Fluid Dynamics and Viscosity
4.7 Newton's Formula for Viscosity
Definition of Viscosity: Viscosity is the property of a fluid by virtue of which it opposes the relative motion between its adjacent layers. It is the internal friction in fluids.
When a fluid flows, different layers move with different velocities. The layer in contact with a solid surface is at rest (no-slip condition), while layers farther away move faster. This velocity difference creates friction between layers.
Newton's Law of Viscosity
Newton's Law of Viscosity Statement:
"The viscous force (F) acting on a layer of fluid is directly proportional to the area (A) of the layer and the velocity gradient (dv/dx) perpendicular to the direction of flow."
F = η A (dv/dx)
Or in terms of stress and strain rate:
τ = η (dv/dx)
where τ = F/A (shear stress)
Where:
- F = Viscous force (N)
- η (eta) = Coefficient of viscosity (Pa·s or N·s/m²)
- A = Area of the layer (m²)
- dv/dx = Velocity gradient (s⁻¹)
Coefficient of Viscosity (η)
Definition: The coefficient of viscosity is defined as the tangential force per unit area required to maintain a unit velocity gradient between two parallel layers of fluid.
Unit: Pa·s (Pascal-second) or N·s/m² or kg/(m·s)
CGS unit: Poise (P) = 0.1 Pa·s
Common unit: Centipoise (cP) = 0.001 Pa·s
Viscosity of Common Fluids
| Fluid | Viscosity (Pa·s) at 20°C | Viscosity (cP) |
|---|---|---|
| Air | 0.000018 | 0.018 |
| Water | 0.001 | 1.0 |
| Blood (37°C) | 0.003 - 0.004 | 3 - 4 |
| Olive Oil | 0.084 | 84 |
| Glycerine | 1.5 | 1500 |
| Honey | 2 - 10 | 2000 - 10000 |
| Motor Oil (SAE 30) | 0.2 - 0.5 | 200 - 500 |
Effect of Temperature on Viscosity
For Liquids:
Temperature increases → Viscosity decreases
- Higher temperature increases molecular kinetic energy
- Molecules move more freely, reducing intermolecular forces
- Flow resistance decreases
- Example: Honey flows more easily when heated
For Gases:
Temperature increases → Viscosity increases
- Higher temperature increases molecular collisions
- More momentum transfer between layers
- Flow resistance increases
- Example: Hot air is more viscous than cold air
Newtonian vs Non-Newtonian Fluids:
Newtonian Fluids: Obey Newton's law of viscosity. Viscosity remains constant regardless of shear rate. Examples: Water, air, most gases, simple liquids.
Non-Newtonian Fluids: Do not obey Newton's law. Viscosity changes with shear rate. Examples:
- Shear-thinning: Ketchup, blood, paint (become less viscous under stress)
- Shear-thickening: Cornstarch in water (become more viscous under stress)
- Bingham plastics: Toothpaste (require threshold stress to flow)
4.8 Laminar vs Turbulent Flow & Reynolds Number
Types of Fluid Flow
Comparison Table
| Property | Laminar Flow | Turbulent Flow |
|---|---|---|
| Flow pattern | Smooth, parallel layers | Chaotic, irregular, with eddies |
| Mixing | No mixing between layers | Vigorous mixing |
| Velocity | Low to moderate | High |
| Energy loss | Low (minimal friction) | High (significant friction) |
| Reynolds Number | Re < 2000 (pipes) | Re > 4000 (pipes) |
| Predictability | Highly predictable | Difficult to predict precisely |
| Examples | Blood in capillaries, slow river flow, oil in pipelines | Fast river rapids, air around airplane wings, water from faucet at high speed |
Reynolds Number
Definition: Reynolds number (Re) is a dimensionless quantity that predicts whether fluid flow will be laminar or turbulent. It represents the ratio of inertial forces to viscous forces.
Re = (ρvD) / η
Or equivalently: Re = (vD) / ν
where ν = η/ρ (kinematic viscosity)
Where:
- ρ = Density of fluid (kg/m³)
- v = Flow velocity (m/s)
- D = Characteristic length (pipe diameter for pipe flow) (m)
- η = Dynamic viscosity (Pa·s)
- ν = Kinematic viscosity (m²/s)
Critical Reynolds Numbers
For Flow in Pipes:
- Re < 2000: Laminar flow (smooth, predictable)
- 2000 < Re < 4000: Transition region (unstable, may switch between laminar and turbulent)
- Re > 4000: Turbulent flow (chaotic, with eddies)
For Flow Around Objects:
- Re < 1: Creeping flow (Stokes' flow) - viscous forces dominate
- 1 < Re < 100: Laminar flow with separation
- Re > 1000: Fully turbulent flow with wake formation
Physical Interpretation
What Reynolds Number Tells Us:
Low Re (Re << 1):
- Viscous forces dominate
- Flow is smooth and laminar
- Fluid "sticks" together
- Example: Honey flowing slowly, bacteria swimming
High Re (Re >> 1):
- Inertial forces dominate
- Flow becomes turbulent
- Eddies and vortices form
- Example: Water from fire hose, air around fast-moving car
Why it matters:
- Engineers use Re to design efficient pipes and channels
- Determines heat transfer rates in fluids
- Affects drag on vehicles and aircraft
- Critical in blood flow analysis (medical applications)
Applications of Reynolds Number
1. Pipeline Design:
- Engineers maintain laminar flow (low Re) to minimize energy loss
- Larger pipes, lower velocities, or more viscous fluids reduce Re
2. Blood Flow:
- Normal blood flow in arteries: Re ≈ 100-400 (laminar)
- Turbulent blood flow causes audible sounds (heart murmurs)
- Atherosclerosis (plaque buildup) increases local velocity → higher Re → turbulence
3. Airplane Wing Design:
- Smooth laminar flow over wing reduces drag
- Transition to turbulence increases drag significantly
- Wing design aims to delay turbulence transition
4. Mixing in Chemical Reactors:
- High Re (turbulent) promotes mixing of reactants
- Low Re (laminar) keeps components separate
Important for NEB Exams:
- Remember the formula: Re = ρvD/η
- Know critical values: Re < 2000 (laminar), Re > 4000 (turbulent) for pipes
- Understand that Re is dimensionless (no units)
- High Re → turbulent; Low Re → laminar
- Increasing velocity, pipe diameter, or density increases Re
- Increasing viscosity decreases Re
4.9 Poiseuille's Formula
Poiseuille's Law describes the flow of viscous fluid through a cylindrical pipe under laminar flow conditions. It relates the volume flow rate to the pressure difference, pipe dimensions, and fluid viscosity.
Poiseuille's Formula:
The volume of liquid flowing per second through a horizontal capillary tube is directly proportional to the pressure difference and the fourth power of the radius, and inversely proportional to the coefficient of viscosity and length of the tube.
Q = (πr⁴ΔP) / (8ηL)
Or in terms of velocity:
vavg = (r²ΔP) / (8ηL)
Where:
- Q = Volume flow rate (m³/s)
- r = Radius of pipe (m)
- ΔP = Pressure difference (P₁ - P₂) (Pa)
- η = Coefficient of viscosity (Pa·s)
- L = Length of pipe (m)
- vavg = Average flow velocity (m/s)
Key Features of Poiseuille's Law
Important Relationships:
1. Fourth Power of Radius (Q ∝ r⁴):
- Doubling the radius increases flow rate by 2⁴ = 16 times!
- Halving the radius decreases flow rate by (1/2)⁴ = 1/16
- Small changes in radius have HUGE effects on flow
- Medical significance: Arterial plaque reducing diameter by 50% reduces blood flow to 6.25% of normal!
2. Inversely Proportional to Viscosity (Q ∝ 1/η):
- More viscous fluids flow more slowly
- Honey (high η) flows much slower than water (low η)
3. Inversely Proportional to Length (Q ∝ 1/L):
- Longer pipes → slower flow (more friction)
- Doubling length halves flow rate
4. Directly Proportional to Pressure Difference (Q ∝ ΔP):
- Greater pressure difference → faster flow
- This is why heart must pump harder when blood vessels narrow
Conditions for Validity
Poiseuille's formula is valid only under these conditions:
- Laminar flow: Reynolds number Re < 2000
- Steady flow: Flow rate constant with time
- Incompressible fluid: Density remains constant
- Newtonian fluid: Viscosity independent of shear rate
- Rigid pipe: Pipe walls don't expand or contract
- No-slip condition: Fluid velocity at wall is zero
- Fully developed flow: Velocity profile doesn't change along pipe
- Horizontal or gravity-compensated: No height difference, or ΔP includes gravitational effects
Applications of Poiseuille's Law
1. Blood Flow in Circulatory System
Normal arteries: Blood flows smoothly with low resistance
Atherosclerosis: Plaque buildup reduces radius
- If radius reduced to 75% (r → 0.75r), flow becomes (0.75)⁴ ≈ 0.32 = 32% of normal
- Heart must increase pressure significantly to maintain adequate flow
- This leads to hypertension (high blood pressure)
2. Intravenous (IV) Drips
Flow rate depends on:
- Needle radius (r⁴ dependence) - larger needles → faster flow
- Height of IV bag (creates pressure difference ΔP = ρgh)
- Length of tubing (longer tubing → slower flow)
3. Oil Pipelines
- Heating oil reduces viscosity → increases flow rate
- Larger diameter pipes dramatically increase throughput
- Pumping stations maintain pressure difference along pipeline
4. Capillary Viscometers
- Measure viscosity by timing fluid flow through capillary tube
- Rearrange Poiseuille's equation to solve for η
- Used in laboratories and industry
Derivation Outline (For Reference)
Starting assumptions:
1. Consider cylindrical shell of fluid at radius x with thickness dx
2. At steady state, viscous force equals pressure force
3. Viscous force: F = η(2πxL)(dv/dx)
4. Pressure force: F = (P₁ - P₂)πx²
Equating forces:
η(2πxL)(dv/dx) = ΔP(πx²)
dv/dx = (ΔP·x)/(2ηL)
Integrating to get velocity profile:
v(x) = (ΔP/(4ηL))(r² - x²)
This is a parabolic velocity profile with vmax at center (x=0)
Integrating velocity over cross-section to get flow rate:
Q = ∫v(x)dA = (πr⁴ΔP)/(8ηL)
Important for NEB Exams:
- Remember the r⁴ dependence - this is the most critical feature
- Understand physical meaning: larger radius → exponentially faster flow
- Know conditions for validity (laminar flow, Newtonian fluid, etc.)
- Be able to solve numerical problems involving Q, r, ΔP, η, L
- Medical applications (blood flow) are frequently asked
4.10 Stoke's Law
Stoke's Law describes the viscous drag force experienced by a small spherical object moving through a viscous fluid at low Reynolds number (creeping flow or laminar flow around the sphere).
Stoke's Law Statement:
"When a small sphere moves through a viscous fluid, it experiences a retarding force (viscous drag) that is directly proportional to its radius, velocity, and the coefficient of viscosity of the fluid."
F = 6πηrv
Where:
F = Viscous drag force (N)
η = Coefficient of viscosity (Pa·s)
r = Radius of sphere (m)
v = Velocity of sphere relative to fluid (m/s)
Terminal Velocity
When a sphere falls through a viscous fluid, three forces act on it:
- Weight (W): W = mg = (4/3)πr³ρsg (downward)
- Buoyant Force (Fb): Fb = (4/3)πr³ρfg (upward)
- Viscous Drag (Fd): Fd = 6πηrv (upward, opposes motion)
Initially, weight exceeds upward forces, so sphere accelerates downward. As velocity increases, drag force increases (Fd ∝ v). Eventually, the three forces balance, and the sphere reaches constant velocity called terminal velocity (vt).
Derivation of Terminal Velocity
At terminal velocity, net force = 0:
W = Fb + Fd
Substituting expressions:
(4/3)πr³ρsg = (4/3)πr³ρfg + 6πηrvt
Simplifying:
(4/3)πr³g(ρs - ρf) = 6πηrvt
Solving for vt:
vt = (2r²g(ρs - ρf)) / (9η)
vt = (2r²g(ρs - ρf)) / (9η)
Where:
vt = Terminal velocity (m/s)
r = Radius of sphere (m)
g = Acceleration due to gravity (9.8 m/s²)
ρs = Density of sphere (kg/m³)
ρf = Density of fluid (kg/m³)
η = Coefficient of viscosity of fluid (Pa·s)
Determination of Coefficient of Viscosity Using Stoke's Law
Experimental Method: Measure terminal velocity of a sphere falling through fluid, then calculate η.
Procedure:
- Setup: Fill a tall graduated cylinder with the test liquid (e.g., glycerine)
- Drop sphere: Drop a small steel ball bearing into the liquid
- Measure terminal velocity:
- Mark two points on cylinder separated by known distance (d)
- Measure time (t) for ball to travel between marks
- Terminal velocity: vt = d/t
- Ensure ball has reached terminal velocity before first mark
- Measure sphere properties:
- Radius (r) using vernier calipers or micrometer
- Density (ρs) = mass / volume
- Calculate viscosity: Rearrange terminal velocity formula
Viscosity Calculation:
From vt = (2r²g(ρs - ρf)) / (9η)
Rearranging:
Important Observations from Terminal Velocity Formula
1. vt ∝ r²:
- Larger spheres fall faster
- Doubling radius increases terminal velocity by 4 times
- This is why raindrops (larger) fall faster than fog droplets (tiny)
2. vt ∝ (ρs - ρf):
- Denser spheres fall faster
- If ρs < ρf, terminal velocity is negative (sphere rises!)
- Example: Air bubbles in water rise because ρair < ρwater
3. vt ∝ 1/η:
- More viscous fluids slow down falling objects
- Ball bearing falls faster in water than in honey
4. vt ∝ g:
- Stronger gravitational field → faster terminal velocity
- On moon (g = 1.6 m/s²), terminal velocities would be lower
Applications of Stoke's Law
1. Measurement of Viscosity:
- Falling ball viscometer uses Stoke's law
- Simple, accurate method for Newtonian fluids
2. Cloud Formation and Rain:
- Tiny water droplets in clouds (r ~ 10 μm) have very small vt ~ 1 cm/s
- They remain suspended in air currents
- When droplets coalesce into larger drops (r ~ 1 mm), vt ~ 6 m/s
- Large drops overcome updrafts and fall as rain
3. Sedimentation:
- Separation of particles by size/density in centrifuges
- Blood cells settle in plasma (used in medical tests)
- Soil particles settle in water
4. Millikan's Oil Drop Experiment:
- Used Stoke's law to determine charge of electron
- Measured terminal velocity of charged oil droplets in electric field
- Nobel Prize-winning experiment (1923)
5. Parachute Design:
- Increase drag to reduce terminal velocity
- Larger surface area → more air resistance
6. Pharmaceutical Industry:
- Drug particle size affects settling rate in suspensions
- Uniformity of suspensions depends on Stoke's law
Limitations of Stoke's Law
- Low Reynolds Number: Valid only for Re < 1 (creeping flow)
- Small spheres: Works best for spheres with radius less than 0.1 mm
- Laminar flow: Breaks down when flow becomes turbulent
- Infinite fluid: Assumes fluid extends infinitely (walls should be far from sphere)
- Rigid sphere: Doesn't account for deformable objects
- No-slip condition: Assumes fluid sticks to sphere surface
Important for NEB Exams:
- Remember Stoke's law: F = 6πηrv
- Remember terminal velocity: vt = (2r²g(ρs - ρf)) / (9η)
- Derivation of terminal velocity is important (6-8 marks)
- Understand experimental determination of viscosity
- vt ∝ r² is crucial for applications
- Know applications: rain formation, sedimentation, viscosity measurement
4.11 Equation of Continuity
The equation of continuity is a fundamental principle in fluid dynamics that expresses the conservation of mass for flowing fluids. It states that for an incompressible fluid in steady flow, the mass flow rate remains constant throughout the flow.
Derivation of Equation of Continuity
Consider a pipe with varying cross-sectional area:
Section 1: Area = A₁, velocity = v₁
Section 2: Area = A₂, velocity = v₂
Step 1: In time Δt, volume of fluid entering at section 1:
ΔV₁ = A₁ × (v₁Δt)
Step 2: Mass of fluid entering at section 1:
Δm₁ = ρ₁ΔV₁ = ρ₁A₁v₁Δt
Step 3: Similarly, mass leaving at section 2:
Δm₂ = ρ₂A₂v₂Δt
Step 4: By conservation of mass (steady flow):
Δm₁ = Δm₂
ρ₁A₁v₁Δt = ρ₂A₂v₂Δt
Step 5: For incompressible fluids (ρ₁ = ρ₂ = ρ):
A₁v₁ = A₂v₂
For incompressible fluids:
A₁v₁ = A₂v₂ = constant
Or: Av = constant
General form (compressible fluids):
ρ₁A₁v₁ = ρ₂A₂v₂ = constant
This is equivalent to:
Volume flow rate Q = Av = constant
Physical Interpretation
What the Equation Means:
A₁v₁ = A₂v₂ tells us:
- When area decreases → velocity increases
- When area increases → velocity decreases
- Product Av remains constant throughout the flow
- This ensures mass is conserved (what flows in must flow out)
Example: If area is halved (A₂ = A₁/2), then velocity doubles (v₂ = 2v₁)
This is why water flows faster when you partially block a hose with your thumb—you reduce the cross-sectional area!
Applications of Equation of Continuity
1. Garden Hose / Water Tap:
- Blocking hose outlet with thumb reduces area
- Water velocity increases dramatically
- Creates a strong jet of water
2. Rivers and Streams:
- Where river narrows (gorge), water flows faster
- Where river widens, water slows down
- Explains why currents are stronger in narrow sections
3. Blood Flow in Arteries:
- Blood flows faster in aorta (large diameter)
- Slows down in capillaries (total cross-sectional area is huge due to billions of capillaries)
- Allows time for oxygen/nutrient exchange in capillaries
4. Venturi Meter (Flow Rate Measurement):
- Uses continuity equation to measure flow rate
- Fluid speeds up in constriction, pressure drops (Bernoulli)
- Pressure difference is measured to determine flow rate
5. Traffic Flow:
- Analogous to fluid flow: cars = fluid particles
- When highway narrows (lanes reduce), traffic must slow down to maintain constant "flow rate"
- Or cars speed up if density remains constant
6. Spray Bottles and Atomizers:
- Nozzle has very small opening
- Fluid accelerates to high velocity
- Creates fine spray or mist
Relation to Volume Flow Rate
The equation of continuity can also be expressed in terms of volume flow rate (Q):
Where:
Q = Volume flow rate (m³/s)
A = Cross-sectional area (m²)
v = Flow velocity (m/s)
This means the volume of fluid passing through any cross-section per unit time is the same throughout the pipe.
Important Assumptions
The equation of continuity assumes:
- Steady flow: Velocity at any point doesn't change with time
- Incompressible fluid: Density remains constant (ρ₁ = ρ₂)
- No leaks: Fluid doesn't enter or leave except at inlet and outlet
- Single stream: No branching of flow (or apply separately to each branch)
Important for NEB Exams:
- Equation of continuity: A₁v₁ = A₂v₂ (remember this!)
- Derivation based on conservation of mass
- When area decreases, velocity increases (inversely proportional)
- Applies to incompressible fluids in steady flow
- Frequently combined with Bernoulli's equation in problems
- Q = Av = constant (volume flow rate)
4.12 Bernoulli's Equation
Bernoulli's equation is one of the most important equations in fluid dynamics. It relates pressure, velocity, and height for a flowing fluid, expressing the principle of conservation of energy for ideal fluids.
Bernoulli's Principle:
"For an ideal fluid undergoing steady flow along a streamline, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant."
P + ½ρv² + ρgh = constant
Or between two points:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Terms:
P = Pressure energy per unit volume (Pa)
½ρv² = Kinetic energy per unit volume (Pa)
ρgh = Potential energy per unit volume (Pa)
Derivation of Bernoulli's Equation
Derivation using Work-Energy Theorem:
Consider fluid element moving from point 1 to point 2.
Work done BY pressure forces:
Wpressure = P₁A₁Δx₁ - P₂A₂Δx₂
Since A₁Δx₁ = A₂Δx₂ = ΔV (volume of fluid element):
Wpressure = (P₁ - P₂)ΔV
Change in kinetic energy:
ΔKE = ½Δm(v₂² - v₁²) = ½ρΔV(v₂² - v₁²)
Change in potential energy:
ΔPE = ΔmgΔh = ρΔVg(h₂ - h₁)
By work-energy theorem:
Wpressure = ΔKE + ΔPE
(P₁ - P₂)ΔV = ½ρΔV(v₂² - v₁²) + ρΔVg(h₂ - h₁)
Dividing by ΔV and rearranging:
P₁ - P₂ = ½ρ(v₂² - v₁²) + ρg(h₂ - h₁)
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Understanding Each Term
| Term | Name | Physical Meaning |
|---|---|---|
| P | Static Pressure | Pressure energy per unit volume due to molecular collisions |
| ½ρv² | Dynamic Pressure | Kinetic energy per unit volume due to bulk motion |
| ρgh | Hydrostatic Pressure | Potential energy per unit volume due to elevation |
Special Cases and Applications
Case 1: Horizontal Flow (h₁ = h₂)
Bernoulli's equation simplifies to:
Implication: Where velocity increases, pressure decreases!
Example: Airplane wing lift, Venturi tube
Case 2: Fluid at Rest (v₁ = v₂ = 0)
Bernoulli's equation becomes:
Or: P₂ - P₁ = ρg(h₁ - h₂)
This is the hydrostatic pressure formula!
Case 3: Free Jet from Tank (Torricelli's Theorem)
Water flowing from hole at depth h below surface:
Efflux velocity equals velocity of free fall from height h!
Applications of Bernoulli's Equation
1. Airplane Wing Lift (Airfoil)
- Air flows faster over curved upper surface than flat lower surface
- By Bernoulli: faster flow → lower pressure on top
- Pressure difference creates upward lift force
- Lift = ½ρ(vtop² - vbottom²) × wing area
2. Venturi Meter (Flow Measurement)
- Pipe narrows at throat → velocity increases
- Pressure drops at throat (Bernoulli's principle)
- Pressure difference measured with manometer
- Calculate flow rate from pressure drop
3. Atomizer / Spray Bottle
- Squeeze bulb forces air through narrow tube
- Fast-moving air creates low pressure region
- Atmospheric pressure pushes liquid up tube
- Liquid mixes with air stream, creating fine spray
4. Bunsen Burner
- Gas flows through narrow jet
- Creates low pressure region
- Draws in air through side holes
- Proper gas-air mixture for clean blue flame
5. Curved Ball in Sports (Magnus Effect)
- Spinning ball drags air around it
- On one side, air speed increases (spin + flight direction)
- On other side, air speed decreases (spin opposes flight)
- Pressure difference causes ball to curve
- Used in soccer, baseball, tennis
6. Chimney Draft
- Hot gases rise through chimney
- Wind blowing across top creates low pressure
- Enhances updraft, improving ventilation
- Taller chimneys have better draft
7. Hydraulic Jump
- Fast shallow water suddenly slows and deepens
- Kinetic energy converts to potential energy
- Seen at base of waterfalls, spillways
8. Blood Pressure and Aneurysm
- Aneurysm (bulge in blood vessel) has larger diameter
- Blood slows in aneurysm (continuity equation)
- Pressure increases (Bernoulli)
- Can cause rupture - medical emergency!
Limitations and Assumptions
Bernoulli's equation assumes:
- Incompressible fluid: Density constant (valid for liquids, low-speed gases)
- Inviscid (ideal) fluid: No viscosity, no energy loss due to friction
- Steady flow: Velocity at any point doesn't change with time
- Flow along streamline: Equation applies along a single streamline
- No external work: No pumps or turbines adding/removing energy
- No heat transfer: Energy is only mechanical (not thermal)
Real fluids have:
- Viscosity → energy losses due to friction
- Turbulence → chaotic mixing, energy dissipation
- Compressibility (for high-speed gases)
- Modified Bernoulli's equation includes loss terms for real applications
Important for NEB Exams:
- Bernoulli's equation: P + ½ρv² + ρgh = constant
- Derivation using work-energy theorem is important (8-10 marks)
- Key principle: High velocity → Low pressure (and vice versa)
- Combined with continuity equation: Narrow pipe → High velocity → Low pressure
- Know applications: airplane lift, Venturi meter, spray bottles
- For horizontal flow: P + ½ρv² = constant
- Torricelli's theorem: v = √(2gh) for efflux velocity
4.13 Numerical Problems and Conceptual Questions
This section contains solved numerical problems and conceptual questions covering all topics in fluid statics. These problems are representative of NEB exam patterns and difficulty levels.
Problem Set 1: Archimedes' Principle & Buoyancy
Problem 1:
A piece of metal weighs 46 g in air. When immersed in a liquid of density 1.24 g/cm³, it weighs 30 g. Find the relative density of the metal.
Given:
Weight in air (Wair) = 46 g
Weight in liquid (Wliquid) = 30 g
Density of liquid (ρliquid) = 1.24 g/cm³
Solution:
Step 1: Calculate loss in weight (buoyant force)
Loss in weight = Wair - Wliquid = 46 - 30 = 16 g
Step 2: Loss in weight = Weight of liquid displaced
Weight of liquid displaced = V × ρliquid × g = 16 g
Therefore: V × 1.24 = 16
Volume of metal: V = 16/1.24 = 12.9 cm³
Step 3: Calculate density of metal
ρmetal = mass/volume = 46/12.9 = 3.57 g/cm³
Step 4: Relative density
Relative density = ρmetal / ρwater = 3.57/1 = 3.57
Problem 2:
An iceberg of density 920 kg/m³ floats in seawater of density 1025 kg/m³. What fraction of the iceberg is above the water surface?
Given:
ρice = 920 kg/m³
ρseawater = 1025 kg/m³
Solution:
By law of floatation: Weight of iceberg = Weight of water displaced
ρice × Viceberg × g = ρwater × Vsubmerged × g
Fraction submerged = Vsubmerged/Viceberg = ρice/ρwater
= 920/1025 = 0.8976 = 89.76%
Fraction above water = 1 - 0.8976 = 0.1024 = 10.24%
Problem Set 2: Pressure in Fluids
Problem 3:
Calculate the pressure at a depth of 100 m below the surface of seawater. Take density of seawater = 1025 kg/m³, atmospheric pressure = 1.01 × 10⁵ Pa, g = 9.8 m/s².
Given:
Depth (h) = 100 m
ρ = 1025 kg/m³
P₀ = 1.01 × 10⁵ Pa
g = 9.8 m/s²
Solution:
Pressure at depth h: P = P₀ + ρgh
P = 1.01 × 10⁵ + (1025)(9.8)(100)
P = 1.01 × 10⁵ + 1.0045 × 10⁶
P = 0.101 × 10⁶ + 1.0045 × 10⁶
P = 1.1055 × 10⁶ Pa
P ≈ 11.06 × 10⁵ Pa = 11 atm (approximately)
Problem Set 3: Surface Tension & Capillarity
Problem 4:
Calculate the height to which water will rise in a capillary tube of radius 0.5 mm. Given: surface tension of water = 0.073 N/m, density of water = 1000 kg/m³, angle of contact = 0°, g = 9.8 m/s².
Given:
r = 0.5 mm = 0.5 × 10⁻³ m
T = 0.073 N/m
ρ = 1000 kg/m³
θ = 0° (cos 0° = 1)
g = 9.8 m/s²
Solution:
Ascent formula: h = (2T cos θ)/(rρg)
h = (2 × 0.073 × 1) / (0.5 × 10⁻³ × 1000 × 9.8)
h = 0.146 / (0.5 × 10⁻³ × 9800)
h = 0.146 / 4.9
h = 0.0298 m = 29.8 mm ≈ 3 cm
Problem Set 4: Viscosity & Stoke's Law
Problem 5:
A steel ball of radius 2 mm falls through glycerine with a terminal velocity of 0.05 m/s. Calculate the coefficient of viscosity of glycerine. Given: density of steel = 7800 kg/m³, density of glycerine = 1260 kg/m³, g = 9.8 m/s².
Given:
r = 2 mm = 2 × 10⁻³ m
vt = 0.05 m/s
ρs = 7800 kg/m³
ρf = 1260 kg/m³
g = 9.8 m/s²
Solution:
Terminal velocity formula: vt = (2r²g(ρs - ρf)) / (9η)
Rearranging for η:
η = (2r²g(ρs - ρf)) / (9vt)
η = [2 × (2×10⁻³)² × 9.8 × (7800 - 1260)] / (9 × 0.05)
η = [2 × 4×10⁻⁶ × 9.8 × 6540] / 0.45
η = (512.544 × 10⁻³) / 0.45
η = 1.139 Pa·s
Problem Set 5: Equation of Continuity
Problem 6:
Water flows through a horizontal pipe of cross-sectional area 48 cm² at one end and 12 cm² at the other end at a rate of 3000 cm³/s. Find the velocity of water at each end.
Given:
A₁ = 48 cm²
A₂ = 12 cm²
Q = 3000 cm³/s
Solution:
Volume flow rate: Q = A₁v₁ = A₂v₂
Velocity at end 1:
v₁ = Q/A₁ = 3000/48 = 62.5 cm/s
Velocity at end 2:
v₂ = Q/A₂ = 3000/12 = 250 cm/s
Check using continuity: A₁v₁ = 48 × 62.5 = 3000 ✓
A₂v₂ = 12 × 250 = 3000 ✓
Note: Velocity quadruples when area is reduced to 1/4
Problem Set 6: Bernoulli's Equation
Problem 7:
Water flows through a horizontal pipe. At one point, the pressure is 200 kPa and velocity is 2 m/s. At another point, the velocity is 8 m/s. Find the pressure at the second point. (Density of water = 1000 kg/m³)
Given:
P₁ = 200 kPa = 2 × 10⁵ Pa
v₁ = 2 m/s
v₂ = 8 m/s
ρ = 1000 kg/m³
Horizontal flow (h₁ = h₂)
Solution:
For horizontal flow, Bernoulli's equation:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
Solving for P₂:
P₂ = P₁ + ½ρ(v₁² - v₂²)
P₂ = 2 × 10⁵ + ½ × 1000 × (2² - 8²)
P₂ = 2 × 10⁵ + 500 × (4 - 64)
P₂ = 2 × 10⁵ + 500 × (-60)
P₂ = 2 × 10⁵ - 3 × 10⁴
P₂ = 200,000 - 30,000
P₂ = 170,000 Pa = 170 kPa
Note: Pressure decreased by 30 kPa as velocity increased from 2 to 8 m/s
Problem 8: (Torricelli's Theorem)
A large tank is filled with water to a height of 5 m. A small hole is made at a depth of 3 m below the water surface. Calculate the velocity of efflux and the horizontal distance where the water strikes the ground. (g = 10 m/s²)
Given:
Total height of water = 5 m
Depth of hole below surface (h) = 3 m
Height of hole above ground (H) = 5 - 3 = 2 m
g = 10 m/s²
Solution:
Part (a): Velocity of efflux
Using Torricelli's theorem: v = √(2gh)
v = √(2 × 10 × 3)
v = √60 = 7.75 m/s
Part (b): Horizontal range
Water exits horizontally with velocity v and falls from height H
Time to fall height H:
H = ½gt²
2 = ½ × 10 × t²
t² = 0.4
t = 0.632 s
Horizontal distance (range):
R = v × t = 7.75 × 0.632 = 4.9 m
(b) Horizontal range = 4.9 m ≈ 5 m
Conceptual Questions
Question 1: Why do icebergs float with most of their volume underwater?
Answer: Icebergs float according to the law of floatation. The density of ice (≈917 kg/m³) is about 0.92 times the density of seawater (≈1025 kg/m³). By the floatation principle:
Fraction submerged = ρice/ρseawater ≈ 0.9
Therefore, about 90% of the iceberg's volume is underwater, with only 10% visible above the surface. This is why icebergs are dangerous to ships—the vast majority of their mass is hidden below the waterline.
Question 2: Explain why it's easier to swim in seawater than in freshwater.
Answer: It's easier to swim in seawater because of greater buoyancy. Seawater has higher density (≈1025 kg/m³) than freshwater (1000 kg/m³) due to dissolved salts. According to Archimedes' principle:
Buoyant force = ρfluid × Vdisplaced × g
For the same volume displaced, seawater exerts a greater upward buoyant force. This makes floating easier and requires less effort to stay afloat, making swimming less tiring.
Question 3: Why does water rise in a capillary tube but mercury gets depressed?
Answer: This behavior depends on the angle of contact (θ) between the liquid and the tube material:
Water in glass: θ ≈ 0° (acute), cos θ ≈ 1 (positive)
- Adhesive forces (water-glass) > Cohesive forces (water-water)
- Water wets the glass
- Concave meniscus forms
- h = (2T cos θ)/(rρg) is positive → capillary rise
Mercury in glass: θ ≈ 138° (obtuse), cos θ < 0 (negative)
- Cohesive forces (mercury-mercury) > Adhesive forces (mercury-glass)
- Mercury doesn't wet glass
- Convex meniscus forms
- h is negative → capillary depression
Question 4: Explain how an airplane generates lift using Bernoulli's principle.
Answer: Airplane wings (airfoils) are designed with a curved upper surface and flatter lower surface. When the airplane moves:
- Air flows faster over the curved upper surface than the flat lower surface
- By continuity equation, same mass of air must pass both surfaces in same time
- By Bernoulli's equation: P + ½ρv² = constant (horizontal flow)
- Higher velocity on top → Lower pressure on top
- Lower velocity on bottom → Higher pressure on bottom
- Pressure difference creates net upward force (lift)
Lift force = ½ρ(vbottom² - vtop²) × wing area
This upward lift overcomes the airplane's weight, allowing it to fly.
Question 5: Why does blood pressure decrease in capillaries despite their small diameter?
Answer: This seems paradoxical but is explained by considering the total cross-sectional area:
- Though individual capillaries are tiny (7-10 μm diameter)
- There are billions of capillaries in parallel
- Total cross-sectional area of all capillaries combined is much larger than that of the aorta
- By continuity equation (A₁v₁ = A₂v₂): larger total area → slower velocity
- By Bernoulli's equation: slower velocity → higher pressure would be expected
- However, viscous resistance (friction) in tiny capillaries causes pressure drop
- Net effect: blood pressure is lowest in capillaries (≈20 mmHg vs ≈120 mmHg in aorta)
The slow flow in capillaries is beneficial—allows time for oxygen/nutrient exchange with tissues.
📝 Exam Tips for Fluid Statics Chapter:
- Derivations (8-10 marks each): Be thorough with derivations of terminal velocity, Bernoulli's equation, ascent formula, Poiseuille's formula
- Diagrams: Always draw neat, labeled diagrams for questions on Archimedes' principle, capillarity, Bernoulli's equation
- Units: Pay attention to unit conversions (mm to m, cm³ to m³, kPa to Pa, cP to Pa·s)
- Sign conventions: Height above reference is positive, below is negative; upward forces positive, downward negative
- Combined problems: Many problems combine continuity equation + Bernoulli's equation—practice these!
- Key formulas to memorize:
- Archimedes: Fb = ρfluidVg
- Capillarity: h = (2T cos θ)/(rρg)
- Stoke's law: F = 6πηrv
- Terminal velocity: vt = (2r²g(ρs - ρf))/(9η)
- Continuity: A₁v₁ = A₂v₂
- Bernoulli: P + ½ρv² + ρgh = constant
- Poiseuille: Q = (πr⁴ΔP)/(8ηL)
- Common mistakes to avoid:
- Forgetting to square velocity in Bernoulli's equation (½ρv²)
- Confusing pressure with pressure energy per unit volume
- Missing the factor of 2 in surface tension problems (soap film has two surfaces)
- Not checking if angle of contact is acute or obtuse in capillarity problems
🎯 Quick Revision Summary:
Archimedes' Principle: Upthrust = Weight of fluid displaced = ρfluidVg
Pascal's Law: Pressure transmitted undiminished; F₂/F₁ = A₂/A₁
Surface Tension: T = F/L = E/A (minimizes surface area)
Capillarity: Rise/depression h = (2T cos θ)/(rρg); h ∝ 1/r
Viscosity: Newton's law F = ηA(dv/dx); η decreases with temperature for liquids
Reynolds Number: Re = ρvD/η; Re < 2000 laminar, Re > 4000 turbulent
Stoke's Law: Fdrag = 6πηrv; valid for Re < 1
Poiseuille's Law: Q = (πr⁴ΔP)/(8ηL); Q ∝ r⁴ (very sensitive!)
Continuity: A₁v₁ = A₂v₂; smaller area → higher velocity
Bernoulli: P + ½ρv² + ρgh = constant; high velocity → low pressure