Bernoulli's Principle: How Does Water Flow in a Tower with Multiple Holes?

AI Thread Summary
Water flows in a tower with multiple holes, maintaining a constant height due to the balance between inflow and outflow. The pressure at both the top and the point of exit is considered to be atmospheric pressure, as it is the primary influence in this scenario. To determine the velocity of water exiting the holes, Bernoulli's equation can be simplified by focusing on the height differences between the holes. The hole at the bottom, being the lowest, is predicted to project water the farthest due to the greater pressure difference compared to the higher holes. Understanding these principles allows for accurate predictions of water flow dynamics in the tower.
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Homework Statement



Examine the schematic of the tower. Water flows in at the top to balance the water flowing out through the holes, so the height of the water in the tower stays fixed.

Take state 1 at the top of the water in the tower and state 2 in the water just as it flows out of the hole.

for this part there is a picture that shows a water tower there are three holes one on the top one in the middle and one at the bottom.


a. What is the pressure at states 1 and 2?Explain.

c. If the holes are at heights of 3.0 cm, 13cm and 23 cm above the surface of the water in the lower basin and the top hole is 3.0 cm below the top of the water in the tower, predict which one you believe will go farthest (top, middle or bottom). Explain your reasoning.

Homework Equations



P1+1/2(m/V)v12+(m/V)gy1=P2+(m/V)v22+(m/V)gy2


density = m/V


The Attempt at a Solution



a) I would say that the pressure is the same as the atmospheric pressure because that is the only pressure that i found it could affect in this problem in this case then pressure at both states are the same

b) If P1 and P2 then somehow i can manipulate the bernoulli equation to get a velocity but i don't know how

please help me
 
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You can get rid of the 1/2(m/V)v's in your Bernoulli's principle equation since that is irrelevant. You need to use the equation without 1/2(m/V)v's to figure out the pressure of the water at each height and then add the air pressure to get the absolute pressure at each height.
 
So i would use this

P1 +m/V)gy1= P2 +(m/V)gy2

use water's density and g=10 and then for y1 would I use zero and for y2 I would use what height? what about P1?
 
P1 +m/V)gy1= P2 +(m/V)gy2

Then move things around so it's

P1-P2=(m/V)gy2-(m/V)gy1

Then that changes to

Delta P = (m/V)g(y2-y1)

Delta P = (m/V)*g*Delta h

Delta P is the pressure of the water aka what you are looking for, and Delta h is the different heights of the holes 3cm,13cm,23cm or .03m, .13m, .23m (in SI units)

After you get the three different answers to Delta P add the pressure of the atmosphere to get the absolute pressures.
 
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