Dynamics of Machine (ME-503 (B)) - Unit 2 Short Notes

UNIT 2: Dynamics of Machines - Advanced Topics

1. Flywheels and Energy Management in Engines

Turning Moment Diagrams

• Definition: Graph of turning moment (torque) vs. crank angle for one complete cycle.

• Four-Stroke Cycle IC Engine:

  • Power stroke (expansion): Torque > Mean torque.

  • Compression, exhaust, intake strokes: Torque < Mean torque.

  • Only one power stroke per two revolutions (720°).

  • Diagram shows large positive area (work output) and negative areas (work input).

• Multi-Cylinder Engine: Overlapping power strokes result in a smoother, more uniform torque curve with smaller fluctuations.

• Scales & Interpretation:

  • Vertical scale: 1 mm = k N-m (k given).

  • Horizontal scale: 1 mm = β degrees (β given).

  • Area under curve between two crank angles represents work done in that interval: Work = Area × (Vertical scale × Horizontal scale).

  • Intercepted areas above/below the mean torque line represent excess/deficit of energy.

Fluctuation of Energy

• Definition: The maximum excess or maximum deficit of energy in the flywheel during a cycle, relative to the mean energy level. Denoted by ΔE.

• Significance: Determines the size (mass) of the flywheel needed to limit speed fluctuations.

• Coefficient of Fluctuation of Energy (Kₑ):

$$\boxed{K_e = \frac{\Delta E}{\text{Mean energy per cycle}}}$$

It is a dimensionless measure of energy variation.

Fluctuation of Speed

• Definition: The difference between the maximum speed (N₁) and minimum speed (N₂) of the flywheel during a cycle. Denoted by ΔN = N₁ - N₂.

• Significance: Directly indicates the smoothness of operation.

• Coefficient of Fluctuation of Speed (Kₙ):

$$\boxed{K_n = \frac{\Delta N}{N_{mean}}}$$

Where `N_mean` is the mean rotational speed (rpm or rps). `K_n` is typically small (e.g., 0.01 to 0.05).

Maximum and Minimum Speeds

• Relationship: N₁ = N_mean + ΔN/2, N₂ = N_mean - ΔN/2.

• Calculation from Energy Fluctuation:

  The maximum fluctuation of energy ΔE is stored/released by the flywheel as its kinetic energy changes between N₁ and N₂.

$$\Delta E = \frac{1}{2} I (\omega_1^2 - \omega_2^2)$$

Where `I = m k²` (mass `m`, radius of gyration `k`), `ω = 2πN/60`.

For small `K_n`, approximation:

$$\boxed{\Delta E \approx I \omega_{mean}^2 \cdot K_n}$$

> [!TIP] **Exam Focus:** You will be given `ΔE`, `m`, `k`, `N_mean` and asked for `N₁`, `N₂`. Use the exact formula `ΔE = ½ I (ω₁² - ω₂²)` and solve the quadratic in `ω`.

Mass and Radius of Gyration

• Energy-based relation: I = m k² is the key parameter.

• Given ΔE, N_mean, and K_n (or ΔN), the required moment of inertia is:

$$I = \frac{\Delta E}{\omega_{mean}^2 \cdot K_n}$$

Then `m` or `k` can be found if the other is known.

Mean Torque and Energy Storage

• Mean Torque (Tₘ): The constant torque that would do the same net work as the actual variable torque over one cycle.

$$T_m = \frac{\text{Net work per cycle}}{2\pi} \quad \text{(for radian measure)}$$

• The flywheel stores energy when T_actual > T_m (speed increases) and releases energy when T_actual < T_m (speed decreases).

2. Governors for Speed Regulation

Function and Classification

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

• Classification:

  1. Centrifugal Governors: (Watt, Porter, Proell) - Rotating masses move radially outward with speed.

  2. Inertia Governors: (e.g., swinging pendulum) - Use inertia of a mass to sense acceleration/deceleration.

Watt Governor

• Derivation of Height (h):

  For a ball of mass m, arm length l, rotating at ω rad/s with sleeve height h.

  Forces: Centrifugal F_c = m r ω², Weight W = mg.

  Geometry: r = l sinθ, h = l cosθ.

  Equilibrium: tanθ = F_c / W = (r ω²)/g.

$$\boxed{h = \frac{g}{\omega^2}}$$

> [!TIP] **Proof:** From `tanθ = rω²/g` and `r = l sinθ`, `h = l cosθ`. For small `θ`, `tanθ ≈ sinθ ≈ θ`. Then `θ ≈ rω²/g`, and `h ≈ l - lθ²/2`. Dominant term: `h ≈ g/ω²`. **Height ∝ 1/Speed²** (`N ∝ ω`).

Porter Governor

• Derivation (with arms and sleeve mass):

  Let:

  • m = mass of each ball

  • M = mass of sleeve

  • l = length of each arm (upper & lower)

  • r = radius of rotation of balls

  • h = height of governor (sleeve lift from lowest position)

  • f = constant frictional force on sleeve (given as force equivalent).

  Forces on one ball: Centrifugal m r ω² outward, Tension T along lower arm, Weight mg down.

  For sleeve: 2T cosθ upward, (2m + M)g + f downward.

  Equilibrium:

$$\frac{T \cos\theta}{T \sin\theta} = \frac{(2m + M)g + f}{m r \omega^2}$$

But `tanθ = r / (h - a)` where `a` is the distance from pivot to sleeve axis (often `a=0` if pivoted on axis).

Solving:

$$\boxed{h = \frac{(2m + M)g + f}{m \omega^2} + a}$$

> [!TIP] **Friction Effect:** Friction `f` **increases** the effective downward force, making `h` larger for a given `ω`. This **reduces sensitiveness**.

Proell Governor

• Derivation (equal arms, extensions):

  Let:

  • l = length of upper arm (hinged on axis)

  • b = length of lower arm/extension from pivot to ball.

  • a = distance from axis to lower arm pivot.

  • At minimum speed, extensions are parallel to axis (r_min = a + b).

  Forces on ball: m r ω², mg, Tension T in lower arm.

  For sleeve: 2T cosφ upward, (2m + M)g down.

  Geometry: tanφ = (r - a)/b.

  Equilibrium:

$$\frac{2T \cos\phi}{2T \sin\phi} = \frac{(2m + M)g}{m r \omega^2} \implies \cot\phi = \frac{(2m + M)g}{m r \omega^2}$$

$$\boxed{\omega^2 = \frac{(2m + M)g \cdot b}{m (r - a)(r - a + b)}}$$

**Minimum Speed:** When `r = r_min = a + b` (extensions parallel).

$$\boxed{\omega_{min}^2 = \frac{(2m + M)g \cdot b}{m \cdot b \cdot (a + b)} = \frac{(2m + M)g}{m (a + b)}}$$

**Equilibrium Speed for given `r`:** Use the general formula above.

Governor Performance Terms

Controlling Force Diagram

• Plot of controlling force (F_c = T cosφ) vs. radius of rotation (r).

• Stable Governor: F_c increases with r (dF_c/dr > 0). Curve lies above the line F_c = m ω² r (isochronous line).

• Unstable Governor: F_c decreases with r (dF_c/dr < 0). Curve lies below the isochronous line.

• Isochronous Governor: F_c ∝ r (F_c = m ω² r). Straight line through origin.

  [!TIP] Stability Condition: For a stable governor, the slope of the F_c curve must be greater than the slope of the isochronous line at the equilibrium point.

Comparison: Porter vs. Proell

• Porter: Lower arms pivot on the axis (a=0). Height h ∝ (2m+M)g/(mω²).

• Proell: Lower arms pivot away from axis (a>0). For same r, ω is smaller (more sensitive).

• Proof of Sensitiveness: For same m, M, l, r, Proell's ω² denominator (r-a)(r-a+b) is smaller than Porter's (r)(r+l), hence ω is smaller at min speed and larger at max speed → larger speed range → greater sensitiveness.

Coefficient of Insensitiveness

• Definition: A measure of how much a governor's speed deviates from its nominal setting due to friction.

$$\boxed{C_i = \frac{\text{Frictional force}}{\text{Controlling force at mean radius}}}$$

It quantifies the **loss of sensitiveness** caused by friction. Higher `C_i` means less sensitive governor.

3. Balancing of Rotating and Reciprocating Masses

Primary and Secondary Balancing

• Rotating Masses: Produce centrifugal forces m r ω². Balanced by placing equal masses diametrically opposite.

• Reciprocating Masses:

  • Primary Force: m r ω² cosθ (in line with crank). Partially balanced by adding a revolving mass m_b at radius r_b such that m_b r_b = α m r (α = fraction balanced, usually 2/3 or 100% for high-speed engines).

  • Secondary Force: m r ω² (r/l) cos2θ (due to obliquity of connecting rod). Requires separate secondary balance masses rotating at 2ω. Often unbalanced in in-line engines.

Unbalance Effects in Engines

Balancing of Multi-Cylinder Engines

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

  • Primary forces: Can be completely balanced if m r same for all and crank angles 120° apart (vector sum zero).

  • Primary couples: Also balance if symmetric about center.

  • Secondary forces/couples: May not balance completely.

• In-line Engines:

  • Primary forces: Can be balanced by revolving mass (partial balance).

  • Primary couples: Cannot be balanced completely for even number of cylinders (unless crank throws are symmetrically arranged). For odd n, couples can balance if n ≥ 3 and cranks equally spaced.

  • Complete balance impossible for in-line 4-stroke engines with more than 2 cylinders due to conflicting requirements for force and couple balance.

• V-Engines:

  • Primary forces: Balance each other if V-angle = 90° for 90° crank separation (e.g., V8). Generally, forces balance if cylinder banks have equal m r and appropriate crank angles.

  • Primary couples: May need separate balance shafts.

Balancing Masses Calculation

1. Revolving Parts: For each crank, m_b r_b must be placed diametrically opposite to m r.

2. Reciprocating Parts (Partial): Balance fraction α (e.g., 2/3). For each cylinder, add a revolving mass m_b such that m_b r_b = α m r at the same crank angle as that cylinder's crank.

3. Given Crank Positions: Resolve all unbalanced primary forces (after partial balancing) into horizontal/vertical components at various crank angles. Use vector/couple diagrams to find resultant.

Dynamically Equivalent System

• Definition: A two-mass system (or sometimes three) that produces identical kinetic energy and resultant inertia forces/torques as the original distributed mass system at a given instant.

• Significance: Simplifies dynamic analysis of complex linkages (e.g., connecting rod) by replacing it with equivalent point masses.

• For a connecting rod (mass m_c, length l, c.g. at distance l₁ from small end, radius of gyration k about c.g.):

  • Mass at c.g.: m_c (to preserve total mass).

  • Two equal masses m₁ = m_c (k²)/(l₁ l₂) at both ends (to preserve moment of inertia about c.g. and kinetic energy).

  • Often simplified to one mass at c.g. plus a revolving mass at crank pin.

4. Friction in Bearings and Power Transmission Devices

Friction Circle

• Definition: A circle of radius r_f = r μ (where r = journal radius, μ = coefficient of friction) used to determine the direction of total friction force on a journal bearing.

• Expression: The total reaction R from the bearing lies within the friction circle. The angle between R and the normal is the angle of friction φ (tanφ = μ). The radius of the friction circle is r_f = r tanφ ≈ r μ for small φ.

$$\boxed{r_f = r \cdot \mu}$$

Conical Pivot Bearings

• Assumptions:

  1. Uniform Pressure: Pressure p constant over bearing surface.

  2. Uniform Wear: p r constant (pressure inversely proportional to radius).

• Given: Load W, cone angle 2α (semi-angle α), outer radius R, inner radius r, speed N, μ.

• Power Loss (P): P = Friction Torque × ω.

  Friction Torque (T_f):

  • Uniform Pressure: p = W / (π (R² - r²) sinα)

$$T_f = \frac{2}{3} \mu W \cdot \frac{R^3 - r^3}{R^2 - r^2} \cdot \frac{1}{\sin\alpha}$$

*   **Uniform Wear:** `p r = constant = W / (2π R r sinα)`

$$T_f = \frac{1}{2} \mu W \cdot \frac{R^2 - r^2}{R r} \cdot \frac{1}{\sin\alpha}$$

> [!TIP] **Exam Pattern:** You will be given `W, α, N, μ` and either `R/r` ratio or `R` and `r`. Compute `T_f` using the correct assumption, then `P = (2πN/60) × T_f`.

Collar Bearings

• Pressure Distribution: Usually assumed uniform pressure (p = constant).

• Friction Torque:

  For a collar with outer radius R, inner radius r, axial load W:

$$T_f = \mu W \cdot \frac{R + r}{2}$$

(Mean radius `R_m = (R+r)/2`).

• Number of Collars: If multiple collars on a shaft share the total load W, friction torque per collar is calculated and summed. Often used to increase friction surface.

Plate Clutches

• Assumptions:

  1. Uniform Pressure: p = constant = F / (π (R² - r²)) where F = axial force.

  2. Uniform Wear: p r = constant. Pressure max at inner radius r, min at outer R. p_max = F / (2π r (R - r)), p_min = F / (2π R (R - r)).

• Torque Transmission (T): For an elemental ring at radius x of width dx:

  • Uniform Pressure: dT = μ p (2π x dx) x = 2π μ p x² dx

$$\boxed{T = \frac{2}{3} \mu F \cdot \frac{R^3 - r^3}{R^2 - r^2}}$$

*   **Uniform Wear:** `dT = μ (k/x) (2π x dx) x = 2π μ k x dx`

$$\boxed{T = \frac{1}{2} \mu F \cdot \frac{R^2 - r^2}{R r}}$$

• Mean Radius (R_m): Radius at which the entire torque could be assumed to act.

  • Uniform Pressure: R_m = (2/3) × (R³ - r³)/(R² - r²)

  • Uniform Wear: R_m = (2/3) × (R² - r²)/(R r) × R? Actually, for uniform wear, R_m = (2/3) × (R² - r²)/(R r) × ? Better: T = μ F R_m, so R_m = T/(μ F).

• Face Width (b): b = R - r. Given R_m / b ratio, solve for R and r.

Conical Clutches

• Working Principle: Friction on conical surface. Axial force F creates normal pressure p and friction.

• Torque Calculation:

  For cone angle 2α (semi-angle α), mean radius R_m.

  Normal force on conical surface: F_n = F / sinα.

  Friction torque: T = μ F_n × R_m = μ F R_m / sinα.

$$\boxed{T = \frac{\mu F R_m}{\sin\alpha}}$$

> [!TIP] `R_m` is typically the **mean radius** of the conical friction surface: `R_m = (R + r)/2`.

5. Brakes

Band Brakes

• Types: Simple (one end fixed), Differential (lever attached to one end, different tensions).

• Tension Ratio (T₁/T₂): T₁/T₂ = e^{μθ} where θ = angle of wrap (radians), μ = coefficient of friction.

• Simple Band Brake: Braking torque T_b = (T₁ - T₂) × r_drum.

  If force F applied on lever at distance l from fulcrum, and band attached at distance x from fulcrum:

  T₁ × x = F × l (moment about fulcrum). Solve for T₁, then T₂ = T₁ / e^{μθ}, then T_b.

• Differential Band Brake: Two tensions act on lever at different radii. Condition for self-energizing: T₁ on longer arm, T₂ on shorter. T_b larger.

Internal Expanding Brakes (Shoe Brakes)

• Working: Two shoes inside a drum. Spring pulls shoes together (friction lining against drum). Hydraulic/pneumatic/mechanical force pushes shoes outward against drum.

• Shoe Force (P): For a given braking torque T_b, find force P on each shoe (assume equal).

  For a shoe with contact angle 2θ (radians), radius R:

  T_b = 2 × (Friction force from one shoe) × R.

  Friction force from one shoe = μ × ∫ p R dθ (pressure distribution assumed uniform or linear).

  Uniform Pressure: T_b = 2 μ P R θ (if P is total normal force on one shoe, and pressure uniform → p = P/(2Rθ)? Actually, for uniform pressure, dN = p R dθ, dF = μ p R dθ, dT = dF × R = μ p R² dθ. Integrate: T_one_shoe = μ p R² (2θ). But p = P / (2 R θ)? Total normal force N = ∫ p R dθ = p R (2θ) = P. So p = P/(2Rθ). Then T_one_shoe = μ (P/(2Rθ)) R² (2θ) = μ P R. So total T_b = 2 μ P R.

  Uniform Wear: Pressure p ∝ 1/cosφ? Actually, for internal shoe, pressure often varies. Common formula: T_b = μ P R (sinθ₁ + sinθ₂) for symmetric shoes? Better to use standard result: For a brake shoe with uniform pressure, T_b = μ P R (θ - sinθ cosθ)? I need to recall standard formula.

  Standard Approach: For a brake shoe with contact angle 2θ (from center line), and assuming uniform pressure, the braking torque is:

$$\boxed{T_b = \mu P R \left( \theta - \frac{\sin 2\theta}{2} \right)}$$

where `P` is the total normal force on that shoe.

• Width of Brake Shoes: Given maximum allowable pressure p_max, and total normal force P on one shoe:

  P = p_max × (contact area) = p_max × (b × arc length) = p_max × b × (2R θ).

  Solve for b.

6. Dynamometers

Classification

Torsion Dynamometers

• Working Principle: Measure the angle of twist (φ) in a shaft under torque T. T = (G J / L) φ, where G = shear modulus, J = polar moment of inertia, L = length over which twist is measured.

• Types:

  1. Strain Gauge Type: Strain gauges on shaft measure strain → torque.

  2. Epicyclic Gear Train Type: Fixed carrier, input shaft to sun gear, output to planet carrier. Torque on fixed member measured.

• Power Calculation: P = T × ω, where ω = angular velocity of shaft.

Types Overview

• Prony Brake: Band around drum, lever with weights. T_b = W × l. P = (2πN/60) × T_b.

• Rope Brake: Rope on drum, spring balance readings T₁, T₂. T_b = (T₁ - T₂) × R_drum.

• Hydraulic Dynamometer: Water-filled casing, impeller. Torque from reaction vanes. T = K (N² - N₁²).

• Epicyclic Gear Train Dynamometer: As above.

Key Formulas Box:

1. Flywheel Energy: $$\displaystyle \Delta E = \frac{1}{2} I (\omega_1^2 - \omega_2^2) $$

2. Watt Governor Height: $$\displaystyle h = \frac{g}{\omega^2} $$

3. Porter Governor Height: $$\displaystyle h = \frac{(2m+M)g + f}{m \omega^2} + a $$

4. Proell Governor: $$\displaystyle \omega^2 = \frac{(2m+M)g \cdot b}{m (r-a)(r-a+b)} $$

5. Friction Circle Radius: $$\displaystyle r_f = r \mu $$

6. Conical Pivot (Uniform Pressure): $$\displaystyle T_f = \frac{2}{3} \mu W \frac{R^3 - r^3}{R^2 - r^2} \csc\alpha $$

7. Plate Clutch (Uniform Wear): $$\displaystyle T = \frac{1}{2} \mu F \frac{R^2 - r^2}{R r} $$

8. Conical Clutch: $$\displaystyle T = \frac{\mu F R_m}{\sin\alpha} $$

9. Band Brake Tension: $$\displaystyle \frac{T_1}{T_2} = e^{\mu\theta} $$