Order of Magnitude Difference: Solving with Torricelli & Bernoulli

AI Thread Summary
The discussion centers on the discrepancy between velocities calculated using Torricelli's and Bernoulli's equations, with Torricelli yielding Vt = 6x10^-3 m/s and Bernoulli giving Vt = 6x10^-4 m/s. Participants analyze the application of Bernoulli's principle, noting the simplification steps taken, such as removing terms related to pressure and density. The calculations reveal that the ratio of cross-sectional areas significantly impacts the final velocity result. The importance of careful mathematical handling and the need to consider instantaneous velocity when measuring height are emphasized. Overall, the conversation highlights the complexities of fluid dynamics and the nuances in applying different equations.
maxolina
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Homework Statement
A cylindrical open tank is filled with water, has a depth of 20 m and a hole at the bottom. The tank cross-sectional area is Atank=10 m^2 and the hole's area is Ahole=3x10^-3 m^2. Find the velocity of the fluid at the top surface.
Relevant Equations
EQ. OF CONTINUITY: A*V=constant
BERNOULLI: P+1/2*ro*v^2+ro*g*h=constant
TORRICELLI: V bottom = sqrt ( 2*g*h)
Solving with Torricelli I get Vt = 6x10^-3 m/s

Solving with Bernoulli I get Vt = 6x10^-4 m/s, a whole order of magnitude smaller.

How can it be correct? I know that Torricelli is an approximation, but the solution given by the book uses Torricelli which doesn't seem right to me.
 
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maxolina said:
Solving with Bernoulli I get Vt = 6x10^-4 m/s, a whole order of magnitude smaller.
Exactly how did you apply this to get the number? If I use subscript ##T## for "Top" and ##B## for "Bottom", I get $$p_T+\frac{1}{2}\rho~v_T^2+\rho ~g~h_T=p_B+\frac{1}{2}\rho~v_B^2+\rho ~g~h_B.$$Then I note that at the top and at the bottom (exit point) the pressures are the same and equal to atmospheric, ##p_T=p_B##, and the equation becomes $$\frac{1}{2}\rho~v_T^2+\rho ~g~h_T=\frac{1}{2}\rho~v_B^2+\rho ~g~h_B.$$ Now what do you think should be done to get a number?
 
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Ok here is what I did after your equation:

- delete ro*g*hb on the right side since Hb = 0
- remove all the ro (density)

I'm left with Vb^2 - Vt^2 = 2 g h

Using Eq. of continuity: AtVt = AbVb -> Vb = At/Ab * Vt

Then substitute inside Bernoulli and solve for Vt.

(At/Ab)^2 * Vt^2 - Vt^2 = 2 g h

Collect Vt:

Vt^2((At/Ab)^2 -1) = 2 g h

Solve for Vt:

Vt = sqrt ( 2 g h / (At/Ab)^2 -1) = sqrt ( 2*9.8*20 / (10/3*10^-3)^2-1)

Ignore the -1:

Vt = sqrt (400 / ((3.3*10^3)^2)) = 20 / 3 * 10^3 = 6 * 10^-4
 
##AtVt=AbVb=Ab\sqrt(2gh)##
##Vt=(Ab/At)\sqrt(2gh)##

When h is measured, Vt must be considered instantaneous; otherwise, h will decrease, as well as both velocities, as it actually happens.
 
maxolina said:
Ok here is what I did after your equation:

- delete ro*g*hb on the right side since Hb = 0
- remove all the ro (density)

I'm left with Vb^2 - Vt^2 = 2 g h

Using Eq. of continuity: AtVt = AbVb -> Vb = At/Ab * Vt

Then substitute inside Bernoulli and solve for Vt.

(At/Ab)^2 * Vt^2 - Vt^2 = 2 g h

Collect Vt:

Vt^2((At/Ab)^2 -1) = 2 g h

Solve for Vt:

Vt = sqrt ( 2 g h / (At/Ab)^2 -1) = sqrt ( 2*9.8*20 / (10/3*10^-3)^2-1)

Ignore the -1:

Vt = sqrt (400 / ((3.3*10^3)^2)) = 20 / 3 * 10^3 = 6 * 10^-4
You messed up the math in the end. If you ignore the -1, you get
$$v_T\approx \sqrt{2gh}\frac{A_B}{A_T}.$$ Now ##\sqrt{2gh}=20~##m/s. What is the ratio of the areas?
 
kuruman said:
You messed up the math in the end. If you ignore the -1, you get
$$v_T\approx \sqrt{2gh}\frac{A_B}{A_T}.$$ Now ##\sqrt{2gh}=20~##m/s. What is the ratio of the areas?

Oh my god I lost more than an hour over this.
Thank you!
 
maxolina said:
Oh my god I lost more than an hour over this.
Thank you!
Putting in the numbers at the very end is a good habit.
 
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