At two different points along a streamline:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

Three Terms:

1. $P$ - Pressure (static pressure)
2. $\frac{1}{2}\rho v^2$ - Kinetic energy per unit volume (dynamic pressure)
3. $\rho g h$ - Potential energy per unit volume

💡 Key Insight: As velocity increases, pressure decreases (and vice versa), assuming height is constant.

Energy Interpretation

Bernoulli's equation represents energy per unit volume:

$\frac{\text{Pressure energy}}{\text{Volume}} + \frac{\text{Kinetic energy}}{\text{Volume}} + \frac{\text{Potential energy}}{\text{Volume}} = \text{constant}$

Pressure energy: Work done by pressure forces Kinetic energy: $\frac{1}{2}\rho v^2 = \frac{1}{2}\frac{m}{V}v^2 = \frac{KE}{V}$ Potential energy: $\rho g h = \frac{m}{V}gh = \frac{PE}{V}$

Special Cases

Horizontal Flow (h₁ = h₂)

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

• Faster flow → lower pressure
• Slower flow → higher pressure

Static Fluid (v₁ = v₂ = 0)

$P_1 + \rho g h_1 = P_2 + \rho g h_2$ $P_2 - P_1 = -\rho g (h_2 - h_1) = \rho g h$

This reduces to our pressure-depth equation from fluid statics! ✓

Applications

Venturi Effect

In a horizontal pipe that narrows:

• By continuity: velocity increases in narrow section
• By Bernoulli: pressure decreases in narrow section

Used in:

• Carburetors (draws fuel into air stream)
• Venturi meters (measure flow rate)
• Atomizers and spray bottles

Airplane Wings (Lift)

• Air flows faster over curved top surface
• Faster flow → lower pressure (Bernoulli)
• Higher pressure below wing pushes up
• Net upward force = lift

(Note: This is simplified; actual lift is more complex)

Torricelli's Theorem

Water flowing from a hole at depth $h$ below surface:

$v = \sqrt{2gh}$

Same as free fall velocity! Water exits as if it fell from height $h$.

Chimney Draft

• Hot air rises in chimney
• Creates low pressure at bottom
• Outside air rushes in to replace it
• Maintains fire combustion

Limitations and Assumptions

Bernoulli's equation assumes:

1. Incompressible fluid (liquids, not gases at high speeds)
2. Non-viscous (no internal friction/resistance)
3. Steady flow (no time variation)
4. Along a streamline (single flow path)

Real fluids have viscosity (resistance to flow), which causes:

• Energy loss (heat)
• Pressure drop along pipes
• Deviation from ideal Bernoulli predictions

Problem-Solving Strategy

1. Identify two points along the flow
2. Write Bernoulli's equation: $P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$
3. Simplify based on conditions:
   • Same height? Drop $\rho gh$ terms
   • Open to atmosphere? Set $P = P_{atm}$
   • Static point? Set $v = 0$
4. Use continuity if needed: $A_1v_1 = A_2v_2$
5. Solve for unknown
6. Check units and reasonableness

Bernoulli + Continuity

Many problems require both equations:

Continuity: $A_1 v_1 = A_2 v_2$ (relates velocities) Bernoulli: $P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$ (relates pressures)

Common Mistakes

❌ Forgetting to use gauge pressure (subtract $P_{atm}$ when appropriate) ❌ Mixing up which point has higher/lower pressure ❌ Not using continuity to find second velocity ❌ Forgetting the $\frac{1}{2}$ in kinetic term ❌ Using Bernoulli for viscous fluids or turbulent flow

Find: Pressure $P_2$

Solution:

Since horizontal, Bernoulli simplifies: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

Solve for $P_2$: $P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2)$ $P_2 = 2.0 \times 10^5 + \frac{1}{2}(1000)(2.0^2 - 8.0^2)$ $P_2 = 2.0 \times 10^5 + 500(4 - 64)$ $P_2 = 2.0 \times 10^5 - 30,000$ $P_2 = 1.7 \times 10^5 \text{ Pa} = 170 \text{ kPa}$

Answer: 170 kPa

Velocity quadrupled (×4), so kinetic term increased by 16×. Pressure must decrease to conserve energy.