Torricelli’s Theorem (College Board AP® Physics 1: Algebra-Based): Revision Note

Torricelli’s theorem

• Consider a container filled with fluid

• An opening is cut into the container at a given depth below the fluid's surface

  • This could be in the side of the container or the base, as pressure only depends on depth below the surface, not orientation

• Torricelli's theorem predicts the speed of fluid exiting the hole

  • It assumes that the surface and opening are both open to the same environment and therefore have matching pressures

Derived equation

• The speed of fluid leaving an opening in a container is described by the equation:

$v=\sqrt{2g\mathrm{\Delta }y}$

• Where:

  • $v$ = the fluid speed leaving the opening, measured in $\mathrm{m}/\mathrm{s}$

  • $g$ = acceleration due to gravity, measured in $\mathrm{m}/{\mathrm{s}}^2$

  • $\mathrm{\Delta }y$ = distance from the opening to the fluid's surface, measured in $\mathrm{m}$

• Interestingly, this equation shows that the speed depends only on depth

Container with fluid leaving an opening

Derivation

Step 1: Identify the fundamental equations

• Recall Bernoulli's equation:

$P_1+\rho g{y}_1+\frac{1}{2}\rho {v_1}^2=P_2+\rho g{y}_2+\frac{1}{2}\rho {v_2}^2$

• Where:

  • $P_1$ = pressure at point 1, measured in $\mathrm{Pa}$

  • $\rho$ = fluid density, measured in $\mathrm{kg}/{\mathrm{m}}^3$

  • $y_1$ = height of point 1 above reference level, measured in $\mathrm{m}$

  • $v_1$ = fluid speed at point 1, measured in $\mathrm{m}/\mathrm{s}$

  • $P_2$ = pressure at point 2, measured in $\mathrm{Pa}$

  • $y_2$ = height of point 2 above reference level, measured in $\mathrm{m}$

  • $v_2$ = fluid speed at point 2, measured in $\mathrm{m}/\mathrm{s}$

• This compares quantities at the surface (point 1) and the opening (point 2)

Step 2: Apply the specific conditions

• At the surface, speed is zero, as no fluid is passing through that point, therefore, it can be inferred that:

$v_1=0\mathrm{m}/\mathrm{s}$

• The reference height can be defined at the same level as the opening, therefore, it can be inferred that:

$y_1=\mathrm{\Delta }y$

$y_2=0\mathrm{m}$

• Assume that both the fluid's surface and the opening are open to the same atmosphere, therefore, it can be inferred that:

$P_1=P_2=P_{atm}$

Step 3: Combine the specific conditions

• Applying these conditions to Bernoulli's equation gives:

${\cancel{P}}_{atm}+\rho g\mathrm{\Delta }y+\frac{1}{2}\rho \times 0^2={\cancel{P}}_{atm}+\rho g\times 0+\frac{1}{2}\rho {v_2}^2$

$\rho g\mathrm{\Delta }y=\frac{1}{2}\rho {v_2}^2$

• The density terms can be canceled out on each side:

$g\mathrm{\Delta }y=\frac{1}{2}{v_2}^2$

• Rename $v_2$ as $v$ and rearrange for fluid speed at the opening:

$v=\sqrt{2g\mathrm{\Delta }y}$

• This is Torricelli's theorem