# TBR Physics: Fluid Pressure and Column Height

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TBR Physics Book II page 88 #22

Correct Answer: C “ A lower fluid velocity at point B, but an unchanged fluid height in Column 1”

The TBR answer explanation acknowledges that velocity in the lower tube will slow down if the liquid is more viscous. According to Bernoulli’s equation, a decreased velocity should leads to INCREASED pressure.

Why isn’t this true?

Because if TBR’s answer is correct, where is the extra energy (from the velocity being slower) going?

The problem I attached (also from TBR - answer is D) shows that indeed, when velocity is highest (where cross-sectional area is smallest), the pressure is the smallest. Similarly then, when velocity is smallest, pressure should be GREATER.

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I do not think you should be using Bernoulli's equation here because you are looking at a non-ideal fluid. Instead, you should be using Poiseuille's equation:

The pressure gradient remains unchanged. The (rho)(g)(h) term remains constant because the density of the new fluid is the same as the old. Therefore (rho)(g)(h) + Patm on the left remains unchanged as does Patm on the exit end of the tube. So you can see how delta P remains unchanged.

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@t5Nitro

Sorry I seem not to be understanding your post completely. Where are you getting the "rho*g*H" term from?

I know that's in Bernoulli's but you said not to use it here?

.

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@DrknoSDN

Wait so if pressure is greater in the reservoir, how does that affect the reservoir (does the fluid level increase)?

I thought the pressure pushing the fluid up the right-column came just from the velocity (and that any conclusions about this fluid height had to be based on velocity).

@justadream After looking at the books description it is just using a simple approximation

@t5Nitro has an accurate interpretation of the pressure gradient decreasing to atmospheric pressure at point B. Based on that the pressure gradient is the same for both fluids changing from (rho gh) at point A to atmospheric at point B. If the gradients are the same then the point directly below column 1 would have the same pressure in both scenarios.

There is a lot of information out there that would disagree with the way TBR solves this but after reading the solution it is probably the most "MCAT style" way of finding an answer. (This should be emphasized)
Real world modeling of Torricelli's law often uses an additional constant to handle changes due to viscosity and a real world calculation may find a pressure gradient change between fluids but probably not worth debating.

I do get what you are suggesting, that the rate of flow in the pipe changes and so under extremes a change in the pressure gradient could be seen.

Crazy Hypothetical to validate that argument: (does not apply to MCAT world)
If you imagine a scenario where the reservoir is very tall and narrow with a large area for point A, the fluid can drain from the reservoir extremely rapidly. At some point it is draining at near freefall velocity. At that point the pressure is indeed changed... If you took a 1,000 meter tall column of water, the pressure at the bottom would be greater on earth but if the same column was in freefall the pressure would be uniform. And if the column was stationary but you cut the bottom off, pressure would be nearly uniform from the top of the draining fluid to the bottom.

--> Technically a less viscous liquid would have less pressure at point A because the reservoir level is decreasing at some velocity at the time of measurement.
To read it presented in a different way: http://seniorphysics.com/physics/pinhole_water_flow.pdf

TBR is simply ignoring any flow in the reservoir itself, and assuming point A is limiting flow. Which is probably a good idea to do. Especially on the MCAT.

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@DrknoSDN

How do you know for MCAT purposes to just focus on the pressure differences here (and ignore the effects of velocity)?

Hypothetically, since we are only considering pressure difference, does that mean that if I increased the tube's width by 100X (or decreased it by 100X), the answer would still be "same column height" since again, we're not considering velocity effects here (and just focusing on pressure difference)?

@justadream Honestly I'm not sure. Don't make the problem too complicated I suppose?

does that mean that if I increased the tube's width by 100X (or decreased it by 100X), the answer would still be "same column height"...?
If you are talking about changing the tubes cross sectional area along the entire length then I don't think it would matter. The solution description only talks about the pressure difference from the start to end of the pipe and that wouldn't change by changing the diameter of the pipe. The flow rate would change but speed would not change because speed/velocity is determined by the height of the water in the reservoir and not the size of the tube. (Torricelli's Law). Moderately changing in the tube's width seems like it would have no effect on the column height. But like I said before, extremes are not presented on the MCAT because extreme situations cause breakdowns of normal "predictable" behavior.

@DrknoSDN

Well now that you say flow rate (volume flow rate) can change if the width of the pipe changes, then it makes more sense to me.

I was confused before because it is not compatible to have constant flow rate, changing cross-sectional area, and non-changing velocity (at least 2 of these vars have to change).

I realize it's been awhile, but I'll go ahead and answer this since I just had the same problem myself.

The above explanation that there is no pressure change does justify that there is no column height change. P at A and P at B are the same, so pressure did not change.

However, I was still baffled that decreased velocity would have the same pressure. After much thought, here is the answer. (n = viscosity in all equation)

v = (change in pressure/4*n*L)*r^2. This equation shows that if we increase our viscosity, n, we decrease velocity. Flow rate (Q) = vA. Thus by increasing viscosity, we have decreased the flow rate proportionally.

Now look at Poiseuille's principle. Q = vA = [(pi*r^2)/(8*n*L)] * (P1-P2). If Q (and thus velocity) has already changed proportionally to viscosity, then there can be no change in pressure.

Edit: I also realized you shot not apply the continuity equation here at all. The continuity equation states the flow in is the same as the flow out (Qin=Qout so a1V1=a2V2). In this problem, this is not the case. Fluid is only flowing out with no input.

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