If water flows at 2 m/s, with pressure 90000 N/m^2, through a tube whose radius is .07 m, then what will be the pressure after the tube narrows to radius .035 m?
When the tube narrows, v increases. Since h is not mentioned, it is implicitly assumed not to change significantly. So, if .5 `rho v^2+`rho g h +P is to remain constant, P must decrease to compensate for the increase in v.
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 areas are `pi( .07 m)^2 and `pi( .035 m)^2, so the proportion is
The new speed is
Thus, .5 `rho v^2 will change from its original value .5(1000 kg/m^3)( 2 m/s)^2 = 2000 N/m^2 to .5(1000 kg/m^3)( 8 m/s)^2 = 32000 N/m^2. This represents an increase of 30000 N/m^2 in the value of .5 `rho v^2.
This increase must be matched by a decrease in P, so the new pressure will be
The quantity .5 `rho v^2 + `rho g h + P remains constant. If the altitude h remains constant (or undergoes a negligible change) the sum .5 `rho v^2 + P must remain constant. If fluid velocity changes from v1 to v2, then we have
so the pressure change is
The continuity equation tells us that the water velocity will change by factor A1 / A2 = (d1/ d2) ^ 2, from v1 to v2 = v1 ( d1 / d2)^2. Thus
The figure below indicates that the diameter ratio d2 / d1 implies area ratio (d2 / d1) ^ 2. Since velocity is inversely proportional to area, this implies velocity ratio (d1 / d2) ^ 2 so that the final velocity is (d1 / d2) ^ 2 * v1.
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