Types of Fluid Flow

Laminar Flow

• Smooth, orderly flow in parallel layers
• No mixing between layers
• Occurs at low velocities
• Example: Honey pouring slowly

Turbulent Flow

• Chaotic, irregular flow with eddies
• Significant mixing
• Occurs at high velocities
• Example: Rapids in a river

Steady Flow

• Velocity at any point doesn't change with time
• Fluid properties constant at each location
• Most problems assume steady flow

Flow Rate

Volume flow rate (Q) is the volume of fluid passing a point per unit time:

$Q = \frac{V}{t}$

Units: m³/s (or L/s, gallons/min)

For flow through a pipe:

$Q = A \cdot v$

where:

• $A$ = cross-sectional area (m²)
• $v$ = flow speed (m/s)

Mass Flow Rate

$\frac{dm}{dt} = \rho \cdot Q = \rho A v$

For incompressible fluids (liquids), density is constant.

Equation of Continuity

For an incompressible fluid in steady flow, mass is conserved. This leads to the continuity equation:

$A_1 v_1 = A_2 v_2$

or equivalently:

$Q_1 = Q_2$

Key Insight: Volume flow rate is constant throughout the pipe.

What This Means:

• Wide pipe (large A) → slow flow (small v)
• Narrow pipe (small A) → fast flow (large v)

This is why:

• Water shoots faster from a partially covered hose
• Rivers flow faster through narrow sections
• Blood flows faster through capillaries (collectively larger area than arteries)

Derivation of Continuity

Consider fluid flowing through a pipe that changes diameter:

In time Δt:

• Volume entering at point 1: $V_1 = A_1 v_1 \Delta t$
• Volume leaving at point 2: $V_2 = A_2 v_2 \Delta t$

Conservation of mass (incompressible fluid): $V_1 = V_2$ $A_1 v_1 \Delta t = A_2 v_2 \Delta t$ $A_1 v_1 = A_2 v_2$ ✓

Applications

Garden Hose

When you cover part of the opening:

• Area decreases → velocity increases
• Water sprays farther

Blood Flow

• Aorta: large area, slower velocity
• Capillaries (total): larger total area, slower velocity
• Individual capillary: tiny area, moderate velocity

River Dynamics

• Wide sections: slow, deep flow
• Narrow sections: fast, shallow flow
• Total flow rate constant

Problem-Solving Strategy

1. Identify the two cross-sections where you know/need information
2. Write the continuity equation: $A_1 v_1 = A_2 v_2$
3. Express areas:
   • Circle: $A = \pi r^2$
   • Rectangle: $A = w \times h$
4. Solve for unknown (usually velocity or diameter)
5. Check units and reasonableness

Common Mistakes

❌ Using diameter instead of radius in area formula ❌ Forgetting to square the radius: $A = \pi r^2$ not $\pi r$ ❌ Assuming velocity is constant (only flow rate Q is constant) ❌ Applying to compressible fluids (gases) without accounting for density changes ❌ Confusing cross-sectional area with surface area

Find: Volume flow rate $Q$

Solution:

$Q = A \cdot v = (0.50)(2.0) = 1.0 \text{ m}^3\text{/s}$

Answer: 1.0 m³/s (or 1000 L/s)

This is a lot of water - equivalent to filling a cubic meter container every second!

Find: Final velocity $v_2$

Solution:

Step 1: Calculate areas. $A_1 = \pi r_1^2 = \pi (0.040)^2 = 5.03 \times 10^{-3} \text{ m}^2$ $A_2 = \pi r_2^2 = \pi (0.020)^2 = 1.26 \times 10^{-3} \text{ m}^2$

Step 2: Apply continuity equation. $A_1 v_1 = A_2 v_2$ $v_2 = v_1 \frac{A_1}{A_2}$

Step 3: Calculate ratio of areas. $\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{(0.040)^2}{(0.020)^2} = 4$

Step 4: Find final velocity. $v_2 = 3.0 \times 4 = 12 \text{ m/s}$

Answer: 12 m/s

The diameter halved, so the area became 1/4 as large. Therefore velocity must quadruple to maintain constant flow rate.

Find: Final velocity $v_2$

Solution:

Step 1: Calculate areas. $A_1 = \pi r_1^2 = \pi (0.040)^2 = 5.03 \times 10^{-3} \text{ m}^2$ $A_2 = \pi r_2^2 = \pi (0.020)^2 = 1.26 \times 10^{-3} \text{ m}^2$

Step 2: Apply continuity equation. $A_1 v_1 = A_2 v_2$ $v_2 = v_1 \frac{A_1}{A_2}$

Step 3: Calculate ratio of areas. $\frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{(0.040)^2}{(0.020)^2} = 4$

Step 4: Find final velocity. $v_2 = 3.0 \times 4 = 12 \text{ m/s}$

Answer: 12 m/s

The diameter halved, so the area became 1/4 as large. Therefore velocity must quadruple to maintain constant flow rate.

Solution:

Part (a): Speed in hose

Step 1: Calculate hose area. $A_1 = \pi r_1^2 = \pi (0.010)^2 = 3.14 \times 10^{-4} \text{ m}^2$

Step 2: Use $Q = A v$. $v_1 = \frac{Q}{A_1} = \frac{6.0 \times 10^{-4}}{3.14 \times 10^{-4}} = 1.91 \text{ m/s}$

Part (b): Exit speed from nozzle

Step 1: Calculate nozzle area. $A_2 = \pi r_2^2 = \pi (0.0025)^2 = 1.96 \times 10^{-5} \text{ m}^2$

Step 2: Use continuity or $Q = A v$. $v_2 = \frac{Q}{A_2} = \frac{6.0 \times 10^{-4}}{1.96 \times 10^{-5}} = 30.6 \text{ m/s}$

Part (c): Speed ratio

$\frac{v_2}{v_1} = \frac{30.6}{1.91} = 16$

Alternative for (c): Using area ratio. $\frac{v_2}{v_1} = \frac{A_1}{A_2} = \frac{r_1^2}{r_2^2} = \left(\frac{2.0}{0.50}\right)^2 = 16$ ✓

Answer:

• (a) Speed in hose: 1.91 m/s
• (b) Exit speed: 30.6 m/s (about 68 mph!)
• (c) Water exits 16 times faster than in the hose

Solution:

Part (a): Speed in hose

Step 1: Calculate hose area. $A_1 = \pi r_1^2 = \pi (0.010)^2 = 3.14 \times 10^{-4} \text{ m}^2$

Step 2: Use $Q = A v$. $v_1 = \frac{Q}{A_1} = \frac{6.0 \times 10^{-4}}{3.14 \times 10^{-4}} = 1.91 \text{ m/s}$

Part (b): Exit speed from nozzle

Step 1: Calculate nozzle area. $A_2 = \pi r_2^2 = \pi (0.0025)^2 = 1.96 \times 10^{-5} \text{ m}^2$

Step 2: Use continuity or $Q = A v$. $v_2 = \frac{Q}{A_2} = \frac{6.0 \times 10^{-4}}{1.96 \times 10^{-5}} = 30.6 \text{ m/s}$

Part (c): Speed ratio

$\frac{v_2}{v_1} = \frac{30.6}{1.91} = 16$

Alternative for (c): Using area ratio. $\frac{v_2}{v_1} = \frac{A_1}{A_2} = \frac{r_1^2}{r_2^2} = \left(\frac{2.0}{0.50}\right)^2 = 16$ ✓

Answer:

• (a) Speed in hose: 1.91 m/s
• (b) Exit speed: 30.6 m/s (about 68 mph!)
• (c) Water exits 16 times faster than in the hose