Solving Bernoulli Problem: Find Water Tank Height

[Redstar2]

Hey guys, I've got a pressure problem that I've been stuck on for a while now and was just wondering if somebody could give me some guidance in solving it. The problem is as follows:

"A water tank open to the atmosphere at the top has two small holes punched in its side, one above the other. The holes 2.08 cm and 14.7 cm above the ground.
What is the height of the water in the tank if the two streams of water hit the ground at the same place? Answer in units of cm."

I'm pretty sure that it has something to do with Bernoulli's equation and the conservation of energy but I don't know how to find the variables needed.


Any help would very much appreciated, thank you!

 

[catkin]

http://www.google.co.in/search?as_q=bernoulli+tank+hole

 

[Redstar2]

Right...so can anyone put that into terms that I can understand? :P

 

[catkin]

Fair response. Most of those are a bit obscure!

[rant mode]
Bernoulli is more complex than most other things taught at the same time. To make it worse it is usually explained in a way that is technically correct but does not bring out its essentials. Thus many students never fully understand it; some become textbook writers and go on to bamboozle later generations of students. Many examples of Bernoulli in textbooks depend on other effects and this causes further confusion.
[\rant mode]

Bernoulli is essentially a statement of conservation of energy in a dynamic fluid. It is expressed as "energy per unit volume" for:

• Gravitational potential energy, GPE
• Kinetic energy, KE
• Pressure energy

The first two are familiar from mechanics where they are mgh and ½mv^2. The third is simply pressure P.

Fluid dynamics is a gruesomely complex part of Physics. Bernoulli simplifies it by imposing several conditions; if these are not met (or not nearly enough met) Bernoulli cannot be used.

For Bernoulli to apply:

• Streamline flow (no turbulence)
• Constant density (no compression)
• Steady state (no transients, unchanging with time)
• Lossless (no friction)

Under these conditions a tiny volume of fluid entering the system goes through various changes of height, velocity and pressure; its energy content does not change. Expressed mathematically:

$${\rho}gh +0.5{\rho}v^2 + p = k$$
where $\rho$ is density, p is pressure and k is the total energy per unity volume.

Often Bernoulli is used to compare the state of the fluid in two places, as in this problem. The value of k is irrelevant and a more useful Bernoulli equation is.
$${\rho}gh_1 + 0.5{\rho}v_{1}^2 + p_1 = {\rho}gh_2 + 0.5{\rho}v_{2}^2 + p_2$$

To find the velocity of water coming out of a hole, consider two points on a streamline flow from the surface of the water (1) to just outside the hole (2).

• GPE: using point 2 as datum (h = 0), GPE(1) is ρgh and GPE(2) is 0.
• KE: KE(1) is 0 (the movement is negligable); KE(2) is ½ρv^2.
• Pressure: at points 1 and 2 the pressure is atmospheric.

Using the second form of Bernoulli's equation:
$${\rho}gh_1 + 0.5{\rho}v_{1}^2 + p_1 = {\rho}gh_2 + 0.5{\rho}v_{2}^2 + p_2$$
$${\rho}gh_1 + 0.5{\rho}0^2 + atmospheric = {\rho}g0 + 0.5{\rho}v_{2}^2 + atmospheric$$
Subtracting atmospheric pressure from both sides, removing the zero terms and dividing by $\rho$
$$gh_1 = 0.5v_{2}^2$$

So now you can use the depths of the holes to find the velocity of the water coming out of them ...

 

[Redstar2]

I understand much better now, thanks for the input!

However, I have a question pertaining to the depths of the holes. How would I correctly represent them relative to the surface of the water rather than the ground?