Calculate Flow Velocity and Hole Diameter in Water Tank Problem

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Homework Help Overview

The problem involves calculating the flow velocity and hole diameter of water exiting a tank through a small hole located below the water surface. The context is fluid dynamics, specifically dealing with pressure and flow rates in an open tank system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Bernoulli's equation and question how to set up the equation correctly given the conditions of the tank and hole. There are attempts to clarify the terms and conditions of the equation, including the velocities and pressures at different points.

Discussion Status

The discussion is ongoing with participants exploring the setup of Bernoulli's equation. Some have provided guidance on the relevant formulas and assumptions, while others are seeking clarification on specific terms and the interpretation of the equation. There is no explicit consensus yet, but productive dialogue is occurring.

Contextual Notes

Participants are working under the assumption that the tank is open to the atmosphere, affecting the pressure calculations. There is also mention of the velocities being zero inside the tank and at the water surface, which is a point of discussion.

rawimpact
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1. A large tank isopen to the atmosphere and is filled with water. A small hole is 16m below the surface of the water. If the flow rate of the water is out of the hole is 2.5e^3 m^3/min, calculate: a)the velocity of the water as it leaves the hole and b) the diameter of the hole.
2. Let's say the pool is 1 and the hole is 2, P = pressure and P is density, i was given the equation: P1 + Pv1^2 + 1/2PY1 = P2 + 1/2PV2^2 + PY23. I have no clue how to solve this equation, can someone help me out please?
 
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Check the relevant formula given in 2.
Inside the pool velocity is zero. And outside the pool Y2 is zero.
 
rl.bhat said:
Check the relevant formula given in 2.
Inside the pool velocity is zero. And outside the pool Y2 is zero.

So the equation is:

P1 + 1/2PY1 = P2 + 1/2PV2^2

Correct?

Where do i go from here?
 
Again check the left hand side of the equation.
 
Can you please explain the equation, i do not understand it so i really do not know what the velocities are
 
OK. At the opening what is the total pressure inside the tank?
 
rl.bhat said:
OK. At the opening what is the total pressure inside the tank?

Well since it is exposed to the atmosphere, isn't it atmospheric pressure? 1.01x10^5 Pa?
 
rawimpact said:
Well since it is exposed to the atmosphere, isn't it atmospheric pressure? 1.01x10^5 Pa?
Plus the pressure due to the water at the depth Y1
 
Ok, I've figured it out, thank you for all of your help. I have another problem, should i continue that here or start another thread?
 
  • #10
You can continue here.
 

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