# A venturi meter problem!

#### sinedine

these are the following data:

diameter(pipe)=0.1m
diameter(throat)=0.05m
difference in pressure=55kpa
Cd=0.98
question requires me to find the volume flow rate.

1)i know i need to find h(diff in pressure/density*gravity)
i have a doubt here where can i just leave the pressure in kpa or must i change it to pa??

2)from there i know i can find volume flow rate=sq root of [ (2gh)/ (A1/A2)^2 -1],where A1=surface of pipe and A2=surface area of throat(this formula is where my doubt lies in cause i have seen other variations!!is this formula correct in the 1st place)

3)finally Q (actual)= Q (theor)*Cd

i am not quite sure of what i am doin in step 2 is correct because i have come across another formula in which Q= Cd*surface area(throat)*sq root of (2gh)!!but while i use this formula my ans are not the same!!can someone help me with doubt.

Related Introductory Physics Homework Help News on Phys.org

#### arun_mid

ans to 2) no, you cannot use that formula. It is applicable only to the situation of velocity of efflux, and that too, to a specific case. Do not use it here, it is better to start from Bernoulli's theorem.
ans to 1) You DO NOT need to find h. The Bernoulli's theorem deals directly with the problem. Please try it out using this theorem or try to find out what it is.
ans to 3) What is Cd?!
Don't rely on too many formulae...they usually confuse and cram the head up :). Start from scratch, or if you're prone to mistakes, just slightly more than scratch.

#### sniffer

Cd is the drag coefficient, it will reduce the resulting flow rate because you have bigger pressure drop.

since height of the two sections are the same (is it?) arun mid is correct, i.e. use bernoulli and continuity (incompressible fluid).

P1 + 0.5*rho*(v1)^2 = P2 + 0.5*rho*(v2)^2.

and

(v1)(A1)=(v2)(A2).

solve simulatneously. P1-P2 = 55000 Pascal.

you'll get v1 and v2. Calculate flow rate = rho*A1*v1.

But then if you want to take into account pressure drop due to drag,

drag force = Cd*dynamicPressure*SurfaceArea.

(this is standard aerodynamicist definition)

Move Surface Area to the left,

drag pressure = Cd* dynamic pressure.

drag pressure = Cd * 0.5 * rho * v^2.

So, your delta p will be lower than 55000 Pa.

Substitute the equation above into bernoulli, then solve the quadratic equation.

maybe that helps.???

please tell me if i am wrong.

Sniffer.

#### sinedine

isit possible for me to assume that the height is the same??cause there is a diff in pressure which suppose to mean there is a diff in height as well isnt it?correct me if i am wrong here!actually i wanted to use bernouli's for the start but because w/o velocity of the fluid being given but not knowing what is what is Z1 and Z2 i am still left curious on how this problem can be solved?thanks alot of ur help guys

#### FredGarvin

sniffer said:
Cd is the drag coefficient, it will reduce the resulting flow rate because you have bigger pressure drop.
Cd is not a drag coefficient. It is the coefficient of discharge of the orifice/nozzle. It is a function of the formation of the vena contracta at the throat and the exit conditions of the device due to viscous effects.

The accepted form of nozzle flow equation is of the form:
$$Q_{act} = C_d A_n \sqrt{\frac{2 \Delta P}{\rho (1 - \beta^4)}}$$

Where:
$$Q_{act}$$ = Flow rate
$$C_d$$ = Discharge coefficient
$$A_n$$ = Nozzle area
$$\Delta P$$ = Pressure differential between the inlet and the throat area
$$\rho$$ = density
$$\beta$$ = ratio of nozzle diameter to entrance diameter

Last edited:

#### FredGarvin

sinedine said:
isit possible for me to assume that the height is the same??cause there is a diff in pressure which suppose to mean there is a diff in height as well isnt it?correct me if i am wrong here!actually i wanted to use bernouli's for the start but because w/o velocity of the fluid being given but not knowing what is what is Z1 and Z2 i am still left curious on how this problem can be solved?thanks alot of ur help guys
Unless you have a very large nozzle and the change in elevation is drastic, you can assume that there is no change in elevation. That assumption is very valid. The pressure differential is solely due to the restriction in the flow. Hence why these are called 'differential meters.'

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving