Alternate Automotive Fuels & Emissions (ME-503 (C)) - Unit 5 Short Notes

UNIT 5: ENGINE DYNAMICS & FRICTION DEVICES

I. GOVERNORS

Function & Classification

• Purpose: Automatically controls the speed of an engine by regulating the fuel/steam supply in response to load changes.

• Classification:

  • By Principle: Centrifugal (Watt, Porter, Proell) vs. Inertia.

  • By Structure: Simple (one rotating mass set) vs. Compound (two rotating mass sets).

Watt Governor

• Construction: Two hinged arms with balls at ends, connected to a sleeve on the rotating spindle.

• Working: Centrifugal force on balls makes arms rise, lifting sleeve to close fuel/steam valve.

• Height Derivation:

  Let mass of each ball = m, radius of rotation = r, angular velocity = ω, length of arm = l, height = h.

  Centrifugal force: F_c = mω²r

  Weight component: W cosθ = mg cosθ

  For equilibrium: mω²r = mg tanθ

  But tanθ = r/h and r = l sinθ ≈ l tanθ (for small θ).

  Hence, mω² (l tanθ) = mg tanθ → ω²l = g

  Therefore, height: h = √(l² - r²) ≈ l - r²/(2l).

  For small angles, from ω²l = g, we get h ∝ 1/ω² or h ∝ 1/N².

  [!TIP] Exam often asks to prove h ∝ 1/N². Use N = 60ω/(2π) and show h = (g × 60²)/(4π²N²).

Porter Governor

• Construction: Similar to Watt, but with an additional central load (W) on the sleeve. Upper arms connect balls to sleeve, lower arms connect balls to a fixed pivot on spindle.

• Force Analysis:

  Tensions in arms T1 (upper) and T2 (lower).

  For ball: T1 cosθ1 + T2 cosθ2 = mg

  T1 sinθ1 - T2 sinθ2 = mω²r

  For sleeve: W + 2T1 cosθ1 = 2T2 cosθ2 (if frictionless).

• Speed Range: Solve equilibrium equations for ω at limiting inclinations θ_max and θ_min.

• Effect of Friction: Friction in sleeve/pivots adds a constant resisting force F_f. This reduces sensitiveness and lowers the minimum speed but raises the maximum speed for a given θ range.

Proell Governor

• Construction: Lower arms are hinged at a point on the spindle (not at the axis). Extensions from these hinges carry the balls. When at minimum speed, these extensions are parallel to the spindle axis.

• Minimum Speed (Parallel Extensions):

  Let r = min radius, h = height of governor, l = length of each arm, e = distance from spindle axis to lower arm hinge.

  Geometry: r = e + l sinθ, h = l cosθ.

  Force equation on one ball: mω²r = (T - mg) tanθ, where T is tension in lower arm.

  Sleeve equilibrium: W + 2T cosθ = 2mg (assuming massless arms).

  Solve to get ω_min² = [g (r - e)] / [r (h - e tanθ)]. For parallel position (θ=0), ω_min² = g(r-e)/(rh).

• Comparison with Porter: Proell has higher sensitiveness (larger change in ω for same θ change) because the centrifugal force has a longer lever arm about the lower hinge.

Governor Performance Characteristics

• Sensitiveness (S): Ability to respond to small speed changes.

  S = (ΔN/N) / (ΔF/F) or S = (N_max - N_min)/N_mean.

  High sensitiveness → large speed variation for small load change (poor regulation).

• Isochronism: Zero sensitiveness (S=0). Governor maintains constant speed regardless of load change (ideal but unstable without feedback). Achieved by adding a spring (e.g., Hartnell governor).

• Hunting: Oscillations of governor sleeve about its equilibrium position due to over-sensitivity. Causes speed fluctuations. Mitigation: Use friction (damping), increase sleeve mass, or use isochronous design.

• Stability: Governor returns to new equilibrium after a disturbance.

  • Stable: dF/dr > F/r (controlling force increases faster than F/r).

  • Unstable: dF/dr < F/r.

  • Isochronous: dF/dr = F/r.

  [!TIP] Stability Condition Derivation:

  Controlling force F_c = mω²r. For stability, as r increases (speed up), F_c must increase more than linearly. So d(mω²r)/dr > mω² → mω² + mr d(ω²)/dr > mω² → d(ω²)/dr > 0. But from equilibrium, F_control = f(r). The graphical condition is dF/dr > F/r.

Friction & Insensitiveness

• Effect of Friction: Reduces sensitiveness. Causes dead zone—a range of speeds where governor does not move because friction balances the centrifugal force change.

• Coefficient of Insensitiveness (k): k = (ΔN_friction) / (N_mean), where ΔN_friction is the speed range due to friction alone. k increases with friction.

II. FLYWHEELS & TURNING MOMENT DIAGRAMS

Function & Distinction from Governor

• Flywheel: Stores kinetic energy during power strokes and releases it during idle strokes to smoothen speed fluctuations.

• Governor: Controls/sets the mean speed by regulating energy input. Does not reduce speed fluctuation amplitude.

Fluctuation Concepts

• Fluctuation of Energy (ΔE): Maximum difference between kinetic energy at max and min speeds.

  ΔE = (1/2) I (ω_max² - ω_min²).

• Coefficient of Fluctuation of Energy (C_E): C_E = ΔE / (Mean kinetic energy) = (ω_max² - ω_min²) / (2 ω_mean²).

• Fluctuation of Speed (ΔN): ΔN = N_max - N_min.

• Coefficient of Fluctuation of Speed (C_N): C_N = ΔN / N_mean.

  [!TIP] For small fluctuations, C_N ≈ √(2C_E).

Turning Moment Diagram for IC Engines

• Four-Stroke Cycle: One power stroke per two revolutions. Diagram shows torque vs. crank angle over 720°.

  • Power Stroke (combustion): High positive torque.

  • Other Strokes (intake, compression, exhaust): Negative or low torque (resistance).

• Mean Torque (T_m): Horizontal line such that area above = area below over cycle. T_m = (Work per cycle) / (4π) for 4-stroke.

• Interpretation: Area between actual torque curve and T_m line represents energy surplus/deficit that the flywheel must absorb/supply.

Flywheel Design Calculations

• Energy Equation: ΔE = I ω_mean Δω (for small Δω).

  Where I = moment of inertia, ω_mean = mean angular velocity, Δω = angular speed fluctuation.

• From Area-Speed Limits:

  1. Draw turning moment diagram to scales: 1 mm = S_T N-m (torque), 1 mm = S_θ rad (angle).

  2. Calculate net area A (in mm²) above/below mean torque line for one cycle.

  3. Actual energy fluctuation: ΔE = A × S_T × S_θ (in N-m).

  4. Given ΔN or C_N, find Δω = (2π ΔN)/60.

  5. Mass/Role of Gyration: I = m k² = ΔE / (ω_mean Δω).

  \boxed{m k^2 = \frac{\Delta E}{\omega_{\text{mean}} \Delta \omega}}

  [!TIP] If C_N is given, Δω = ω_mean C_N, so m k² = ΔE / (ω_mean² C_N).

Mean Torque & Speed Relations

• Power (P) = T_mean × ω_mean (in consistent units).

• Work per cycle = T_mean × (angle per cycle).

III. BALANCING OF RECIPROCATING & ROTATING MASSES

Fundamentals

• Primary vs. Secondary:

  • Primary: Forces due to simple harmonic motion of reciprocating mass (inertia force mω²r cosθ). Frequency = crank speed.

  • Secondary: Due to obliquity of connecting rod. Force component mω²r (r/L) cos2θ. Frequency = 2×crank speed. (Neglected if r/L is small).

• Dynamically Equivalent System: Replace a distributed mass by two or three point masses that produce same inertia force and moment about a reference point.

Single Cylinder Engines

• Balancing:

  • Revolving mass (m_r): Can be fully balanced by a counterweight.

  • Reciprocating mass (m_rec): Only partially balanced (typically c × m_rec, where c ≤ 1). Full balance creates vertical dynamic forces in single cylinder.

• Forces (at crank angle θ from IDC):

  • Inertia Force: F_I = m_rec ω² r cosθ (opposite to acceleration).

  • Piston Effort (F_P): Net force on piston due to gas pressure minus inertia force.

  • Side Thrust (F_S): F_P tanφ ≈ F_P (r sinθ)/L (on cylinder walls).

  • Crank Effort (F_T): Tangential component on crank: F_T = (F_P cosθ + F_I sinθ) / cosφ ≈ F_P cosθ + F_I sinθ.

  • Thrust in Connecting Rod (F_R): F_R = F_P / cosφ.

Multi-Cylinder Engines

• In-line Engines:

  • Primary Balance: Possible if ∑ m_r r cosθ = 0 and ∑ m_r r sinθ = 0 (crank angles suitably arranged).

  • Secondary Balance: Requires ∑ m_rec (r²/L) cos2θ = 0 and ∑ m_rec (r²/L) sin2θ = 0.

  • Complete Balance? No. Primary can be balanced, but secondary usually unbalanced. Even if both balanced, residual couples may exist.

• Radial Engines:

  • Cylinders equally spaced (360°/n).

  • Primary and secondary forces always balance (if all reciprocating masses equal).

  • Primary and secondary couples may exist if r/L is considered.

Locomotive Engines (Uncoupled Two-Cylinder)

• Crank Arrangements: Usually 90° apart.

• Derivations (for two cylinders, crank angle θ for 1st cylinder, θ+φ for 2nd):

  • Swaying Couple (M_s): Moment about vertical axis due to unbalanced horizontal forces.

    M_s = (1 - c) m_rec ω² r (L/2) [cosθ + cos(θ+φ)] (approx, primary only).

    Maximum: M_s,max = (1-c) m_rec ω² r L |cos(φ/2)|.

  • Hammer Blow (F_V): Unbalanced vertical force at wheels.

    F_V = (1-c) m_rec ω² r [sinθ + sin(θ+φ)] (primary only).

    Magnitude: F_V,max = (1-c) m_rec ω² r √[2(1+cosφ)].

    Varies with speed as ω².

  • Variation in Tractive Effort (ΔF_T): Variation in horizontal force between wheel and rail.

    ΔF_T = (1-c) m_rec ω² r [cosθ - cos(θ+φ)].

    Maximum: ΔF_T,max = (1-c) m_rec ω² r √[2(1-cosφ)].

  [!TIP] For φ=90°: F_V,max = √2 (1-c) m_rec ω² r, ΔF_T,max = √2 (1-c) m_rec ω² r.

• Balancing Fraction (c): Given max allowable hammer blow F_V,allow at speed ω, solve F_V,max ≤ F_V,allow for c.

Balancing Masses & Residual Forces

• Magnitude & Position: Balance ∑ m_r r vectorially. Place balancing mass m_b at radius r_b such that m_b r_b = ∑ m_r r (magnitude and direction).

• Resultant Unbalance Force: At any crank angle θ, sum all unbalanced inertia forces vectorially.

IV. FRICTION CLUTCHES

Single Plate Clutch

• Uniform Pressure Assumption: Pressure p constant over entire friction surface.

  • Axial force: W = π p (r₂² - r₁²)

  • Torque: T = (2/3) μ π p (r₂³ - r₁³)

  • Mean radius: R_m = (2/3) (r₂³ - r₁³)/(r₂² - r₁²)

• Uniform Wear Assumption: p r = constant. Pressure highest at inner radius.

  • Axial force: W = 2π p_mean (r₂ - r₁) r_mean (where p_mean is pressure at mean radius).

  • Torque: T = μ W R_m (with R_m = (r₂ + r₁)/2).

  • Maximum Pressure: p_max = (W r₂) / [2π r_mean (r₂ - r₁)] at inner radius r₁.

• Design: Given P, N, μ, p_max, and often R_m / b = 4 (where b = r₂ - r₁ is face width). Solve for R_m, r₁, r₂, b.

Conical Clutch

• Working: Friction surface is conical. Axial force W creates normal pressure. Frictional torque T = μ W (r₁ + r₂)/(2 sin α), where α is cone angle.

• Comparison: Higher torque capacity for same axial force than plate clutch (due to 1/sin α factor). Self-energizing possible.

V. BRAKES

Band Brake

• Simple Band with Lever:

  Tensions: T₁ (tight side), T₂ (slack side). T₁/T₂ = e^{μθ} (θ in radians).

  Braking torque: T_b = (T₁ - T₂) r_d.

  Lever mechanics: T₁ = P × (l / d), where P = effort, l = lever arm, d = drum radius to lever attachment point.

  [!TIP] If band attached to fulcrum and pin, use moment balance about fulcrum.

Shoe Brakes (Internal Expanding)

• Double Shoe: Two shoes inside drum, pivoted. Spring provides initial force, hydraulic/mechanical linkage provides operating force.

• Self-energizing: Leading shoe gets additional force from friction drag.

• Design:

  Braking torque T_b = 2 μ p b R (θ - sinθ cosθ) (for each shoe, assuming uniform pressure p over angle θ).

  Spring force F_s calculated from torque and pressure limit p_max.

  Shoe width b from p_max and total normal force.

VI. BEARINGS & FRICTION CIRCLE

Conical Pivot Bearings

• Uniform Pressure: p = constant.

  Frictional torque: T_f = (2/3) μ W (R₂³ - R₁³)/(R₂² - R₁²).

  Power loss: P = (2π N T_f)/60.

• Uniform Wear: p r = constant.

  Frictional torque: T_f = (1/2) μ W (R₂ + R₁).

• Design: Given W, α (cone angle), p_max, R₂/R₁ ratio, find R₁, R₂.

Collar Bearings

• Uniform Pressure:

  Frictional torque: T_f = (μ W π (D + d))/4, where D, d = outer/inner diameters.

  Power: P = (μ W π (D + d) N)/(60 × 1000) (W in N, P in kW).

• Uniform Wear:

  T_f = (μ W π (D² - d²))/(4 D).

• Number of Collars: If total load W shared by n collars, W/n used in above formulas.

Friction Circle

• Definition: In journal bearings, the resultant friction force acts along the tangent to a friction circle of radius r_f.

• Radius Derivation: Friction force F_f = μ N. If φ = angle of friction (tan φ = μ), then r_f = r sin φ, where r = journal radius.

  \boxed{r_f = r \sin \phi}

  [!TIP] Used in viscous damping analysis and Coulomb friction in rotating systems.

VII. DYNAMOMETERS

Classification

• Absorption: Absorbs and dissipates engine power as heat (e.g., Prony brake, Rope brake).

• Transmission: Transmits power to a load while measuring (e.g., Epicyclic, Belt).

• Torsion: Measures torque via shaft twist (e.g., Optical, Electrical with strain gauges).

Absorption vs. Transmission

• Absorption: Engine drives dynamometer which acts as a brake. All power converted to heat. Simple, but wasteful.

• Transmission: Power goes to a load (e.g., generator). Dynamometer measures torque/speed. More efficient.

Torsion Dynamometer

• Construction: Shaft with strain gauges or an optical disc with radial lines.

• Power Calculation:

  • Strain Gauge: Measure strain ε → stress σ = Eε → torque T = (σ J)/r (J = polar moment, r = radius). Power P = (2π N T)/60.

  • Optical: Measure angle of twist θ between two discs. T = (G J θ)/L (G = shear modulus, L = length between discs).

VIII. KINEMATICS OF MECHANISMS

Four-Bar Chain Analysis (Instant Center Method)

• Steps:

  1. Draw configuration to scale.

  2. Locate all instant centers (IC). For 4-bar, 3 ICs besides fixed pivots.

  3. Use ω₂/ω₄ = (IC₁₄₋₂₄)/(IC₁₂₋₂₄) for angular velocities.

  4. Velocity of any point: v = ω × distance to IC.

  5. Acceleration: Use relative acceleration equations or differentiate velocity.

• Given: Link lengths, input ω₂, α₂, configuration → find ω₃, ω₄, α₃, α₄.

Cam & Follower Dynamics (Offset Circular Disc Cam)

• Cam: Circular disc radius R, center offset by e from camshaft axis.

• Follower Lift (s): s = e(1 - cosθ) (if zero lift at θ=0).

• Follower Velocity & Acceleration:

  v = e ω sinθ

  a = e ω² cosθ (radial acceleration of cam center towards follower).

• Lift Condition (Critical Speed): Follower loses contact when dynamic force exceeds spring force.

  Spring force F_s = k s + F₀ (initial compression).

  Dynamic force F_d = m a (inertia force).

  Condition for lift: m e ω² cosθ > k e(1 - cosθ) + F₀.

  Solve for max ω before lift occurs at any θ.

IX. SPECIAL TOPICS & INTEGRATED PROBLEMS

Turning Moment Diagram Applications

1. Calculate ΔE = A × S_T × S_θ (from areas and scales).

2. Given N_mean, C_N or ΔN, find Δω.

3. Compute I = ΔE/(ω_mean Δω).

4. If mass m given, find k = √(I/m).

Governor Design Problems

• Speed Range with Friction: Calculate ω_min and ω_max for Porter/Proell with friction force F_f included in sleeve equilibrium.

• Equilibrium Speed: For given r and θ, solve force equations for ω.

Combined Balancing & Locomotive Problems

• Fraction to Balance (c): Given max F_V (hammer blow) at speed ω, use F_V,max = (1-c) m_rec ω² r √[2(1+cosφ)] to find c.

• Maximum Swaying Couple: M_s,max = (1-c) m_rec ω² r L |cos(φ/2)|.

• Tractive Effort Variation: ΔF_T,max = (1-c) m_rec ω² r √[2(1-cosφ)].

Friction Device Synthesis

• Clutch: Given P, N, μ, p_max, R_m/b → find R_m, r₁, r₂, b using torque and pressure formulas (check both uniform pressure and wear assumptions).

• Brake: Given T_b, μ, θ, p_max, D → find spring force F_s and shoe width b from torque and pressure equations.

• Bearing: Given W, α, p_max, D/d ratio → find d, D from axial load formula, then power loss at given N.