Flow measurement in venturi meter

AI Thread Summary
The discussion revolves around calculating the theoretical flow rate using a venturi meter, where confusion arises due to the constant diameter of the pipe. Participants clarify that the diameters at points 1 and 2 are different, with the throat diameter being smaller, which is essential for the calculations. The pressure difference (P1-P2) is a critical factor, and participants discuss how to derive it using hydrostatic principles. There is a consensus that the correct approach involves understanding pressure at a depth and applying the appropriate formulas, including the coefficient of discharge for real flow rates. The conversation highlights the importance of correctly identifying areas and pressures in the venturi meter setup to avoid mathematical errors.
freshbox
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First I am trying to find the theoretical flow rate but I am not able to get the answer because since diameter of the pipe is 0.03m throughout, when I sub into the formula:

My A1/A2 will be 1, and 1-1 = 0. And I will get maths error whatever is divided by 0.

Am I going in the wrong direction?

Please advice, thanks.
 

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freshbox said:
First I am trying to find the theoretical flow rate but I am not able to get the answer because since diameter of the pipe is 0.03m throughout, when I sub into the formula:

My A1/A2 will be 1, and 1-1 = 0. And I will get maths error whatever is divided by 0.

Am I going in the wrong direction?

Please advice, thanks.

No, the diameter is not 0.03 m throughout. Read the problem statement again.
 
I mean the 2 pipes at point 1 & 2 connecting to the venturi meter
 
freshbox said:
I mean the 2 pipes at point 1 & 2 connecting to the venturi meter

As per the given question, at point 1 it is 0.03 m and at point 2 (throat), its 0.012 m.
 
I am confused. Is A1= Area of Left hand side pipe, A2= Area of Right hand side pipe in the theoretical flow rate formula?

"A venturi-meter is installed in a pipe of 30mm diameter" the pipe the question is talking about is it the 2 vertical pipes or the horizontal pipe?
Thanks.
 
freshbox said:
I am confused. Is A1= Area of Left hand side pipe, A2= Area of Right hand side pipe in the theoretical flow rate formula?

No. Your book defines A1 and A2, maybe on the previous page, no?

Do you know about Bernoulli theorem?

"A venturi-meter is installed in a pipe of 30mm diameter" the pipe the question is talking about is it the 2 vertical pipes or the horizontal pipe?

The 30 mm diameter pipe is the horizontal pipe (isn't that obvious? :biggrin: ).

The vertical pipes are used for calculating pressures at point 1 and 2.
 
Ok I think I got it for the pipes. I am having problem finding P1-P2

Can I ask Since P1=P2

P1=pg(0.045)
P2=pg(0.027)

P1=P2
1000x9.81(0.045)=1000x9.81(0.027)
441.45=264.87 -> Stuck...

There is a note in my book saying "The pressure difference, (P1-P2) is obtained from a U-tube manometer."
Since this is not a u-tube manometer, how do I go about finding P1-P2?Thanks.
 
freshbox said:
Ok I think I got it for the pipes. I am having problem finding P1-P2

Can I ask Since P1=P2

P1=pg(0.045)
P2=pg(0.027)

P1=P2
1000x9.81(0.045)=1000x9.81(0.027)
441.45=264.87 -> Stuck...

There is a note in my book saying "The pressure difference, (P1-P2) is obtained from a U-tube manometer."
Since this is not a u-tube manometer, how do I go about finding P1-P2?


Thanks.

How P1=P2? :confused:

What is the pressure at a depth in a liquid?
 
I take the reference point at P1 and P2. Like how I did for Px and Py in the attached screenshot.What is the pressure at a depth in a liquid? I don't know sorry.
 

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  • #10
freshbox said:
I take the reference point at P1 and P2. Like how I did for Px and Py in the attached screenshot.

Looks good to me. Can't you find P1 and P2 in the problem you have posted?

P1=P_(atm)+(density)*g*(h1).

Similarly find P2.
 
  • #11
What is P_(atm)? Why do I need to take in account into my equation?

So can I still take my reference point at point 1 & 2 like how I did for Px & Py (Px=Py) so P1=P2?Thanks.
 
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  • #12
freshbox said:
What is P_(atm)? Why do I need to take in account into my equation?Thanks.

Sorry, I should have defined it. P_(atm) is the atmospheric pressure. I can't help you further until you learn how to calculate pressure at a depth. Pressure at depth, which is taught in hydro statics should be in your notes or you haven't yet started with it? But its unlikely that you begin with hydrodynamics before studying hydro statics.

For this problem, P1-P2=(density)*g*(difference in heights of liquid in vertical tube).
 
  • #13
Is P1=P2?
 
  • #14
freshbox said:
Is P1=P2?

You should refer your textbook or search online for pressure at depth.
 
  • #15
I have search and don't understand the information online. My teacher said that I should find a reference point and the reference point should have the same liquid.

So I tried to apply post #9 method on my post #1 question which is by taking a reference point. Post #9 I can find the pressure difference between 1 & 2. But for post #1, I am unable to form the equation to "P1-P2=xxxx"

Can you tell me where I go wrong?Thanks.
 
  • #16
freshbox said:
I have search and don't understand the information online. My teacher said that I should find a reference point and the reference point should have the same liquid.

So I tried to apply post #9 method on my post #1 question which is by taking a reference point. Post #9 I can find the pressure difference between 1 & 2. But for post #1, I am unable to form the equation to "P1-P2=xxxx"

Can you tell me where I go wrong?


Thanks.

What you do in post #9 is same as calculating pressure at a depth.

In post #9, you go down, in this problem you go up. :wink:

Think where you should mark X and Y in this case.
 
  • #17
Ok. I have labeled my new Px and Py. But I think it's abit wrong because what do I do for the "?" part of water in Px.

But if it is wrong I thought my reference point should have the same liquid:confused:
 

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  • #18
freshbox said:
Ok. I have labeled my new Px and Py. But I think it's abit wrong because what do I for the "?" part of water in Px.

But if it is wrong I thought my reference point should have the same liquid:confused:

It would much easier if you place Px at the top most point on the liquid inside the vertical tube on the left.
 
  • #19
Ok. How about my Py? I thought they should be on the same level with each other and with the same liquid too?

Because my post #9 referencing points are at the same level and having the same liquid.Please explain, I'm abit confused. thank you sir.
 
  • #20
freshbox said:
Ok. How about my Py? I thought they should be on the same level with each other and with the same liquid too?Thanks.

Yes, take them on the same level.

EDIT: And don't call me sir, I am a student myself. :D
 
  • #21
Ok same referencing point but Py is not having the same liquid as Px.

My lecturer said that referencing point must be at the same level and is having the same liquid.

I'm confused over this part.It's ok, it's a form of respect, if you don't like I won't call you that haha.. Thanks anyway for your help :)
 

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  • #22
Px=9810(0.045)+P1
Py=9810(0.027)+P2+Patm

Since Px=Py

9810(0.045)+P1=9810(0.027)+P2+Patm
P1-P2=264.87-441.45
P1-P2=-176.58 -> Is the answer wrong?
 
  • #23
freshbox said:
Px=9810(0.045)+P1
Py=9810(0.027)+P2+Patm

Since Px=Py

9810(0.045)+P1=9810(0.027)+P2+Patm
P1-P2=264.87-441.45
P1-P2=-176.58 -> Is the answer wrong?

Py is actually the atmospheric pressure. Going up from P1,
P1-9810(0.045)=Px

If we go up from P2,
P2-9810(0.027)=Py.

Pressure above the liquid in the vertical tubes is same everywhere so Px=Py. Solve to find P1-P2.
 
  • #24
Don't understand at all. The questions that I did I will add the pressure together. :cry:
 
  • #25
freshbox said:
Don't understand at all. The questions that I did I will add the pressure together. :cry:

The value you calculated for P1-P2 is correct but the sign is wrong. You can substitute that value if you wish and obtain the answer but the way you calculated is wrong. I don't know how can I explain it to you in simpler terms. We should wait for the experts to join.
 
  • #26
Answer wrong :cry:
 

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  • #27
freshbox said:
Answer wrong :cry:

I can't read what's written in front of the root symbol. Also, you need not plug in the value of g, look at the formula, g cancels out so omit it.
 
  • #28
0.03 is written in front of the root symbol. With or without the g I will still can the same answer right?
 
  • #29
freshbox said:
0.03 is written in front of the root symbol.
Why 0.03? The formula shows that its area in front of root symbol so it should ##\pi(0.015)^2##.
With or without the g I will still can the same answer right?
Yes.
 
  • #30
Yea sorry missed out on that one..hehe.. I calculated as 0.27L/s. Did you get the answer?
 
  • #31
freshbox said:
Yea sorry missed out on that one..hehe.. I calculated as 0.27L/s. Did you get the answer?

I haven't calculated it but isn't that correct?
 
  • #32
No answer is 0.064L/s :cry:

I still don't understand :cry:
 
  • #33
freshbox said:
No answer is 0.064L/s :cry:

Hmm...are you sure it is L/s, not mL/s?
 
  • #34
The answer from post #1 0.064 L/s is written by my lecturer.
 
  • #35
The problem I am having right now is how do I go about finding P1-P2.

Can someone please explain to me, thanks!
 
  • #36
freshbox said:
The answer from post #1 0.064 L/s is written by my lecturer.

Sorry, the given answer is correct.

Show your steps, I can't point out the error in your working unless you show your calculations.
 
  • #37
P1-9810(0.045)=Px
P2-9810(0.027)=Py

P1-P2=176.58

I forget the Square at [A1/A2]

My final answer is 0.27L/s
 

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  • #38
freshbox said:
P1-9810(0.045)=Px
P2-9810(0.027)=Py

P1-P2=176.58

I already said you to omit g. Also, don't forget to multiply with coefficient of discharge.

Check the term outside the radical, It should be ##\pi(0.015)^2##.
 
  • #39
I thought the question is asking to find the theoretical flow rate so why we have to multiply Cd?

And can you explain to me how you get P1-P2 again?

I used to add all the pressure together, but I see that you subtract them.Thanks.
 
  • #40
freshbox said:
I thought the question is asking to find the theoretical flow rate so why we have to multiply Cd?

The question states that we have to find the water flow rate and 0.064 L/s is the actual flow rate. That's why I asked you to multiply with Cd.
And can you explain to me how you get P1-P2 again?
Explaining things is not my strong point, I thought you had some problem in arithmetic but I did not know that you haven't yet learned about calculating pressure at a depth. I would suggest you to ask your lecturer or wait for someone else to join.
 

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