Flow measurement in venturi meter

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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?
 
freshbox said:
No answer is 0.064L/s :cry:

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

Can someone please explain to me, thanks!
 
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.
 
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
 
Attachments
  • Handwritten-style equation: Vth = π(0.03) times the square root of 2 × 2x9.81x176.58 over 9810[(π0.015²)/(π0.006²) − 1], written in red on a white background.
    as.jpg
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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##.
 
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.
 
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.