) create cavitation bubbles, that's true. However, OP asks: "does the boiling point of the water reduce due to the loweing of this air pressure over the surface". So he's not asking for boiling inside the water, but specifically for boiling at the water surface. And yes, dissipating energy into the water can obviously raise the temperature to the boiling point. This would probably take a while, though, and is no pressure effect. :) Mar 31 '17 at 20:24