How Does Atmospheric Pressure Affect Water Stream Shrinkage?

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Homework Help Overview

The discussion revolves around the effects of atmospheric pressure on the behavior of a water stream flowing from a tank. The original poster presents a problem involving the speed of water exiting a hole and the subsequent reduction in stream radius as it falls.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between gravitational acceleration and the shrinking of the water stream, questioning how atmospheric pressure influences this phenomenon. There are attempts to apply conservation principles to calculate velocity and distance.

Discussion Status

The discussion includes various calculations and interpretations of the problem. Some participants have provided calculations based on conservation of flow rate, while others have raised questions about the role of atmospheric pressure in the stream's behavior. There is no explicit consensus, but productive dialogue is ongoing.

Contextual Notes

Participants are navigating the implications of atmospheric pressure on the flow of water and its effects on stream characteristics. The original poster expresses uncertainty about the appropriate formulas to use in this context.

dmk90
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Homework Statement


Water flows through a hole at the bottom of a tank that is filled to a height h=3m. The radius of the hole is r1 = 1.5 cm.
1. What is the speed of the water immediately after it leaves the hole?

2. At what distance d below the bottom of the tank is the radius of the stream reduced to r2 = 1 cm?

Homework Equations

The Attempt at a Solution


I managed to solve the speed for 7.7 m/s. However, I'm not sure what formula to use to address the atmospheric pressure that cause the stream to shrink.
 
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The stream shrinks because it is accelerating gravitationally. As it speeds up, for the flow rate to remain constant, the cross sectional area must decrease.

Chet
 
Based on your clarification, I calculated the velocity at the bottom using the conservation of volumetric flow rate (R = Av = constant), v0 = 17.3. Then g = dv/dt -> dt = dv/g = 0.98 s. Then d = d0 + v0t + 0.5at2 = 12.3, which seems to be the correct result. What do you think?
 
dmk90 said:
Based on your clarification, I calculated the velocity at the bottom using the conservation of volumetric flow rate (R = Av = constant), v0 = 17.3. Then g = dv/dt -> dt = dv/g = 0.98 s. Then d = d0 + v0t + 0.5at2 = 12.3, which seems to be the correct result. What do you think?
I haven't checked your arithmetic, but you seem to have the right idea.
 

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