Hey guys help with Venturi Meter question

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vip_uae
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A venturi meter with a 75 mm diameter throat is installed in a horizontal 150 mm diameter pipeline. The pressure at entry to the meter is 70 kN/m2 gauge and the pressure at the meter throat must not fall below 25 kN/m2 absolute. Calculate the maximum flow for which the meter may be used, given that the density of the flowing fluid is 900 kg/m3 and the coefficient of discharge for the meter is 0.96.


Answer = [0.079 m3/s]

ive been trying this question for the past hour and seriously my head is abt to explode coz i am getting a tottaly different answer i even tried all the formulas on the net i am still gettin my answer

Flow rate (Q) = 0.0456275

and i got this answer
0.0438

soo pleasez guys help me out thanks
 
on Phys.org
can you show your attempt, it would be better if you debug it step by step
 
Thanks for the reply i used this formula

Q= Cd x At
-------- x Sqrt 2 x ( P1 - Pt)
sqrt ( 1 - (At/A1)^2 ) ------- + G
row

Cd drag coff

At Area throat

A1 Area pipe

P1 first pressure

Pt Preasure at throat

G as in gravity

i replaced all the values in and that's it
 
Your entry pressure is in gauge and throat pressure is absolute. Have you considered that?
 
Oh no i haven't noticed that at all... can u explain the difference please
 
any one please explain for me :)
 
Well I've been trying to do what you told me exactly but i still get a lower value please show me the steps for it ...
 
vip_uae said:
Well I've been trying to do what you told me exactly but i still get a lower value please show me the steps for it ...

Since the venturi meter is horizontal and assuming the friction is ignored, the flow rate through the meter (and thus inlet) would be:

[tex]Q = A_2 \cdot v_2 \cdot C_d[/tex]

And the energy balance reduces to:

[tex]\frac{P_1}{\rho} + \frac{v_1^2}{2g} = \frac{P_2}{\rho} + \frac{v_2^2}{2g}[/tex]

Using the Continuity equation to find v_1 in terms of v_2:

[tex]A_1v_1 = A_2v_2[/tex]

[tex]v_1 = \frac{A_2}{A_1} \cdot v_2[/tex]

Using the continuity equation along with an energy balance gives a velocity in the throat of:

[tex]v_2 = \sqrt{\frac{\frac{2g(P_1 - P_2)}{\rho}}{1 - (\frac{A_2}{A_1})^2}}[/tex]

Since P_1 is a gauge pressure, add 101.325 to get 171.325 kPa absolute at the entrance (since 1 kN/m^2 = 1 kPa).

You should be able to plug in the given values now to get the flow rate.

Hope this helps.

CS

PS
You might want to check the math as I ran through that really quick on my scratch pad!
 
Thats Gr8 ... i have found the problem... when i am calculating i forgot to multiply the pressure by 10^3 that's why its giving me a lower value :)

thanks