Module 2: Physics

Module 2: Physics – B1.3 Subsonic Aeroplane Aerodynamics, Structures and Systems

1. Overview

Module 2 of the EASA Part-66 syllabus provides the foundational physics knowledge required for aircraft maintenance certifying staff. This module covers the core principles of mechanics, thermodynamics, optics, and wave motion, and applies them to the operation, maintenance, and troubleshooting of aircraft systems. For the B1.3 (Helicopter) category, the emphasis is on understanding how these physical laws govern helicopter flight, power transmission, hydraulic and pneumatic systems, and engine performance.

The module is structured to build from basic concepts to more complex applications. It begins with statics (forces, moments, equilibrium), progresses through kinematics and kinetics (motion and its causes), and then moves into thermodynamics and fluid dynamics, which are essential for understanding engine cycles, gas laws, and hydraulic systems. The syllabus also includes oscillations, waves, and sound, which are relevant to vibration analysis and rotor dynamics.

The knowledge levels are defined as:

• Level 1: A familiarisation with the principal elements of the subject. The candidate should have a basic overview.
• Level 2: A general knowledge of the theoretical and practical aspects of the subject. The candidate should be able to apply general knowledge to typical maintenance situations.
• Level 3: A detailed knowledge of the theoretical and practical aspects of the subject. The candidate should be able to analyse, diagnose, and solve complex problems.

This study material synthesises the key concepts from the exam questions provided, focusing on the areas most frequently tested and most critical for helicopter maintenance.

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2. Key Concepts Explained in Detail

2.1 Units and Dimensions

All physical quantities consist of a numerical value and a unit. The international system of units (SI) is the standard used in aviation maintenance documentation and calculations.

• Base SI Units:
• Length: metre (m)
• Mass: kilogram (kg)
• Time: second (s)
• Temperature: kelvin (K)
• Electric Current: ampere (A)
• Derived SI Units:
• Force: newton (N) = kg·m/s²
• Pressure: pascal (Pa) = N/m²
• Energy/Work: joule (J) = N·m
• Power: watt (W) = J/s
• Frequency: hertz (Hz) = s⁻¹
• Common Conversions:
• 1 bar = 100,000 Pa = 100 kPa
• 1 psi = 6894.76 Pa ≈ 6.895 kPa
• 1 litre = 0.001 m³
• 1 kg/L = 1000 kg/m³

Example (Q19, Q23, Q47): A hydraulic system pressure of 3000 psi is converted to SI units as follows:

3000 psi × 6894.76 Pa/psi = 20,684,280 Pa ≈ 20.68 MPa.

2.2 Statics: Forces, Moments, and Equilibrium

Statics deals with bodies at rest or moving at constant velocity, where the net force is zero.

• Force: A vector quantity that causes a change in motion. Its SI unit is the newton (N). Weight (W) is the force due to gravity: W = m × g, where g = 9.81 m/s².
• Moment (Torque): The turning effect of a force about a point. It is calculated as the product of the force and the perpendicular distance from the line of action of the force to the pivot point: Moment = Force × Distance. The SI unit is the newton-metre (N·m).
• Equilibrium: A body is in equilibrium when the vector sum of all forces acting on it is zero, and the sum of all moments about any point is zero. This is a direct application of Newton's First Law.
• Stress: The internal resistance of a material to an applied force. It is calculated as Stress = Force / Cross-sectional Area. The SI unit is the pascal (Pa), but for structural materials, megapascals (MPa) are commonly used.

Example (Q7, Q26, Q43): A helicopter climbing or descending at a constant velocity is in equilibrium. The net force is zero. For a climb, the upward lift force (L) must balance the downward weight (W) and any drag component. In a simplified vertical climb, L = W.

Example (Q8): A helicopter hovering with a weight of 20,000 N and a rotor disc area of 50 m² exerts an average pressure on the air of:

Pressure = Force / Area = 20,000 N / 50 m² = 400 Pa.

Example (Q20): A steel bolt with a cross-sectional area of 50 mm² (50 × 10⁻⁶ m²) subjected to a 20 kN tensile force experiences a stress of:

Stress = 20,000 N / (50 × 10⁻⁶ m²) = 4 × 10⁸ Pa = 400 MPa.

Example (Q28): The braking torque of a disc brake is calculated using the friction force and the mean radius:

Friction Force = μ × Normal Force = 0.4 × 5000 N = 2000 N

Torque = Force × Radius = 2000 N × 0.25 m = 500 N·m.

2.3 Kinematics and Kinetics: Motion and Its Causes

Kinematics describes motion without considering its causes, while kinetics relates motion to the forces that cause it.

• Linear Motion:
• Velocity (v) is the rate of change of displacement: v = Δs/Δt (m/s).
• Acceleration (a) is the rate of change of velocity: a = Δv/Δt (m/s²).
• Newton's Second Law: F = m × a.
• Angular Motion:
• Angular velocity (ω) is the rate of change of angular displacement, measured in radians per second (rad/s).
• Conversion from RPM: ω = (RPM × 2π) / 60.
• Linear velocity at a radius (r): v = ω × r.
• Angular momentum (H) is the product of moment of inertia (I) and angular velocity: H = I × ω. The SI unit is kg·m²/s.

Example (Q3, Q5, Q24, Q41): A rotor blade of length 8 m rotating at 400 RPM has a tip speed calculated as:

ω = (400 × 2π) / 60 = 41.87 rad/s

v = ω × r = 41.87 × 8 = 334.9 m/s.

Example (Q12): A rotor blade with a moment of inertia of 50 kg·m² rotating at 300 RPM has angular momentum:

ω = (300 × 2π) / 60 = 31.4 rad/s

H = 50 × 31.4 = 1570.8 kg·m²/s.

Example (Q40): The net thrust of a turbojet engine is calculated from the change in momentum of the air:

Thrust = Mass Flow Rate × (Exit Velocity - Inlet Velocity) = 5 kg/s × (300 - 60) m/s = 1200 N.

Example (Q44): The torque in a tail rotor drive shaft transmitting 200 kW at 6000 RPM is:

ω = (6000 × 2π) / 60 = 628 rad/s

Torque = Power / ω = 200,000 W / 628 rad/s = 318.5 N·m.

2.4 Thermodynamics: Heat, Work, and Gas Laws

Thermodynamics is the study of heat, work, and energy conversion. It is fundamental to understanding gas turbine engines, air conditioning, and hydraulic systems.

• Temperature Scales:
• Kelvin (K) is the SI unit of absolute temperature.
• Conversion: K = °C + 273.15.
• Gas Laws:
• Boyle's Law (constant temperature): P₁V₁ = P₂V₂. Pressure is inversely proportional to volume.
• Charles's Law (constant pressure): V₁/T₁ = V₂/T₂. Volume is directly proportional to absolute temperature.
• Gay-Lussac's Law (constant volume): P₁/T₁ = P₂/T₂. Pressure is directly proportional to absolute temperature.
• Ideal Gas Law: PV = nRT, where P is pressure (Pa), V is volume (m³), n is the number of moles, R is the universal gas constant (8.314 J/(mol·K)), and T is absolute temperature (K). For a given gas, this can be rearranged to find density: ρ = P / (R_specific × T), where R_specific for air is 287 J/(kg·K).

Example (Q2): A nitrogen-filled shock strut at 500 kPa and 20°C (293 K) is heated to 50°C (323 K). Using Gay-Lussac's Law:

P₂ = P₁ × (T₂/T₁) = 500 × (323/293) = 551 kPa.

Example (Q10): An OAT of -20°C is converted to Kelvin: -20 + 273.15 = 253.15 K ≈ 253 K.

Example (Q42): Air density at the compressor inlet with P = 95 kPa and T = 288 K:

ρ = P / (R × T) = 95,000 / (287 × 288) ≈ 1.15 kg/m³.

• Heat Transfer:
• Specific heat capacity (c) is the energy required to raise the temperature of 1 kg of a substance by 1 K. The unit is J/(kg·K).
• Heat transferred: Q = m × c × ΔT, where m is mass (kg) and ΔT is the temperature change (K).

Example (Q16): A heat exchanger cooling oil at a flow rate of 0.5 kg/s from 120°C to 80°C with a specific heat of 2.0 kJ/(kg·K) removes:

Q = 0.5 × 2.0 × (120 - 80) = 40 kJ/s = 40 kW.

• Thermodynamic Processes:
• Adiabatic Process: No heat transfer occurs. For an ideal gas, T₂ = T₁ × (P₂/P₁)^((γ-1)/γ), where γ is the specific heat ratio (1.4 for air).

Example (Q49): Air compressed adiabatically from 1 bar and 300 K to 5 bar:

T₂ = 300 × (5)^(0.4/1.4) = 300 × 1.584 = 475 K.

• Thermodynamic Cycles and Efficiency:
• Carnot Efficiency: The maximum theoretical efficiency of a heat engine operating between two temperatures: η = 1 - (T_cold / T_hot).
• Second Law of Thermodynamics: Heat cannot be completely converted into work in a cyclic process; some heat must be rejected to a lower temperature reservoir.

Example (Q14): A turbine with inlet temperature 1200 K and exhaust temperature 800 K has a Carnot efficiency of:

η = 1 - (800/1200) = 0.333 = 33.3%.

Example (Q51): An engine producing 1000 kW with 30% efficiency requires a heat input of:

Input Power = Output / Efficiency = 1000 / 0.30 = 3333.3 kW.

2.5 Fluid Mechanics and Hydraulics

Fluid mechanics deals with the behaviour of liquids and gases. It is essential for understanding hydraulic systems, pneumatic systems, and aerodynamics.

• Pressure: The force exerted per unit area. P = F / A. In a static fluid, pressure is transmitted equally in all directions (Pascal's Principle).
• Density: Mass per unit volume. ρ = m / V. Specific gravity is the ratio of a substance's density to the density of water (1000 kg/m³).
• Bulk Modulus: A measure of a fluid's resistance to compression. A low bulk modulus means the fluid is more compressible, leading to a "spongy" feel in hydraulic systems.
• Continuity Equation: For an incompressible fluid, the mass flow rate is constant: A₁V₁ = A₂V₂. A reduction in cross-sectional area results in an increase in velocity.
• Bernoulli's Principle: For an incompressible, inviscid fluid, the total pressure (static + dynamic) remains constant along a streamline. Total Pressure = Static Pressure + Dynamic Pressure.
• Hydraulic Power: The power transmitted by a hydraulic system is the product of pressure and flow rate: Power (kW) = (Pressure in bar × Flow in L/min) / 600.

Example (Q4, Q33): A hydraulic actuator with a 50 mm diameter piston and a system pressure of 150 bar (15 MPa) exerts a force of:

Area = π × (0.025 m)² = 0.00196 m²

Force = 15,000,000 Pa × 0.00196 m² = 29,437 N.

Example (Q6): A hydraulic pump delivering 20 L/min against 200 bar has a power output of:

Power = (200 × 20) / 600 = 6.67 kW.

Example (Q13, Q37): A 200-litre fuel tank filled with fuel of density 0.8 kg/L contains:

Mass = 0.8 × 200 = 160 kg.

Example (Q46): If the throat area of a venturi is half the inlet area, the velocity at the throat is double the inlet velocity (A₁V₁ = A₂V₂).

Example (Q27): An increase in airspeed increases dynamic pressure, which increases total pressure as measured by a pitot tube.

2.6 Rotational Dynamics and Gears

Helicopters rely heavily on rotating components and gearboxes to transmit power from the engine to the rotors.

• Gear Ratio: The ratio of the number of teeth on the driven gear to the number of teeth on the driving gear. Gear Ratio = N_driven / N_driver.
• Speed and Torque Relationship: Output speed is inversely proportional to gear ratio. Output torque is directly proportional to gear ratio. Power is conserved (ignoring friction): P = Torque × ω.

Example (Q15, Q25): A gear with 20 teeth driving a gear with 60 teeth has a gear ratio of 3:1. An input speed of 3000 RPM results in an output speed of:

Output Speed = 3000 / 3 = 1000 RPM. The output torque is 3 times the input torque.

Example (Q45): A bevel gear with 30 teeth driven by a pinion with 10 teeth at 6000 RPM results in a gear speed of 2000 RPM and a torque ratio of 3:1.

2.7 Oscillations, Waves, and Vibration

Vibration analysis is critical for rotor track and balance and for monitoring engine and gearbox health.

• Acceleration due to Vibration: Measured in multiples of g (9.81 m/s²) or in m/s².
• Vector Addition: The imbalance in a rotor system is the vector sum of the centrifugal forces of each blade. Adding a balance weight creates a known force vector to cancel the resultant imbalance.

Example (Q48): An accelerometer reading of 5g corresponds to:

5 × 9.81 = 49.05 m/s².

Example (Q38): Rotor track and balance uses the principle of vector addition of centrifugal forces. Each blade's mass creates a force proportional to its mass and radius (F = m × r × ω²). The imbalance is the resultant vector sum.

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3. Important Formulas and Procedures

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4. Common Relationships Between Concepts

• Power, Torque, and Speed: Power is the product of torque and angular velocity. For a given power, an increase in speed results in a decrease in torque, and vice versa. This is fundamental to gearbox design.
• Pressure, Force, and Area: Pressure is the ratio of force to area. A small piston area can generate a large force with a relatively modest pressure, which is the basis of hydraulic actuation.
• Gas Laws and Engine Performance: The ideal gas law (PV = nRT) explains why air density decreases with altitude and temperature. Lower air density reduces engine mass flow, which can lead to higher turbine gas temperatures (TGT) at a constant power setting.
• Thermodynamics and Efficiency: The Second Law of Thermodynamics dictates that no heat engine can be 100% efficient. The Carnot efficiency provides the theoretical maximum, which is always less than 1.
• Kinematics and Kinetics: The linear velocity of a point on a rotating body is directly proportional to its distance from the axis of rotation. The force required to change this motion is described by Newton's Second Law.
• Fluid Dynamics and Airspeed Measurement: Bernoulli's principle relates static and dynamic pressure. The pitot-static system uses this relationship to measure airspeed.

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5. Typical Exam Focus Points

Based on the source questions, the following areas are frequently tested in the EASA Part-66 Module 2 exam for the B1.3 category:

• Unit Conversions: Converting between psi and Pa, bar and Pa, °C and K, and RPM to rad/s. These are essential for interpreting maintenance manuals and performing calculations.
• Newton's Laws of Motion: Understanding equilibrium (net force = 0) and the relationship between force, mass, and acceleration (F = ma). This is applied to hovering, climbing, and descending flight.
• Rotational Motion: Calculating angular velocity, linear tip speed, and angular momentum. This is critical for understanding rotor dynamics.
• Gas Laws: Applying Boyle's, Charles's, and Gay-Lussac's laws, as well as the ideal gas law, to scenarios involving shock struts, pneumatic systems, and engine performance.
• Hydraulic Systems: Calculating force from pressure and area, hydraulic power, and understanding Pascal's principle and the effects of bulk modulus.
• Thermodynamics: Calculating heat transfer, Carnot efficiency, and understanding the Second Law of Thermodynamics and adiabatic processes.
• Gear Ratios: Calculating output speed and torque from gear tooth counts. This is fundamental to understanding main and tail rotor gearboxes.
• Fluid Dynamics: Applying the continuity equation and Bernoulli's principle to venturi flow and airspeed measurement.
• Power and Efficiency: Calculating power from force and velocity, torque and angular velocity, and understanding the relationship between input and output power with efficiency losses.
• Density and Specific Gravity: Calculating mass from volume and density, and converting between specific gravity and density.

To succeed in the exam, candidates should be able to:

1. Identify the relevant physical principle from the question scenario.
2. Select the correct formula.
3. Convert all units to SI before performing calculations.
4. Perform the calculation accurately.
5. Interpret the result in the context of the aircraft system.