Fluid Dynamics: Find the Pipe Radius at Height Increase

[student 1]

Homework Statement

Water is flowing through a pipe of radius 4 cm with a speed of 15 m/s The pipe increases 3 M in height with no change in pressure or speed. What is the radius of the pipe in cm at this new height?

Homework Equations

P1+1/2\rhoV1²+\rhogY1=P2+1/2\rhoV2²+\rhogY2
The rho's are not exponents they just appeared like that.

A1V1=Constant

The Attempt at a Solution

I know since I have the radius and speed I can find the Flow rate using the continuity equation. (.04)²\pi(15 M/s) = .075 m³/s

The Pi should not be an exponent either. Sorry
Since I have the flow rate I do not know what I should do to relate the rise of 3M in height with Bernoulli's equation. I know since they have the same speed density and pressure; most of the factors will cancel out. Am I missing an equation that could help me out??

Homework Statement

Homework Equations

The Attempt at a Solution

 

[minger]

Are you sure that shouldn't read there is no change in pressure or mass/volume flow? If there is a change in elevation then there either must be a change in pressure and/or a change in velocity.

 

[student 1]

No, the problem states that there is no change in pressure or speed. With that being said I assume since the Pipe has changed elevation the Pipe has to change the radius to accompany the fact that there is not a change in the pressure or speed. The Answer is supposedly 4.3 cm, but I have no clue how to get that. I'm sure I have to relate the potential energy of the higher pipe to the original starting location but I don't Know what formula I should use or even if this will help me.

 

[student 1]

Well, I know you are suppose to pretend the pipe does change and find the change in velocity that would occur for the pipe. Then change the pipe's radius to allow for the speed of the water to flow the same as the pipe on the ground, Any help with this problem please!

 

[nucleus]

You need to use the Continuity equation Flow Q = A1v1 = A2v2 = A3v3 ...
You have already found A1v1 (.075 m3/s)
The velocity v1 = v2 = 15. Note this should be call velocity head because overall velocity does change due to change in area.
Now you need to find Z1 – Z2.
Then you can find v2 and hence A2.

 

[minger]

This problem doesn't make any sense. Maybe I'm overthinking it, but you cannot just use continuity. If you use continuity, then the problem is just as simple as
(\rho A V)_{in} = (\rho AV)_{out}

However, this that's assuming that there's only one unknown, A, meaning that the velocity is unchanged (which the problem states). HOWEVER, this means that you can directly solve for A_{out} without taking into account the change in elevation.

Simply put, conversation of energy states that a change in elevation will cause either a change in pressure, velocity, or both.

 

[student 1]

Minger you had it right. You do have to assume that there is a change in velocity since there is a change in the elevation. However, the way the problem is stated it would not seem that you could do that. So, once you have found the velocity of the second spot 3m above you can use the continuity equation to solve for your radius at the elevated spot.

 

[nucleus]

minger and student 1
The way I see this problem is that we already have a change of pressure (p1-p2) that is causing a flow of 15 m/s. Now we are adding a flow due to the change of elevation.
Because this is a volume flow rate the density and specific weights cancel.
Z1-Z2 = 3-.4 = 2.6m
V2 = 15 – 2.6 = 12.4
Then A2 = A1v1/v2 = 4.3cm which is the answer provided by student 1.
student 1 By Bern formula you must have a difference in pressure or elevation, or both in this case, to get flow.

PS I see you have posted while I was writing this but I will post anyway.