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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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