Dynamics of Machine (ME-503 (B)) - Unit 3 Short Notes

UNIT 3: DYNAMICS OF MACHINE

I. FLYWHEELS AND ENERGY FLUCTUATION

Function & Role

• Flywheel: Stores kinetic energy during power stroke and releases during other strokes to smooth out speed fluctuation. It does not control mean speed.

• Governor: Controls mean speed by regulating energy supply. It is a feedback control device.

[!TIP] Key Difference: Flywheel deals with energy fluctuation (cyclic), governor with speed regulation (steady-state).

Fluctuation Concepts

Turning Moment Diagram (Multi-Cylinder Engine)

• Plots crank angle ($\theta$) vs. torque ($T$).

• Mean torque line: Horizontal line at average torque $$\displaystyle T_{mean} $$.

• Areas above line: Excess energy (flywheel absorbs).

• Areas below line: Energy deficit (flywheel releases).

• Scale conversion:

  • Vertical scale: $$\displaystyle 1 \text{ mm} = S_T \text{ N-m} $$

  • Horizontal scale: $$\displaystyle 1 \text{ mm} = S_\theta \text{ degrees} $$

  • Area $$\displaystyle A \text{ mm}^2 $$ corresponds to energy: $$\displaystyle \Delta E = A \times S_T \times \frac{\pi}{180} \times S_\theta $$ (convert $\theta$ to radians).

[!TIP] For four-stroke engine, one cycle = 720° ($4\pi$ rad). Diagram repeats every 720°.

Flywheel Design & Calculations

• Energy stored in flywheel: $$\displaystyle E = \frac{1}{2} I \omega^2 = \frac{1}{2} m k^2 \omega^2 $$

  • $I$: Moment of inertia, $m$: mass, $k$: radius of gyration, $\omega$: angular speed.

• Maximum/Minimum Speeds given $\Delta E$ and $$\displaystyle N_{mean} $$:

$$N_{max}, N_{min} = N_{mean} \left(1 \pm \frac{K_N}{2}\right) \quad \text{or} \quad \Delta E = I \omega_{mean}^2 \cdot \frac{\Delta \omega}{\omega_{mean}}$$

• Minimum radius of gyration from speed limits ($\Delta N$ given):

  \boxed{k = \sqrt{\frac{\Delta E}{\frac{1}{2} m (\omega_{max}^2 - \omega_{min}^2)}}}

  where $$\displaystyle \omega = \frac{2\pi N}{60} $$.

II. GOVERNORS

Function & Classification

• Function: Maintain constant mean speed by controlling fuel/steam supply despite load changes.

• Classification:

  • Centrifugal: Masses move outward with speed (Watt, Porter, Proell).

  • Inertia: Masses oscillate due to inertia forces (e.g., pendulum).

  • Simple: Single rotating mass (Watt).

  • Compound: Two rotating masses (Porter, Proell).

Watt Governor

• Construction: Two balls on arms, hinged to spindle; sleeve moves vertically.

• Derivation of Height ($h$):

  Forces on ball: $mg$ down, centrifugal $$\displaystyle mr\omega^2 $$ out.

$$\tan \phi = \frac{r\omega^2}{g} \approx \phi \text{ (small angle)}$$

Geometry: $$\displaystyle \sin \phi \approx \frac{r}{h} \Rightarrow h \propto \frac{1}{\omega^2} \propto \frac{1}{N^2} $$

\boxed{h = \frac{g}{4\pi^2 N^2}} \quad (N \text{ in rps})

[!TIP] Proof: $$\displaystyle h \propto 1/N^2 $$ shows inverse square relationship.

Porter Governor

• Construction: Upper arms ($$\displaystyle l_1 $$) and lower arms ($$\displaystyle l_2 $$), central load $W$ on sleeve.

• Speed Range with Friction:

  Friction $$\displaystyle F_f $$ acts on sleeve. Effective load = $$\displaystyle W \pm F_f $$.

  For limiting inclinations $$\displaystyle \phi_1, \phi_2 $$:

$$N_{1,2} = \frac{1}{2\pi} \sqrt{\frac{g(W \pm F_f) \cdot (l_1 + l_2)}{(W \pm F_f) l_1 l_2 \cdot \cos^2 \phi_{1,2} \cdot \sin \phi_{1,2}}}$$

[!TIP] Friction increases speed range (sensitivity reduces).

Proell Governor

• Construction: Lower arms have extensions; pivots offset from axis.

• Minimum Speed (extensions parallel to axis at min radius $$\displaystyle r_1 $$):

  \boxed{N_{min} = \frac{1}{2\pi} \sqrt{\frac{g \cdot (m + M)}{m r_1 \cos \phi \cdot (l_1 + l_2 \cos \phi)}}}

  where $\phi$ = angle of upper arm to vertical, $m$ = ball mass, $M$ = sleeve mass.

• Equilibrium Speed for any radius $r$:

$$N = \frac{1}{2\pi} \sqrt{\frac{g (m + M)}{m r \cos \phi \cdot (l_1 + l_2 \cos \phi)}}$$

$\phi$ varies with $r$.

Governor Characteristics

Analysis & Stability

• Controlling Force vs. Radius Diagram:

  • Stable: $$\displaystyle F_c $$ increases with $r$ (positive slope).

  • Unstable: $$\displaystyle F_c $$ decreases with $r$ (negative slope).

  • Isochronous: $$\displaystyle F_c $$ constant with $r$ (horizontal line).

• Condition for Stability: $$\displaystyle \frac{dF_c}{dr} > 0 $$.

• Proell vs. Porter Sensitiveness:

  Proell has higher sensitivity because for same $r$, $\phi$ is smaller → larger change in $r$ for $\Delta N$.

III. BALANCING OF ROTATING AND RECIPROCATING MASSES

Fundamentals

• Dynamically Equivalent System: Replaces actual mass distribution with point masses at specific locations to produce same inertia forces/moments.

• Primary vs. Secondary:

  • Primary: Due to simple harmonic component of reciprocating inertia force ($$\displaystyle mr\omega^2 \cos \theta $$).

  • Secondary: Due to quadratic component ($$\displaystyle mr\omega^2 \frac{r}{l} \cos 2\theta $$), where $l$ = connecting rod length.

Reciprocating Engines: Inertia Forces

• Inertia Force ($$\displaystyle F_I $$):

$$F_I = m \cdot a = m r \omega^2 \left( \cos \theta + \frac{r}{l} \cos 2\theta \right)$$

(neglecting piston rod area; $\theta$ = crank angle from IDC).

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

$$F_P = P \cdot A - F_I - F_f$$

where $P$ = gas pressure, $A$ = piston area, $$\displaystyle F_f $$ = friction.

• Thrust on Cylinder Walls ($$\displaystyle F_T $$):

$$F_T = \frac{F_P}{\tan \phi} \approx F_P \cdot \frac{r \sin \theta}{l \cos \theta}$$

$\phi$ = angle of connecting rod with piston axis.

• Thrust in Connecting Rod ($$\displaystyle F_R $$):

$$F_R = \frac{F_P}{\cos \phi}$$

• Crank Effort ($$\displaystyle F_T $$ on crank pin):

$$F_T = F_R \sin(\theta + \phi) = F_P \tan(\theta + \phi)$$

(or from torque: $$\displaystyle T = F_T \cdot r $$).

Partial Balancing

• Why only part? Balancing all reciprocating mass (primary) would cause unbalanced vertical force at high speed (hammer blow in locomotives). Hence, only a fraction ($1 - c$) is balanced by revolving mass.

• In-line Engines: Can achieve complete primary balance if cylinders are phased appropriately (e.g., 4-stroke inline-4: 180° crank interval). Secondary balance is incomplete unless special arrangements (e.g., balance shafts).

Multi-Cylinder Engines

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

  • Primary forces cancel if $$\displaystyle m_r $$ equal and crank angles $0°, 120°, 240°$.

  • Secondary forces also cancel for symmetric radial engines.

Locomotive Dynamics (Uncoupled Two-Cylinder)

• Assumptions: Crank angle $\theta$, cylinders at $0°$ and $$\displaystyle \theta_0 $$ (usually $90°$ for 4-stroke).

• Primary Unbalanced Force ($$\displaystyle F_P $$):

$$F_P = m_r r \omega^2 \left[ \cos \theta + \cos(\theta + \theta_0) \right]$$

• Secondary Unbalanced Force ($$\displaystyle F_S $$):

$$F_S = m_r r \omega^2 \frac{r}{l} \left[ \cos 2\theta + \cos 2(\theta + \theta_0) \right]$$

• Swaying Couple ($$\displaystyle M_S $$): Unbalanced couple about vertical axis.

$$M_S = m_r r \omega^2 \frac{r}{l} \cdot d \cdot \sin 2\theta \quad (d = \text{distance between cylinder centerlines})$$

• Hammer Blow ($$\displaystyle F_V $$): Unbalanced vertical force on wheel.

$$F_V = \text{Primary vertical component} + \text{Secondary vertical component}$$

For two-cylinder $90°$:

$$F_V = \sqrt{2} m_r r \omega^2 \left( \cos \theta + \frac{r}{l} \cos 2\theta \right)$$

• Variation in Tractive Effort ($$\displaystyle F_T $$):

$$F_T = \frac{2\pi T}{d_w} \pm \text{unbalanced horizontal force}$$

where $$\displaystyle d_w $$ = wheel diameter.

• Balancing Fraction ($c$): Fraction of $$\displaystyle m_r $$ balanced by revolving mass.

  Limit hammer blow: $$\displaystyle F_{V,max} \leq F_{allow} $$ → solve for $c$.

Balancing Mass Calculation

• Magnitude ($$\displaystyle m_b $$) for partial balance:

$$m_b r_b = (1 - c) m_r r$$

• Positioning: Opposite to crank at radius $$\displaystyle r_b $$.

• Resultant Residual Unbalance Force:

$$\vec{F}_{res} = \vec{F}_{unbalanced} + \vec{F}_{secondary} + \text{any remaining primary}$$

IV. FRICTION CLUTCHES AND BRAKES

Friction Clutches

• Single Dry Plate Clutch:

  • Assumptions:

    • Uniform Pressure: $$\displaystyle p = \frac{F}{\pi (r_o^2 - r_i^2)} $$ → $$\displaystyle T = \mu F \cdot \frac{2}{3} \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} $$

    • Uniform Wear: $$\displaystyle p r = \text{const} $$ → $$\displaystyle T = \mu F \cdot \frac{r_o + r_i}{2} $$

  • Mean Radius ($$\displaystyle R_m $$):

    • Uniform pressure: $$\displaystyle R_m = \frac{2}{3} \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} $$

    • Uniform wear: $$\displaystyle R_m = \frac{r_o + r_i}{2} $$

  • Face width ($b$): Given $$\displaystyle R_m / b = 4 $$ (typical), and $$\displaystyle T = \mu F R_m $$.

[!TIP] Uniform wear is more realistic for clutches; uniform pressure for brakes.

• Conical Clutch:

  • Torque: $$\displaystyle T = \mu F \cdot \frac{r_o + r_i}{2 \sin \alpha} \cdot \frac{1}{\sqrt{1 + \left( \frac{r_o - r_i}{2h} \right)^2}} \approx \frac{\mu F (r_o + r_i)}{2 \sin \alpha} $$

  where $\alpha$ = cone angle, $h$ = axial width.

Friction Brakes

• Internal Expanding Brake (Double Shoe):

  • Braking torque: $$\displaystyle T = 2 \mu F_s R \cdot \frac{\sin \theta}{\cos \phi} $$ (simplified), where $\theta$ = wrap angle per shoe, $\phi$ = angle of force.

  • Spring force ($$\displaystyle F_s $$) for given $T$: $$\displaystyle F_s = \frac{T}{2 \mu R \sin \theta} $$ (approx).

  • Shoe width ($b$): From pressure limit $$\displaystyle p_{max} $$:

$$p = \frac{F_s}{b \cdot R \cdot \theta} \leq p_{max} \Rightarrow b \geq \frac{F_s}{R \theta p_{max}}$$

• Simple Band Brake:

  • Braking torque: $$\displaystyle T = (T_1 - T_2) R = T_1 R \left(1 - \frac{1}{e^{\mu \theta}}\right) $$

    where $$\displaystyle T_1 $$ = tight side tension, $\theta$ = embrace angle (radians).

  • Lever force ($P$): From moment about fulcrum.

$$P \cdot l = T_1 \cdot a - T_2 \cdot b \quad \text{(geometry dependent)}$$

Friction Circle

• Definition: Circle of radius $$\displaystyle r_f = \mu R $$ representing maximum frictional moment for a journal bearing.

• Radius:

  \boxed{r_f = R \cdot \tan \phi \approx R \cdot \mu}

  where $R$ = journal radius, $\phi$ = friction angle ($$\displaystyle \tan \phi = \mu $$).

V. BEARINGS (FRICTION AND POWER LOSS)

Cone Pivot Bearings

• Construction: Conical surface, shaft rotates, load axial.

• Power Lost in Friction:

  • Uniform Pressure:

$$P = \frac{2}{3} \mu W \omega \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2}$$

• Uniform Wear:

$$P = \frac{1}{2} \mu W \omega \frac{r_o^2 - r_i^2}{r_o - r_i} = \frac{1}{2} \mu W \omega (r_o + r_i)$$

• Design (given $W$, $$\displaystyle p_{max} $$, $$\displaystyle \frac{r_o}{r_i} = k $$):

  From $$\displaystyle W = p \cdot \pi (r_o^2 - r_i^2) \cdot \frac{1}{\sin \alpha} $$ (uniform pressure) → solve for $$\displaystyle r_i, r_o $$.

Collar Bearings

• Power Absorption (uniform pressure):

$$P = \mu W \omega \frac{r_o^2 + r_i^2}{2(r_o - r_i)}$$

• Number of Collars ($n$): Total projected area $$\displaystyle n \cdot \pi (r_o^2 - r_i^2) \geq \frac{W}{p_{max}} $$.

VI. DYNAMOMETERS

Classification

Torsion Dynamometers

• Construction: Shaft with torque meter (e.g., spring-loaded lever, dead weights).

• Working: Torque $T$ causes angular twist $\theta$; measured by dial/scale.

• Power Calculation:

  \boxed{P = T \cdot \omega = \frac{2\pi N T}{60}}

  where $T$ = torque (N-m), $N$ = rpm.

Comparison

• Absorption: Simple, but wastes energy; used for engine testing.

• Transmission: In-line, no energy loss; used for in-situ measurement.

VII. KINEMATICS OF MECHANISMS

Four-Bar Mechanism

• Instantaneous Center (IC) Method:

  1. Find ICs ($$\displaystyle I_{12}, I_{23}, I_{34}, I_{14} $$).

  2. Velocity ratio: $$\displaystyle \frac{\omega_2}{\omega_4} = \frac{I_{14}I_{41}}{I_{12}I_{21}} $$ (sign by direction).

  3. Acceleration: Use relative acceleration equation:

$$\vec{a}_B = \vec{a}_A + \vec{a}_{B/A}^t + \vec{a}_{B/A}^n$$

• Angular Acceleration ($\alpha$):

$$\alpha_4 = \frac{a_t}{I_{14}I_{41}} \quad \text{(tangential component)}$$

• Midpoint of Coupler: Velocity = $$\displaystyle \omega_4 \times \text{distance from } I_{14} $$.

Cams and Followers

• Eccentric Cam (offset $e$):

  • Lift $$\displaystyle L = e(1 - \cos \theta) $$ for radial follower.

  • Acceleration: $$\displaystyle a = e \omega^2 \cos \theta $$.

• Follower Dynamics:

  • Spring force $$\displaystyle F_s = k \cdot \text{deflection} $$.

  • Lift-off speed: When $$\displaystyle F_s = 0 $$ (follower loses contact). Condition:

$$m a \geq F_{spring} + \text{inertia effects}$$

Solve for $\omega$.

VIII. SPECIAL APPLICATIONS AND TERMINOLOGY

Engine Balancing Terminology

Governor Terminology (Detailed)

Comparative Studies

> [!IMPORTANT] EXAM STRATEGY

1. Flywheel: Always convert area to energy using scales; use $$\displaystyle \Delta E = I \omega_{mean} \Delta \omega $$.

2. Governors: Draw force diagram; include friction as $$\displaystyle \pm F_f $$ on sleeve.

3. Balancing: For locomotives, remember $$\displaystyle F_{hammer} \propto \omega^2 $$; balance fraction $c$ reduces unbalanced mass.

4. Friction Devices: Identify assumption (uniform pressure/wear) before applying torque formula.

5. Four-Bar: IC method is fastest for velocity; acceleration needs relative motion equations.

6. Cam: Lift-off when spring force = 0 → $$\displaystyle m a = -k \cdot \text{deflection} $$ at that point.