Bernoulli's Principle Problem/Energy Conservation

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SUMMARY

The discussion centers on applying Bernoulli's Principle and conservation of energy to determine the velocity of water exiting a nozzle and the height it reaches. The correct expression for the exit velocity, Vexit, is derived as Vexit = (2[(Pres - Patm)/ρ])^(1/2), where Pres is the pressure in the reservoir, Patm is atmospheric pressure, and ρ is the density of water. The confusion arises regarding the neglect of the ρgh term, which is justified since there is no change in potential energy from the reservoir to the nozzle exit, as the pressure already accounts for the height difference.

PREREQUISITES
  • Understanding of Bernoulli's equation
  • Knowledge of conservation of energy principles
  • Familiarity with fluid dynamics terminology
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study Bernoulli's equation in detail, focusing on its applications in fluid mechanics.
  • Explore the concept of pressure in fluids, particularly how it relates to potential energy.
  • Learn about the implications of neglecting certain terms in fluid dynamics problems.
  • Investigate real-world applications of nozzle flow and pressure systems in engineering.
USEFUL FOR

Students studying fluid mechanics, engineers working with hydraulic systems, and anyone interested in the practical applications of Bernoulli's Principle in real-world scenarios.

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Homework Statement


Hi! The problem states: Water through a certain sprinkler system flows trhough a level hose connected to a nozzle which is directed directly upwards. The water leaves the nozzle and shoots to a height, h, before falling back down again into a pool.
The hose is connected to a reservoir which maintains the water there at a pressure, Pres Assume no viscosity of flow, until water exits the nozzle. Use the following notation:
Anoz= cross sectional area of nozzle
Lhose=length of hose
Patm=atmospheric pressure
ρ=density of water

What is the best expression for the velocity of the water just after it leaves the nozzle? (A)
What is the best expression for the height, h? (B)



Homework Equations


Bernoulli's equation, and conservation of energy (Initial kinetic energy of water=final potential energy of water at height, h)


The Attempt at a Solution



So, I actually was fine on the calculation of this problem, but I am having difficulty understanding the rationale behind (A). I calculated Vexit= (2[(Pres-Patm)/ρ])^(1/2)
However, to do this, I ignored the velocity of the water in the reservoir. That makes sense, it is hardly moving compared to the velocity of the novel because the regions have equal flow rates. However, I also ignored the ρgh term for the water in the reservior. I guess that I am confused as to why this term doesn't have to be taken into account.

If anyone could explain, that would be great!

 
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erinfalk said:

Homework Statement


Hi! The problem states: Water through a certain sprinkler system flows trhough a level hose connected to a nozzle which is directed directly upwards. The water leaves the nozzle and shoots to a height, h, before falling back down again into a pool.
The hose is connected to a reservoir which maintains the water there at a pressure, Pres Assume no viscosity of flow, until water exits the nozzle. Use the following notation:
Anoz= cross sectional area of nozzle
Lhose=length of hose
Patm=atmospheric pressure
ρ=density of water

What is the best expression for the velocity of the water just after it leaves the nozzle? (A)
What is the best expression for the height, h? (B)



Homework Equations


Bernoulli's equation, and conservation of energy (Initial kinetic energy of water=final potential energy of water at height, h)


The Attempt at a Solution



So, I actually was fine on the calculation of this problem, but I am having difficulty understanding the rationale behind (A). I calculated Vexit= (2[(Pres-Patm)/ρ])^(1/2)
However, to do this, I ignored the velocity of the water in the reservoir. That makes sense, it is hardly moving compared to the velocity of the novel because the regions have equal flow rates. However, I also ignored the ρgh term for the water in the reservior. I guess that I am confused as to why this term doesn't have to be taken into account.

If anyone could explain, that would be great!

Can you draw the diagram of the problem please ? Looks like you are confused with wordings.
 
erinfalk said:
However, I also ignored the ρgh term for the water in the reservior. I guess that I am confused as to why this term doesn't have to be taken into account.

If anyone could explain, that would be great!

Homework Statement





You were given the pressure in the reservoir at the point where the hose connects to it. So there is no change in potential energyfrom that point until the water squirts out of the nozzle & climbs.

Had the problem stated 'the hose is connected to a point h below the surface of the water', then you would have had to use p = patm + rho g h. The result would have been the same.

The problem statement did not make clear whether pres included patm. You might want to check on that.
 

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