Or, between two points:

$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g y_1 \;=\; P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g y_2$

💡 Key Idea: Where the fluid moves faster, the pressure is lower (at the same height). This is why airplane wings lift, shower curtains pull inward, and roofs blow off in hurricanes.

Special Cases

Horizontal flow ($y_1 = y_2$)

$P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$

Static fluid ($v_1 = v_2 = 0$)

$P_1 + \rho g y_1 = P_2 + \rho g y_2 \;\;\Rightarrow\;\; P_2 - P_1 = \rho g (y_1 - y_2)$

This recovers $P = P_0 + \rho g h$.

Torricelli's Theorem (open tank)

For a small hole a depth $h$ below the open top of a tank: $v = \sqrt{2 g h}$

Same as a freely-falling object dropped from height $h$ — fluid speed depends on depth, not on the tank's cross-section.

Combining With Continuity

Most AP problems use both $A_1 v_1 = A_2 v_2$ and Bernoulli's equation. Use continuity to find an unknown speed, then plug into Bernoulli to find the pressure.

⚠️ Common Mistakes

1. Forgetting the height term. Always check whether $y_1$ and $y_2$ differ.
2. Misreading the relation. Pressure does not decrease everywhere a fluid moves — only along the same streamline (and at the same height).
3. Using gauge vs. absolute pressure inconsistently. Use the same convention on both sides.
4. Applying Bernoulli through a fan or pump. Bernoulli is only valid where no work is added or removed (and viscous losses are negligible).

Key Formulas

Bernoulli (horizontal): $P_2 = P_1 + \tfrac12\rho(v_1^2 - v_2^2) = 2.0\times10^5 + \tfrac12(1000)(2.25 - 20.25)$ $= 2.0\times10^5 - 9000 = 1.91\times10^{5}\text{ Pa}.$