Speed of Water Emerging from Large Tank: [itex] \sqrt{2gh} [/tex]

  • Thread starter Thread starter tandoorichicken
  • Start date Start date
  • Tags Tags
    Speed Tank Water
Click For Summary

Homework Help Overview

The problem involves determining the speed of water emerging from a hole in a large storage tank, specifically exploring the relationship expressed by the equation \( v = \sqrt{2gh} \), where \( h \) is the depth of the hole below the water surface. The subject area pertains to fluid dynamics and the application of Bernoulli's principle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Bernoulli's equation and its components, questioning how to set up the equation for both inside and outside the tank. There are attempts to manipulate the equation to derive the speed of the water, with some participants expressing confusion over the resulting expressions.

Discussion Status

The discussion is active, with participants providing hints and exploring the implications of Bernoulli's theorem. Some guidance has been offered regarding the setup of the equation, but there is no explicit consensus on the correct interpretation or outcome yet.

Contextual Notes

Participants are working under the assumption that viscosity can be neglected, and there is a focus on the implications of gravitational forces on the fluid. There is also a mention of the density of water, which may influence the calculations being discussed.

tandoorichicken
Messages
245
Reaction score
0
A large storage tank is filled with water. Neglecting viscosity, show that the speed of water emerging through a hole in the side a distance h below the surface is [itex]v = \sqrt{2gh} [/tex].[/itex]
 
Physics news on Phys.org
Please show us what you have to work with and what you have tried on this problem.
 
Originally posted by tandoorichicken
A large storage tank is filled with water. Neglecting viscosity, show that the speed of water emerging through a hole in the side a distance h below the surface is [itex]v = \sqrt{2gh} [/tex].[/itex][itex][/itex]
[itex] <br /> Square the whole thing and it starts to look familiar.<br /> <br /> [tex]V^2 = 2gh[/tex]<br /> <br /> Remember that gravity applies a downward force on the fluid and that a force on any side of a fluid is applied to ALL sides of the fluid.[/itex]
 
Good! A good solid hint without completely solving the problem!
 
Write the Bernoulli's Theorem at the Point just inside the tank and just outside the tank.
 
ummmmm...
If I use Bernoulli's Equation, one side of it says
[tex]P+ \rho gh + \frac{1}{2}\rho v^2[/tex].
What does the other side of the equation look like?

If the other side is 0,
then
[tex]P = \rho gh[/tex]
[tex]2\rho gh = -\frac{1}{2}\rho v^2[/tex]
But since h is going below the surface the negative is expected and can be ignored. Then-
[tex]2\rho gh = \frac{1}{2}\rho v^2[/tex]
The density of water is 1 g/cm^3.
[tex]2gh = \frac{1}{2} v^2[/tex]
[tex]4gh = v^2[/tex]
But then I get
[tex]v = 2\sqrt{gh}[/tex], not [tex]v = \sqrt{2gh}[/tex]
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
Replies
2
Views
10K
Replies
11
Views
5K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
12
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K