Fluid Mechanics-I (CE-501) Unit 3 Notes - RGPV

UNIT 3: FLUID MECHANICS-I (Exam-Focused Short Notes)

I. FLUID PROPERTIES & PRESSURE FUNDAMENTALS

A. Basic Properties

Property	Symbol	Definition	SI Unit	Relation
Density	$\rho$	Mass per unit volume	kg/m³	$$\displaystyle \rho = \frac{m}{V} $$
Specific Weight	$\gamma$	Weight per unit volume	N/m³	$$\displaystyle \gamma = \rho g $$
Specific Volume	$v$	Volume per unit mass	m³/kg	$$\displaystyle v = \frac{1}{\rho} $$
Specific Gravity	$SG$	Ratio of density to water density at 4°C	-	$$\displaystyle SG = \frac{\rho}{\rho_{water}} $$

Viscosity

Dynamic Viscosity ($\mu$): Measure of internal resistance to flow (shear stress/velocity gradient).

$$\tau = \mu \frac{du}{dy}$$

Unit: N·s/m² (Pa·s).

Kinematic Viscosity ($\nu$): Dynamic viscosity divided by density.

$$\nu = \frac{\mu}{\rho}$$

Unit: m²/s (Stoke: 1 St = 10⁻⁴ m²/s).

[!TIP] Temperature Variation

Liquids: Viscosity ↓ with temperature ↑ (molecular bonds weaken).

Gases: Viscosity ↑ with temperature ↑ (molecular momentum transfer increases).

Common Pitfall: Confusing trends for liquids vs. gases.

Surface Tension ($\sigma$)

Force per unit length acting tangentially on surface.
Pressure difference across curved interface (Young-Laplace):

$$\Delta P = \sigma \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$

For spherical droplet/bubble:

Droplet: $$\displaystyle \Delta P = \frac{2\sigma}{R} $$

Soap bubble: $$\displaystyle \Delta P = \frac{4\sigma}{R} $$ (two interfaces).

Bulk Modulus of Elasticity ($K$)

Measure of fluid compressibility:

$$K = -V \frac{dP}{dV} = \rho \frac{dP}{d\rho}$$

For isentropic process: $$\displaystyle K = \rho c^2 $$, where $c$ = speed of sound.

B. Pressure Concepts

Absolute Pressure: Measured relative to perfect vacuum.
Gauge Pressure: Relative to atmospheric pressure ($$\displaystyle P_{atm} $$).

$$P_{abs} = P_{gauge} + P_{atm}$$

Vacuum Pressure: When $$\displaystyle P_{abs} < P_{atm} $$, $$\displaystyle P_{vac} = P_{atm} - P_{abs} $$.

Pascal’s Law

Pressure at a point in a static fluid is transmitted equally in all directions.
Intensity of Pressure: $$\displaystyle P = \frac{F}{A} $$ (normal force per unit area).

Buoyant Force

Archimedes’ Principle: $$\displaystyle F_b = \gamma \cdot V_{disp} $$ (weight of displaced fluid).
Center of Buoyancy: Centroid of displaced volume.
Stokes’ Law context: Drag force on sphere at low Re ($$\displaystyle Re < 0.1 $$):

$$F_d = 3\pi \mu d U$$

Used in settling velocity calculations.

II. PRESSURE MEASUREMENT

A. Manometers

Simple Manometer

Measures gauge pressure at a point.
Formula for U-tube with different liquids:

$$P_1 + \rho_1 g h_1 = P_2 + \rho_2 g h_2$$

Differential Manometer

Measures pressure difference between two points.
U-tube differential:

$$P_A - P_B = (\rho_m - \rho) g h$$

($$\displaystyle \rho_m $$ = manometric fluid density, $\rho$ = fluid density, $h$ = mercury height difference).

Inverted U-tube: Used for low pressure differences; sensitive to gas flows.

Micromanometer

U-tube with enlarged ends; measures very small $\Delta P$.

$$\Delta P = 2 \sigma \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$$

(if meniscus curvature considered).

[!TIP] Typical Problem

Given: Two pipes with liquids of sp.gr. 1.5 and 0.9, pressures $$\displaystyle P_A = 1 $$ kgf/cm², $$\displaystyle P_B = 1.8 $$ kgf/cm², mercury manometer (sp.gr. 13.6). Find $h$.

Solution: Convert to consistent units (e.g., N/m² or m of water). Use:

$$P_A + \rho_A g h_A = P_B + \rho_B g h_B + \rho_{Hg} g h$$

With $$\displaystyle h_A = h_B $$ if pipe centers at same level. Solve for $h$.

B. Pressure Distribution on Surfaces

Plane Surface

Total force: $$\displaystyle F = P_{c} \cdot A $$, where $$\displaystyle P_c = \gamma \bar{h} $$ (pressure at centroid).
Center of Pressure (CoP):

$$h_{cp} = \bar{h} + \frac{I_{xx}}{A \bar{h}^2}$$

$$\displaystyle I_{xx} $$ = second moment of area about centroidal axis.

Curved Surface

Horizontal component = force on projected vertical area.
Vertical component = weight of fluid above surface.
Resultant force magnitude & direction from components.

III. FLOW KINEMATICS & VISUALIZATION

A. Flow Description

Type	Basis	Example
Steady	Properties independent of time	$$\displaystyle \frac{\partial}{\partial t}(\cdot) = 0 $$
Unsteady	Properties vary with time	Startup flow
Uniform	Properties constant along flow direction	Fully developed pipe flow
Non-uniform	Properties vary along flow	Converging duct
Laminar	Smooth, ordered motion	$$\displaystyle Re < 2000 $$ (pipe)
Turbulent	Chaotic, mixing	$$\displaystyle Re > 4000 $$ (pipe)
Rotational	Vorticity $\omega \neq 0$	Boundary layers
Irrotational	$$\displaystyle \omega = 0 $$	Potential flow

Reynolds Number ($Re$):

$$Re = \frac{\rho U L}{\mu} = \frac{U L}{\nu}$$

Criterion for transition (pipe: $Re \approx 2300$).

B. Flow Lines & Functions

Line	Definition	Key Property
Pathline	Actual trajectory of a fluid particle	Time-dependent
Streakline	Locus of particles passing through a fixed point	Injection point
Streamline	Line tangent to velocity vector at instant	$$\displaystyle \frac{dx}{u} = \frac{dy}{v} = \frac{dz}{w} $$; no cross-flow

Stream Function ($\psi$)

Exists for incompressible flow.
Properties:
1. $$\displaystyle \psi = \text{constant} $$ ⇒ streamline.
2. Flow between two streamlines = $\Delta \psi$ (discharge per unit depth).
3. $$\displaystyle u = \frac{\partial \psi}{\partial y},\; v = -\frac{\partial \psi}{\partial x} $$ (2D).
4. Satisfies continuity automatically.

Velocity Potential ($\phi$)

Exists for irrotational flow: $$\displaystyle \vec{V} = \nabla \phi $$.
Properties:
1. $$\displaystyle \phi = \text{constant} $$ ⇒ equipotential line.
2. Streamlines and equipotentials are orthogonal.
3. Satisfies Laplace’s equation: $$\displaystyle \nabla^2 \phi = 0 $$.

Flow Net

Grid of intersecting streamlines and equipotentials (orthogonal).
Used in 2D irrotational flow (e.g., groundwater, airfoils).
Method of Drawing:
1. Draw boundary streamlines ($$\displaystyle \psi = 0, \psi = Q $$).
2. Sketch equipotentials orthogonal to boundaries.
3. Adjust until squares (equal $\Delta \psi$, $\Delta \phi$).
Each “square” represents equal discharge and potential drop.

C. Continuity Equation

General 3D (Cartesian):

$$\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0$$

Derivation: Apply conservation of mass to infinitesimal control volume ($dx\,dy\,dz$). Net outflow = rate of mass decrease.

1D Flow: $$\displaystyle A_1 V_1 = A_2 V_2 $$ (discharge constant).
2D Stream Function: Automatically satisfies continuity.

IV. FLOW DYNAMICS: ENERGY & MOMENTUM

A. Bernoulli’s Equation

Statement (along a streamline for inviscid, steady, incompressible flow):

$$P + \frac{1}{2}\rho V^2 + \rho g z = \text{constant}$$

Terms: Pressure head, Velocity head, Elevation head.

Derivation from Euler’s equation:

$$-\frac{dP}{\rho} + g \, dz + V \, dV = 0$$

Integrate along streamline ⇒ Bernoulli.

Assumptions:

Steady flow
Incompressible ($$\displaystyle \rho = \text{const} $$)
Inviscid ($$\displaystyle \mu = 0 $$)
Along streamline
No shaft work/heat transfer

Limitations:

Not valid for viscous flows, rotational flows, or across shocks.
Cannot predict pressure recovery in diffusers (separation).

Modified Bernoulli (with losses):

$$P_1 + \frac{1}{2}\rho V_1^2 + \rho g z_1 = P_2 + \frac{1}{2}\rho V_2^2 + \rho g z_2 + h_L \rho g$$

$$\displaystyle h_L $$ = head loss (major + minor).

B. Momentum Equation

General Form (Control Volume):

$$\sum \vec{F} = \frac{d}{dt} \int_{CV} \rho \vec{V} \, dV + \int_{CS} \rho \vec{V} (\vec{V} \cdot \vec{n}) \, dA$$

For steady flow: $$\displaystyle \sum \vec{F} = \int_{CS} \rho \vec{V} (\vec{V} \cdot \vec{n}) \, dA $$.

Applications:

Force on pipe bend/nozzle:

$$F_x = \dot{m}(V_{out,x} - V_{in,x}) + (P_{out}A_{out} - P_{in}A_{in}) \cos\theta$$

Force on vane/plate (jet deflection).

C. Energy Equation (First Law)

For steady flow:

$$\dot{Q} - \dot{W}_s = \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1) \right)$$

$\dot{Q}$: heat transfer rate
$$\displaystyle \dot{W}_s $$: shaft work (pump/turbine)

Pump Power:

$$P_{pump} = \dot{m} \cdot \frac{\Delta P}{\rho} / \eta_{pump}$$

Turbine Power:

$$P_{turbine} = \dot{m} \cdot g \Delta h \cdot \eta_{turbine}$$

V. FLOW MEASUREMENT & ORIFICES

A. Flow Measurement Principles

Pitot Tube

Measures stagnation pressure $$\displaystyle P_0 = P + \frac{1}{2}\rho V^2 $$.
Velocity: $$\displaystyle V = \sqrt{\frac{2(P_0 - P)}{\rho}} $$ (if static pressure $P$ known).
Used for velocity profiling.

Venturi Meter

Converging section, throat, diverging diffuser.
Derivation (Bernoulli + continuity):

$$Q_{theo} = A_1 A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho (A_1^2 - A_2^2)}}$$

Actual discharge: $$\displaystyle Q = C_d Q_{theo} $$, $$\displaystyle C_d \approx 0.98 $$ (high accuracy, low loss).

Flow Nozzle

Converging nozzle only (no diffuser).
$$\displaystyle C_d \approx 0.96 $$; higher loss than venturi.

Orifice Meter

Thin plate with orifice.
Derivation:

$$Q_{theo} = A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho}}$$

Actual: $$\displaystyle Q = C_d A_2 \sqrt{\frac{2(P_1 - P_2)}{\rho}} $$

$$\displaystyle C_d = C_c C_v $$ (contraction × velocity coefficients).

$$\displaystyle C_d \approx 0.6–0.7 $$; high energy loss due to vena contracta.

B. Comparison of Devices

Feature	Venturi	Flow Nozzle	Orifice
Energy Loss	Lowest	Moderate	Highest
Cost	High	Medium	Low
Accuracy	High (±1%)	Medium (±2%)	Low (±5–10%)
$$\displaystyle C_d $$	0.98	0.96	0.6–0.7
Applications	Large pipes, permanent	Steam, high velocity	Low cost, temporary

C. Notches & Weirs

Rectangular Weir

Discharge: $$\displaystyle Q = C_d \frac{2}{3} L \sqrt{2g} \, H^{3/2} $$

$L$ = length, $H$ = head over crest.

Triangular (V-notch)

$$\displaystyle Q = C_d \frac{8}{15} \tan(\theta/2) \sqrt{2g} \, H^{5/2} $$

Trapezoidal (Cipoletti): Combination of rectangular + triangular.

VI. DIMENSIONAL ANALYSIS & SIMILITUDE

A. Fundamental Concepts

Dimensions: $[M], [L], [T]$ (mass, length, time).
Units: Numerical values (kg, m, s).
Dimensional Homogeneity: Equation must have same dimensions on both sides.

B. Buckingham Pi Theorem

Statement: If $n$ variables in a problem with $r$ fundamental dimensions, then $(n - r)$ independent $\Pi$ groups exist.

Procedure:

List all variables ($n$).
Identify repeating variables ($r$; usually $\rho, V, D$).
Form $\Pi$ groups: $$\displaystyle \Pi = \text{var} \cdot (\text{repeating})^a $$.
Solve exponents by dimensional homogeneity.
Write functional relation: $$\displaystyle F(\Pi_1, \Pi_2, ...) = 0 $$.

Example: Drag Force $R$ on body

Variables: $R, \rho, V, D, \mu, g$ → $$\displaystyle n=6 $$, $$\displaystyle r=3 $$ ($M,L,T$) → 3 $\Pi$ groups.

Choose $\rho, V, D$ as repeating.

$$\displaystyle \Pi_1 = \frac{R}{\rho V^2 D^2} $$ (Force coefficient)
$$\displaystyle \Pi_2 = \frac{\rho V D}{\mu} = Re $$ (Reynolds)
$$\displaystyle \Pi_3 = \frac{V^2}{g D} = Fr $$ (Froude)

Thus: $$\displaystyle R = \rho V^2 D^2 \cdot f(Re, Fr) $$.

C. Model Testing

Similarities:

Geometric: Same shape ($$\displaystyle L_m/L_p = \text{constant} $$).
Kinematic: Same motion ($$\displaystyle V_m/V_p = \text{constant} $$).
Dynamic: Same force ratios (same $\Pi$ groups).

Model Laws:

Reynolds Law: $$\displaystyle Re_m = Re_p $$ (viscous dominated). Hard to achieve full similarity due to different $\nu$.
Froude Law: $$\displaystyle Fr_m = Fr_p $$ (gravity dominated, free surface). Common in ships, open channels.
Euler Law: $$\displaystyle Eu_m = Eu_p $$ (pressure forces).
Mach Law: $$\displaystyle Ma_m = Ma_p $$ (compressibility).

Attainability:

Usually partial similarity; prioritize dominant forces.
Example: Ship model → Froude similarity (gravity), ignore viscous (Re not matched).

VII. PIPE FLOW & FRICTION LOSSES

A. Laminar Flow (Hagen-Poiseuille)

Assumptions: Steady, laminar, Newtonian, horizontal, circular pipe, no slip.

Velocity Distribution (parabolic):

$$u(r) = \frac{\Delta P}{4\mu L} (R^2 - r^2)$$

Max velocity at center: $$\displaystyle u_{max} = \frac{\Delta P R^2}{4\mu L} $$.

Shear Stress Distribution (linear):

$$\tau(r) = \frac{r}{2} \frac{\Delta P}{L}$$

Hagen-Poiseuille Equation (discharge):

$$Q = \frac{\pi R^4 \Delta P}{8\mu L} = \frac{\pi D^4 \Delta P}{128\mu L}$$

[!TIP] Numerical Problem

Given: $$\displaystyle D=150 $$ mm, at $$\displaystyle r=20 $$ mm, $$\displaystyle u=0.4 $$ m/s (laminar). Find $$\displaystyle u_{max} $$, $$\displaystyle V_{mean} $$, $Q$.

Solution:

$$\displaystyle u_{max} = u(r) + \frac{r^2}{2} \frac{d^2u}{dr^2} $$? Actually, use parabolic profile: $$\displaystyle u = u_{max}(1 - r^2/R^2) $$.

So $$\displaystyle 0.4 = u_{max}(1 - (0.02)^2/(0.075)^2) \Rightarrow u_{max} \approx 0.4 / (1 - 0.0711) = 0.431 $$ m/s.

$$\displaystyle V_{mean} = \frac{u_{max}}{2} = 0.2155 $$ m/s.

$$\displaystyle Q = V_{mean} \cdot A = 0.2155 \times \pi (0.075)^2 \approx 0.012 $$ m³/s = 12 lps.

B. Turbulent Flow & Friction

Darcy-Weisbach Equation:

$$h_f = f \frac{L}{D} \frac{V^2}{2g}$$

where $f$ = friction factor (dimensionless).

Friction Factor $f$:

Laminar: $$\displaystyle f = \frac{64}{Re} $$.
Turbulent: Depends on $Re$ and relative roughness $\varepsilon/D$.

Use Blasius (smooth pipes, $$\displaystyle Re < 10^5 $$): $$\displaystyle f = 0.316 / Re^{0.25} $$.

Moody Chart: Graphical solution for all regimes.

Minor Losses:
$$\displaystyle h_m = K \frac{V^2}{2g} $$, where $K$ = loss coefficient (tabulated).

Entrance: $K \approx 0.5$ (sharp), 0.04 (well-rounded).
Exit: $$\displaystyle K = 1 $$.
Bends/valves: Manufacturer data.

C. Pipes in Series & Parallel

Series: Same $Q$, total $$\displaystyle h_f = \sum h_{f,i} $$.

Equivalent length method: $$\displaystyle L_{eq} = \sum L_i \frac{f_i D_i}{f D} $$.

Parallel: Same $$\displaystyle h_f $$, $$\displaystyle Q_{total} = \sum Q_i $$.

For two pipes:

$$\frac{Q_1}{Q_2} = \left( \frac{D_1}{D_2} \right)^{2.63} \left( \frac{L_2}{L_1} \right)^{0.54} \quad \text{(approx, turbulent)}$$

Or solve: $$\displaystyle h_f = f_1 \frac{L_1}{D_1} \frac{V_1^2}{2g} = f_2 \frac{L_2}{D_2} \frac{V_2^2}{2g} $$ with $$\displaystyle Q = Q_1 + Q_2 $$.

[!TIP] Common Pitfall: In parallel pipes, pressure drop is same, not flow rate. Use iterative method if $f$ depends on $Re$.

VIII. KEY DEFINITIONS & SHORT NOTES (3-Mark Questions)