# Bernoulii's Principle and Equation of Continuity help

1. Feb 23, 2007

### Robbomb116

1. The problem statement, all variables and given/known data
A liquid with a specific gravity of 0.9 is stored in a pressurized, closed storage tank to a height of 7 m. The pressure in the tank above the liquid is 8700 Pa. What is the intial velocity of the fluid when a 5 cm valve is opened at a point 0.5 m from the bottom of the tank? Assume the gage pressure at the opening is 0 Pa (Atmospheric pressure)

2. Relevant equations
[tex]\rho_1{A_1}{v_1}=\rho_2{A_2}{v_2}[\tex] ?
or
[tex]{A_1}{v_1}={A_2}{v_2}[\tex] ?

and
[tex]{P_1}+{.5{\rho}{v_1}^2}+{{\rho}gh_1}={P_2}+{.5{\rho}{v_2}^2}+{{/rho}gh_2}[\tex]

3. The attempt at a solution
I tried simply rearanging the 3rd equation and solving for v2 but i dont no [tex]\rho[\tex] so that was just a mess that didnt work....I cant figure out where I'm supposed to use the diameter of the hole...I've been working on this for hours and everytime i think i've frigured it out i second guess myself and think i did something I couln't. Could someone please poin me in the right direction...to top it all off i think i did that latex thing wrong
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 23, 2007

### arildno

1. Remember that Bernoulli's principle is only valid in its standard form if we may assume the whole inviscid fluid field to be STATIONARY.

2. In a REAL tank problem, remember that the upper surface will gradually move DOWNWARDS (due to emptying of the tank from the bottom!); but this means that since the upper surface is time-VARYING in its position, the fluid spatial domain is therefore time-varying as well, and hence, we cannot regard the fluid velocity at a fixed point in space to remain the same throughout the emptying process!!
Thus, strictly speaking, the necessary condition for the applicability of Bernoulli's principle is not present throughout time..

3. Therefore, in order to use Bernoulli's principle, this necessitates that we make a SIMPLIFICATION or rough approximation that assumes the error we introduce is tiny with respect to the value we get out in real life, and what we get out by making a sleazy use of Bernoulli's principle!

4. The first thing you should note, is that you haven't been given the cross-section of the tank, but only the radius of the valve.
That means the continuity equation has THREE unknowns, the velocity at the top-surface of the fluid (where the pressure is 8700 Pa), the velocity at the valve (where the pressure is 0 Pa), and the unkown cross-section at the top surface.
Along with Bernoulli's equation, you therefore have only two equations to determine three unknowns, which is IMPOSSIBLE!

5. However, if we make the reasonable ASSUMPTION that the top surface cross-section is A LOT LARGER THAN the valve cross section, then it follows from the continuity equation that the velocity at the top surface is WHOLLY NEGLIGBLE, compared to whatever velocity is generated at the valve opening.
That is, by the assumption of a much larger tank cross section than valve cross-section, it is justified to regard the velocity of the fluid at the top surface as 0.

6. Now, information of pressure changes propagates through a fluid with the velocity of sound, and for any velocity change, Newton's laws demands some passage of time in order for the velocity change to actually occur.
This tiny time interval is frightfully difficult to model exactly, so we make yet another simplification:
Bernoulli's equation, as measured along a streamline connecting the top surface of the fluid and the valve opening HOLDS, with a non-zero "initial" velocity being present at the valve opening, and the pressures at each cross-section being equal to the pressures there just prior to the opening of the valve.

7. Thus, at the top surface, you now know the pressure&velocity, whereas at the valve you know the pressure value (i.e, 0Pa) and the height difference to the top surface.

Thus, you may solve for the velocity of the fluid at the valve opening, by use of Bernoulli's equation.

END NOTE:
The fluid's DENSITY (and hence it's specific gravity) is the same wherever in the fluid domain!!

Last edited: Feb 23, 2007