Water is confined to flow through a tube whose cross-sectional radius decreases from .14 m to radius .028 m. If the flow speed at the .14 m cross-section is 3 m/s, what is the flow speed at the .028 m cross-section?
If the area through which water flows decreases, and if the water has nowhere to escape, then since water is effectively incompressible at ordinary pressures, its speed must increase in proportion to the decrease in cross-sectional area. This is the essence of the continuity equation.
The areas are `pi ( .14 m)^2 and `pi( .028 m)^2, so the proportion is
which simplifies to
The new speed is new speed: `vbl1( 3 m/s) = 74.99999 m/s.
The figure shows a side view of water falling in a pipe which narrows from cross-sectional area A1 to cross-sectional area A2.
Since the volume of fluid passing a point in time `dt is A1 v1 `dt in the first section of pipe and A2 v2 `dt in the second section, and since the fluid passing one point has to displace an equal volume past the second point, the two volumes are equal.
The result is the continuity equation:
It follows that v2 = (A1 / A2) * v1
If the radii of the pipe are r1 and r2, then A1 = `pi * r1^2 and A2 = `pi * r2^2. The area ratio A1 / A2 is therefore
and v2 is
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