Question

A 6.3 cm diameter horizontal pipe gradually narrows to 4.5 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.0 kPa and 26.0 kPa , respectively. What is the volume rate of flow?

Answer 1

Apply the Bernouilli equation -

means we have -

31000 + rho*v1^2 = 26000 + rho*v2^2 --------------------------------------(i)

and the continuity equation
v1*A1=v2*A2
or v1=(d2/d1)^2*v2= (4.5 / 6.3)*v2 = 0.714*v2

now put this value in (i) -

rho = 1000 kkg/m^3

31000 + 1000*(0.714*v2)^2 = 26000 + 1000*v2^2

=> 5000 = 490*v2^2

=> v2 = 3.19 m/s

and A2 = pi*(d2/2)^2 = 3.141*(0.0225)^2 = 0.00159 m^2

Therefore, the volume flow rate = A2*v2 = 0.00159*3.19 = 0.0050721 m^3/s = 5.0721 liter/s.