If the pressure in the pipe is 8830 N/m2 and in the constricted section is 6622.5 N/m2 , what is the rate at which oil is flowing? Answer in units of m^3/s.

Question

A horizontal pipe of diameter 0.769 m has a smooth constriction to a section of diameter 0.4614 m . The density of oil flowing in the pipe is 821 kg/m ^3 . If the pressure in the pipe is 8830 N/m2 and in the constricted section is 6622.5 N/m2 , what is the rate at which oil is flowing? Answer in units of m^3/s.

Answer 1

We can use Bernoulli’s equation which is P1 + ½ x ρ x V1^2 = P2 + ½ x ρ x V2^2.

 Density = ρ = 821 kg/m^3

P1 = 8830 N/m^2

P2 = 6622.5 N/m^2

It can be shown that Velocity = Flow / Area and flow rate is the same for both sections of the pipe ( V1 = flow / Area1 and V2 = flow / Area2.

So, rearranging Bernoulli’s equation and substituting, gives

8830 – 6622.5 = ½ x 821 x flow^2 x(1/A2^2 – 1/A1^2)

A1 = 3.14 x (0.769/2)^2   A2 = 3.14 x (0.4614/2)^2

A1^2 = 0.464 m^2   A1^2 = 0.167 m^2

Now, 2207.5 = 410.5 x Flow^2 x (1/0.167^2 – 1/0.464^2)

2207.5 = 410.5 x Flow^2 x (4.645)

Flow^2 = 2207.5 /(410.5 x 4.645) = 1.158

Flow = √ 1.158= 1.08 m^3/s                                                                                            

Check all math because I haven’t.