Fluid Dynamics and Continuity

Flow rate, continuity equation, and fluid motion in pipes

🌊 Fluid Dynamics and Continuity

Introduction

So far we've studied fluids at rest (fluid statics). Now we examine fluids in motion (fluid dynamics). Understanding how fluids flow is essential for everything from plumbing to blood circulation to airplane design.

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

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

📚 Practice Problems

1Problem 1easy

❓ Question:

Water flows through a pipe with a cross-sectional area of 0.50 m² at a velocity of 2.0 m/s. What is the volume flow rate?

💡 Show Solution

Given:

Area: $A = 0.50$ m²
Velocity: $v = 2.0$ m/s

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!

2Problem 2easy

❓ Question:

Water flows through a pipe with a cross-sectional area of 0.50 m² at a velocity of 2.0 m/s. What is the volume flow rate?

💡 Show Solution

Given:

Area: $A = 0.50$ m²
Velocity: $v = 2.0$ m/s

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!

3Problem 3medium

❓ Question:

Water flows through a pipe at 3.0 m/s. The pipe narrows from a diameter of 8.0 cm to 4.0 cm. What is the water velocity in the narrow section?

💡 Show Solution

Given:

Initial velocity: $v_1 = 3.0$ m/s
Initial diameter: $d_1 = 8.0$ cm $= 0.080$ m
Final diameter: $d_2 = 4.0$ cm $= 0.040$ m

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.

4Problem 4medium

❓ Question:

Water flows through a pipe at 3.0 m/s. The pipe narrows from a diameter of 8.0 cm to 4.0 cm. What is the water velocity in the narrow section?

💡 Show Solution

Given:

Initial velocity: $v_1 = 3.0$ m/s
Initial diameter: $d_1 = 8.0$ cm $= 0.080$ m
Final diameter: $d_2 = 4.0$ cm $= 0.040$ m

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.

5Problem 5hard

❓ Question:

A garden hose (diameter 2.0 cm) delivers water at 0.60 L/s. (a) What is the water speed in the hose? (b) A nozzle reduces the diameter to 0.50 cm. What is the exit speed? (c) How much faster does water exit compared to the hose?

💡 Show Solution

Given:

Hose diameter: $d_1 = 2.0$ cm $= 0.020$ m
Flow rate: $Q = 0.60$ L/s $= 6.0 \times 10^{-4}$ m³/s
Nozzle diameter: $d_2 = 0.50$ cm $= 0.0050$ m

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

6Problem 6hard

❓ Question:

💡 Show Solution

Given:

Hose diameter: $d_1 = 2.0$ cm $= 0.020$ m
Flow rate: $Q = 0.60$ L/s $= 6.0 \times 10^{-4}$ m³/s
Nozzle diameter: $d_2 = 0.50$ cm $= 0.0050$ m

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

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