Question

A 6.7 cm diameter horizontal pipe gradually narrows to 5.3 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.5 kPa and 23.4 kPa , respectively.What is the volume rate of flow?

Answer 1

Bernoulli's law says that energy is conserved in the flow (no viscous or turbulent losses):
P + 1/2 d v^2 + dgh = constant

where d is the density of the fluid, P is pressure, v is linear velocity, and h is the height of the flow. If the pipe is horizontal, h is the same at both points, so
P1 + 1/2 d v1^2 + dgh = P2 + 1/2 d v2^2 + dgh
P2 - P1 + 1/2 d (v2^2 - v1^2) = 0

The volume velocity u is linear velocity times cross-sectional area. If the fluid is incompressible, then the volume velocity is constant.
u = v1 (pi r1^2) = v2 (pi r2^2)
P2 - P1 + 1/2 d (v2^2 - v1^2) = 0

P2 - P1 + 1/2 d (u / (pi r2^2) )^2 - (u / (pi r1^2) )^2 =0
u^2 = - (P2 - P1) / { 1/2 d (1/r2^4 - 1/r1^4) / pi^2 }
= - (31.5e3 N/m^2 - 23.4e3 N/m^2) / { 1/2 (1e3 kg/m^3) (1/(0.067/2 m)^4 - 1/(0.053/2 m)^4) / pi^2 }
=12.9e-5 m^4/s^2
u = 11.364e-3 m^3/s^2
x (1000 L / 1 m^3) = 11.364 L/s