# Water flow venturi meter problem! Heeeelp please

1. Feb 21, 2005

### fcukniles

water flow venturi meter problem! Heeeelp please :)

Hi,
The objects of my experiment where
1) to find out the theoretical and experimental pressures in a venturi tube,
2) To determin the coefficient of discharge for the venturi

here is the data i collectec http://cniles0.tripod.com/work020.jpg

im having problems with my two objectives,
i have i have been playing arround with the following formula's but with no luck,
.
m = A2 [sqrt (2 x delta p x viscosity)/1-k^2

K=A1/A2

therefore K = 2.63998

.
V = 201.1 x sqrt [2 x delta p x 1 (waters viscosity coeffient)]/[1 - 2.64^2]

Here is the hand out i got, giving some formula's and it gives cross sectional values for the venturi tube but not sure what do do with these?

http://cniles0.tripod.com/venturi_water.doc

i have worked out the experimental rh but i dont know how to find out the therotical rh, and im not sure what Cb (show on results table is).

thanks for any help ppl
Kris

2. Feb 21, 2005

### fcukniles

any help would be really apricated thanks

3. Feb 21, 2005

### minger

I'm not sure what you are measuring in your experiment....velocity I assume. Here's what I would think you would be doing.

You have a given Q (flow rate), and you are looking for coefficient of discharge. Well based on continutity, you know what the flow rate at any point along your tube must be the same. So given the area upstream of the venturi, you can find pressure and velocity using Bernoulli's Equation, and then Q=VA. Now using that last equation, you can see that if flow rate stays constant (which in your case it does), any change in the area will yield a proportional change in the velocity. With your new velocity, plug that back into Bernoulli's Equation to solve for a theoretical pressure. Now take your actual measured velocity and find your actual pressure. Divide these to get your coefficient.

4. Feb 21, 2005

### FredGarvin

I am not quite sure what your question refers to. What is "rh?" What is "Cb?" Do you mean Cd (which is the discharge coefficient)? Also, I am having a lot of troubles reading your results page that you scanned in.

So...let's talk in general.

The theoretical flow through the venturi is given by the equation you have already mentioned:

$$M = A_{2} \sqrt{\frac{2 \Delta P \rho}{1-k^2}}$$

That is the theoretical flow (in a perfect world). With the measurements you have you should be able to calculate both the theoretical and what actually happened. There will be a difference. That difference is the discharge coefficient of the venturi:

$$M_{actual} = M_{theoretical} C_{d}$$

I guess the first big question is do you understand what the data you took is for and where it fits into the venturi equation?

5. Feb 21, 2005

### fcukniles

yeh how do i work out the coefficient of discharge?
thanks

6. Feb 21, 2005

### minger

I think I explained it decently well, but let me try again.

Your coefficent of discharge is a fraction that tells you how much flow you will actual get based on theoretical. Qactual = Qtheo*C

To get your theoretical flow, you will use a combination of Bernoulli's Equation, and the equation for flow, flow simply being Q=VA.

Since you have your areas and initial velocity, finding your theoretical flow through the venturi should be simple. It seems to me that you are measuring velocity in your experiment so...take your measured velocity and multiply by area to get your actual flow. Divide your theoretical by your actual to get the coefficient.

7. Feb 22, 2005

### fcukniles

im not quite sure how to find out the theoretical and experimental pressures in a venturi tube is it something to do with the thickness of the tube and the high difference?

8. Feb 22, 2005

### fcukniles

heres what am getting at the minute and it doesnt seam right:

A1=201.1 (area of crossection)
A2=530.9
K=0.37879
DeltaP= 195

Putting these numbers into the equation i get 17912.11 which is nothing like any of my other values any tell me what this means or where i have gone wrong?
thanks

9. Feb 22, 2005

### FredGarvin

What are the units of your $$\Delta P$$? That large of a number looks like possibly inHg or inH2O. You must use the proper units.

Also, you are not finding the pressures. You measured them in your experiment! You are calculating the theoretical flow using your measured pressures and comparing it to what you actually measured in the bucket.

10. Feb 22, 2005

### fcukniles

i measured the height difference in each of the tubes, these values are in mm how do i work out the pressure change? thanks for help :)

11. Feb 22, 2005

### FredGarvin

millimeters of what? Water I am assuming. You need to use consistent units with your calculations. Look at the units you have for density and area.

Look for conversions from mmH2O to psi or Lbf/ft^2.

12. Feb 22, 2005

### Gamma

He is measuring the mass flow rate. So the

Volume flow rate = 1/density * Mass flow rate

Isn't this the experimental flow rate?

So he is left with finding the theoretical flow rate.

fcukniles:
Do you have a diagram of your experimental set up? What are p1, p2 .p11? Difficult to read your data.

13. Feb 22, 2005

### FredGarvin

Since there seems to be some confusion...here's what I see as confusing you:

The mass flow in the tube at any point along the line has to be the same (conservation of mass). Therefore, if you take any two points where you did the measurements, you should calculate the same theoretical mass flow rate. So, for example, take station #1 (the entrance to the venturi) and station #4 (the throat section, i.e. smallest area):

- You measured a mass flow rate of .424 kg/sec (trial #2)

- The $$\Delta P$$ is the absolute value of the difference between two pressure readings. In your case, P1=222 mmH2O and P2=20 mmH2O. You HAVE TO convert these readings to the proper units. Since this is in metric, that would be into Pascals or N/m^2 (the unit of Newtons can be further broken down into kg*m/s^2. That leaves you with an overall pressure unit of (kg*m)/(s^2*m^2))

- The area ratio, k (what is normally called the "beta" ratio) is the ratio of the smaller area to the larger area, or this case, 201.1/530.9. NOTE: Since this is a ratio, you don't have to worry about the units here. No matter what units you use, the answer will be the same.

- Since I didn't see it anywhere, I am assuming you used a value for the density of water of 1000 kg/m^3. Again, you have to use the proper units!

NOW you can calculate the THEORETICAL mass flow through the venturi. Once you get that value, compare that value to the measured mass flow rat to calculate the Cd.

I did run the numbers that you had for this trial and they came out pretty good. The Cd I calculated is along the lines of what I would expect for a venturi meter.

Last edited: Feb 22, 2005
14. Feb 24, 2005

### fcukniles

how do i convert them? i've been playing around with the numbers but no luck, nothing that is close to 0.450

15. Feb 24, 2005

### fcukniles

ok im lost...

me = wild sheep in new york

16. Feb 24, 2005

### fcukniles

baaaaaaaaaa

17. Feb 24, 2005

### Staff: Mentor

The equivalence for pressure at the some height of water can be found from the relationship P = $\rho$gh. Units have to be consistent.

However,

1 atm = 1.013 bar = 1.01325E+5 Pa = 760 mmHg = 1.01325E+4 mmH2O = 1.033 kg/cm2 = 14.696 psi

1 Pa = 1E-5 bar = 9.869E-6 atm = 0.0075 mmHg = 0.1 mmH2O = 1.02E-5 kg/cm2=0.000145 psi

Calculator for pressure - http://www.lenntech.com/unit-conversion-calculator/pressure.htm

Just put a number in one of the boxes and hit calculate.

18. Feb 24, 2005

### fcukniles

ok so if:
A1 = 30mm => 300 N/m^2
A2 = 195mm => 1950 N/m^2
k= 0.15385
so puttin these into the formula:

1950 x [sqrt (2x1650x1000)/1-0.15385^2]

3569835.251 !?!

pretty sure this aint right!

any help?

19. Feb 24, 2005

### FredGarvin

The "A" variables are the AREAS of the sections you are considering. How are you going from A in 'mm ' to A in N/m^2?

Short of actually doing it for you, I would suggest you post the work you have done. Every step. Then we can show you where you are taking the wrong steps.

20. Feb 24, 2005

### fcukniles

ok right here is what i think im doing:
if A1= 221.7mm^2 A2= 530.9mm^2 need to convert these into N/m^2
so i divide these by 1000. Giving me A1 = 0.2217N/m^2 A2= 0.5309N/m^2.
so k = 0.4176. Putting the Change in height 165 mm into this http://www.lenntech.com/unit-conversion-calculator/pressure.htm I get Delta P to =1650.
So... putting these numbers into the formula I get...

0.5309N/m^2 x sqrt([2 x 1650 x 1000]/ 1-0.4176^2])

Ans 1061.4 ?