Mechatronics (ME-503 (A)) - Unit 5 Short Notes

Flywheels and Energy Fluctuation

Turning Moment Diagram

• Construction & Interpretation: A graphical representation of the net turning moment (torque) on the crankshaft versus crank angle for one complete cycle (typically two revolutions for a 4-stroke engine).

• Scales:

  • Vertical (Torque): 1 mm = X N-m (given in problem).

  • Horizontal (Crank Angle): 1 mm = Y degrees (given in problem).

• Mean Torque Line: Horizontal line representing the average torque over the cycle. Its height = (Area under actual torque curve) / (Total crank angle, i.e., 720° for 4-stroke).

• Areas Above/Below: The areas between the actual torque curve and the mean torque line represent the excess energy (above line) and energy deficit (below line) during different parts of the cycle.

  [!TIP] The maximum area (either above or below) corresponds to the maximum fluctuation of energy (ΔE) for the cycle. This is the key value for flywheel design.

Fluctuation of Energy and Speed

• Fluctuation of Energy (ΔE): The maximum difference between the energy stored and the mean energy during a cycle. It is the maximum area (in N-m or J) between the turning moment diagram and the mean torque line.

  Significance: Governs the size (mass) of the flywheel. Larger ΔE requires a larger flywheel to limit speed fluctuations.

• Fluctuation of Speed (ΔN): The difference between maximum and minimum speeds (N_max - N_min) of the flywheel/engine shaft.

• Relationship: ΔE is stored/released from the flywheel's kinetic energy change.

$$\Delta E = \frac{1}{2} I \left( \omega_{max}^2 - \omega_{min}^2 \right)$$

Where `I` = mass moment of inertia, `ω` = angular velocity.

For small fluctuations, `ΔN << N_mean`:

$$\Delta E \approx I \cdot \omega_{mean}^2 \cdot \frac{\Delta N}{N_{mean}}$$

> **Key Insight:** `ΔE` causes `ΔN`. The flywheel's inertia `I` resists this change.

Coefficients of Fluctuation

• Coefficient of Fluctuation of Energy (K_E):

$$K_E = \frac{\Delta E}{E_{mean}} = \frac{\Delta E}{\frac{1}{2} I \omega_{mean}^2}$$

*   `E_mean` = mean kinetic energy.

*   **Indicates relative energy variation.**

• Coefficient of Fluctuation of Speed (K_N or δ):

$$K_N = \frac{\Delta N}{N_{mean}}$$

*   **Indicates relative speed variation.**

*   For small `K_N`, `K_E ≈ 2 K_N`.

Flywheel Design and Analysis

• Mass & Radius of Gyration (k):

  • Mass moment of inertia: I = M * k²

  • k = radius of gyration (distance from axis where mass M can be concentrated to give same I).

• Energy Stored:

$$E = \frac{1}{2} I \omega^2 = \frac{1}{2} M k^2 \omega^2$$

• Determination of Minimum Mass from Turning Moment Diagram:

  1. Find ΔE (max area) from diagram using given scales.

  2. Given N_mean, N_max, N_min (or ΔN), calculate ω_mean, ω_max, ω_min.

  3. Use energy equation:

$$\Delta E = \frac{1}{2} M k^2 \left( \omega_{max}^2 - \omega_{min}^2 \right)$$

4.  Solve for `M` (if `k` given) or `k` (if `M` given).

> [!TIP] Ensure consistent units: Convert `N-m` to `J` (1 N-m = 1 J), `rpm` to `rad/s`: $$\displaystyle \omega = \frac{2\pi N}{60} $$.

• Max/Min Speeds Given ΔE & N_mean:

  From ΔE = I ω_mean² (K_E) and K_N ≈ K_E/2:

$$N_{max} = N_{mean} \left(1 + \frac{K_N}{2}\right), \quad N_{min} = N_{mean} \left(1 - \frac{K_N}{2}\right)$$

Or directly solve:

$$\omega_{max,min} = \omega_{mean} \pm \frac{\Delta E}{2 I \omega_{mean}}$$

Governors

Function and Classification

• Function: To regulate the mean speed of an engine/machine under varying load by automatically adjusting the fuel/steam supply.

• Classification:

  | Type | Principle | Examples | | :--- | :--- | :--- | | Centrifugal | Centrifugal force on rotating masses | Watt, Porter, Proell, Hartnell | | Inertia | Inertia of masses | Pendulum, shaft governors |

• Key Terms:

  • Sensitiveness: Ability to respond to small speed changes. Sensitiveness ∝ 1/(N_max - N_min).

  • Isochronism: N_max = N_min (zero speed fluctuation). Requires F_c/r = constant.

  • Hunting: Oscillations (speed rise/fall) about mean speed due to over-sensitivity.

  • Stability: Governor returns to new equilibrium without excessive hunting. Requires dF_c/dr > 0 (controlling force curve steeper than centrifugal force curve).

Watt Governor

• Construction: Two arms (l) with balls (m) hinged to crankpin (not directly on shaft). Arms connected to sleeve via links.

• Height Expression:

$$h = \frac{g}{\omega^2} \quad \text{or} \quad h \propto \frac{1}{N^2}$$

**Proof:** For equilibrium, `Centrifugal force = Tension component`:

$$m r \omega^2 = T \sin\theta \approx T \frac{r}{h} \Rightarrow T = \frac{m r \omega^2 h}{r} = m \omega^2 h$$

Also, `Weight = T cosθ ≈ T`:

$$mg = T = m \omega^2 h \Rightarrow h = \frac{g}{\omega^2}$$

• Controlling Force Diagram: Plot F_c = m r ω² vs r. For Watt, F_c/r = mω² (constant). Stable if slope of F_c curve > slope of mω² line.

Porter Governor

• Construction: Upper arms (l₁) hinged to shaft, lower arms (l₂) connected to sleeve. Central load W on sleeve.

• Speed Range (with friction F_f):

  Let θ = inclination of upper arm.

  Minimum Speed (sleeve about to descend): Friction opposes downward motion.

$$\frac{(W + 2mg) h}{r} = 2 m (g + a) \omega_{min}^2 \quad \text{where } a = \frac{F_f}{2m}$$

**Maximum Speed (sleeve about to rise):** Friction opposes upward motion.

$$\frac{(W + 2mg) h}{r} = 2 m (g - a) \omega_{max}^2$$

Where `h = l₁ cosθ`.

• Comparison with Watt: More sensitive (greater change in r for same Δω) due to added central load W. Height h is constant for given θ, not ∝ 1/ω².

Proell Governor

• Construction: Upper arms (l) hinged to shaft, lower arms (l) pivoted on extensions from shaft axis. Balls at end of lower arms. No central load on sleeve.

• Speed Range:

  Minimum Speed:

$$\omega_{min}^2 = \frac{(W + 2mg) (l + r) \cos\theta}{2 m r (l \cos\theta + r \sin\theta)}$$

**Maximum Speed:** Replace `(W + 2mg)` with `(W - 2mg)` if `W > 2mg`.

Where `r` = radius of rotation, `θ` = inclination of lower arm.

• Comparison with Porter: More sensitive because centrifugal force acts on lower arm pivot, increasing effective moment arm. ω_min is lower for same r.

Governor Characteristics & Stability

• Stability Conditions:

  1. Stable: dF_c/dr > dF_c/r (controlling force curve steeper). N_max > N_min.

  2. Unstable: dF_c/dr < dF_c/r. N_max < N_min (not practical).

  3. Isochronous: dF_c/dr = dF_c/r and F_c/r = constant. N_max = N_min.

• Effect of Friction:

  • Increases ω_min, decreases ω_max → reduces sensitiveness.

  • Coefficient of Insensitiveness (K_i):

$$K_i = \frac{\text{Increase in } \omega_{min}}{\omega_{mean}} = \frac{\text{Decrease in } \omega_{max}}{\omega_{mean}}$$

Balancing of Engines

Fundamentals

• Static Balance: Center of mass lies on axis of rotation. For rotating masses, Σ m r = 0 and Σ m r θ = 0 (in polar coordinates).

• Dynamic Balance: Requires both static balance and balancing of couples (Σ m r² = 0).

• Primary vs. Secondary Balancing (Reciprocating):

  • Primary: Balance due to m ω² r cosθ (unbalanced force).

  • Secondary: Balance due to m ω² r (r/l) cos2θ (due to obliquity, r/l ratio).

• Hammer Blow: Vertical dynamic force transmitted to rails due to unbalanced primary force of reciprocating masses. P = (1 - c) m_r ω² r, where c = fraction balanced.

• Swaying Couple: Horizontal couple about the center of gravity due to unbalanced primary forces in multi-cylinder engines. C = (1 - c) m_r ω² r * d, where d = distance between cylinder centerlines.

Balancing of Single-Cylinder Engines

• Partial Balancing: Only a fraction (c) of reciprocating mass is balanced by a revolving mass at crank radius r.

  • Balancing mass m_b = c * m_r * (r / r_b) placed opposite to crank.

  • Reason: Complete balancing would produce large vertical forces (hammer blow) at high speeds.

• Residual Unbalanced Force:

$$F_{res} = (1 - c) m_r \omega^2 r \cos\theta + m_r \omega^2 \frac{r^2}{l} \cos 2\theta$$

(Primary + Secondary terms)

Balancing of Multi-Cylinder Engines

• In-line Engines:

  • Complete Primary Balance: Requires Σ m_r r cosθ_i = 0 and Σ m_r r sinθ_i = 0 (crank angles θ_i).

  • Complete Secondary Balance: Requires Σ m_r r² cos2θ_i = 0 and Σ m_r r² sin2θ_i = 0.

  • Practical: Often only primary partial balance is done. Results in swaying couple and variation in tractive effort.

• V-type Engines: Cylinders in two banks at angle α. Primary forces can be balanced if θ_B = θ_A ± 180° and m_A r_A = m_B r_B. Secondary forces generally unbalanced.

• Radial Engines (e.g., 3-cylinder at 120°):

  • Primary: Σ m_r r cosθ_i = 0 and Σ m_r r sinθ_i = 0 if θ_i = 0°, 120°, 240° and m_r r equal.

  • Secondary: Σ m_r r² cos2θ_i = 0 and Σ m_r r² sin2θ_i = 0 if θ_i spaced 60° apart (not satisfied for 120° spacing). So secondary unbalanced.

Unbalanced Forces & Couples (Two-Cylinder Locomotive)

• Assumptions: Crank radius r, reciprocating mass per cylinder m_r, crank angle θ, distance between cylinder centers d, wheel radius R_w.

• Swaying Couple (Derivation):

  Unbalanced primary forces: F_{1} = (1-c) m_r ω² r cosθ, F_{2} = (1-c) m_r ω² r cos(θ ± φ) where φ = crank angle difference.

  Couple about CG: C = F_1 * (d/2) - F_2 * (d/2).

  For φ = 90° (typical):

$$C = \frac{(1-c) m_r \omega^2 r d}{2} \left( \cos\theta \mp \sin\theta \right)$$

**Maximum magnitude:** `C_max = \frac{(1-c) m_r \omega^2 r d}{\sqrt{2}}`.

• Variation in Tractive Effort: Net horizontal force on wheels = (1-c) m_r ω² r (cosθ ± cos(θ±φ)). Causes cyclic variation in drawbar pull.

• Hammer Blow (at speed v):

$$P = (1-c) m_r \omega^2 r = (1-c) m_r \left( \frac{v}{R_w} \right)^2 r$$

Given max allowable `P_max`, solve for fraction `c` to be balanced:

$$c = 1 - \frac{P_max R_w^2}{m_r v^2 r}$$

Friction Clutches and Brakes

Friction Clutches

• Single Plate Clutch:

  • Pressure Distribution:

    • Uniform Pressure (p constant): p = F / (π (r_o² - r_i²))

    • Uniform Wear (p r = constant): p_i r_i = p_o r_o, F = 2π p_i r_i (r_o - r_i)

  • Torque Transmission:

    • Uniform Pressure: T = μ F \frac{r_o^3 - r_i^3}{3(r_o^2 - r_i^2)}

    • Uniform Wear: T = μ F \frac{r_o^2 + r_i^2}{2 r_o + 2 r_i} (Simpler, often used)

  • Design: Given P (power), N (rpm), μ, p_max, find T = P / ω, then F, then r_o, r_i (often r_o/r_i = 1.2 to 1.5), face width b ≈ (r_o - r_i)/2.

• Multi-plate Clutch: T_total = n * T_single_plate (n = number of friction surfaces).

• Conical Clutch:

  • Torque: T = μ F \frac{R ( \sin\alpha + \mu \cos\alpha )}{\cos\alpha - \mu \sin\alpha} where α = cone angle, R = mean radius.

  • Advantages: Higher torque for same F (self-energizing effect), compact.

Friction Brakes

• Band Brake:

  • Simple Band: T_b = (T_1 - T_2) R = T_1 R \left(1 - e^{-μθ}\right) where θ = wrap angle (radians).

  • Differential Band: Two ends attached to lever at different radii. T_b = (T_1 - T_2) R, T_1/T_2 = e^{μθ}. Lever arm ratio l_1/l_2 gives mechanical advantage.

  • Self-locking: If μθ ≥ 1 (i.e., θ ≥ 1/μ rad), brake holds without force F (dangerous).

• Block Brake: Pivoted block on drum. T_b = μ F R for simple case (ignoring moment of F).

• Internal Expanding Shoe Brake (Automotive Drum Brake):

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

  • Spring Force F_s for Torque T: For two shoes (one leading, one trailing):

$$T = 2 μ F_s R \frac{\sin\alpha + μ \cos\alpha}{\cos\alpha - μ \sin\alpha}$$

    where `α` = angle of shoe tip from vertical.

*   **Brake Shoe Width `b`:** From pressure limit `p_max`:

$$F_s = p_{avg} * (2 R b) \quad \Rightarrow \quad b = \frac{F_s}{2 R p_{avg}}$$

    (Assuming uniform pressure over shoe area `2R*b`).

Friction Circle

• Definition: Circle of radius r_f = μ R in journal bearing, representing locus of reaction force between journal and bearing for constant μ.

• Derivation: For a journal radius R, friction force F_f = μ N. The resultant reaction R must lie within a circle of radius μR (friction circle) centered on the normal force line.

  Application: In journal bearings, the frictional torque T_f = F_f * R = μ N R. The friction circle helps visualize the direction of R and F_f for different loading conditions.

Bearings and Friction Losses

Pivot Bearings (Conical)

• Assumptions: Cone angle 2α (so half-angle α), shaft radius R, load W, speed N rpm, μ.

• Uniform Pressure (p constant):

  • Pressure: p = W / (π R^2 ( \cosec α - cot α ))

  • Friction Torque: T_f = \frac{2}{3} μ W R \cosec α

  • Power Loss: P_f = T_f ω = \frac{2}{3} μ W R \cosec α * \frac{2πN}{60}

• Uniform Wear (p r = constant):

  • Pressure: p_i R = constant, W = π p_i R^2 ( \cosec α - cot α )

  • Friction Torque: T_f = \frac{1}{2} μ W R \cosec α

  • Power Loss: P_f = \frac{1}{2} μ W R \cosec α * ω

  [!TIP] Uniform wear gives lower torque/power loss than uniform pressure for same W, R, α.

Collar Bearings

• Construction: Shaft with integral collar(s), bearing surface on flat annular area.

• Power Absorbed (Uniform Pressure):

$$P_f = \frac{μ W \pi (D_o + D_i) N}{60}$$

Where `D_o`, `D_i` = outer/inner diameters.

• Number of Collars: If pressure limit p_max:

$$n = \frac{W}{p_{avg} * π (D_o^2 - D_i^2) / 4} \quad \text{(for one collar)}$$

Use `p_avg` based on uniform pressure or wear.

Journal Bearings

• Friction Circle Concept: As above, r_f = μ R.

• Power Loss: P_f = μ W R ω (for full journal bearing, assuming μ constant).

Dynamometers

Kinematics of Mechanisms (Selected)

Four-Bar Mechanism (Instantaneous Center Method)

• Velocity: v_B = ω_AB * r_AB (perpendicular). ω_BC = v_B / r_IC_{BC} where IC_{BC} is instantaneous center of link BC.

• Acceleration: Use relative acceleration method:

$$\vec{a}_C = \vec{a}_B + \vec{a}_{C/B}$$

where `a_B` known (from `α_AB`), `a_{C/B}` has tangential (`α_BC * r_BC`) and radial (`ω_BC² * r_BC`) components.

Cam Dynamics (Offset Cam, Flat-faced Follower)

• Follower Acceleration a_f in terms of cam angle θ:

  For offset e, base circle radius r_b, follower lift s (function of θ):

$$a_f = -e \omega^2 \sin\theta + \omega^2 \frac{d^2 s}{d\theta^2}$$

(Derivation from kinematics of oscillating follower).

• Critical Speed for Lift-off:

  Lift-off occurs when spring force F_s = k (s_0 - s) ≤ inertia force m_f a_f (downwards).

  Critical ω_cr found from m_f a_f(ω_cr, θ) = F_s(θ) at worst-case θ.

Advanced Concepts from Recent Papers

Dynamically Equivalent System

• Definition: A two-mass system (or point mass) that produces identical kinetic energy and identical velocity/acceleration at a specified point (usually the center of gravity) as the original distributed mass system.

• Significance: Simplifies dynamic analysis of complex bodies (like connecting rods) by replacing them with equivalent masses at crankpin and along the connecting rod.

Piston Effort and Crank Effort in Reciprocating Engines

• Given: Crank angle θ, gas pressure P (on piston), masses: m_r (reciprocating), m_R (revolving at radius r), dimensions: r (crank radius), l (connecting rod length), ω (crank speed).

• Inertia Force: F_I = m_r ω² r \left( \cos\theta + \frac{r}{l} \cos 2\theta \right) (along cylinder axis).

• Piston Effort (F_P): Net force on piston.

$$F_P = P \cdot A - F_I - F_f \quad \text{(for horizontal engine)}$$

Where `A` = piston area, `F_f` = friction (often neglected).

• Thrust on Cylinder Walls (F_C): F_C = \frac{F_P}{\tan\phi}, where φ = angle of connecting rod with vertical (sinφ ≈ (r/l) sinθ).

• Connecting Rod Thrust (F_T): F_T = F_P / \cos\phi.

• Crank Effort (F_T'): Tangential component on crankpin.

$$F_T' = F_T \sin(\theta + \phi) \approx F_T \sin\theta \quad \text{(for } l >> r\text{)}$$

This is the **driving torque** on crankshaft: `T = F_T' * r`.

Frictional Couple in Locomotive Engines

• Uncoupled Two-Cylinder Four-Stroke:

  • Assumptions: Crank radius r, cylinder bore D, piston speed V_p, μ (friction coefficient), n (rpm), A = πD²/4.

  • Frictional Power Loss per cylinder: P_f = (μ P_{mean} A V_p) / 2 (for 4-stroke, mean pressure P_{mean}).

  • Frictional Torque: T_f = P_f / ω.

  • Total Frictional Couple: C_f = 2 T_f (for two cylinders, assuming no phase cancellation).

  Note: This is a simplified average value. Actual friction torque varies with crank angle.