Flow Rate Examkrackers Physics Question 603

[rowjimmy]

2 one liter containers are filled with the same viscous fluid via different size funnels. The cross-sectional area of the stem on the first funnel is twice as large as the cross-sectional area of the stem on the second funnel. If the first container is filled in 24 seconds, which of the following could be the time required to fill the second container?

A. 24 s
B. 42 s
C. 48 s
D. 54 s

Answer: D, 54 s.

I understand that with viscous fluids the velocity of the fluid decreases as the radius of the vessel decreases because the rate of volume change is greater than the rate of area change so you end up with a fluid exposed to a greater amount of area which increases drag.

However, if this was an ideal fluid would the time required to fill both containers be the same because the volume flow rate would be the same? I ask this because in the answer explanation it states: "Ignoring viscosity, the flow rate would be doubled for the first container: Q = Av. The viscosity has a greater effect on the narrower funnel so the second container requires more than twice the time to fill as the first."

This explanation makes sense assuming the first sentence of the explanation is true, but it suggests that doubling the area of a vessel would double the flow rate, but why isn't Q constant assuming no viscosity (remember the answer explanation states "ignoring viscosity" so is that the same as saying an ideal fluid--and isn't Q constant in an ideal fluid?).

In other words, if the fluid isn't viscous and the area of a pipe is doubled then the rate at which volume flows should still be constant because the velocity would decrease by half. But because this is a viscous fluid the flow rate will be affected by drag and the flow rate will be slower. However, the narrower the pipe then the greater effect of drag but how do we know that it will take more than twice as long to fill and not somewhere between 1.x-2 times as long?

Can Poiseuilles' Law be used to answer my question? If Q is proportional to r^4 and the area of funnel 1 is twice the area of funnel 2 then the radius is greater by sqrt(2), thus Q of funnel 1 is 4 times greater so funnel 2 should take 4 times longer to fill. Please correct me if I'm wrong. That explanation of doubling the area somehow doubles the flow rate if viscosity is ignored doesn't make logical sense to me because I would've assumed that if viscosity is ignored then the flow rates should be equal because the fluid is the same and if you double the area the the velocity would halve but now I'm just repeating myself so I hope you get the gist of my question(s).

Thanks!!

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[milski]

Here is what the explanations is trying to do: For ideal fluid, the decrease in flow will be exactly twice and the answer would be 48s. Adding viscosity will make the flow decrease a bit more than the ideal case, so the answer will be something slightly more than 48 - the only choice being 54.

For an ideal fluid the translational velocity of the fluid will be proportional to the pressure difference. If you increase the area, you are increasing the volume of fluid moving at this velocity and as a result the flow rate of the fluid.

What they want you to realize here is that when you start decreasing the radius the viscosity will have more effect on the flow.

 

[rowjimmy]

For an ideal fluid the translational velocity of the fluid will be proportional to the pressure difference. If you increase the area, you are increasing the volume of fluid moving at this velocity and as a result the flow rate of the fluid.QUOTE]

I thought Q stays constant for an ideal fluid when A increases because the velocity would decrease proportionately.

 

[milski]

I thought Q stays constant for an ideal fluid when A increases because the velocity would decrease proportionately.

The translational velocity stays the same, not Q. Two quick ways to convince yourself:

- if there is no friction, what would slow the flow down when the area is increased? No viscosity means that the particles don't experience any forces from the other particles around them.

- what you're saying is the equivalent of saying that you can decrease the flow out of a tank by widening the hole in it. That's not supported by experiments.


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[SaintJude]

I thought Q stays constant for an ideal fluid when A increases because the velocity would decrease proportionately.

In an ideal fluid, the flow rate Q is constant and the changes you're referring to would theoretically happen.Ideal fluids also no viscosity. So since this question indicated that these fluids have viscosity, it must be real fluids.

Thanks for posting this good question that made me concretely see how the application of theories from ideal fluids would affect predicting results about changes in a real fluid. I posted a question about this like a day ago.

 

[chiddler]

I am very glad you asked this question because I had apparently made the same mistake that you have.

thanks for the helpful explanations milski and stjude

 

[Class1P]

So just to clarify..

Ideal fluids with no viscosity = when you decrease output pore size (cross sectional area of tube) the velocity increases proportionally and the flow stays the same..

Non-ideal viscous fluid with friction = decrease velocity when tube cross-sectional area is decreased due to friction and viscous forces and thus a large decrease in fluid flow.



Also good to note, the Velocity is not directly inverse-proportional to the radius

since Q = A*V

and Area = pi*(r^2)

The cross sectional area increases proportionally with the square of the radius, thus wouldn't a radial increase of a factor of 2 result in an increase of the area by 4? and thus an decrease in velocity by a factor of 4?

 

[milski]

So just to clarify..

Ideal fluids with no viscosity = when you decrease output pore size (cross sectional area of tube) the velocity increases proportionally and the flow stays the same..

Non-ideal viscous fluid with friction = decrease velocity when tube cross-sectional area is decreased due to friction and viscous forces and thus a large decrease in fluid flow.



Also good to note, the Velocity is not directly inverse-proportional to the radius

since Q = A*V

and Area = pi*(r^2)

The cross sectional area increases proportionally with the square of the radius, thus wouldn't a radial increase of a factor of 2 result in an increase of the area by 4? and thus an decrease in velocity by a factor of 4?

For non-viscous fluid the velocity is constant. That means that increased area increases the flow. All this translates to the fairly easily observable fact that a bigger hole makes for a bigger leak.


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[rowjimmy]

Here is what the explanations is trying to do: For ideal fluid, the decrease in flow will be exactly twice and the answer would be 48s. Adding viscosity will make the flow decrease a bit more than the ideal case, so the answer will be something slightly more than 48 - the only choice being 54.

For an ideal fluid the translational velocity of the fluid will be proportional to the pressure difference. If you increase the area, you are increasing the volume of fluid moving at this velocity and as a result the flow rate of the fluid.

What they want you to realize here is that when you start decreasing the radius the viscosity will have more effect on the flow.

Let me just first say thank you for trying to explain this and I apologizefor beating this thing into the ground. But I don't understand how youcan say that V stays constant for an IDEAL fluid. I understand your explanationregarding V staying constant for real fluids per your very helpful reference towater emptying faster with a larger hole.

However, the explanation for this answer choice involves the assumption ofthe flow rate of an IDEAL FLUID (no viscosity) increasing by 2 if the areadecreases by 2 (and then the explanation factors in the additional resistancedue to viscosity). But one thing that I am sure of is that Q is constant in anideal fluid. If you've ever gone kayaking or tubing down a long river that haswide areas and narrow areas you are familiar with the phenomena of increasingflow rate when the river narrows and a decreasing flow rate when the riverwidens. As to what is responsible for this phenomenon, I would look toBernoulli's Equation, the conservation of fluid mass and/or the incompressibilityof ideal fluids (or low-viscosity fluids) as potential answers. Randomtranslational motion increases when uniform translational motion decreases, theformer being responsible for greater fluid pressure. Per Bernoulli’s equation, greaterfluid pressure results in slower fluid velocity. This explains why wider tubeshave slower linear velocity of fluid—because the random translational motion isgreater which increases fluid pressure. Another way of looking at it is thatideal fluids are incompressible so the rate at which a given volume of fluidflows by a certain point must be the same at all points of that fluid (this isalso explained by the conservation of energy.) I think you are confusing thelinear velocity of fluid with the volumetric flow of a fluid—the latter is whatis stays constant and the former is what changes. FOR IDEAL FLUIDS, VOLUME FLOWRATE IS ALWAYS A CONSTANT. A thumb over a hose that increases the LINEAR fluidvelocity, windows that pop out during a tornado, and the lift of an airplaneare all examples of the volume flow rate and Bernoulli’s equation working inconcert.

So, again sorry if I'm repeating myself, if there is no viscosity (aka idealfluid?) and the flow area decreases by 2 then the linear velocity—and not Q—shouldincrease by 2.

 

[milski]

Bernoulli's law is applicable only in case when you have constant flow. All your examples (the river, the hose) are an already constant flow which moves through a pipe with different area. That is not the same as comparing two different setups with two different areas of the hole in the tank.

For an ideal fluid the only forces acting on a parcel of the fluid at the boundary between the tank and the 'outside' are from the difference in pressure and gravity. The net force is the same for any parcel with equal size (or more precisely, equal area, assuming equal thickness). That results in a constant velocity of the ideal fluid leaving the tank.

 

[rowjimmy]

Bernoulli's law is applicable only in case when you have constant flow. All your examples (the river, the hose) are an already constant flow which moves through a pipe with different area. That is not the same as comparing two different setups with two different areas of the hole in the tank.

For an ideal fluid the only forces acting on a parcel of the fluid at the boundary between the tank and the 'outside' are from the difference in pressure and gravity. The net force is the same for any parcel with equal size (or more precisely, equal area, assuming equal thickness). That results in a constant velocity of the ideal fluid leaving the tank.

What is the linear velocity of the fluid dependent on? Won't linear velocity of a fluid be affected by a change in the area of the tube it flows through? P=F/A and the area changes the pressure change that results at that point will cause a change in the linear velocity of the fluid (bernouli's equation)?

 

[milski]

What is the linear velocity of the fluid dependent on? Won't linear velocity of a fluid be affected by a change in the area of the tube it flows through? P=F/A and the area changes the pressure change that results at that point will cause a change in the linear velocity of the fluid (bernouli's equation)?

Density of the fluid, gravity and pressure difference between the inside and the outside of the tank. You are correct that P=F/A or F=PA. Consider the bottom layer of water with height δ. A parcel with area A of this layer has a mass of Vρ=δAρ. The force for that parcel is going to be PA+mg=PA+δΑρg. From a=F/m, we have a=(PA+δAρg)/(δΑ&#96=(P+δρg)/(δ&#96=g+P/δρ

From this you can conclude that the linear velocity increases when gravity or the pressure difference increases and decreases when the density decreases.

Bernoulli's law cannot be used to compare the two cases - you can apply it only to a flow which is already constant, not to compare two different flows. It's the same difference as saying that momentum does not change in a system with no external forces and expecting all systems with no external forces to have the same momentum as each other.

 

[rowjimmy]

Density of the fluid, gravity and pressure difference between the inside and the outside of the tank. You are correct that P=F/A or F=PA. Consider the bottom layer of water with height δ. A parcel with area A of this layer has a mass of Vρ=δAρ. The force for that parcel is going to be PA+mg=PA+δΑρg. From a=F/m, we have a=(PA+δAρg)/(δΑ&#96=(P+δρg)/(δ&#96=g+P/δρ

From this you can conclude that the linear velocity increases when gravity or the pressure difference increases and decreases when the density decreases.

Bernoulli's law cannot be used to compare the two cases - you can apply it only to a flow which is already constant, not to compare two different flows. It's the same difference as saying that momentum does not change in a system with no external forces and expecting all systems with no external forces to have the same momentum as each other.

Thanks I think I finally get it but don't you mean acceleration "increases when density decreases" because if density decreases (in denominator) then acceleration will increase a=g+P/δρ and thus change in velocity would increase. I think what really made me understand this question was realizing that these are two seperate systems and with that distinction are two seperate Q values--the Q values for two seperate bodies of fluid will differ and not be constant if V or A change. This wasn't clear to me before because all you really read about is how flow rate is always constant in ideal fluids, so this caught me off guard for the obvious reason of Q not being constant. Got it got it thank you milski!

 

[milski]

Thanks I think I finally get it but don't you mean acceleration "increases when density decreases" because if density decreases (in denominator) then acceleration will increase a=g+P/δρ and thus change in velocity would increase. I think what really made me understand this question was realizing that these are two seperate systems and with that distinction are two seperate Q values--the Q values for two seperate bodies of fluid will differ and not be constant if V or A change. This wasn't clear to me before because all you really read about is how flow rate is always constant in ideal fluids, so this caught me off guard for the obvious reason of Q not being constant. Got it got it thank you milski!

Yes, you're right, my bad - too many decreases and increases in the same sentence. I know what you mean - the consequences of Bernoulli's equation are not exactly intuitive so everyone keep harping on them and then we tend to use them even when they don't apply.

One thing I was thinking as we were discussing this: Ideal behavior is not that unusual. Most of the time you can treat non-extreme real life events as ideal and the laws for ideal behavior are a good starting point. It's only when it comes down to being more precise with the numbers where adjustments have to be made.