Back to Course
Fluid Flow and Viscosity
Ideal vs. Real Fluids and Types of Flow
Fluid flow is categorized as Streamline (laminar) or Turbulent based on the stability of particle paths and velocity consistency.
$$R_e = \frac{\rho v D}{\eta}$$
The Reynolds Number predicts flow regime transitions.
$R_e$=Reynolds Number(dimensionless)
$\rho$=Fluid density(kg/m³)
$v$=Flow velocity(m/s)
$D$=Characteristic length (diameter)(m)
$\eta$=Dynamic viscosity(Pa·s)
$R_e < 2000$
→Laminar flow — smooth, predictable streamlines
$2000 < R_e < 3000$
→Transition region — flow is unstable
$R_e > 3000$
→Turbulent flow — chaotic eddies and mixing
Steady Flow: Particles passing a point follow identical paths, ensuring constant velocity over time at that location. Streamlines never cross in steady flow.
Turbulence: Occurs at high velocities or low viscosities, characterized by chaotic eddies and vortices. Exact particle paths become unpredictable.
Ideal Fluid: Assumed to be Incompressible, non-viscous, and irrotational for simplified modeling. Real fluids always have some viscosity.
Dimensional Check: $R_e = \frac{[\text{kg/m}^3][\text{m/s}][\text{m}]}{[\text{Pa·s}]}$ — all units cancel, confirming it is dimensionless.
The Equation of Continuity
The Equation of Continuity expresses the Conservation of Mass for an Incompressible fluid moving through a channel of varying width.
$$A_1 v_1 = A_2 v_2$$
The volume flow rate remains constant throughout a closed system.
$A$=Cross-sectional area(m²)
$v$=Fluid velocity(m/s)
$Av$=Volume flow rate(m³/s)
$A_2 = A_1/2$
→$v_2 = 2v_1$ — halving the area doubles the speed
$A_2 = A_1/4$
→$v_2 = 4v_1$ — quartering area quadruples speed (diameter halved)
Flow Rate: The product $Av$ represents the volume of fluid passing a section per unit time (m³/s). This remains constant at every cross-section.
Incompressibility: Assumes density $\rho$ is uniform, making mass conservation ($\rho_1 A_1 v_1 = \rho_2 A_2 v_2$) simplify to volume conservation.
Falling Water Stream: Water from a tap narrows as it falls — gravity increases its speed $v$, so continuity requires the cross-sectional area $A$ to decrease.
Bernoulli's Principle
Bernoulli's Principle relates pressure, speed, and height, fundamentally stating that an increase in fluid speed coincides with a decrease in Static Pressure. It applies to ideal (non-viscous, incompressible, steady) fluid flow.
$$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$$
An expression of energy conservation for flowing fluids — each term has units of pressure (Pa).
$P$=Static Pressure(Pa)
$\frac{1}{2}\rho v^2$=Dynamic Pressure (KE per unit volume)(Pa)
$\rho gh$=Hydrostatic Pressure (PE per unit volume)(Pa)
$h_1 = h_2$ (horizontal flow)
→$P + \frac{1}{2}\rho v^2 = \text{const}$ — pressure and speed trade off directly
$v_1 = v_2$ (uniform pipe)
→$P + \rho gh = \text{const}$ — reduces to hydrostatic pressure equation
Venturi Effect: Fluid pressure drops in a constricted section of a pipe where velocity is higher. Used in Venturi meters, carburetors, and atomizers.
Energy Components: Each term represents energy per unit volume — static pressure is internal energy, $\frac{1}{2}\rho v^2$ is kinetic, $\rho gh$ is gravitational potential.
Dimensional Check: All three terms have dimensions of $[\text{Pa}] = [\text{kg·m}^{-1}\text{·s}^{-2}]$, confirming dimensional consistency.
Viscosity and Stokes' Law
Viscosity ($\eta$) represents internal friction between adjacent fluid layers. It measures how much force is needed to slide one layer of fluid over another.
High Viscosity: Thick fluids like honey and glycerin ($\eta \approx 630 \times 10^{-3}$ Pa·s) resist flow strongly.
Low Viscosity: Water ($\eta \approx 0.8 \times 10^{-3}$ Pa·s) and air ($\eta \approx 0.019 \times 10^{-3}$ Pa·s) flow easily.
Temperature Dependence: Viscosity decreases with temperature in liquids (molecules move apart) but increases in gases (more molecular collisions).
Units: The SI unit of $\eta$ is Pa·s (= kg·m⁻¹·s⁻¹ = N·s·m⁻²). All are equivalent.
Stokes' Law quantifies the Drag Force on a sphere moving slowly through a viscosity-dominated medium under laminar conditions.
$$F_d = 6\pi \eta r v$$
Calculates the retarding force on a small sphere in laminar (low Reynolds number) conditions.
$F_d$=Viscous Drag Force(N)
$\eta$=Coefficient of dynamic viscosity(Pa·s)
$r$=Radius of sphere(m)
$v$=Velocity of sphere relative to fluid(m/s)
$v \to 0$
→$F_d \to 0$ — no drag when stationary
High speed ($R_e \gg 1$)
→Stokes' Law breaks down — drag becomes proportional to $v^2$
Linear Dependence: Unlike air resistance at high speeds ($\propto v^2$), Stokes' drag is proportional to $v$ — valid only for slow, laminar flow around the sphere.
Dimensional Check: $[F_d] = [\text{Pa·s}][\text{m}][\text{m/s}] = [\text{kg·m}^{-1}\text{·s}^{-1}][\text{m}][\text{m·s}^{-1}] = [\text{kg·m·s}^{-2}] = [\text{N}]$ ✓
Terminal Velocity
Terminal Velocity is the constant speed a falling object reaches when the net downward force becomes zero — the weight is exactly balanced by the upward drag force (and buoyancy in a fluid).
$$v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}$$
Derived by setting $mg = F_d + F_b$ (weight = Stokes' drag + buoyancy) and solving for $v$.
$v_t$=Terminal velocity(m/s)
$r$=Radius of sphere(m)
$\rho_s$=Density of sphere(kg/m³)
$\rho_f$=Density of fluid(kg/m³)
$g$=Gravitational acceleration(m/s²)
$\eta$=Fluid viscosity(Pa·s)
$\rho_s = \rho_f$ (neutral buoyancy)
→$v_t = 0$ — object floats, no net downward force
$\rho_f \approx 0$ (falling through air)
→Simplifies to $v_t = \frac{2r^2 \rho_s g}{9\eta}$ (buoyancy negligible)
$\eta \to 0$ (no viscosity)
→$v_t \to \infty$ — object accelerates indefinitely (free fall)
Derivation: At terminal velocity: $\frac{4}{3}\pi r^3 \rho_s g = 6\pi \eta r v_t + \frac{4}{3}\pi r^3 \rho_f g$. Rearranging gives the formula above.
Approach to Terminal Velocity: A dropped sphere initially accelerates at $g$, then acceleration decreases exponentially as drag builds up, asymptotically reaching $v_t$. It never truly reaches $v_t$ but gets arbitrarily close.
Application Strategy
Worked Examples
Problem Statement
Calculate the terminal velocity of a steel ball (radius 2 mm, density 8000 kg/m³) falling through glycerin (density 1200 kg/m³, viscosity 0.83 Pa·s).
Terminal VelocityStokes' LawBuoyancy
Strategic Toolkit
$v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}$
Step 1
Identify Variables
Extract all given constants and convert to SI units.
$r = 2$ mm $= 0.002$ m, $\rho_s = 8000$ kg/m³, $\rho_f = 1200$ kg/m³, $\eta = 0.83$ Pa·s, $g = 9.8$ m/s²
Step 2
Calculate Density Difference
Find the effective density driving the motion (sphere density minus fluid density).
$\Delta\rho = \rho_s - \rho_f = 8000 - 1200 = 6800$ kg/m³
Step 3
Substitute Known Values
Plug values into the terminal velocity formula, noting $r^2 = (0.002)^2 = 4 \times 10^{-6}$ m².
$v_t = \frac{2 \times (0.002)^2 \times 6800 \times 9.8}{9 \times 0.83} = \frac{2 \times 4 \times 10^{-6} \times 66640}{7.47}$
Step 4
Calculate
Evaluate the numerator and divide:
$v_t = \frac{0.5331}{7.47} \approx 0.071$ m/s
Problem Statement
A venturi meter has a pipe diameter of 4 cm and a throat diameter of 2 cm. If the velocity in the pipe is 2 m/s, find the velocity in the throat.
Equation of ContinuityArea Ratios
Strategic Toolkit
$A_1 v_1 = A_2 v_2$
Step 1
Area Ratio from Diameters
Since $A = \pi r^2 = \pi(d/2)^2$, the ratio of areas equals the square of the ratio of diameters.
$\frac{A_1}{A_2} = \left(\frac{d_1}{d_2}\right)^2 = \left(\frac{4}{2}\right)^2 = 4$
Step 2
Apply Continuity
Solve for the throat velocity $v_2$.
$v_2 = v_1 \times \frac{A_1}{A_2} = 2 \times 4 = 8$ m/s
Problem Statement
Water ($\rho = 1000$ kg/m³) flows horizontally. Its speed increases from 3 m/s to 5 m/s. Find the pressure drop.
Bernoulli's PrincipleDynamic Pressure
Strategic Toolkit
$\Delta P = \frac{1}{2}\rho(v_2^2 - v_1^2)$
Step 1
Identify the Simplification
Horizontal flow means $h_1 = h_2$, so the $\rho gh$ terms cancel. Bernoulli reduces to $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$.
$\Delta P = P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)$
Step 2
Substitute Values
Square the velocities first, then subtract.
$\Delta P = \frac{1}{2} \times 1000 \times (5^2 - 3^2) = 500 \times (25 - 9) = 500 \times 16 = 8000$ Pa
Problem Statement
A tiny water droplet of radius 0.010 cm descends through air. Calculate its terminal velocity. ($\eta_{\text{air}} = 19 \times 10^{-6}$ Pa·s, $\rho_{\text{water}} = 1000$ kg/m³)
Terminal VelocityUnit Conversion
Strategic Toolkit
$v_t = \frac{2r^2 \rho g}{9\eta}$
Step 1
Convert Units
Convert radius from cm to metres carefully.
$r = 0.010$ cm $= 1.0 \times 10^{-4}$ m
Step 2
Write Simplified Formula
Air density is negligible compared to water ($\rho_f \approx 0$), so buoyancy is ignored:
$v_t = \frac{2r^2 \rho g}{9\eta}$
Step 3
Substitute Known Values
With $r = 1.0 \times 10^{-4}$ m, $\rho = 1000$ kg/m³, $g = 9.8$ m/s², $\eta = 19 \times 10^{-6}$ Pa·s:
$v_t = \frac{2 \times (1.0 \times 10^{-4})^2 \times 1000 \times 9.8}{9 \times 19 \times 10^{-6}} = \frac{1.96 \times 10^{-4}}{1.71 \times 10^{-4}}$
Step 4
Calculate
Evaluate the fraction:
$v_t \approx 1.1$ m/s
Memorize formulas
Equation of Continuity
Flow rate ($Av$) stays constant; smaller area means higher speed.
Formula
$A_1 v_1 = A_2 v_2$
Bernoulli's Equation
Energy conservation for fluids: Pressure + Kinetic + Potential = Constant.
Formula
$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$
Stokes' Law
Drag force on a slow sphere depends on viscosity, radius, and speed.
Formula
$F_d = 6\pi \eta r v$
Terminal Velocity
Gravity balanced by drag and buoyancy results in constant falling speed.
Formula
$v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}$
Review
1
A main water pipe of cross-sectional area 12 cm² splits into three identical smaller pipes. If water flows at 2 m/s in the main pipe and the flow divides equally, what is the velocity in each smaller pipe if each has an area of 2 cm²?
Ans
4 m/s
Review
Tap to reveal
Apply continuity to the split flow
The total volume flow rate must be conserved, so the flow in the main pipe equals the combined flow through the three identical branches. Since the flow divides equally, each branch carries one-third of the total.
Solve for the branch speed
Use $A_1v_1 = 3A_2v_2$ and substitute the given areas and speed.
$$12 \times 2 = 3 \times 2 \times v_2 \Rightarrow v_2 = 4\ \text{m/s}$$
2
Water ($\rho = 1000$ kg/m³) flows at 3 m/s through a ground-floor pipe where the pressure is 200,000 Pa. The pipe rises to a height of 5 m where the velocity increases to 5 m/s. The pressure in the upper pipe is: ($g = 10$ m/s²)
Ans
142,000 Pa
Review
Tap to reveal
Write Bernoulli between the two points
Use the pressure, kinetic, and gravitational terms at the lower and upper pipe positions. Take the ground floor as the zero-height reference.
Substitute and isolate the unknown pressure
Insert the given values and move the known energy terms to the other side.
$$200000 + \frac{1}{2}(1000)(3^2) = P_2 + \frac{1}{2}(1000)(5^2) + (1000)(10)(5)$$
$$P_2 = 142000\ \text{Pa}$$
3
A steel sphere ($\rho_s = 7800$ kg/m³) reaches terminal velocity $v_t$ in a viscous liquid of density 1300 kg/m³. An aluminium sphere ($\rho_a = 2600$ kg/m³) of the same radius is dropped in the same liquid. Its terminal velocity is:
Ans
$\frac{v_t}{5}$
Review
Tap to reveal
Use the density-difference dependence
For spheres of the same radius in the same fluid, terminal velocity is proportional to the excess density $(\rho_s - \rho_f)$, not the raw object density.
Compare steel and aluminium
Compute the two excess densities and take their ratio.
$$\frac{v_{\text{Al}}}{v_{\text{steel}}} = \frac{2600-1300}{7800-1300} = \frac{1300}{6500} = \frac{1}{5}$$
4
A sphere of radius $r$ falls through a viscous fluid with terminal velocity $v_t$. If the radius is doubled and the viscosity of the fluid is also doubled, the new terminal velocity is:
Ans
$2v_t$
Review
Tap to reveal
Use the Stokes-law scaling
For terminal motion in a viscous fluid, $v_t$ is proportional to $r^2/\eta$ when the fluid and sphere densities are unchanged.
Apply the doubled-radius and doubled-viscosity changes
Doubling the radius gives a factor of $2^2$, while doubling the viscosity gives a factor of $1/2$.
$$v_t' = v_t \times \frac{(2r)^2}{2\eta} \times \frac{\eta}{r^2} = 2v_t$$
5
In Bernoulli's equation $P + \frac{1}{2}\rho v^2 + \rho gh = \text{const}$, the dimension of each term is:
Ans
kg·m⁻¹·s⁻²
Review
Tap to reveal
Check the pressure dimension
The Bernoulli equation sums pressure-like terms, so each term must have the same unit as pressure.
Verify the derived units
Pressure is force per unit area, which in base SI units is $\text{kg·m}^{-1}\text{·s}^{-2}$; the other terms reduce to the same unit.
$$[P] = [\rho v^2] = [\rho gh] = \text{kg·m}^{-1}\text{·s}^{-2}$$
6
The drag force on a sphere moving through a viscous fluid is given by Stokes' Law as $F = 6\pi\eta rv$. If both the radius and speed are doubled, the drag force becomes:
Ans
4 times
Review
Tap to reveal
Use the proportionality in Stokes' law
The drag force is directly proportional to both radius and speed, so changing either one rescales the force multiplicatively.
Double both factors
If radius doubles and speed doubles, the force is multiplied by $2 \times 2$.
$$F' = 6\pi\eta(2r)(2v) = 4F$$
7
A tiny water droplet of radius $1.0 \times 10^{-4}$ m falls through air ($\eta = 1.9 \times 10^{-5}$ Pa·s). Taking $\rho_{\text{water}} = 1000$ kg/m³ and $g = 9.8$ m/s², its terminal velocity is approximately:
Ans
1.1 m/s
Review
Tap to reveal
Use the terminal-velocity formula
For a small sphere in a viscous medium, the terminal speed is $v_t = \frac{2r^2\rho g}{9\eta}$, and the air buoyancy is negligible here.
Substitute the values and simplify
Insert the radius, density, gravity, and viscosity, then evaluate the powers of ten carefully.
$$v_t = \frac{2(10^{-4})^2(1000)(9.8)}{9(1.9\times10^{-5})} \approx 1.1\ \text{m/s}$$
8
A water hose with an internal diameter of 20 mm discharges 30 kg of water in 60 s. The water speed at the outlet is: ($\rho = 1000$ kg/m³)
Ans
1.6 m/s
Review
Tap to reveal
Find the mass flow rate and outlet area
Convert the discharged mass per time into $\dot{m}$, and use the hose radius, not the diameter, in the circular area formula.
Use $\dot{m} = \rho Av$
Solve for the speed after substituting the given density and area.
$$\dot{m} = \frac{30}{60} = 0.5\ \text{kg/s},\quad A = \pi(0.01)^2 = 3.14\times10^{-4}\ \text{m}^2$$
$$v = \frac{0.5}{1000\times3.14\times10^{-4}} \approx 1.6\ \text{m/s}$$
9
As water falls freely from a tap, the stream gets narrower. This is best explained by:
Ans
The equation of continuity — speed increases so area must decrease
Review
Tap to reveal
Identify the mechanism controlling stream thickness
As the water falls, gravity increases its speed. For steady flow, the volume flow rate must stay constant along the stream.
Apply continuity
If speed increases, the cross-sectional area must decrease so that $Av$ stays constant.
$$A_1v_1 = A_2v_2 \Rightarrow v_2>v_1 \; \text{so} \; A_2<A_1$$
10
A sphere of density $\rho_s$ is dropped into a liquid of density $\rho_f$ and viscosity $\eta$. If $\rho_s < \rho_f$, the sphere will:
Ans
Rise upward and reach a terminal velocity
Review
Tap to reveal
Compare buoyancy and weight
If the sphere is less dense than the fluid, the buoyant force is larger than its weight, so the net force is upward.
Recognize the terminal-state direction
As it rises, viscous drag acts downward and grows until it balances the upward net force, giving an upward terminal speed.
$$\rho_s < \rho_f \Rightarrow v_t < 0 \text{ in the usual downward sign convention}$$
Question 1 of 10
0:00
1
A main water pipe of cross-sectional area 12 cm² splits into three identical smaller pipes. If water flows at 2 m/s in the main pipe and the flow divides equally, what is the velocity in each smaller pipe if each has an area of 2 cm²?