Question

A 7.2 cm diameter horizontal pipe gradually narrows to 4.8 cm . When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 33.0 kPa and 26.0 kPa , respectively.

What is the volume rate of flow?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Given,

d1 = 6.3 cm ; d2 = 3.4 cm ; P1 = 32 k Pa and P2 = 22.6 kPa

We need to find the volume rate flow.

We know that,

A1v1 = A2v2

v2 = (A1/A1) v1 = (d1/d2)² v1

We know from Bernaulli's equation that,

P1 + 1/2 rho v1² + h1 rho g = P2 +  1/2 rho  v2² + h2 rho  g

substituting the value of v2 we get

P1 + 1/2  rho v1² + 0 = P2 +  1/2  rho  [(d1/d2)² x v1] ² + 0 =  P2 +  1/2  rho  (d1/d2)⁴ v1² +

solving for v1² we get

v1² = 2 (P1 - P2) / [  rho (d1/d2)⁴ - 1 ]

putting in the values:

v1² = 2 ( 33 - 26 ) x 10³ / [ 1000 x (7.2 x 10^-2 / 4.8 x 10^-2)⁴ - 1 ]

v1 = 1.66 m/s

So the volume rate flow will be:

A1v1 = pi (d1/2)² x v1 = 3.14 x 7.2 x 7.2 x 10^-4 x 1.66 / 4 = 6.76 x 10^-3 m³/s (aprox)

Hence, Volume flow rate = 6.76 x 10^-3 m³/s (aprox)