My Torricelli solution is one order of magnitude off compared to my Bernoulli solution

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  • #1
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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  • #2
kuruman
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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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  • #3
maxolina
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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
 
  • #4
Lnewqban
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##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.
 
  • #5
kuruman
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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
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?
 
  • #6
maxolina
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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!
 
  • #7
kuruman
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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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