Consider a fluid flow through a pipe from left to right. The pipe is cylindrical

Question

Consider a fluid flow through a pipe from left to right. The pipe is cylindrical and consists of two sections with different radii. In the wider section, the radius is R1 , the fluid flow speed is V1 , and the gauge pressure of the fluid is P1. In the narrower section, the radius, the fluid flow speed, and the gauge pressure of the fluid are R2 , V2 , and P2 . The fluid is incompressible.

(a) If R1 = 3 cm, R2 = 1 cm, and the fluid flow speed in the wider section is V1 = 18.2 cm/s, what is the flow rate at the narrower section of the pipe?
(b) If the fluid is water with a mass density of 1000 kg/m3, what is P1 - P2 , the pressure difference between the wider and the narrower section of the pipe?

(c) For the given pipe dimensions of R1 = 3 cm, R2 = 1 cm, in order to have the fluid flow speed in the narrower section of the pipe V2 = 189.2 cm/s, what should be the fluid flow speed in the wider section V1?

Explanation / Answer

a) A1 = pi*R1^2 = pi*0.03^2 = 2.83*10^-3 m^2

A2 = pi*R2^2 = pi*0.01^2 = 3.14*10^-4 m^2

Apply Continuty equation

A2*V2 = A1*V1

V2 = V1*(A1/A2)

= 18.2*(2.83*10^-3/(3.14*10^-4))

= 164 cm/s

= 1.64 m/s

volume flow rate = A2*V2

= 3.14*10^-4*1.64

= 5.15*10^-4 m^3/s

b) Apply Bernoulli's equation

P1 + 0.5*rho*v1^2 = P2 + 0.5*rho*v2^2

P1 - P2 = 0.5*rho*(v2^2 - v1^2)

= 0.5*1000*(1.64^2 - 0.182^2)

= 1328 Pa

c) Apply, A1*V1 = A2*V2

V1 = V2*(A2/A1)

= 189.2*(3.14*10^-4/(2.83*10^-3))

= 21 cm/s

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