Flow Rate Questions

[rowjimmy]

Would the flow rate and translational velocity of fluid through spigot A be lesser, greater, or the same as spigot B. Why? Assume spigots A and B are at the same height and area of spigot B is twice that of spigot A.

Would the flow rate and translational velocity of the river at Q1 be lesser, greater, or the same as Q2? Why? Assume the river splits evenly and the areas on both sides of the split is the same.

Assume fluid in both problems is ideal.

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[MedPR]

For the spigot question.

If spigot B is larger, its volume flow rate will be greater. Q=V*A. According to the continuity equation, spigot A should have a greater velocity than spigot B. V1A1=V2A2. Bigger area = smaller velocity.

I'm less sure about the river one, but I would assume both would be less for Q2. If Q2 and Q3 have the same area, and the water splits evenly, then all values for Q1 must be equal to Q2+Q3.

 

[milski]

There is no continuity between spigot A and spigot B. They're both going to have the same linear velocity, since that depends on the hydrostatic pressure (or more precisely on the pressure difference between the inside and outside of the spigot). Higher area of spigot B will give you higher flow there.

In the river case you have Q1=Q2+Q3. If the areas at Q2 and Q3 the same, Q2=Q3 and Q2=1/2Q1 and A1=A2+A3, A2=A3=A1/2. Since you halved the area, you'll have to maintain the same linear speed to maintain the same flow.

You cannot apply Bernoulli's law in either case where the flow splits - the split itself is not a continuous flow.

 

[rowjimmy]

There is no continuity between spigot A and spigot B. They're both going to have the same linear velocity, since that depends on the hydrostatic pressure (or more precisely on the pressure difference between the inside and outside of the spigot). Higher area of spigot B will give you higher flow there.

In the river case you have Q1=Q2+Q3. If the areas at Q2 and Q3 the same, Q2=Q3 and Q2=1/2Q1 and A1=A2+A3, A2=A3=A1/2. Since you halved the area, you'll have to maintain the same linear speed to maintain the same flow.

You cannot apply Bernoulli's law in either case where the flow splits - the split itself is not a continuous flow.

Ok so this is exactly what I thought. The spigot question I got from examkrackers 595 and the other question I just made up because the spigot question got me thinking. I had a problem understanding how having two different flow rates for the same ideal fluid didn't violate the rule that "volume flow rate is the same at all points in a conduit of ideal fluid." So if the flow rate of spigot b is twice that of a there seems to be an inconsistency, until you look at the flow rate of the whole body. Would I be wrong if I said that the body of fluid above the spigots flows at a certain flow rate, let's just call it Qc. Is Qc like Q1 in the river problem? In other words, are spigots a and b like river splits Q2 and Q3-- will the sum of the flow rates of Qa and Qb add to Qc? Even if this may be wrong it helps me see how the rule for ideal flow rate still applys. If my observation isn't correct then how else would volume flow rate be the same for this ideal fluid? (Note I know that this rule does hold in this scenario because the explanation in the back of the book says that it is NOT violates but doesn't explain in detail how)

Thanks!!

 

[milski]

Yes, you can say that Qc=Qa+Qb or you can even generalize it as ΣQi=0 for all flows in the tank. Violating that would mean that you're getting/losing additional fluid from somewhere. Note that having the linear velocities be the same is only a special case here. You can have two spigots at different heights - they'll have different linear velocities and different flows but the sum will still stay the same.

 

[rowjimmy]

Yes, you can say that Qc=Qa+Qb or you can even generalize it as ΣQi=0 for all flows in the tank. Violating that would mean that you're getting/losing additional fluid from somewhere. Note that having the linear velocities be the same is only a special case here. You can have two spigots at different heights - they'll have different linear velocities and different flows but the sum will still stay the same.

So if the spigots (a and b) were at different heights theyll both have different Q, V, and A values? I think I get it.

 

[Class1P]

do you spend all day on these forums Milski? You're everywhere.

 

[milski]

do you spend all day on these forums Milski? You're everywhere.

Recently - yes. It's my way of keeping the pre-reqs fresh in my mind. Right now I have plenty of free time on my schedule to do that but obviously that cannot last forever.


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[rowjimmy]

So if the spigots (a and b) were at different heights theyll both have different Q, V, and A values? I think I get it.

That was a question to you milski! Thanks.

 

[milski]

So if the spigots (a and b) were at different heights theyll both have different Q, V, and A values? I think I get it.

Different heights will be different V. A does not really depend on it - they'll be as big or as small as you make them. Q will depend on both the height/velocity and the area. In general they will be different, although you could make a higher spigot with larger area have the same flow as a lower spigot with smaller area.

 

[MedPR]

Different heights will be different V. A does not really depend on it - they'll be as big or as small as you make them. Q will depend on both the height/velocity and the area. In general they will be different, although you could make a higher spigot with larger area have the same flow as a lower spigot with smaller area.

And if you put two spigots of identical area at two different heights, the one nearer to the bottom of the container will have a greater flow velocity through it.

 

[milski]

And if you put two spigots of identical area at two different heights, the one nearer to the bottom of the container will have a greater flow velocity through it.

Yes. Both flow and linear velocity.

 

[johnwandering]

Hey guys, a few questions:

For the question where the river splits

You cannot apply Bernoulli's law in either case where the flow splits - the split itself is not a continuous flow.

I was wondering why exactly this was the case because the two rivers still flow continuously from the first body. Is this just because there are 2 moving water systems now and hence the equation has to be modified to hold two system values?






And for the spigot question

They're both going to have the same linear velocity, since that depends on the hydrostatic pressure (or more precisely on the pressure difference between the inside and outside of the spigot).

I was under the belief that radius had a great effect on velocity.

v= (r^2)(P1-P2)/(4x(viscosity constant)xL)

Considering Pressure is equal for both of these spigots, the one with the greater radius should have the greater velocity?

What am I missing here?

 

[milski]

Hey guys, a few questions:

For the question where the river splits

I was wondering why exactly this was the case because the two rivers still flow continuously from the first body. Is this just because there are 2 moving water systems now and hence the equation has to be modified to hold two system values?

Continuous means that all the fluid flows in the same system without disturbances, adding or losing fluid. Splitting the fluid in two separate systems does not count for that.

And for the spigot question


I was under the belief that radius had a great effect on velocity.

v= (r^2)(P1-P2)/(4x(viscosity constant)xL)

Considering Pressure is equal for both of these spigots, the one with the greater radius should have the greater velocity?

What am I missing here?

That's for fluids with some viscosity. For ideal fluids (like this question) there is no pressure loss and the radius won't matter. In reality the spigot is short enough that for any reasonably big opening you can still treat fluids like water as ideal fluids.

The velocity that you have above is the flow velocity which for ideal fluid is proportional to the area. That seems to agree with the formula above, which makes it proportional to r^2.

 

[johnwandering]

Thanks



If the fluid is ideal, would the "proportional" formula go as:

v= (r^2)(P1-P2)/(4x(viscosity constant)xL)
v=(r^2)(P1-P2)
Hydrostatic Pressure= v/(r^2)

The velocity that you have above is the flow velocity which for ideal fluid is proportional to the area. That seems to agree with the formula above, which makes it proportional to r^2.

Even then I was wondering why the linear velocities of the two spigots were identical.
It seems that the one with the larger A should also have a greater velocity?


I am aware of the Av=Av formula which seems to equate the values to Flow Rate Q.
But considering we don't know flow rate (because the spigots are not continuous), and only know that the pressures are equal, wouldn't it be more accurate to set the relationship between the two spigots to be

Pressure= v/(r^2)
Pressure= v/A
v/A=v/A


So the spigot with the greater Area has the higher velocity?

 

[milski]

The Poiseuille's formula which you are quoting is applicable only for viscous fluids. For non-viscous ideal fluids there is no friction with the walls and no pressure loss, making P1-P2=0. Viscosity is also 0, so the right side becomes 0/0 and you cannot get anything meaningful for the velocity.

There are two popular ways to derive the linear velocity in that case - either use Bernoulli's law for the whole container and the spigot (that requires a few assumptions and is not very intuitive) or just treat it as an energy problem. Both lead to the same result, v=sqrt(2gh).

Having the velocity depend on the radius would imply that the fluid interacts in some way with the walls of the spigot which cannot happen for non-viscous fluid.

 

[johnwandering]

Oh wow thanks a million

 

[hellocubed]

I can't seem to find in the BR a mention that "pressure is the sole cause of velocity in ideal fluids." (unless someone could point it out, which would be very much appreciated.)


Is there any equation that can represent this?

Perhaps:
Pressure= (1/2)(density)(Volume)(velocity^2)