Question

A 5.1 cm diameter horizontal pipe gradually narrows to 2.5 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 35.0 kPa and 21.8 kPa , respectively. What is the volume rate of flow?

Answer 1

Use Bernoulli's theorem to solve this problem.

The expression of the theorem is given as -  

P + 1/2 d v^2 + dgh = k (constant)

where d is the density of the fluid, P is pressure, v is linear velocity, and h is the height of the flow.

Now, assume that 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.

This means -

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

u^2 = - (P2 - P1) / { 1/2 d (1/r2^4 - 1/r1^4) / pi^2 }

Given that,

r1 = 5.1/2 cm = 0.0255 m, r2 = 2.5/2 cm = 0.0125 m, P1 = 35000 Pa = 35000 N/m^2, P2 = 21800 N/m^2

So -

u^2 = - (21800 - 35000) / { 1/2 (1000) (1/(0.0125)^4 - 1/(0.0255)^4) / pi^2 }
= 13200 / {500*(40960000 - 2365044) / pi^2]

= 13200 / 1955981224

=> v = 2.60 x 10^-3 m^3/s

Convert this in Liter / s -

v = 2.60 x 10^-3 x 1000 L/s = 2.60 L/s (Answer)