Calculating Discharge Rate of Fluid in Circular Area

[WhiteWolf98]

Homework Statement

As shown in Figure. 12.5, consider a case where fluid discharges from the origin (point O) at quantity $q$ per unit time. Putting velocity in the radial direction on a circle of radius $r$ to $v_r$, the discharge $q$ per unit thickness is $2\pi r v_r =constant$.

Relevant Equations

$q=2\pi r v_r =constant$
$Q=Av$

This is actually right at the start of another derivation, but I can't understand how the author gets the formula for $q$. So the discharge per unit thickness is the circumference of the circle, multiplied by the velocity at that point (at $r$)? I thought the formula for flow rate was $Q=Av$, so I just can't work this out in my head.

Thank you

 

[Twigg]

$Q = Av$ works fine in 3D. For this 2D problem, try thinking along the lines of $Q = Lv$ where L is the length of the curve that is normal the flow.

 

[haruspex]

The description is sloppily inconsistent. It says the discharge is q per unit time, which would mean q is a volumetric flow rate, dimensionally L³T^-1. But then it refers to "discharge q per unit thickness", implying dimension L², then the equation gives q the dimension L²T^-1.
If we take the last as correct (i.e. the discharge volume per unit time per unit thickness is q) and set the thickness to be D then the volume per unit time is qD and this should equal Av = 2πrDv. Cancelling, q=2πrv.

 

[WhiteWolf98]

Thanks for your replies, but I apologise since I'm still not getting it. If it's 2D, what exactly is the thickness of? And when we talk about volume flow rate here, is this how much fluid the source is discharging at anyone time? It's just more difficult to visualise this in 2D than it is in 3D for some reason. For example, the formula $Q=Av$: when I think about this formula, it makes sense to me when you think about fluid flowing out of a cross-section (for example, a pipe). But here... how can it even be 2D? Is the water just on some flat plate or something? How can I more easily think about the amount of water being discharged per unit time in 2D? Circumference multiplied by velocity just isn't making sense to me

 

[Twigg]

I hear you, and I'm going to second haruspex that this problem could've been written much more clearly. If you want a real-life situation to visualize the problem, imagine if you took two sheets of glass, say 1mm apart, drilled a hole in the middle of one sheet, and poured water in. The quantity Q that you're calculating is essentially the true volumetric flow rate divided by that 1mm separation. That's why they call q "discharge per unit thickness".

 

[WhiteWolf98]

Apologies for the late response, and thanks for the replies so far. That description you gave Twigg, helped a lot in picturing what's happening, but I'm unsure how the units come out to be $L^2~T^{-1}$ as haruspex said. The circumference, multiplied by the depth and the velocity should give $L^3~T^{-1}$. Is there anything wrong with proceeding forward like this? Assuming a depth of just $1$, it would still give me the formula $q=2 \pi r v_r$.

 

[haruspex]

how the units come out to be $L^2~T^{-1}$ as haruspex said. Assuming a depth of just $1$, it would still give me the formula $q=2 \pi r v_r$.

Depth has dimension, it cannot just be "1".
In $q=2 \pi r v_r$, r has dimension L, $v_r$ has dimension LT^-1, so q has dimension L²T^-1.
That makes sense if we consider q to be the volumetric flow rate per unit thickness, and that is a reasonable way to describe what is effectively a 2D flow.
As I wrote, the two text descriptions, "quantity per unit time" and "discharge per unit thickness" mismatch, but in combination suggest that "quantity per unit time per unit thickness" is intended.

 

[WhiteWolf98]

Depth has dimension, it cannot just be "1".
In $q=2 \pi r v_r$, r has dimension L, $v_r$ has dimension LT^-1, so q has dimension L²T^-1.
That makes sense if we consider q to be the volumetric flow rate per unit thickness, and that is a reasonable way to describe what is effectively a 2D flow.
As I wrote, the two text descriptions, "quantity per unit time" and "discharge per unit thickness" mismatch, but in combination suggest that "quantity per unit time per unit thickness" is intended.

So if it was, 'quantity per unit time, per unit thickness', would $L^3~T^{-1}$ be correct? I did mention thickness as $1$; however, if we give it a unit, then we'd get the units of $L^3~T^{-1}$. What I'm trying to say, maybe, is that it only makes sense to me if those are the units, as it also agrees with what Twigg said.

 

[Twigg]

So if it was, 'quantity per unit time, per unit thickness', $L^3 T^{-1}$ would be correct?

$q$ is a measure of volumetric flow per unit time, per unit thickness. Volume has dimensions of $L^3$, time has dimensions of $T^{-1}$ and thickness has dimensions of $L$. So $q$ has dimensions of $(L^3 / T) / L = L^2 T^{-1}$

 

[haruspex]

I did mention thickness as 1;

And I already responded that that makes no sense. You cannot get rid of an inconvenient physical dimension by saying it is "1". 1 what?
As @Twigg posted above, "per unit thickness" means 1/L, so makes the dimension of q equal to L²T^-1 , not L³T^-1 .

 

[WhiteWolf98]

$q$ is a measure of volumetric flow per unit time, per unit thickness. Volume has dimensions of $L^3$, time has dimensions of $T^{-1}$ and thickness has dimensions of $L$. So $q$ has dimensions of $(L^3 / T) / L = L^2 T^{-1}$

That, combined with the example you've given I think clears it up; doesn't really get better than that. And yeah, not assigning a dimension was silly on my part, just wanted to move on with the question. I don't think the dimensions mattered too much in the next steps, but it's quite a problem proceeding with a false understanding. Thank you anyway both of you for helping me understand what's going on, much appreciated.