Determine the rate of flow of water through a pipe of 300mm dia placed in an inclined position where a venturimeter is inserted, having a throat dia of 150mm.The difference of pressure between the main throat is measured by a liquid of specific gravity 0.7 in an inverted u-tube which gives a reading of 260mm.The loss of head the main and throat is 0.3 times the kinetic head of the pipe.

 Determine the rate of flow of water through a pipe of 300mm dia placed in an inclined position where a venturimeter is inserted, having a throat dia of 150mm.The difference of pressure between the main throat is measured by a liquid of specific gravity 0.7 in an inverted u-tube which gives a reading of 260mm.The loss of head the main and throat is 0.3 times the kinetic head of the pipe. 

Solution: 

 Given: 

Dia of inlet, 

D1=300mm=0.3m 

Therefore area of inlet, A1= π/4*(0.3)2 =0.07m2 www.getmyuni.com 42 

Throat dia, D2=150mm=0.15m 

Therefore area of throat, A2= π/4*(0.15)2 =0.01767m2 

 Specific gravity of lighter liquid (u-tube) sl =0.7 

Specific gravity of liquid (water) flowing through pipe, 

Reading of differential manometer, x=260mm=0.26m 

Difference of pressure head, h is given by 

((p1/ρg) +z1) - (p2/ρg) +z2) =h 

Also, h= x (1- sl/sw) =0.26(1-0.7/1.0)

 =0.078m of H2O 

Loss of head, hL=0.3*kinetic head of pipe = 0.3 * v1 2 /2g 

Now applying Bernoulli’s equation at section ‘1’ and ‘2’, 

We get, (p1/ρg) + z1 + (v1 2 /2g) = (p2/ρg) +z2 + (v2 2 /2g ) + hL 

[(p1/ρg) +z1) - (p2/ρg) +z2 ]+ [(v1 2 /2g) - (v2 2 /2g)] = hL 

But [(p1/ρg) +z1) - (p2/ρg) +z2 ] =0.078 m of H2O 

And hL =0.3*(v1 2 /2g) 

 therefore, 0.078 + [(v1 2 /2g) - (v2 2 /2g)] = 0.3*(v1 2 /2g) 

0.078 +0.7(v1 2 /2g) - (v2 2 /2g) = 0 ---- (1) 

Applying continuity equation on section (1) and (2) , 

we get A1v1=A2v2 v1= A2v2/A1 = v2/4 

Substitute ‘v1’ in equation (1), we get 

0.078 + (0.7(v2 2 /4))/2g - (v2 2 /2g) 

= 0 0.078 + (v2 2 /2g) ((0.7/16)-1) = 0

 (v2 2 /2g) * (-0.956) = - 0.078 

v2 2 =0.078*2*9.81/0.956=1.6 

 v2 =1.26m/s 

Rate of flow Q= A2v2=0.01767*1.26 

 Q=0.0222m3 /s