# How long a straw could Superman use?

To suck water through a straw, you create a partial vacuum in your lungs. Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure. This corresponds to drinking water through a straw about ten meters long at maximum.

By taping several straws together, a friend and I drank through a $3.07m$ straw. I think we may have had some leaking preventing us going higher. Also, we were about to empty the red cup into the straw completely. My question is about what would happen if Superman were to drink through a straw by creating a complete vacuum in the straw. The water would rise to ten meters in the steady state, but if he created the vacuum suddenly, would the water's inertia carry it higher? What would the motion of water up the straw be? What is the highest height he could drink from?

Ignore thermodynamic effects like evaporation and assume the straw is stationary relative to the water and that there is no friction.

• The "Fluid Dynamics" tag makes this question more intimidating Dec 21 '10 at 8:37
• +1 for the picture, cause we all know: pics or liar! ;p Dec 21 '10 at 20:55
• Stupid nit-picky comment - but you actually make the vacuum with your mouth, not your lungs. Do an experiment to prove it... start by breathing in and then while still breathing in place the straw in your mouth and close your lips. See if anything different happens than normal. Also have the phone handy to dial 911. Dec 22 '10 at 6:12
• @John Berryman: It depends how much of a vacuum you want to create. The mouth can create so much of a pressure differential, which is usually enough, but the lungs need to be used to create a larger one. Jan 19 '11 at 20:45
• @MarkEichenlaub I may be late but Hagen Poisseulle? Oct 22 '12 at 18:46

I think we can most easily consider the problem from the perspective of energy. For a unit area column the external energy put in equals the volume of the column times the air pressure. The gravitational energy is the mass of water raised times the average height of the water. So as we pull the water up, we the water is gaining kinetic energy until the water reaches the static limit (roughly 10meters), but at this point the average water in the tube has only risen by half that amount, so the rest is kinetic energy of (upward) water motion. So the vacuum energy in dimensionless units is $h$, while the gravitational energy is $\frac{1}{2}h^2$. The solution is $h=2$. When we reach 2 times the static limit (20 meters) then the gravitational energy in the water matches the "vacuum" energy we put in, so that would represent the high point of the oscillation. So I think we would get 2 times the static limit. The water velocity will be messy to solve for, as the amount of water moving in the column depends upon height, so just look only at net energy as a function of water height... Of course he will only get a sip of water, then the column would start to fall........

• @Omega lol - 17 hours after the question is asked we posted the same solution within two minutes Dec 21 '10 at 22:24
• @Martin: what? How is discussing energy balance philosophy? And what use is it doing the long-winded calculations as you did if there is mistake in your approach in the first place? I wish I could give -1 to your comment... Dec 22 '10 at 8:01
• @Marek: So where is a mistake in my solution? Dec 22 '10 at 8:17
• @Martin: I am not saying there necessarily is one. Just that there is a possibility. I argued against your stance of doing the math first. Thinking and physics comes first, math only second. By the way, are you able to find a flaw in Omega's and Mark's solutions? I am not, because they are conceptually very simple. But I am not so sure about your solution because it deals with gory details and it's not clear that you didn't forget to account for something (like the Bernoulli equation Mark has mentioned). Dec 22 '10 at 8:24
• @Martin: call it Euler equation if it helps you. By making the flow unsteady you are not getting rid of the phenomena found in Bernoulli. Just introducing other, more complicated ones. Dec 22 '10 at 8:46

I have an argument that the water in the straw will rise to twice the equilibrium height.

David and Martin's answers consider the system of water in the straw. I will consider the system of the water in the straw plus the water in the reservoir.

As water goes into the straw, the water level in the reservoir drops, and the atmosphere does work on the system. If a volume $V$ of water enters the straw, the work done on the system is $PV$, with $P$ the atmospheric pressure. Assume that the reservoir has a large surface area so that the level the reservoir drops is negligible.

When the water is at its peak in the straw, the kinetic energy of the system is zero, so the potential energy is $PV$. The potential energy is also $\rho g V h/2$. So the maximum height of the water is

$$h = \frac{2P}{\rho g}$$

This answer is different from Martin and David's. I think this might be because when the water starts moving, the pressure at the entrance to the straw may not be $P$ any more.

• Heh, nice argument. Hard to find a flaw in this. Dec 21 '10 at 22:32
• @Mark: If you are right then the pressure at the entrance to the straw must be greater than $P_0$. How do you explain this? Dec 22 '10 at 8:06
• @Mark: If we follow Bernoulli's equation then the pressure at the entrance to the straw must be even lower than $P_0$. Dec 22 '10 at 10:12
• @Mark: True! Marek would have had to write this comment. Dec 22 '10 at 11:08
• I think this makes perfect sense. Also, if you imagine an oscillating system without friction, oscillating around the stable height, and starting from h=0, then the maximum height also turns out to be twice the stable height. In reality I think this oscillation is damped and so the water won't reach the full height and will very quickly settle to the stable height. Dec 22 '10 at 20:56

I went back and took a more careful look at this. I'm still not convinced it's correct, but I'm hoping this is at least better than what I had before.

Let $h$ be the height of the column of water inside the straw. As this height rises by an amount $\delta h$, the work done on the column is

$$\delta W = P A \delta h$$

where $A$ is the cross-sectional area. The change in potential energy is

$$\delta U = \rho A \delta h g h$$

so by conservation of energy,

$$\delta K = \delta W - \delta U = \left(P - \rho g h\right) A \delta h$$

This excess kinetic energy comes from two contributions: the added mass,

$$\delta K_1 = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A \delta h)\dot{h}^2$$

and any change in speed of the column of water,

$$\delta K_2 = \frac{1}{2}m(2v\delta v) = (\rho A h)\dot{h} \delta\dot{h}$$

Putting it all together, we get

$$P A \delta h - \rho A g h \delta h = \frac{1}{2}\rho A \dot{h}^2 \delta h + \rho A h\dot{h}\delta\dot{h}$$

If you assume (or prove) that $P$ is dependent on $h$ through Bernoulli's theorem,

$$P + \frac{1}{2}\rho\dot{h}^2 = P_0$$

Substituting in (and canceling the common factor of $A$), you get

$$P_0 \delta h - \rho g h \delta h = \rho \dot{h}^2 \delta h + \rho h \dot{h}\delta\dot{h}$$

which at least accounts for the mysterious factor of $\frac{1}{2}$ that appeared in previous versions of my answer.

Now, I don't think we can simply assume that $\delta h \neq 0$ and divide it out, because if we do that, we get a factor of $\frac{\delta\dot{h}}{\delta h}$ which is undefined at the initial and maximum heights. (Roughly speaking, the variation $\delta h$ is second-order at those points whereas the variation $\delta\dot{h}$ is still first-order.) Instead, I'll divide by $\delta t$, which certainly should not produce any singularities, to get

$$P_0 \dot{h} - \rho g h \dot{h} = \rho \dot{h}^3 + \rho h \dot{h}\ddot{h}$$

At the initial and maximum heights, $\dot{h} = 0$, so the equation is trivially satisfied there. But consider the situation when displaced from either initial or maximum height by an arbitrarily small amount, such that $\dot{h}\neq 0$. Here we can cancel out $\dot{h}$ to get

$$P_0 - \rho g h = \rho \dot{h}^2 + \rho h \ddot{h}$$

Since $\dot{h}$ will be infinitesimally small around the maximum height, we can neglect the first term on the right, but not the second. So we're left with

$$P_0 - \rho g h = \rho h \ddot{h}$$

Note that this agrees with a simple analysis using Newton's second law. (The forces acting on the column of water at its maximum height are the pressure force $P_0 A$ acting upwards and gravity $\rho Ahg$ acting downwards, and the difference is equal to $ma = \rho Ah\ddot{h}$.) So the differential equation passes at least one basic consistency test.

Anyway, this equation no longer admits the solution $h = \frac{P_0}{\rho g}$. Instead we have

$$h = \frac{P_0}{\rho (g + \ddot{h})}$$

Unfortunately I can't think of a way to determine $\ddot{h}$ at maximum without solving the equation, so for now I'm limited to a numerical solution.

For a quick estimation, I plugged the full differential equation from above into Mathematica's NDSolve function. With boundary conditions $h(0) = 0$ and $\dot{h}(0) = 0$, it complained about undefined expressions, so I used boundary conditions at a nonzero time,

$$h(\epsilon_t) = \epsilon_h$$

and

$$\dot{h}(\epsilon_t) = \epsilon_{\dot{h}}$$

for values of the various $\epsilon$ constants ranging from $10^{-3}$ to $10^{-8}$. In my tests, I get this graph, seemingly independently of the values of $\{\epsilon\}$ or the ratios between them: Mathematica indicates that the graph peaks at $15.5\,\mathrm{m}$, so if this analysis is correct, that would be the maximum height. (FWIW I am still very suspicious of this calculation though)

• I can't dispute your argument, but it seems to me (and is implied in the OP) that the maximum height will be higher than the balance height you found. Dec 21 '10 at 21:10
• @David: why would $\delta h$ need to be zero at the maximum? For me it's just a testing parameter that helps you determine the equation. In the maximum it amounts to putting the system a little out of equilibrium, so it's a principle of virtual work, right? As for that $1 \over 2$ factor, it's also bugging me. I can't understand why yours and Martin's derivations differ :-) Dec 21 '10 at 22:00
• @Sklivvz: yeah, that's what I thought too. I was kind of surprised to see $P_0/\rho g$ pop out of the equation at the end (and that's part of the reason I'm a little suspicious of this). Dec 22 '10 at 2:24
• @Marek: I guess that works. When I first wrote this up I used $\delta t$ instead, and I had stuck in my head the fact that, at the maximum, $\delta h \equiv \dot{h}\delta t = 0$ to first order. Dec 22 '10 at 2:27
• @David:There is a fundamental error in your analysis:Bernoulli's equation is applicable only to the steady flow of a fluid. Dec 22 '10 at 8:46

If I follow up on keenan pepper's suggestion, if the water is deep, and especially if you can mess with the topology of the straw you can go to almost unlimited height!

Consider a straw that is stuck very deeply into the ocean. Then coil the straw around at great depth many many times. I then blow very hard (I am superman afterall), and create a huge volume of airfilled straw at great depth. This configuration has a great deal of potential energy, so if we simply stop blowing we have the pressure at the great depth of the bottom of the straw accelerating water into and up the straw. Since by coiling the straw at great depth I can obtain an unlimited ratio of volume of the straw underwater to volume above water, the energy analysis allows me to reach an unlimited height. So the issue becomes if there is some other sort of limit. Can we get cavitation of water trying to enter the straw or something if the velocity gets too high? But, in any case you should be able to get really high, tens or hundreds of times the static limit, by preenergizing the system in this way. In the real world friction will limit how far you can take it.

• -1) Show this mathematically! Dec 22 '10 at 7:02
• @Martin I think it is sufficiently clear from energy considerations that what Omega said makes sense. Of course, you'd be very welcome to submit a more mathematical analysis of the same ideas if you want. Dec 22 '10 at 9:27
• @Mark:This must be done by Omega. Without any math the whole discussion goes dead. Why not add at least some back-of-the-envelope calculations to confirm their claims. This is an excellent quantitative problem. Dec 22 '10 at 10:34
• The math request seems unneeded. The energy stored by depressing the water level is simply the integral of d volume times depth under the surface. The energy needed above water is the same integral above the water level. Equate the two. If you can make the first integral grow unboundedly then it is solved. Dec 22 '10 at 14:25

Trick question, he'd use his super strength to bend the straw into an Archimedes' screw, then hold it at an angle to the surface of the water and rotate it about the axis. This lets him draw it up to any height, and then he can drain the world's oceans to prove a point or do whatever other superdickery he's trying to do.

I am not fully convinced by this argument, but can't find a flaw in it.

Let's analyse a similar experiment, which I believe to be equivalent. Assume that initially, the water is already at the stable level $H=\frac{P_{atm}}{\rho g}$. The vacuum is already present.

Now, by some unimportant means, we lower the water to $h=0$, and then let it go up freely.

How much energy are we storing in the system by lowering the water? We can find out by calculating the work done. The work is done against pressure and in favour of gravity.

$$W= sP_{atm}H + \int^0_H{m(h)g\ \mathrm{d}h}$$

Substituting $H=\frac{P_{atm}}{\rho g}$, $m(h)=\rho s h$

$$W= \frac{sP_{atm}^2}{\rho g} + s\rho g\int^0_H{h\ \mathrm{d}h}$$

$$W= \frac{sP_{atm}^2}{\rho g} - \frac{1}{2}s\rho gH^2$$

$$W= \frac{sP_{atm}^2}{2\rho g}$$

Now when the water is released and allowed to rise, all this energy will used to make the water rise. No energy is assumed to be wasted on friction.

At the highest point, all the energy will be converted in gravitational potential energy. This can be expressed through the following formula:

$$U(h)=\frac{1}{2}s\rho gh^2$$

Therefore, at the top point, $W=U(h)$

$$\frac{sP_{atm}^2}{2\rho g} = \frac{1}{2}s\rho gh^2$$

Solving for $h$:

$$h^2=\frac{P_{atm}^2}{g^2\rho^2}$$

$$h = \frac{P_{atm}}{g\rho} = H$$

Therefore, the water will raise up to $H$, which it will reach with zero velocity.

• @Sklivvz Your expression for the work done is correct, but the total energy is the work done plus the original potential energy. When this is accounted for, your method shows that the water rises to twice its original height. Dec 23 '10 at 21:04
• @Mark Eichenlaub: I don't understand - the original potential energy is the second term in the RHS of the first equation. Do I need to add it twice? Dec 24 '10 at 9:40
• @Sklivvz The system starts out with some energy. Call it $E_0$. Next, you add energy to the system. That's the work you calculated, $W$. Now the total energy of the system is $E_0 + W$. Finally, you want to turn all the energy into potential energy at the highest point. Thus, the potential energy is $U(h) = E_0 + W$, not $U(h) = W$ as you have it now. As written, your calculation says that you start with the water at equilibrium height, then you add energy and that energy doesn't dissipate, and nonetheless it goes back to equilibrium height. If so, where is all the energy you added? Dec 24 '10 at 9:47
• @Mark Eichenlaub: My line of thought is actually a bit different. Initially the system has potential energy $E_0$. To lower the level we do some work, but less than expected because we use the potential energy as well - the water goes down, the potential energy goes down. So when the water is at the bottom level, there is no potential energy left. This makes sense to me, because if we actually started in that position, where would the potential energy come from? Dec 24 '10 at 9:54
• @Mark, I think I am getting it now... :-) I basically need to redo the answer from scratch :-/ Dec 27 '10 at 14:08

I can not comment yet, so I will put it as an answer :-( I like the energy approach, but why not to use Archimedes principle? First, substitute the air atmosphere by an extra ten meters of water around the straw, so that now the initial conditions are a vacuum straw inserted a lenght h in a fluid. The energy to produce such vacuum in the fluid you can see by Archimedes; and it is h/2 times g times the mass of the removed fluid. Let it move, and it can go up until filling a column of lenght h above the level, because the energy (now gravitational) of this column is, again, h/2 times its mass times g.

• Actually, should it be possible to answer without using a reference to the energy? Old pal Archimedes had not this concept. The equivalent problem is, I have a closed barrel half filled of water, floating. I push it until it is exactly covered, then I release it. Can I prove, using old knowledge, that it jumps until exactly out of the water? Jan 16 '11 at 1:00
• You have some interesting ideas, but it very difficult to parse your writing. I am unsure exactly what you mean by Archimedes' principle, and in general I am only partially confident I understand what points you're trying to make. Jan 16 '11 at 1:20
• Archimedes principle is that a body sumerged in a fluid suffers a force equal to its volume times the density of the fluid. The principle of equality between work and variation of energy implies that the energy to move down an empty body (say a crystal sphere with vacuum inside) in a fluid to a depth h is force times h. In this way you can calculate the energy needed to do a vacuum hole drilled in water, without using explicitly the concept of pressure. Jan 16 '11 at 3:06
• @Mark, glad to serve. I hope also that the edits will do it more readable. Of course, it is still to be proved that hydrodinamically all the stored energy can be used to move the column up. If it can not, the argument gives only an upper bound. Jan 16 '11 at 3:41

Superman is going to manipulate the air pressure in the straw. To get the water to go up, he must provide a reduction in the pressure. It's clear that if this reduction is provided at a very slow rate, then he will not be able to significantly exceed 10 meters or so (as limited by atmospheric pressure).

On the other hand, if he reduces the pressure quickly, it's at least possible that the water could reach a higher height. How high can the water go under this assumption? The idea is to use the momentum of the water to get the water higher, so the figure of merit will be the maximum speed of the water at surface level.

Reducing the pressure cannot move the water faster than the air and the air speed is limited by the speed of the gas molecules in the air or about 330 meters per second. By equating kinetic energy $0.5 m v^2$ with potential energy $mgh$, water with that initial speed can reach a height of $h = 0.5 v^2/g =$ 2775 meters. The height is small enough to justify the assumption that $g$ is a constant. Maybe you should add 10 meters for the usual vacuum effect.

Hmmm. Ah, what the heck, I ought to just do the fracking calculation for how high the water goes in a wide straw when a vacuum is applied to it.

• I'm not sure I understand. How is he getting the water up to this speed? Jan 17 '11 at 21:33
• Er, is your Superman blowing into the ocean surface to increase atmospheric pressure? Or are you simply planning some method to store energy in the straw using forced oscilations? Jan 18 '11 at 1:10
• Mark, Alejandro: Okay, I'll edit the answer to give an idea why this sort of calculation comes to mind. Jan 19 '11 at 4:44
• I see your point, but the water needs to be accelerated by the pressure from the water below it, so I guess that's a good upper bound, but I don't think the water picks up nearly the velocity you mentioned. Jan 19 '11 at 5:12
• @Mark @Carl I could see the point if he uses a really deep straw say 10 or 11 kilometers. First he blows air inside, up to one thousand atmospheres, then he releases, no need to suck at all... Jan 19 '11 at 17:52

If we are looking at this from a purely suction related problem then superman the maximum height sumperman could lift water in a straw would be equal to the pressure being exerted on the water he is drinking. If drinking from sea level then he could lift or suck the water about 10 m. Theoretically he could create a complete vacuum in his mouth then the amount of lift is merely the differential pressures. We run into this limit all the time with vacuum pumps. However, if he is able to suck really quickly then the velocity of the air could allow for water entrainment beyond the maximum lift. He could not suck the water as one big slug but rather as droplets carried up the straw due to the wind velocity within the straw. In a more practicle case, vacuum pumps have been shown to lift water from over a couple of hundred feet above static water level using this method. But the rate of recharge within the well must be suffuciently low to ensure that the water does not "clog" your suction pipe. To put another way you need to have far more air than water going up your straw.

Unless I'm missing something, this is simply the height of a water-based barometer, since it is really the atmospheric pressure that is pushing the water up the straw. at STP, the answer is 33 1/2 feet. If he were sucking Hg up the straw (not recommended for non-superheros!), the height would be ~30 inches.

• You are missing something. Everybody knows the height of water in a water barometer is about 10 meters. It even says so in the question. The question is what would happen if Superman created a vacuum really suddenly, or perhaps alternately sucked and blew. Dec 21 '10 at 20:29

If you immerse the staw in 10m of water, while holding the end closed, and then release it, the water will accelerate upwards past the level point and it will overshoot up to a height of about 6m. No sucking needed. Add some suction to this and you can go higher. A square section straw should minimize the friction loss allowing for better "spring back". Overall you can help the suction, by immersing the straw deeper and deeper.

Note that it takes work do immerse the straw (displacing the water) and that is the energy conveted into ponetial energy that allows the water to rise. Close the end when the water reaches the maximum height and you can measure how high you can reach.

• -1) Show the math! Dec 22 '10 at 7:43
• @Martin There's nothing innately wrong with a qualitative answer. In this case, jalexiou was trying to point out a way of thinking about the problem that differed from the way I described in the original question. The point of the answer was this new physical insight, and so especially for this circumstance a detailed calculation is unnecessary. (I didn't upvote, though, because the answer restates one posted a few hours earlier.) Dec 22 '10 at 9:31
• @Mark:The problem is that jalexiou gives a quantitative argument: "the water will accelerate upwards past the level point and it will overshoot up to a height of about 6m" without any calculation. Dec 22 '10 at 10:56
• @Martin Well, I agree with you there. Dec 22 '10 at 11:07
• @Mark, I am an engineer, and such I am allowed to guestimate results based on experience. Yes, in reality the bounce back will not be 6.0000m but somewhere near there. I estimate the damping ratio of a typical flow through pipe as between 0.1 and 0.6 and from my exprience with 2nd order damped systems the overshoot (bounce) is about 60% of the initial excitation. If I show the math will you remove the (-1) ? Addionally, I have perfomed this experiment various times with actual drinking staws and water and I stand by my guestimate. Dec 23 '10 at 16:21

You can manipulate the vacuum suction limitation of maximum height of 33.9 feet or 10 meter (14.7 psia or 0.1 Mpa) by using oscillating blow and suct. Use longer straw submerged deep enough into the water, blow it until the air almost reach the bottom of the straw then suck it! You will get height boost!!

• Welcome to physics SE! Physics lives through discussion and coherent reason. -1 At least mention a physical law (Newton's $F=m\cdot a$). This height boost already is mentioned as intertia in OP's question. Apr 25 '13 at 19:54