This device measures the flow speed of incompressible fluid which are flowing through the pipe of different cross sectional area. This device is also called flow meter or venturi tube.
This device consists of a horizontal tube or pipe with having wider opening of cross sectional area A1 and a narrow neck of cross section A2 , this two regions of the horizontal pipe are connected to a manometer which contains a liquid of density σ . At the wider opening the velocity of fluid is v1 and pressure p1 and at the narrower opening the velocity of fluid is v2 and fluid pressure is p2. And the density of fluid which are flowing through the pipe is ρ. For reference see the diagram below:
Now we are going to deriving the formula for speed of fluid , which are flowing through the tube or pipe.
DERIVATION FOR FLOW RATE
If fluid is flowing through the tube then according to the equation of continuity , we get:
A1v1 = A2v2. Or. (A1/A2) = (v2/v1)
Now using Bernoulli principle for horizontal flow of liquid in the pipe with density ρ.
p1 + (1/2)ρv1^2 = p2 + (1/2)ρv2^2
Then we get;
p1 – p2 = (1/2)ρ[v2^2 -v1^2]
Taking common v1^2 then we get;
p1 – p2 = (1/2)ρv1^2[(v2^2/v1^2) -1]
We know that ; ( v2^2/v1^2) = ( A1^2/A2^2)
If v2/v1 get squared then A1/A2 also get squared.
Then equation become;
p1 – p2 = (1/2)ρv1^2[(A1^2/A2^2) – 1]
This pressure difference in the pipe also creates a difference in height in the liquid which are filled in the manometer.
So the difference in height h of two arms of the manometer measure the pressure difference , which is given as.
p1 – p2 = σgh
So putting σgh in the above equation in place of p1 – p2 then ,
σgh = (1/2)ρv1^2[(A1^2/A2^2) – 1]
So from the above equation
Velocity of fluid is
v1 = √(2hσg/ρ)×[(A1^2/A2^2) -1]^-1/2
Here second term is out of square root.
It is the speed of liquid in the wider opening.
Then the volume of the liquid flowing per second through the wider opening is given as follows;
According to the equation of continuity ,
V = A1v1 then putting the value of v1, we get ;
A1v1 = A1√(2hσg/ρ)×[(A1^2/A2^2) -1]^-1/2
On simplifying we get;
V = A1A2√[2hσg/ρ(A1^2 – A2^2)]
And we know that hσg is equal to the p1 -p2
Then after putting the value of σgh in the above volume equation , then we get the final formula for volume flowing per second in the tube.
V = A1A2√[2(p1-p2)/ρ(A1^2 – A2^2)]
Watch this video for reference.