Physics, flow rate with viscosity?
I don't understand how you get 2.1 (m^3?, l?) per second here.
Normally I would need to know the pressure difference at the beginning and end of the horizontal pipe:
P1 should be pgh = 1.5m * g * 1260kg/m³ ≈ 18540.9 Pa (the pressure at a given depth in the vertical vessel).
The dynamic viscosity of glycerol should be 1.48 Pa*s. But I'll never get 2.1/s if I simply take ∆p = 18540.9 Pa (i.e., assume P2 = 0).
So I have P1, but how the hell do I get P2???
(3 votes)
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Δh and Δp are proportional between the ends of the horizontal tube.
Flow rate:
It may have been expected to have another η. Often, in such tasks the viscosity is calculated beforehand by sinking (and then something else comes out as η = 1.45)
So your p2 is just 0 to get to ∆p= 18.5409Pa? I don’t understand
See Supplement at the beginning of the answer.
In the horizontal tube, the pressure drop occurs because of the viscosity.
But thanks why P2 = 0 Pa? The sketch of Hyperphysics says F = (P1-P2)/R in the * horizontal* tube. The use ΔP here belongs to the vertical part.