The equivalent problem is, I have a closed barrel half filled of water, floating. 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. This is an excellent quantitative problem. One end of the straw is placed in a water-filled glass and at the other end the water is sucked up into the mouth.
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! Announcing the Stacks Editor Beta release! 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.
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. Superman is going to manipulate the air pressure in the straw. I am not fully convinced by this argument, but can't find a flaw in it. The energy stored by depressing the water level is simply the integral of d volume times depth under the surface. ), the height would be ~30 inches. 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). 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. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The height is small enough to justify the assumption that $g$ is a constant.
@Mark @Carl I could see the point if he uses a really deep straw say 10 or 11 kilometers. In this way, the water is pushed upwards by the greater external ambient pressure against the lower pressure inside the drinking straw.
All of our straws are easy to clean, dishwasher safe food grade and BPA free. 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. 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. Our shortest reusable straw is the Short Cocktail Straw, which measures 5 long. Instead, there would be a near-vacuum in the upper part of the straw that could make the water boil at room temperature.
When the water is at its peak in the straw, the kinetic energy of the system is zero, so the potential energy is $PV$.
How is it possible for tall trees to pull water to heights more than 10m? The resultant upward force Fp results from the difference between the force F0 at the lower end of the water column which pushes the water upwards by the ambient pressure p0 and the force F1 at the upper end of the water column which pushes the water downwards by the internal drinking straw pressure p1. This answer is different from Martin and David's. If you can make the first integral grow unboundedly then it is solved. How high can the water go under this assumption? Why does OpenGL use counterclockwise order to determine a triangle's front face by default? This creates a lower counterforce so that the force of the external ambient pressure is able to push the water upwards. Let $h$ be the height of the column of water inside the straw.
I think this makes perfect sense. The water column would simply stop at this height.
It only takes a minute to sign up. With a longer drinking straw, the ambient pressure can no longer push the water column upwards.
How is he getting the water up to this speed?
The weight increases with increasing height.
This height boost already is mentioned as intertia in OP's question. First imagine a glass of water. and any change in speed of the column of water, $$\delta K_2 = \frac{1}{2}m(2v\delta v) = (\rho A h)\dot{h} \delta\dot{h}$$, $$P A \delta h - \rho A g h \delta h = \frac{1}{2}\rho A \dot{h}^2 \delta h + \rho A h\dot{h}\delta\dot{h}$$. In a state with the common law definition of theft, can you force a store to take cash by "pretending" to steal?
@Mark, I am an engineer, and such I am allowed to guestimate results based on experience. $$W= sP_{atm}H + \int^0_H{m(h)g\ \mathrm{d}h}$$, Substituting $H=\frac{P_{atm}}{\rho g}$, $m(h)=\rho s h$, $$W= \frac{sP_{atm}^2}{\rho g} + s\rho g\int^0_H{h\ \mathrm{d}h}$$, $$W= \frac{sP_{atm}^2}{\rho g} - \frac{1}{2}s\rho gH^2$$.
Derivation of the Navier-Stokes equations, Derivation of the Euler equation of motion (conservation of momentum), Derivation of the continuity equation (conservation of mass). I hope also that the edits will do it more readable. As the water level in the straw increases, the weight of the water column increases. 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. Junior employee has made really slow progress. Some people prefer bent reusable smoothie straws, while others like the straight reusable smoothie straws. This short straw is available only in a .24 diameter. Archimedes principle is that a body sumerged in a fluid suffers a force equal to its volume times the density of the fluid. To get the water to go up, he must provide a reduction in the pressure. At given pressures this suction height h is determined as follows: \begin{align}\require{cancel}F_p &\overset{! }{=} F_g \\[5px] \left(p_0 p_1 \right) \cdot \bcancel{A} & = \rho gh \cdot \bcancel{A} \\[5px]p_0 p_1 &=\rho h g \\[5px]\end{align}, \begin{align}&\underline{h = \frac{p_0 p_1}{\rho g} } \\[5px]\end{align}. Whether your reusing stainless steel drinking straws at a restaurant or home setting cleaning, maintaining and reusing them is easy.
Do I need to add it twice?
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). It features a .24 width, and 1-3/8 round spoon attached.
So when the water is at the bottom level, there is no potential energy left. As long as your consent is not given, no ads will be displayed. My question is about what would happen if Superman were to drink through a straw by creating a complete vacuum in the straw. How is discussing energy balance philosophy? I think we may have had some leaking preventing us going higher.
Like their plastic counterparts, our steel Boba Straws are 8.5 long with a .36 diameter. This is by far the most popular straw size, and it will work for most applications. Note that this agrees with a simple analysis using Newton's second law.
Thinking and physics comes first, math only second. I will consider the system of the water in the straw plus the water in the reservoir. 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. If the plate is pressed downwards, the water is pressed upwards through the straw. No energy is assumed to be wasted on friction. (Roughly speaking, the variation $\delta h$ is second-order at those points whereas the variation $\delta\dot{h}$ is still first-order.) Please additionally consider local health department protocols, which can vary by location and state.
Welcome to physics SE! 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, $$\dot{h}(\epsilon_t) = \epsilon_{\dot{h}}$$. As for that $1 \over 2$ factor, it's also bugging me.
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. 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.
However, if he is able to suck really quickly then the velocity of the air could allow for water entrainment beyond the maximum lift.
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 only way to get higher would be for the air pressure (the thing actually pushing the liquid up) to be higher.
Initially the system has potential energy $E_0$.
If I show the math will you remove the (-1) ? In the same way the ambient pressure on the water surface is able to push the water upwards through the drinking straw. The weight of the water column to be pushed up depends on the water level in the straw. So the issue becomes if there is some other sort of limit.
Of course, it is still to be proved that hydrodinamically all the stored energy can be used to move the column up.
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https://www.youtube.com/watch?v=HUmZrtiXDik. But the rate of recharge within the well must be suffuciently low to ensure that the water does not "clog" your suction pipe.
Another popular straw for cocktails is our 8 long reusable spoon straw. This does not have to do with a too weakly developed mouth musculature but has a natural physical cause.
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. So the maximum height of the water is.
Sucking creates a vacuum in the drinking straw, so that the greater ambient pressure pushes the beverage through the drinking straw. It has already been mentioned that the immersion depth of the straw or the hoses of suction pumps has no influence on the maximum suction lift. Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure. 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)!
@MarkEichenlaub I may be late but Hagen Poisseulle? 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.
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. If he were sucking Hg up the straw (not recommended for non-superheros! I argued against your stance of doing the math first. At this level, the ambient pressure p0 also acts in the water itself (since there is no hydrostatic pressure at this point). Let's analyse a similar experiment, which I believe to be equivalent. 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.
by Steely Straws | Feb 19, 2019 | Wholesale Reusable Drinking Straws. I am not, because they are conceptually very simple. The standard drinking straw diameter is .24 wide. For this reason, the maximum drinking height is 10 metres. Boba drinks which are infused with sweet, thick tapioca pearls are difficult to consume without an extra wide boba drinking 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? Finally, you want to turn all the energy into potential energy at the highest point. You create a vacuum at the top of the straw. The immersion depth of the drinking straw plays no role (more about this later). @Mark:This must be done by Omega.
The resulting suction lift h is determined as follows: \begin{align}\require{cancel}F_p &\overset{! I push it until it is exactly covered, then I release it.
Without any math the whole discussion goes dead. @Sklivvz The system starts out with some energy. At some point the weight Fg will eventually become as large as the upwards acting force Fp. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. Eventually it will no longer be possible to drink water from such a drinking straw from a height of about 10 metres. I wish I could give -1 to your comment @Marek: So where is a mistake in my solution? which at least accounts for the mysterious factor of $\frac{1}{2}$ that appeared in previous versions of my answer. 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. On the other hand, if he reduces the pressure quickly, it's at least possible that the water could reach a higher height. Anyway, this equation no longer admits the solution $h = \frac{P_0}{\rho g}$.
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. In the state of equilibrium, the downwards acting weight Fg must be equal in magnitude to the upwards acting force Fp. We run into this limit all the time with vacuum pumps. For a unit area column the external energy put in equals the volume of the column times the air pressure.
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. Either way, reusable smoothie straws work great for beverages like smoothies, powder shakes, super food blends, or yogurt. Therefore, the water will raise up to $H$, which it will reach with zero velocity.
If the pressure becomes too low at some point, the liquid in the straw begins to vaporize and the gaseous molecules build up a pressure themselves (vapour pressure). Practice shows, however, that with increasing height it becomes more and more difficult to get the liquid through the straw. at STP, the answer is 33 1/2 feet. Or are you simply planning some method to store energy in the straw using forced oscilations?
The hydrostatic pressure is determined by the density of the liquid , the gravitational acceleration g and the depth below the water surface hd. The differential is enough to force water, or soda, up into your mouth. I have an argument that the water in the straw will rise to twice the equilibrium height.
The work is done against pressure and in favour of gravity. @David:There is a fundamental error in your analysis:Bernoulli's equation is applicable only to the steady flow of a fluid. They are meant to give general tips and practical guidelines on both household and commercial straw cleaning applications. 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. This maximum suction height will be derived mathematically in the following. Then coil the straw around at great depth many many times. Straws work on air pressure. Would it be possible to use Animate Objects as an energy source? What you wuld suck from the straw then would be cold steam. @Martin: I am not saying there necessarily is one. The only question you have is how do you choose the right length and width for your reusable straw. I'm having trouble explaining this to my friends as I do not know the mechanisms fully. A standard smoothie straw is 8.5 long but features a wider .36 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.
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.
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.
Overall you can help the suction, by immersing the straw deeper and deeper. \begin{align}&F_p = F_0 F_1 \\[5px] &F_p = p_0 \cdot A p_1 \cdot A \\[5px] \label{ff}&\underline{F_p = \left(p_0 p_1 \right) \cdot A} \\[5px]\end{align}. 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. What is the highest height he could drink from? @John Berryman: It depends how much of a vacuum you want to create. It even says so in the question.
David and Martin's answers consider the system of water in the straw.
Heh, nice argument.
By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. The vacuum is already present. 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 force with which this water column is effectively pushed upwards again results from the difference between the forces at the lower and upper end of the water column: \begin{align}&F_p = F_2 F_1 \\[5px] &F_p = p_2 \cdot A p_1 \cdot A \\[5px] \label{f}&F_p = \left(p_2 p_1 \right) \cdot A \\[5px]\end{align}. Water vs Milkshake being sucked through a straw. @Mark, I think I am getting it now :-) I basically need to redo the answer from scratch :-/. Does China receive billions of dollars of foreign aid and special WTO status for being a "developing country"? 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. How can one check whether tax money is being effectively used by the government for improving a nation? -1 At least mention a physical law (Newton's $F=m\cdot a$). The pressure p2 acting at the lower end of the drinking straw results from the sum of ambient pressure p0 and hydrostatic pressure ph (see article Pressure in liquids). Close the end when the water reaches the maximum height and you can measure how high you can reach. Even with a perfect vacuum, there is a limit to how much water can be moved in this way, which incidentally tops out at around 10 meters.
Nah, you just wouldn't be strong enough to lift it. 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, $$P_0 \delta h - \rho g h \delta h = \rho \dot{h}^2 \delta h + \rho h \dot{h}\delta\dot{h}$$. @Martin: call it Euler equation if it helps you.
Everybody knows the height of water in a water barometer is about 10 meters. @Omega lol - 17 hours after the question is asked we posted the same solution within two minutes. 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.
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. So we're left with. Just introducing other, more complicated ones. Now the total energy of the system is $E_0 + W$. 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. 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.
Check below for more information on cleaning your reusable straws. By making the flow unsteady you are not getting rid of the phenomena found in Bernoulli. This makes sense to me, because if we actually started in that position, where would the potential energy come from? As an original and leading wholesale bulk supplier of reusable stainless steel drinking straws for the U.S. market, specializing in custom engraved drinking straws, Steelys can help you choose the correct length and width of reusable straw for your needs. I then blow very hard (I am superman afterall), and create a huge volume of airfilled straw at great depth. How do you explain this? The standard reusable straw size is 8.5 long x .24 wide.
@Mark: If we follow Bernoulli's equation then the pressure at the entrance to the straw must be even lower than $P_0$. In this case, the maximum suction lift hmax is obtained by creating a vacuum inside the straw: \begin{align}&h_{max} = \frac{p_0 \overbrace{p_1}^{=0}}{\rho \cdot g}\\[5px]&\boxed{h_{max} = \frac{p_0}{\rho \cdot g} }\\[5px]\end{align}. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. The gravitational energy is the mass of water raised times the average height of the water. 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. Of course, you'd be very welcome to submit a more mathematical analysis of the same ideas if you want. And what use is it doing the long-winded calculations as you did if there is mistake in your approach in the first place?
Hmmm. Maybe you should add 10 meters for the usual vacuum effect. If it can not, the argument gives only an upper bound.
One might think that such a drinking straw could in principle be of any length. 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. Why does water boil faster at high altitudes? To suck water through a straw, you create a partial vacuum in your lungs. The air pressure at sea level is very close to one bar, which is equal to the pressure created by a water column of 10 meters.
Eventually, gravity and the pressure of the water inside the straw balances out the air pressure and the water won't go higher even with a partial vacuum still above it. Boba, or bubble tea, is a delicious treat that is growing in popularity globally. @Sklivvz Your expression for the work done is correct, but the total energy is the work done plus the original potential energy. Press question mark to learn the rest of the keyboard shortcuts. Actually, should it be possible to answer without using a reference to the energy?
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