Blowing air through a straw

I, for one, am not always familiar with what may be customary notation or excepted practices when it comes to notation. I had assumed that your detail "The incoming air flow is V=0.51/s." was the velocity of the air flow through the upper 6mm pipe. I took this number to mean 0.51 m/sec air velocity. In your solution I see that you intended it to be air volume that you have expressed in cubic centimeters per second. How do I know the unit of the air volume is intended to be? At best it is a educated guess based on some notation standard.

I do not know, and really don't care, it may be very well accepted practice and commonly understood that V with a mark above is a notation for volume in cc's but I am frustrated by problems that assume that all of the solvers are familiar with arbitrary notation. I am not at all interested in looking up notation or learning what some symbol may be commonly used to represent when the information could easily be put in the problems wording without any ambiguity.

It seams to me that to be clear the "The incoming air flow" detail should have the units included i.e. V=0.51 cc/s

It was a surprise to me that the effect would be to increase the water level in the straw (I just visualized the action, what the gas would do, and answered based on that). I see that the equations applied in the solution don't differentiate between creating the air flow via under-pressure on the outflow end, or creating it via over-pressure on the inflow end. Also, maybe the solution is correct for the stated numbers, but surely with a high enough over-pressure (blowing), like hurricane wind, some of that incoming pressure would be diverted down -- and the equations appear to not cover that either. I.e. there is evidently some simplifying assumption here (linear flow, perhaps?), that breaks down for the more general case?

I'm not sure about this, because your equation for Delta P compares two distinct points in time and under two distinct flow regimes -- before the flow is established versus after. Bernoulli is really only good for steady state flow.

I think another comment summarises the issue -- a differential pressure needs to be established in order to cause the force that causes the gas to flow. I thought in terms suggested by the diagram -- the flow is suggested to be due to positive pressure applied at the inlet (left) because that's where the arrow and v-dot are. In this case I think the pipe is pressurised, and the gas pressure difference will cause the water level to drop by 1.6 cm. But if the air is sucked out of the pipe from the right, you depressurise the pipe and the water will rise by the same amount. But the arrow is on the left, not on the right, so I think the implication is that the differential pressure is positive. The water will drop slightly in reaction to the pressure that needs to be applied to establish the flow.

I decided it would be a simple little experiment to actually try to test the premise of this problem. I set up a test that I believe replicates this problem quite closely. There are no lengths stipulated in the problem for the horizontal tube length so I just used the short length of the teas that I have available. My Teas are 8mm not 6mm in diameter as indicated in the problem I do not believe that the underlining principle will change with that change in the T shaped tube size.

In the comments, several people have questioned the accepted answer to this question. Ban Yan and Alf Steinbach apear to have some very valid points. . I am now convinced that the accepted answer to this problem is not correct. I cannot produce the results as the problem answer indicates.

Personally, I believe that I have confusing the effects of a ventui effect with that of the Bernoulli effect that is used to calculate the answer. I now believe that with a constant size in the horizontal tube no Bernoulli effect can be demonstrated.

If the answer is correct could someone explain the results of my experiment. What have I done incorrectly? What should I change to get results that would indicated the accepted answer to the problem?

My original intention was to demonstrate the basic principle that would apply in the calculation of the answer to this problem, instead I appear to have demonstrated that under the conditions indicated in this problem there is no negative pressure produced in the vertical tube. In fact there is an positive pressure produced in the vertical tube. As the volume of air is increased thru the horizontal tube the level of the water in the vertical tube become lower until air is bubbling out of the bottom of the vertical tube in the water. I have not attempted to estimate or measure the volume of air moving true the horizontal tube. Am certain that as I slowly increase the air flow from a static condition to a very high speed rush of compressed air I never see any indication of a negative pressure in the standing vertical tube. presumably, at some point in the slowly increasing volume of air flow, the velocity of the air flow should approximate the air flow velocity used for this problem.

Something else to consider that I did not include in my video. When I produce air flow using a large diameter hose end at the T shaped tube I only see positive pressure. But! if I produce air flow using the small diameter orifice at the end of the blow gun I see negative pressure in the horizontal tube. I discovered that it is not necessary to place the end of the blow gun into the T shaped tube. I can blow high velocity air directly into the open end of the tube from a position several centimeter away from the tube. Under these conditions there is no question I get negative pressure in the vertical tube that can be controlled nicely by simply controlling the volume of air coming out from the blow gun orifice.

Therefore it seams apparent to me that the answer to this problem is very dependent on a number of factors that are not stipulated in the problem. I suspect that factors like air will compress and expand to fill a volume which effects the pressure, the length of the horizontal tube both before and after the vertical pipe all play an important role . By adjusting the speed of the entering air I could possibly make this solution work if I blow just the correct amount of air at the correct velocity into the open vertical tube. Keep in mind both ends of the tube remains open even as a high speed air stream is directed into the open tube from some distance away, ambient air could and would enter or exit out of the open tube ends at any time depending on pressure differences. It is not a simple mater of the volume of air moving thru the horizontal tube, it is a matter of how fast that air is moving. It may seam that the volume of air should be directly proportional to the air velocity in the tube, but if that is the case why is there such a pronounced difference in the pressure in the vertical tube that appears to be completely dependent on the manner that the air flow is produced. I think there is a lot more to this question then this problem is considering.

Anyone that may be interested can click to view a short video

The answer to this question as set up is incorrect as Darryl has adaqueatly demonstrated. If we examine Bernoulli's Principle along a streamline the faultly logic becomes apparent. In order for the fluid in the vertical straw to lift there must be vacuum ( relative to ambient ) pressure created in the straw. with the left and right sides labled as 1 and 2 respectively:

$\displaystyle \frac{P_1}{\gamma}+ z_1 + \frac{ {v_1}^2}{2g} = \frac{P_2}{\gamma}+ z_2 + \frac{ {v_2}^2}{2g}$

Start simplifying:

The elevations of 1 and 2 are equal, thus $\displaystyle z_1 = z_2$ and the terms cancel.

The crossectional areas of the inlet and outlet are equal, flow is invicid and incompressible, thus continuity applies and the velocities of 1 and 2 are equal: $\displaystyle v_1 = v_2 \rightarrow \frac{ {v_1}^2}{2g} = \frac{ {v_2}^2}{2g}$ and the terms cancel.

As a consequence, we are left with:

$\displaystyle P_1 = P_2 \rightarrow P_1 - P_2 = 0$ meaning there is no differential pressure across the tube for any flow rate. When we neglect viscous disappation, the model states any flow rate desired can be obtained at zero differential, and why not...there is no resistance! However unrealistic, this in itslelf is not the killer, it could be that the whole tube is under vaccum relative to ambient pressure excerted on the water surface? Again, No. Bernoulli's Principle makes no claim about what happens across streamlines, only what happen on a single streamline. Thus, it is not a valid argument to talk about what happens inside and outside the straw using the principle.

What we can say using a variation of the principle ( conservation of energy) accounting using viscosity is that the pressure gradient decreases in the direction of flow, as a conseqence of the Second Law of Thermodynamics. Meaning if the flow is from left to right, the pressure in the tube is decreasing from left to right. This can happen in two ways, presurizing the system above ambient on the left, driving the fluid surface down, or pressurizing below ambient on the right, pulling the fluid up. The direction of flow is the same in either case, but the effect given as "correct" by the poster in not what would happen when you blow on the left side. It would bubble, not lift.