-1 At least mention a physical law (Newton's $F=m\cdot a$). The ambient pressure is therefore only responsible for pushing the water up from the surrounding water level. Why does the air we blow/exhale out from our mouths change from hot to cold depending on the size of the opening we make with our mouth? 
In fact, it is not the generation of negative pressure during suction that is the limiting factor, but the ambient pressure! The point of the answer was this new physical insight, and so especially for this circumstance a detailed calculation is unnecessary. Archimedes principle is that a body sumerged in a fluid suffers a force equal to its volume times the density of the fluid. Thus it is the atmospheric pressure that makes it possible for the water in the drinking straw to be pushed upwards. 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. ;p. Stupid nit-picky comment - but you actually make the vacuum with your mouth, not your lungs.
@Martin I think it is sufficiently clear from energy considerations that what Omega said makes sense. By the way, are you able to find a flaw in Omega's and Mark's solutions? Yes, in reality the bounce back will not be 6.0000m but somewhere near there.
reusable bent stainless steel drinking straws, Choosing the right length and width for your reusable straws. What is the highest height he could drink from? This depth hd corresponds to the immersion depth of the drinking straw. 
This excess kinetic energy comes from two contributions: the added mass, $$\delta K_1 = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A \delta h)\dot{h}^2$$. The force Fp with which the water can be effectively pushed upwards is the difference between the forces acting at the lower end (F0) and the upper end of the water column (F1). at STP, the answer is 33 1/2 feet. 
By creating a vacuum in your mouth, you create an imbalance: the pressure inside your mouth is lower than the air pressure pressing on the drink in your cup. The water column would simply stop at this height. 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. 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). @Sklivvz The system starts out with some energy. Our customers love these straws when making the perfect drink. I then blow very hard (I am superman afterall), and create a huge volume of airfilled straw at great depth. @Mark: If we follow Bernoulli's equation then the pressure at the entrance to the straw must be even lower than $P_0$.
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. No energy is assumed to be wasted on friction. Standard 8.5 long straws may not work for extra tall tumblers or popular retail beverage containers with extra-large beverage capacity, like the 30-oz YETI Rambler. This force Fp must obviously be so large that it is able to push the water column with the weight Fg=mg upwards. How come water in a cup won't shoot out of a straw placed inside it? @MarkEichenlaub I may be late but Hagen Poisseulle? So I think we would get 2 times the static limit. A square section straw should minimize the friction loss allowing for better "spring back".
In the state of equilibrium, the downwards acting weight Fg must be equal in magnitude to the upwards acting force Fp. However, if he is able to suck really quickly then the velocity of the air could allow for water entrainment beyond the maximum lift. 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. Explain Like I'm Five is the best forum and archive on the internet for layperson-friendly explanations. Either way, reusable smoothie straws work great for beverages like smoothies, powder shakes, super food blends, or yogurt. As for that $1 \over 2$ factor, it's also bugging me. So we're left with.
Finally, you want to turn all the energy into potential energy at the highest point. This short straw is available only in a .24 diameter. The problem is when you go to extremes - eventually the weight of the water excedes the ability of the atmospheric pressure to move it.
@Sklivvz: yeah, that's what I thought too. Or are you simply planning some method to store energy in the straw using forced oscilations? The energy stored by depressing the water level is simply the integral of d volume times depth under the surface. For a unit area column the external energy put in equals the volume of the column times the air pressure. What would the motion of water up the straw be? Close the end when the water reaches the maximum height and you can measure how high you can reach. For the maximum possible force with which the water can be pushed upwards, a vacuum must be created inside the straw (p1=0), so that the entire ambient pressure p0 can push the water upwards inside the straw without counterpressure. (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 $ma = \rho Ah\ddot{h}$.)
Whether your reusing stainless steel drinking straws at a restaurant or home setting cleaning, maintaining and reusing them is easy. Note that when the simulation is re-run with that $\frac12$ the answer corresponds to the energy conservation answers given of just over $20m$.
How do you explain this? There is also a clear explanation for this, because due to the hydrostatic pressure, the water in the open straw is pushed to a uniform level anyway. The lack of air pressure inside the straw means that the air pushing down on the surface of the liquid outside the straw forces it up the straw. All of our straws are easy to clean, dishwasher safe food grade and BPA free. We offer the standard 8.5 x .24 reusable straw in either straight or as a reusable bent stainless steel drinking straws profiles, either standard polished silver or in anodized colors that wont scratch, chip, peel or fade. Let's analyse a similar experiment, which I believe to be equivalent. For this purpose, the entire water column inside the straw, which is pushed upwards, is considered. You have some interesting ideas, but it very difficult to parse your writing. 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. Nah, you just wouldn't be strong enough to lift it. 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! Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure. 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. Our tips for cleaning reusable drinking straws were compiled from on-site visits and interviews with real Steelys customers who use our straws either at home or in commercial settings. And what use is it doing the long-winded calculations as you did if there is mistake in your approach in the first place? How to tell reviewers that I can't update my results. Overall you can help the suction, by immersing the straw deeper and deeper. Instead, there would be a near-vacuum in the upper part of the straw that could make the water boil at room temperature. Then coil the straw around at great depth many many times. 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. 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. Now when the water is released and allowed to rise, all this energy will used to make the water rise. This creates a lower counterforce so that the force of the external ambient pressure is able to push the water upwards. If you prefer a wider straw for thicker beverages, you should consider a smoothie straw. 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. Also have the phone handy to dial 911. How is discussing energy balance philosophy? The standard reusable straw size is 8.5 long x .24 wide. Our reusable travel straw sets give you reusable straw options so you can enjoy any drink in your favorite tumblers, no matter where you are so you can always live up to your commitment to cut single-use plastic waste. It only takes a minute to sign up. As the water level in the straw increases, the weight of the water column increases.
for values of the various $\epsilon$ constants ranging from $10^{-3}$ to $10^{-8}$. Are Banksy's 2018 Paris murals still visible in Paris and if so, where? Therefore, the water will raise up to $H$, which it will reach with zero velocity. For these products, we also offer extra long 9.5 reusable stainless steel straws in boxed sets, and we can upgrade any straw order to extra-long length upon request, just let us know what you need. The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, The "Fluid Dynamics" tag makes this question more intimidating, +1 for the picture, cause we all know: pics or liar! What makes the water flow up the straw, however, is not some kind of sucking force created by the vacuum, it is the ambient air pressure pressing down on the water and pushing it up the straw to equalize the vacuum. The standard length and width of single-use disposable straws that youll find at most bars and restaurants are 8.5 long. In the maximum it amounts to putting the system a little out of equilibrium, so it's a principle of virtual work, right?
If you assume (or prove) that $P$ is dependent on $h$ through Bernoulli's theorem, Substituting in (and canceling the common factor of $A$), you get, $$P_0 \delta h - \rho g h \delta h = \rho \dot{h}^2 \delta h + \rho h \dot{h}\delta\dot{h}$$. The only question you have is how do you choose the right length and width for your reusable straw. 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. In order to overcome any large differences in height, one would only have to suck on it powerfully enough. Even the most powerful vacuum pump in the world would not be able to pump water to such a height! Would it be possible to use Animate Objects as an energy source? $$W= sP_{atm}H + \int^0_H{m(h)g\ \mathrm{d}h}$$, Substituting $H=\frac{P_{atm}}{\rho g}$, $m(h)=\rho s h$, $$W= \frac{sP_{atm}^2}{\rho g} + s\rho g\int^0_H{h\ \mathrm{d}h}$$, $$W= \frac{sP_{atm}^2}{\rho g} - \frac{1}{2}s\rho gH^2$$. If you continue to use this website, we will assume your consent and we will only use personalized ads that may be of interest to you.
At an ambient pressure of 1 bar, a liquid density of 1000 kg/m and a gravitational acceleration of 10 N/kg, the maximum suction lift is about 10 Meters for water: \begin{align}&\underline{h_{max}} = \frac{p_0}{\rho \cdot g} \approx \frac{10^5 \frac{\text{N}}{\text{m}}}{1000 \frac{\text{kg}}{\text{m}} \cdot 10 \frac{\text{N}}{\text{kg}}} = \underline{10 \text{ m}} \\[5px]\end{align}. The hydrostatic pressure is determined by the density of the liquid , the gravitational acceleration g and the depth below the water surface hd. Boba drinks which are infused with sweet, thick tapioca pearls are difficult to consume without an extra wide boba drinking straw.
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. Everybody knows the height of water in a water barometer is about 10 meters. Well cover that in this article, and we have also produced a useful straw size chart, which visually details the various straw sizes.
Consider a straw that is stuck very deeply into the ocean. }{=} F_g \\[5px]\left(p_0 + \rho g h_d p_1 \right) \cdot \bcancel{A} & = \left(\rho g h_d+ \rho g h\right) \cdot \bcancel{A} \\[5px]p_0 + \bcancel{\rho g h_d} p_1 &= \bcancel{\rho g h_d}+ \rho g h \\[5px]p_0 p_1 &=\rho h g \\[5px]\end{align}. To put another way you need to have far more air than water going up your straw. Initially the system has potential energy $E_0$. Some. The math request seems unneeded. Just insert the brush into the straw with warm soapy water, scrub, and rinse clean. Even the most powerful suction pump in the world would not be able to overcome a height difference of more than 10 metres if a perfect vacuum were created. Practice shows, however, that with increasing height it becomes more and more difficult to get the liquid through the straw. 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. @Mark Eichenlaub: I don't understand - the original potential energy is the second term in the RHS of the first equation. Eventually it will no longer be possible to drink water from such a drinking straw from a height of about 10 metres. For this reason it is not possible to drink a glass of water with the help of a straw even in free space (apart from the fact that water would immediately become gaseous anyway due to the vacuum)! This does not have to do with a too weakly developed mouth musculature but has a natural physical cause. Why does water boil faster at high altitudes? In accordance with the principle of communicating vessels, the water level inside the straw will be the same as outside. A great eco-friendly accessory for a zero-waste lifestyle. For convenience, we also sell Steelys Straw Cleaning Brushes to make it easy to clean and maintain any reusable drinking straw.
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. This can be expressed through the following formula: $$\frac{sP_{atm}^2}{2\rho g} = \frac{1}{2}s\rho gh^2$$. How is it possible for tall trees to pull water to heights more than 10m? @Martin There's nothing innately wrong with a qualitative answer. 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. Marek would have had to write this comment. If it can not, the argument gives only an upper bound. If you had a strong enough thumb, the water in the straw would start boiling even at room temperature to fill the vacuum (to a point).
Heh, nice argument. Undefined behavior (according to clang -fsanitize=integer) on libstdc++ std::random due to negative index on Mersenne Twister engine. Even taking into account the immersion depth and the hydrostatic pressure acting there, the same formula is obtained, so that the suction lift is actually not dependent on the immersion depth itself! Welcome to physics SE! By taping several straws together, a friend and I drank through a $3.07m$ straw. Just introducing other, more complicated ones. I wish I could give -1 to your comment @Marek: So where is a mistake in my solution? At the highest point, all the energy will be converted in gravitational potential energy. ), the height would be ~30 inches. You will get height boost!! Hard to find a flaw in this. First he blows air inside, up to one thousand atmospheres, then he releases, no need to suck at all You are missing something. I push it until it is exactly covered, then I release it. Also, what would happen if you had a very long straw, submerged it under the ocean's surface, and pulled it out with your thumb covering the top? Thinking and physics comes first, math only second. The energy needed above water is the same integral above the water level. 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). Let $h$ be the height of the column of water inside the straw. It features a .24 width, and 1-3/8 round spoon attached.