Expression for rate of flow through venturimeter :

 Expression for rate of flow through venturimeter : 

Consider a venturimeter fixed in a horizontal pipe through 

which a fluid is flowing (say water) as shown in figure 4. 

 Let d1 = diameter at inlet or at section 1 

 p1 =Pressure at section 1

 v1 = velocity of fluid at section 1 

 a1 = area at section 1= (π/4)*d1 2 

And d2, p2, v2, a2 are corresponding values at section 2

 Applying Bernoulli’s equation at section 1 & 2 we get 

 (p1/ρg) + (v1 2 /2g) + (z1) = (p2/ρg) + (v2 2 /2g) + (z2)  

As pipe is horizontal, hence z1=z2 

(p1/ρg) + (v1 2 /2g) = (p2/ρg) + (v2 2 /2g) or 

 (p1-p2)/ρg = (v2 2 /2g) - (v1 2 /2g) ---- (1) 

But (p1-p2)/ρg, is the difference of pressure head at sections 1 & 2 and it is equal to ‘h’ or 

(p1-p2)/ρg = h 

Substituting the value of (p1-p2)/ρg in the above eqn. (1) we 

Get, h = (v2 2 /2g) - (v1 2 /2g) ---- (2) 

now applying continuous equation at sections 1 & 2 a

1v1= a2v2 or v1 = (a2v2)/ a1 

substitute the value of v1 in equation (2)

 h= (v2 2 /2g) - [(a2v2/ a1) 2 /2g] = (v2 2 /2g)[1-(a2 2 /a1 2 )] 

 = (v2 2 /2g)[(a1 2 - a2 2 )/ a1 2 ] 

v2 2 =2gh [a1 2 /( a1 2 - a2 2 )]  

Therefore v2 =√[2gh {a1 2 /( a1 2 - a2 2 )}] 

 v2 = [a1 /√( a1 2 - a2 2 )]* √(2gh) 

Discharge Q = a2v2 

Qth = a2*[a1 /√( a1 2 - a2 2 )]* √(2gh) ---- (3) 

Equation (3) gives the discharge under ideal conditions and is called theoretical discharge. Actual discharge will be less than theoretical discharge. 

 Qact = Cd[a1a2 /√( a1 2 - a2 2 )]*√(2gh)] 

Where Cd is coefficient of venturimeter and its value is less than 1.