Fluid Mechanics — Fluid Statics and Kinematics
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
Fluid Statics
- Pressure at depth: p = p₀ + ρgh
- Pascal’s law: Pressure applied anywhere in a confined fluid transmits equally in all directions
- Manometers: p_A = p_atm + ρgh (vertical) or p_A − p_B = ρg(h_A − h_B)
- Buoyancy: F_B = ρ_f × V_d × g (Archimedes’ principle); Body floats if ρ_body < ρ_fluid
Fluid Kinematics
- Continuity equation: A₁V₁ = A₂V₂ (Q = constant)
- Bernoulli’s equation: p/ρg + V²/2g + z = constant (along a streamline)
- Stream function ψ: Lines of constant ψ = streamlines; Δψ = discharge between two streamlines
Key Formulas Table
| Concept | Formula |
|---|---|
| Hydrostatic Pressure | p = p₀ + ρgh |
| Buoyancy Force | F_B = ρVg |
| Continuity | A₁V₁ = A₂V₂ |
| Bernoulli | p/ρ + V²/2 + gz = const |
| Stream Function | dψ = V·dr (2D flow) |
⚡ GATE Tip: Bernoulli is for inviscid, steady, incompressible flow along a streamline. If viscosity or rotation is significant, Bernoulli does NOT apply directly.
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Fluid Statics
Hydrostatic Pressure Distribution
Pressure in a static fluid increases linearly with depth:
p = p₀ + ρgh
where ρ is fluid density, g is acceleration due to gravity, and h is depth below the free surface. This means pressure is the same at all points at the same depth — it acts perpendicularly on any surface (Pascal’s law).
Absolute vs Gauge Pressure:
- Gauge pressure: p_g = ρgh (relative to atmospheric)
- Absolute pressure: p_abs = p_atm + ρgh
⚡ GATE Watch: Many problems use gauge pressure implicitly. Watch for atmospheric pressure being explicitly stated or omitted.
Manometers
Manometers measure pressure difference using fluid column heights.
Simple U-tube manometer:
p_A − p_B = ρ_m·g·h
For a differential manometer with two fluids of different densities:
p_A − p_B = g(ρ₂h₂ − ρ₁h₁)
Where the denser fluid goes on the heavier side. If both limbs have the same fluid, the pressure difference equals the weight of the column difference.
Buoyancy and Archimedes’ Principle
A body submerged or partially submerged in a fluid experiences an upward buoyant force equal to the weight of displaced fluid:
F_B = ρ_fluid × V_displaced × g
Floating vs Submerged:
- Fully submerged: F_B = ρ_fluid × V_body × g; Body sinks if ρ_body > ρ_fluid
- Floating (partially submerged): F_B = ρ_fluid × V_submerged × g = W_body = ρ_body × V_total × g
The metacentre and centre of buoyancy determine stability of floating bodies. A floating body is stable if the metacentre is above the centre of gravity.
Pascal’s Law and Its Applications
Pascal’s law states that pressure applied to an enclosed fluid transmits undiminished to all portions of the fluid and the container walls. This principle powers:
- Hydraulic jacks: Small force on small piston → large force on large piston (F₁/A₁ = F₂/A₂)
- Hydraulic brakes: Brake pedal force multiplied through master/slave cylinder system
Fluid Kinematics
Continuity Equation
For incompressible flow (liquid or low-speed gas):
A₁V₁ = A₂V₂ or Q = AV (volume flow rate constant)
For compressible flow (high-speed gas): ρ₁A₁V₁ = ρ₂A₂V₂
The continuity equation is simply mass conservation applied to a flow tube.
Stream Function
In two-dimensional incompressible flow, the stream function ψ(x,y) satisfies:
u = ∂ψ/∂y, v = −∂ψ/∂x
where u and v are velocity components in x and y directions. Lines of constant ψ are streamlines. The volume flow rate between two streamlines equals the difference in their ψ values:
Q = ψ₂ − ψ₁
Bernoulli’s Equation
Derived from Newton’s second law along a streamline for inviscid, steady, incompressible flow:
p/ρ + V²/2 + gz = constant
Or in head form:
p/(ρg) + V²/(2g) + z = H (total head)
⚡ Critical Reminder: Bernoulli’s equation CANNOT be used when:
- Viscous effects are significant (boundary layers, pipes with friction)
- There is a pump or turbine (add/remove energy)
- Flow is rotational (vorticity present)
- Across a shock wave
Example Problem
Water flows through a horizontal pipe that reduces from 300 mm diameter to 150 mm diameter. Velocity in the 300 mm section is 2 m/s. Find velocity in the 150 mm section and pressure difference if the pressure in the larger section is 150 kPa.
Solution: A₁ = π(0.3)²/4 = 0.0707 m²; A₂ = π(0.15)²/4 = 0.01767 m² V₂ = V₁(A₁/A₂) = 2×(0.0707/0.01767) = 8 m/s Applying Bernoulli (p₁/ρ + V₁²/2 = p₂/ρ + V₂²/2, since z₁ = z₂ = 0): p₂ = p₁ + ρ(V₁²−V₂²)/2 = 150×10³ + 1000(4−64)/2 = 150000 − 30000 = 120 kPa Δp = p₁ − p₂ = 30 kPa
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Potential Flow Theory
Velocity Potential Function
For irrotational flow, a velocity potential φ exists such that:
u = ∂φ/∂x, v = ∂φ/∂y, w = ∂φ/∂z
The Laplace equation governs potential flow:
∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z² = 0
A function is harmonic if it satisfies Laplace’s equation. Both φ (potential) and ψ (stream function) are harmonic in irrotational, incompressible 2D flow.
Combination of Basic Flows
Complex potential flow patterns can be built from superposition of elementary solutions:
| Flow Type | φ (potential) | ψ (stream function) |
|---|---|---|
| Uniform flow | φ = Ux | ψ = Uy |
| Source/Sink | φ = (Q/2π) ln r | ψ = (Q/2π) θ |
| Free vortex | φ = −(Γ/2π) θ | ψ = (Γ/2π) ln r |
| Doublet | φ = −K y/(x²+y²) | ψ = K x/(x²+y²) |
Doublet and Cylinder Flow
Superposition of uniform flow + source + sink (brought infinitely close) produces flow around a cylinder:
- Velocity at cylinder surface: V_θ = 2U sin θ
- Maximum velocity: 2U (at θ = 90° and 270°)
- Pressure distribution from Bernoulli gives d’Alembert’s paradox: Zero drag force on a cylinder in inviscid flow
Stagnation Points
Where V = 0 (from Bernoulli, p = p₀ + ρV²/2 = p₀ + ρ×0, so maximum pressure). On a cylinder in uniform flow, stagnation points occur at θ = 0° and 180°.
Free Surface Flows
For flows with a free surface (open channel, weir flow), the Euler’s equation in the vertical direction under hydrostatic assumption gives:
p/ρg + z = constant (piezometric head)
This is used in Weir formulas and orifice flow with free discharge.
Rotational vs Irrotational Flow
- Irrotational: No vorticity (∇ × V = 0); potential flow theory applies; Bernoulli valid along streamlines
- Rotational: Vorticity present (e.g., flow in boundary layers, downstream of bluff bodies, free vortices); Bernoulli NOT valid between arbitrary points
⚡ GATE Trap: Students often apply Bernoulli between two points on different streamlines in rotational flow — this is incorrect.
Hydrostatic Forces on Surfaces
For a submerged vertical plane surface of width b at depth h:
F_R = ρg·b·h·h̄ (h̄ = centroid depth = h/2) F_R = ρg·Ā·h̄ (Ā = area)
Line of action (center of pressure):
h_cp = I_G/Āh̄ + h̄ (about free surface)
The center of pressure is always below the centroid for a vertical surface (deeper = higher pressure).
For a horizontal surface, pressure is uniform → F_R = p·A at that depth.
Example — Hydrostatic Force on a Gate
A rectangular gate 3 m wide and 4 m deep is vertical, submerged in water with its top 2 m below the water surface. Find resultant force and its location.
Solution: Depth of centroid h̄ = 2 + 4/2 = 4 m Area A = 3×4 = 12 m² F_R = ρg·A·h̄ = 1000×9.81×12×4 = 471,000 N = 471 kN
I_G about horizontal axis through centroid = bh³/12 = 3×4³/12 = 16 m⁴ h_cp = h̄ + I_G/(Āh̄) = 4 + 16/(12×4) = 4 + 0.333 = 4.333 m below surface
GATE Previous Year Pattern
| Year | Topic | Marks |
|---|---|---|
| 2023 | Bernoulli + continuity (pipe contraction) | 5 |
| 2022 | Hydrostatic force on submerged plane | 2 |
| 2021 | Buoyancy and stability | 2 |
| 2020 | Manometer pressure measurement | 2 |
| 2019 | Potential flow — cylinder | 5 |
⚡ Common Mistakes: (1) Applying Bernoulli in viscous flows, (2) Wrong sign in manometer equation, (3) Confusing stream function ψ with velocity potential φ, (4) Forgetting that center of pressure is below centroid.
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