Bernoulli's Equation

Energy conservation in fluids, applications to lift and flow

✈️ Bernoulli's Equation

Introduction

Bernoulli's Equation is the application of energy conservation to fluid flow. It relates pressure, velocity, and height at different points in a flowing fluid.

Bernoulli's Equation

For an ideal fluid (incompressible, non-viscous) in steady flow:

$P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}$

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:

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

Incompressible fluid (liquids, not gases at high speeds)
Non-viscous (no internal friction/resistance)
Steady flow (no time variation)
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

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

📚 Practice Problems

1Problem 1easy

❓ Question:

Water flows horizontally through a pipe. At point 1, the pressure is 200 kPa and velocity is 2.0 m/s. At point 2, the velocity is 8.0 m/s. What is the pressure at point 2? (ρ_water = 1000 kg/m³)

💡 Show Solution

Given:

Horizontal flow: $h_1 = h_2$
$P_1 = 200$ kPa $= 2.0 \times 10^5$ Pa
$v_1 = 2.0$ m/s, $v_2 = 8.0$ m/s
$\rho = 1000$ kg/m³

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.

2Problem 2easy

❓ Question:

Water flows horizontally through a pipe. At point 1, the pressure is 200 kPa and velocity is 2.0 m/s. At point 2, the velocity is 8.0 m/s. What is the pressure at point 2? (ρ_water = 1000 kg/m³)

💡 Show Solution

Given:

Horizontal flow: $h_1 = h_2$
$P_1 = 200$ kPa $= 2.0 \times 10^5$ Pa
$v_1 = 2.0$ m/s, $v_2 = 8.0$ m/s
$\rho = 1000$ kg/m³

Find: Pressure $P_2$

Solution:

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

Answer: 170 kPa

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

3Problem 3medium

❓ Question:

A large open tank of water has a small hole 5.0 m below the water surface. (a) What is the velocity of water exiting the hole? (b) How far horizontally does the water travel if the hole is 1.0 m above the ground?

💡 Show Solution

Given:

Depth below surface: $h = 5.0$ m
Height above ground: $y = 1.0$ m

Solution:

Part (a): Exit velocity (Torricelli's Theorem)

Apply Bernoulli between surface (point 1) and hole (point 2):

Point 1: $P_1 = P_{atm}$ , $v_1 \approx 0$ (large tank), $h_1 = h$
Point 2: $P_2 = P_{atm}$ , $v_2 = ?$ , $h_2 = 0$

$P_{atm} + 0 + \rho g h = P_{atm} + \frac{1}{2}\rho v_2^2 + 0$ $\rho g h = \frac{1}{2}\rho v_2^2$ $v_2 = \sqrt{2gh} = \sqrt{2(9.8)(5.0)} = \sqrt{98} = 9.9 \text{ m/s}$

Part (b): Horizontal distance (projectile motion)

Horizontal motion: $x = v_2 t$ Vertical motion: $y = \frac{1}{2}gt^2$

Find time to fall 1.0 m: $1.0 = \frac{1}{2}(9.8)t^2$ $t = \sqrt{\frac{2.0}{9.8}} = 0.452 \text{ s}$

Horizontal distance: $x = v_2 t = (9.9)(0.452) = 4.5 \text{ m}$

Answer:

(a) Exit velocity: 9.9 m/s
(b) Horizontal distance: 4.5 m

4Problem 4hard

❓ Question:

Water flows through a horizontal pipe. At point A, the diameter is 10 cm and the pressure is 2.0 × 10⁵ Pa. At point B, the diameter is 5.0 cm. If the speed at A is 2.0 m/s, find (a) the speed at B, (b) the pressure at B. Use ρ = 1000 kg/m³.

💡 Show Solution

Solution:

Given: d_A = 10 cm, P_A = 2.0 × 10⁵ Pa, v_A = 2.0 m/s, d_B = 5.0 cm

(a) Speed at B (Continuity equation): A_A v_A = A_B v_B π(0.05)² × 2.0 = π(0.025)² × v_B (0.05)²(2.0) = (0.025)²v_B v_B = (0.05)²(2.0)/(0.025)² v_B = (2.5 × 10⁻³)(2.0)/(6.25 × 10⁻⁴) v_B = 8.0 m/s

(b) Pressure at B (Bernoulli's equation): P_A + ½ρv_A² = P_B + ½ρv_B² (horizontal pipe, same height)

P_B = P_A + ½ρ(v_A² - v_B²) P_B = 2.0 × 10⁵ + ½(1000)(2.0² - 8.0²) P_B = 2.0 × 10⁵ + 500(4 - 64) P_B = 2.0 × 10⁵ - 30,000 P_B = 1.7 × 10⁵ Pa or 170 kPa

Pressure decreases where speed increases!

5Problem 5medium

❓ Question:

💡 Show Solution

Given:

Depth below surface: $h = 5.0$ m
Height above ground: $y = 1.0$ m

Solution:

Part (a): Exit velocity (Torricelli's Theorem)

Apply Bernoulli between surface (point 1) and hole (point 2):

Point 1: $P_1 = P_{atm}$ , $v_1 \approx 0$ (large tank), $h_1 = h$
Point 2: $P_2 = P_{atm}$ , $v_2 = ?$ , $h_2 = 0$

$P_{atm} + 0 + \rho g h = P_{atm} + \frac{1}{2}\rho v_2^2 + 0$ $\rho g h = \frac{1}{2}\rho v_2^2$ $v_2 = \sqrt{2gh} = \sqrt{2(9.8)(5.0)} = \sqrt{98} = 9.9 \text{ m/s}$

Part (b): Horizontal distance (projectile motion)

Horizontal motion: $x = v_2 t$ Vertical motion: $y = \frac{1}{2}gt^2$

Find time to fall 1.0 m: $1.0 = \frac{1}{2}(9.8)t^2$ $t = \sqrt{\frac{2.0}{9.8}} = 0.452 \text{ s}$

Horizontal distance: $x = v_2 t = (9.9)(0.452) = 4.5 \text{ m}$

Answer:

(a) Exit velocity: 9.9 m/s
(b) Horizontal distance: 4.5 m

6Problem 6hard

❓ Question:

💡 Show Solution

Solution:

Given: d_A = 10 cm, P_A = 2.0 × 10⁵ Pa, v_A = 2.0 m/s, d_B = 5.0 cm

(b) Pressure at B (Bernoulli's equation): P_A + ½ρv_A² = P_B + ½ρv_B² (horizontal pipe, same height)

P_B = P_A + ½ρ(v_A² - v_B²) P_B = 2.0 × 10⁵ + ½(1000)(2.0² - 8.0²) P_B = 2.0 × 10⁵ + 500(4 - 64) P_B = 2.0 × 10⁵ - 30,000 P_B = 1.7 × 10⁵ Pa or 170 kPa

Pressure decreases where speed increases!

7Problem 7hard

❓ Question:

Water flows through a horizontal pipe that changes from 10 cm diameter to 5.0 cm diameter. The pressure in the wide section is 150 kPa and velocity is 2.0 m/s. (a) Find velocity in narrow section. (b) Find pressure in narrow section.

💡 Show Solution

Given:

$d_1 = 10$ cm $= 0.10$ m, $d_2 = 5.0$ cm $= 0.050$ m
$P_1 = 150$ kPa $= 1.5 \times 10^5$ Pa
$v_1 = 2.0$ m/s
$\rho = 1000$ kg/m³

Solution:

Part (a): Velocity in narrow section

Use continuity: $A_1 v_1 = A_2 v_2$

$v_2 = v_1 \frac{A_1}{A_2} = v_1 \frac{r_1^2}{r_2^2} = v_1 \left(\frac{d_1}{d_2}\right)^2$ $v_2 = 2.0 \left(\frac{0.10}{0.050}\right)^2 = 2.0(4) = 8.0 \text{ m/s}$

Part (b): Pressure in narrow section

Use Bernoulli (horizontal): $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

$P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2)$ $P_2 = 1.5 \times 10^5 + \frac{1}{2}(1000)(2.0^2 - 8.0^2)$ $P_2 = 1.5 \times 10^5 + 500(4 - 64)$ $P_2 = 1.5 \times 10^5 - 30,000$ $P_2 = 1.2 \times 10^5 \text{ Pa} = 120 \text{ kPa}$

Answer:

(a) Velocity in narrow section: 8.0 m/s
(b) Pressure in narrow section: 120 kPa

Verification: Diameter halved → area became 1/4 → velocity ×4. Faster flow → lower pressure (Bernoulli) ✓

8Problem 8hard

❓ Question:

💡 Show Solution

Given:

$d_1 = 10$ cm $= 0.10$ m, $d_2 = 5.0$ cm $= 0.050$ m
$P_1 = 150$ kPa $= 1.5 \times 10^5$ Pa
$v_1 = 2.0$ m/s
$\rho = 1000$ kg/m³

Solution:

Part (a): Velocity in narrow section

Use continuity: $A_1 v_1 = A_2 v_2$

$v_2 = v_1 \frac{A_1}{A_2} = v_1 \frac{r_1^2}{r_2^2} = v_1 \left(\frac{d_1}{d_2}\right)^2$ $v_2 = 2.0 \left(\frac{0.10}{0.050}\right)^2 = 2.0(4) = 8.0 \text{ m/s}$

Part (b): Pressure in narrow section

Use Bernoulli (horizontal): $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

Answer:

(a) Velocity in narrow section: 8.0 m/s
(b) Pressure in narrow section: 120 kPa

Verification: Diameter halved → area became 1/4 → velocity ×4. Faster flow → lower pressure (Bernoulli) ✓

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