If water flows at 2 m/s through a tube whose cross-sectional area is .09 m^2, then what will be the flow speed if the tube narrows to cross-sectional area .045 m^2?
a = fnf(2,10,1)dumby = fnf(1,20,1)
b = dumby / 100
dummy=fnf(2,6,1)
c = b / dummy
vbl1 = b / c
cor = vbl1 * a
If the area through which water flows decreases, and if the water has nowhere to escape, then since water is incompressible, its speed must increase in proportion to the decrease in cross-sectional area.
The ratio of areas
is
so that the new speed is flow speed:
The volume of water passing a point where the cross-sectional area is A, in time `dt and with water moving perpendicular to the cross-section at speed v must be equal to that of a cylinder with cross-sectional area A and length v `dt. That is, the volume must be
If the water is confined to a pipe, the amount passing one point in time `dt is equal to the amount passing another in time `dt.
If the pipe narrows from cross-sectional area A1 to cross-sectional area A2, then `dV must be the same at both points for equal values of `dt, and the velocities v1 and v2 must be such that
It easily follows that
This is called the continuity equation.
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:
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