Bernoulii's Principle and Equation of Continuity help

[Robbomb116]

Homework Statement

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)

Homework 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]

The Attempt at a Solution

I tried simply rearanging the 3rd equation and solving for v2 but i don't no [tex]\rho[\tex] so that was just a mess that didnt work...I can't 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

 

[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!

 

[m2003]

Hello,

Bernoulli's Principle and Equation of Continuity are fundamental concepts in fluid mechanics that can help us solve this problem.

First, let's review Bernoulli's Principle, which states that in a steady flow of an incompressible fluid, the sum of the pressure, kinetic energy, and potential energy per unit volume is constant at any two points along a streamline. In equation form, this can be written as:

$$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$$

Where P is the pressure, ρ is the density, v is the velocity, g is the acceleration due to gravity, and h is the height above a reference point.

Next, let's look at the Equation of Continuity, which states that the mass flow rate is constant at any two points along a streamline. In equation form, this can be written as:

$$\rho_1 A_1 v_1 = \rho_2 A_2 v_2$$

Where ρ is the density, A is the cross-sectional area, and v is the velocity.

Now, let's apply these concepts to the problem at hand. We are given that the specific gravity (SG) of the liquid is 0.9, which means that its density is 0.9 times the density of water (1000 kg/m^3). Therefore, the density of the liquid can be calculated as:

ρ = 0.9 x 1000 kg/m^3 = 900 kg/m^3

We are also given that the pressure in the tank above the liquid is 8700 Pa, and the gage pressure at the opening is 0 Pa (atmospheric pressure). This means that the pressure at the bottom of the tank (point 1 in our equations) is 8700 Pa + 0 Pa = 8700 Pa. The pressure at the opening (point 2 in our equations) is 0 Pa (atmospheric pressure).

We are asked to find the initial velocity of the fluid when a 5 cm valve is opened at a point 0.5 m from the bottom of the tank. Let's label this point as point 3 in our equations. Using the Equation of Continuity, we