Answered

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What is the maximum flow rate of water in a smooth pipe 8.0 cm diameter if the flow is to be laminar

Sagot :

Answer:

0.05 m/s

Explanation:

We start by finding the average velocity of water in the pipe. This is done by saying

R(e) = ρv(avg)d/μ

Where,

R(e) = Reynolds number, and that's 2000

ρ = Density of water, 1000 kg/m³

μ = Viscosity of water, 10^-3

d = diameter of pipe

v(avg) = average velocity

Since we're interested in average velocity, we make v(avg) the subject of formula. So that

V(avg) = R(e).μ/ρ.d

V(avg) = 2000 * 10^-3 / 1000 * 0.08

V(avg) = 2 / 80

V(avg) = 0.025 m/s

The maximum flow rate of water in the pipe usually is twice the average velocity, and as such

V(max) = 2 * V(avg)

V(max) = 2 * 0.025

V(max) = 0.05 m/s

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