Flow rate through a submerged orifice

AI Thread Summary
The discussion revolves around applying Bernoulli's equation to analyze flow through a submerged orifice. The user attempts to determine flow velocity by considering pressure differences at two points, but struggles with eliminating pressure variables. Clarifications are made regarding the definitions of pressure at different points and the impact of liquid depth on pressure calculations. The conversation highlights the importance of understanding pressure changes with depth and concludes with the user successfully deriving the flow velocity after equating pressure differences. The exchange emphasizes collaborative problem-solving in fluid dynamics.
pressurised
Messages
22
Reaction score
0

Homework Statement


pawWNIe.jpg


Homework Equations


Bernoulli:
p1+½ρU12+ρgz1=p2+½ρU22+ρgz2

Continuity:
Q=U1A1=U2A2

Area=CC*(¼πd2)

The Attempt at a Solution


[/B]
I attempted to apply the Bernoulli equation, making point 1 the top surface of the left hand side and point 2 the instant at the sharp edge.
However I ended up with; pA+ρgz=p+½ρU2

Normally when it is not submerged, the formula can narrow down to U=√2gH, where H=z in this case, but I'm not sure how to eliminate the pressure in the liquid.

Is liquid also trying to move from the right to the left?
 
Physics news on Phys.org
pressurised said:
trying to move from the right to the left?
No. It goes with the flow :smile:

Sorry for being corny. What are pA, z and p ? How many unknowns are there in your attempt equation ?

"Homework Statement " is rather lacking from my point of view...
 
BvU said:
No. It goes with the flow :smile:

Sorry for being corny. What are pA, z and p ? How many unknowns are there in your attempt equation ?

"Homework Statement " is rather lacking from my point of view...

pA would be atmospheric pressure, z is the distance from the contraction point to the top of the left hand side surface, so 1.2m, and then p would be the pressure at the contraction part where it leaves.

The previous part to the problem, in part a) asks me to write down the Bernoulli equation, in b) it tells me to apply the equation to the flow through the sharp edged orifice shown in the picture and sketch the streamline you select to analyse the flow.
I chose to analyse it from the surface and the contraction point.

I've solved questions where the right hand side is air, so the pA's would normally cancel out, but I'm not sure with this one.

Would it be possible to make point 1 a point of very small distance from the surface so that the two p's are equal?
 
pressurised said:
Would it be possible to make point 1 a point of very small distance from the surface so that the two p's are equal?
Point 1? That sounds like the start of the flow. Did you mean at the end?
What would the pressure be at the right of the orifice if it were blocked up?
 
haruspex said:
Point 1? That sounds like the start of the flow. Did you mean at the end?
What would the pressure be at the right of the orifice if it were blocked up?

I mean point 1 as in right at the top of the 1.8m surface.

And what would it be with the liquid there too or if the liquid hadn't have been there? The pressure would be pA I think if it was blocked up?
 
pressurised said:
I mean point 1 as in right at the top of the 1.8m surface.
I don't see how that changes anything. Aren't you already taking the start of the flow as being there?
pressurised said:
what would it be with the liquid there too or if the liquid hadn't have been there?
What the pressure would be with the liquid levels as shown but the hole blocked.
 
haruspex said:
I don't see how that changes anything. Aren't you already taking the start of the flow as being there?

I meant how normally, if this was the most simple straight edged orifice question with no fluid on the right, point 1 at the surface would take atmospheric pressure, and we also let point 2 at the hole take atmospheric pressure too to cancel the Bernoulli equation down to U2=√2gH
I was thinking if we could make point 1 a very small distance below the surface so that it can share the same pressure as point 2, but now I realize that wouldn't work either as the pressure changes with depth.

What the pressure would be with the liquid levels as shown but the hole blocked.
I see. The pressure would be ρg(1.2) on the left and ρg(0.6) on the right?
 
pressurised said:
I see. The pressure would be ρg(1.2) on the left and ρg(0.6) on the right?
That would seem reasonable.
 
haruspex said:
That would seem reasonable.

Thank you! I continued and equated the difference in pressure with the hole blocked with p1-p2 and that gave me the correct answer for U. :)
 
  • #10
Hi I know this post is old but is there any chance that you still have your working for this question or could give me a step by step process of how you did it? thanks
 
  • #11
Sarahbt99 said:
Hi I know this post is old but is there any chance that you still have your working for this question or could give me a step by step process of how you did it? thanks
That would subvert the principles of this forum. Please post your own attempt/thinking first.
 
Back
Top