Alternate Automotive Fuels & Emissions (ME-503 (C)) - Unit 1 Short Notes

UNIT 1: DYNAMICS OF MACHINES - FUNDAMENTALS & APPLICATIONS

1.0 Governors (Speed Control Devices)

1.1 Function and Classification

• Function: Automatically regulates the speed of an engine by controlling the fuel/steam supply in response to load changes, maintaining speed within prescribed limits.

• Classification:

  • Centrifugal Governors:

    • Pendulum Type: (e.g., Watt Governor) - Sensitive to angular acceleration.

    • Pivot Type: (e.g., Porter, Proell Governors) - Sensitive to centrifugal force.

  • Inertia Governors: (e.g., Cross-Head Governor) - Sensitive to linear acceleration.

  • By Speed Range:

    • Constant Speed Governors: Isochronous (e.g., Hartnell Governor).

    • Variable Speed Governors: Most centrifugal types.

1.2 Key Terminology

• Sensitiveness: Ability to respond to small speed changes. Defined as:

$$ \text{Sensitiveness} = \frac{\text{Range of speed}}{\text{Mean speed}} = \frac{N_2 - N_1}{N} $$

Higher sensitiveness means a larger speed variation for a given sleeve movement.

• Isochronism: Ideal condition where the governor maintains constant speed for all load positions within the working range (sensitiveness = 0). Requires controlling force $$\displaystyle F_c \propto r $$.

• Hunting: Continuous fluctuation of engine speed above and below the mean due to over-sensitivity of the governor.

• Stability: Governor returns to its new equilibrium position after a speed change without hunting or instability.

[!TIP] Exam Focus: Sensitiveness, isochronism, and hunting are frequently asked for definitions and inter-relationships.

1.3 Controlling Force Diagram & Stability Conditions

• Controlling Force ($$\displaystyle F_c $$): Radial force exerted by the governor mechanism on the sleeve.

• Diagram: Plots $$\displaystyle F_c $$ (y-axis) vs. radius of rotation $r$ (x-axis).

• Stability Conditions:

  • Stable Governor: Slope of $$\displaystyle F_c $$ vs. $r$ curve > Slope of centrifugal force ($$\displaystyle mr\omega^2 $$) vs. $r$ curve.

  • Unstable Governor: Slope of $$\displaystyle F_c $$ vs. $r$ curve < Slope of $$\displaystyle mr\omega^2 $$ vs. $r$ curve.

  • Isochronous Governor: Both slopes are equal at all points.

1.4 Watt Governor

• Construction: Two balls on arms, pivoted on spindle. Arms connected to sleeve via links.

• Derivation of Height ($h$):

  For equilibrium in vertical direction: $$\displaystyle mg = 2F_c \cos\theta $$

  Centrifugal force: $$\displaystyle F_c = m r \omega^2 $$

  From geometry: $$\displaystyle r = h \tan\theta $$, $\cos\theta \approx 1$ for small $\theta$.

  Combining: $$\displaystyle mg = 2 m r \omega^2 \Rightarrow g = 2 r \omega^2 $$

$$ \boxed{h = \frac{g}{2 \omega^2} \propto \frac{1}{N^2}} $$

Where $N$ is speed in rpm, $$\displaystyle \omega = \frac{2\pi N}{60} $$.

1.5 Porter Governor

• Construction: Similar to Watt, but with an additional central load $W$ on the sleeve. Lower arms extend to balls.

• Analysis: Force equations on one arm:

  • Vertical: $$\displaystyle (W + mg) = 2P \cos\theta $$ (P = force in lower arm)

  • Horizontal: $$\displaystyle P \sin\theta = m r \omega^2 $$

  • Geometry: $$\displaystyle r = (h + x) \sin\theta $$, where $x$ = distance from pivot to lower arm pivot.

• Speed Range with Friction: Friction $$\displaystyle F_f $$ acts oppositely on sleeve during rise/fall.

  • Maximum Speed (Sleeve about to descend): Friction aids gravity.

$$ N_2 = \frac{1}{2\pi} \sqrt{\frac{g(W + 2mg)}{m[(h_1 + x)\cos\theta_1 - \frac{F_f}{m}]}} $$

*   **Minimum Speed (Sleeve about to rise):** Friction opposes gravity.

$$ N_1 = \frac{1}{2\pi} \sqrt{\frac{g(W + 2mg)}{m[(h_2 + x)\cos\theta_2 + \frac{F_f}{m}]}} $$

Where $$\displaystyle h_1, h_2 $$ and $$\displaystyle \theta_1, \theta_2 $$ correspond to max/min radii.

1.6 Proell Governor

• Construction: Bell-crank levers. Balls on extensions of lower arms. Pivot points on the axis.

• Analysis: Similar to Porter, but geometry differs.

  • $$\displaystyle r = (h + e) \sin\theta $$, where $e$ = distance from axis to lower arm pivot.

  • Force equations yield:

$$ m r \omega^2 = \frac{(W + mg)(h + e)\sin^2\theta}{e \cos\theta} $$

• Speed Range with Friction: Similar approach as Porter, but with Proell geometry. Friction $$\displaystyle F_f $$ at sleeve.

1.7 Comparative Analysis: Porter vs. Proell Governor

• Sensitiveness Comparison:

  For same $m, W, h, r, \theta$:

  • Porter: $$\displaystyle m r \omega^2 = \frac{(W+mg)(h+x)\sin^2\theta}{x \cos\theta} $$

  • Proell: $$\displaystyle m r \omega^2 = \frac{(W+mg)(h+e)\sin^2\theta}{e \cos\theta} $$

  Since $$\displaystyle e < x $$ (Porter's lower pivot farther from axis), for same $r, \theta$, Proell requires higher $\omega$.

  Hence, for same speed range, Proell has larger change in $r$, meaning higher sensitiveness.

$$ \boxed{\text{Proell Governor is more sensitive than Porter Governor.}} $$

1.8 Governor Performance: Coefficient of Insensitiveness

• Definition: Quantifies the loss of sensitiveness due to friction.

$$ \text{Coefficient of Insensitiveness} (K) = \frac{N_2 - N_1}{N} \bigg|_{\text{with friction}} - \frac{N_2 - N_1}{N} \bigg|_{\text{without friction}} $$

Alternatively: $$\displaystyle K = \frac{2F_f}{W + 2mg} $$ (for simple governors).

• Significance: Lower $K$ indicates better performance (less sensitivity loss to friction).

2.0 Flywheels (Energy Fluctuation Control)

2.1 Function and Difference from a Governor

2.2 Fluctuation of Energy & Speed

• Fluctuation of Energy ($\Delta E$): Difference between maximum and minimum kinetic energy of the flywheel during a cycle.

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

• Fluctuation of Speed ($\Delta N$ or $\Delta \omega$): Difference between maximum and minimum angular speeds.

• Significance: $\Delta E$ is determined from the turning moment diagram. Flywheel size is designed to limit $\Delta N$ (or $\Delta \omega$) within permissible limits (e.g., 1% for engines).

2.3 Coefficients

• Coefficient of Fluctuation of Energy ($$\displaystyle \beta_e $$):

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

• Coefficient of Fluctuation of Speed ($$\displaystyle \beta_s $$):

$$ \beta_s = \frac{\Delta \omega}{\omega_{\text{mean}}} \approx \frac{\Delta N}{N} \quad (\text{for small fluctuations}) $$

Relationship: $$\displaystyle \beta_e \approx 2 \beta_s $$ (for small $$\displaystyle \beta_s $$).

2.4 Turning Moment Diagram Analysis

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

• Mean Torque Line: Horizontal line such that area above = area below.

$$ T_{\text{mean}} = \frac{\text{Work done per cycle}}{2\pi} = \frac{\text{Area of diagram}}{2\pi} $$

• Interpretation of Areas:

  • Area above mean line ($$\displaystyle A_+ $$): Excess energy during power stroke → Flywheel stores energy.

  • Area below mean line ($$\displaystyle A_- $$): Deficit during exhaust/compression → Flywheel releases energy.

  • Maximum Energy Stored: Corresponds to point where cumulative area from a reference (e.g., TDC) is maximum.

  • Minimum Energy Stored: Corresponds to point where cumulative area is minimum.

$$ \Delta E = \text{Max cumulative area} - \text{Min cumulative area} $$

2.5 Mass & Radius of Gyration Calculation

• From Energy Fluctuation:

$$ \Delta E = I \omega_{\text{mean}} \Delta \omega \quad \text{or} \quad \Delta E = I \omega_{\text{mean}}^2 \cdot \frac{\Delta N}{N} $$

Where $$\displaystyle I = m k^2 $$ ($m$ = mass, $k$ = radius of gyration).

• Given Permissible Speed Fluctuation ($$\displaystyle \frac{\Delta N}{N} $$):

$$ m k^2 = \frac{\Delta E}{\omega_{\text{mean}}^2 \cdot (\Delta N / N)} $$

*   **Step 1:** Determine $\Delta E$ from turning moment diagram (using given scales).

*   **Step 2:** Compute $$\displaystyle \omega_{\text{mean}} = \frac{2\pi N}{60} $$.

*   **Step 3:** Use formula above to find $$\displaystyle m k^2 $$.

*   **If mass $m$ is given, find $k$; if $k$ is given, find $m$.**

2.6 Energy Fluctuation for Multicylinder Engines

• Principle: The net turning moment diagram is the algebraic sum of individual cylinder diagrams (considering crank angles).

• Effect: More cylinders → smaller $\Delta E$ for same mean torque → smaller flywheel required.

• Calculation: Plot resultant diagram, find max/min cumulative areas to get $\Delta E$.

3.0 Balancing of Rotating & Reciprocating Masses

3.1 Core Concepts

• Primary Balancing: Balancing forces due to unbalanced masses at crank radius (fundamental harmonic).

• Secondary Balancing: Balancing forces due to obliquity of connecting rod ($$\displaystyle \frac{r}{l} $$ term), producing secondary forces at twice crank speed.

• Hammer Blow: Vertical dynamic load on rail due to unbalanced primary forces in locomotive drivers. Maximum at 90° crank angle.

• Swaying Couple: Horizontal couple acting on locomotive frame due to unbalanced primary forces in two-cylinder engines. Maximum at 0° & 180°.

• Variation in Tractive Effort: Cyclic fluctuation in wheel-rail adhesion force due to unbalanced horizontal forces.

3.2 Balancing of Inline Engines

• Reciprocating Masses ($$\displaystyle m_r $$):

  • Primary Force: $$\displaystyle F_{p} = m_r \omega^2 r \cos\theta $$

  • Secondary Force: $$\displaystyle F_{s} = m_r \omega^2 r \frac{r}{l} \cos 2\theta $$

  • Partial Balancing: Only a fraction ($c$) of primary force is balanced (to avoid excessive vertical load on rails). Typically $$\displaystyle c = \frac{2}{3} $$ for locomotives.

  • Balancing Mass ($$\displaystyle m_b $$): Placed opposite crank at radius $$\displaystyle r_b $$ to balance revolving mass and fraction $c$ of reciprocating mass:

$$ m_b r_b = m_r r + c \cdot m_r r = (1+c) m_r r $$

• Residual Unbalanced Force ($$\displaystyle F_U $$): At crank angle $\theta$:

$$ F_U = m_r \omega^2 r \left[ (1-c) \cos\theta + \frac{r}{l} \cos 2\theta \right] $$

3.3 Balancing of Radial Engines (e.g., 3-cylinder, 120° apart)

• Primary Forces: For cylinders at $\theta, \theta+120°, \theta+240°$.

$$ F_{p,\text{total}} = m_r \omega^2 r \left[ \cos\theta + \cos(\theta+120°) + \cos(\theta+240°) \right] = 0 $$

**Complete primary balance** achievable for **3-cylinder** at 120°.

• Secondary Forces:

$$ F_{s,\text{total}} = m_r \omega^2 r \frac{r}{l} \left[ \cos 2\theta + \cos 2(\theta+120°) + \cos 2(\theta+240°) \right] $$

$$ = m_r \omega^2 r \frac{r}{l} \left[ \cos 2\theta + \cos(2\theta+240°) + \cos(2\theta+480°) \right] $$

Since $$\displaystyle \cos(2\theta+480°) = \cos(2\theta+120°) $$, sum = 0.

**Complete secondary balance** also achievable for **3-cylinder** at 120°.

3.4 Locomotive Balancing (Two-Cylinder, Uncoupled, Crank Angle $\alpha$)

• Assumptions: Crank radius $r$, reciprocating mass per cylinder $$\displaystyle m_r $$, wheel radius $$\displaystyle R_w $$, distance between cylinder center lines $d$, distance between wheel centers $L$.

• Primary Unbalanced Force (Horizontal):

$$ F_{p} = m_r \omega^2 r (1 + \cos\alpha) \quad \text{(at crank angle $\theta$)} $$

• Swaying Couple ($$\displaystyle C_s $$): Couple about center of gravity.

$$ C_s = m_r \omega^2 r \cdot d \cdot \sin\alpha \quad \text{(Maximum at $$\displaystyle \theta = 90°, 270° $$)} $$

• Hammer Blow ($H$): Vertical unbalanced force on rail (due to balancing mass $$\displaystyle m_b $$ at radius $R$ on wheel).

$$ H = m_r \omega^2 r (1-c) \cdot \frac{R}{r} - m_b \omega^2 R \quad \text{(at $$\displaystyle \theta=90° $$)} $$

For no hammer blow limit: $$\displaystyle H \leq H_{\text{max}} $$ → determines balancing fraction $c$.

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

$$ F_T = \frac{2 \cdot \text{Indicated mean effective force} \times \text{area}}{2} \pm F_{p} \cos\theta \quad \text{(fluctuates with $\theta$)} $$

• Maximum Swaying Couple: $$\displaystyle C_{s,\max} = m_r \omega^2 r d |\sin\alpha| $$.

4.0 Clutches & Brakes (Friction Power Transmission & Dissipation)

4.1 Single Plate Clutch

• Assumptions:

  • Uniform Pressure ($$\displaystyle p = \text{constant} $$): New clutch. Pressure $$\displaystyle p = \frac{F}{2\pi (r_2^2 - r_1^2)} $$.

  • Uniform Wear ($$\displaystyle p r = \text{constant} $$): Worn clutch. Pressure $$\displaystyle p = \frac{F}{2\pi (r_2 - r_1) r_m} $$.

• Torque Transmission:

  • Uniform Pressure: $$\displaystyle T = \mu F \cdot \frac{r_2^3 - r_1^3}{3(r_2^2 - r_1^2)} $$

  • Uniform Wear: $$\displaystyle T = \mu F \cdot \frac{r_2 + r_1}{2} = \mu F r_m $$

  Where $$\displaystyle r_m = \frac{r_2 + r_1}{2} $$ (mean radius).

• Design from Power & Speed:

$$ T = \frac{P}{\omega} = \frac{P}{2\pi N/60} = \frac{60P}{2\pi N} $$

Given $T, F, \mu$, solve for $$\displaystyle r_2, r_1 $$ with ratio $$\displaystyle r_2/r_1 $$ or face width $$\displaystyle b = (r_2 - r_1) $$.

4.2 Conical Clutch

• Working: Friction on conical surface. Normal force $$\displaystyle N = \frac{F}{\sin\alpha} $$ ($\alpha$ = cone angle).

• Torque: $$\displaystyle T = \mu N \cdot \frac{r_2 + r_1}{2} = \mu F \cdot \frac{r_2 + r_1}{2\sin\alpha} $$.

• Self-Energizing: For $$\displaystyle \alpha < \tan^{-1}\mu $$, clutch self-locks.

4.3 Band & Block Brakes (Simple Band Brake)

• Tension Ratio: $$\displaystyle \frac{T_1}{T_2} = e^{\mu \theta} $$ ($\theta$ in radians).

• Braking Torque: $$\displaystyle T_b = (T_1 - T_2) r_d $$ ($$\displaystyle r_d $$ = drum radius).

• With Lever: $$\displaystyle T_1 = F \cdot \frac{l_1}{l_2} $$ (if lever attached to $$\displaystyle T_1 $$ side). Solve for $$\displaystyle T_2 $$, then $$\displaystyle T_b $$.

4.4 Internal Expanding Shoe Brake (Double Shoe)

• Self-Energizing: Leading shoe gets additional force from drag of $$\displaystyle T_1 $$.

• Spring Force ($$\displaystyle F_s $$) Calculation:

  For each shoe: $$\displaystyle F_s + \mu N = \frac{T_1 - T_2}{r_d} \cdot \frac{l}{b} $$ (moment about pivot).

  With $$\displaystyle T_1/T_2 = e^{\mu \theta} $$, solve for $$\displaystyle F_s $$.

• Shoe Width ($b$) from Pressure:

  $$\displaystyle p = \frac{\text{Force on shoe}}{b \cdot R \cdot \theta} \leq p_{\text{max}} $$.

4.5 Bearing Pressure Limits

• Clutch: $$\displaystyle p \leq 0.1 - 0.2 \ \text{N/mm}^2 $$ (wet), $$\displaystyle 0.2 - 0.4 \ \text{N/mm}^2 $$ (dry).

• Brake Lining: $$\displaystyle p \leq 0.2 - 0.4 \ \text{N/mm}^2 $$.

5.0 Friction in Bearings & Power Loss

5.1 Friction Circle

• Definition: Circle of radius $$\displaystyle r_f = \mu R $$ representing locus of resultant friction force for a journal bearing.

• Radius: $$\displaystyle \boxed{r_f = \mu R} $$, where $R$ = journal radius, $\mu$ = friction coefficient.

• Significance: Used in velocity analysis of mechanisms with sliding contacts.

5.2 Pivot Bearings: Conical Pivot

• Uniform Pressure ($$\displaystyle p = \text{const} $$):

  • Load: $$\displaystyle W = \frac{\pi p (r_2^2 - r_1^2) \sin\alpha}{2} $$

  • Friction Torque: $$\displaystyle T_f = \frac{2}{3} \mu W r_m \csc\alpha $$, $$\displaystyle r_m = \frac{2}{3} \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$

  • Power Loss: $$\displaystyle P = T_f \omega $$

• Uniform Wear ($$\displaystyle p r = \text{const} $$):

  • Load: $$\displaystyle W = \frac{\pi p r_m (r_2 - r_1) \sin\alpha}{2} $$

  • Friction Torque: $$\displaystyle T_f = \frac{1}{2} \mu W r_m \csc\alpha $$

  • Power Loss: $$\displaystyle P = T_f \omega $$

• Bearing Dimensioning: Given $$\displaystyle W, p_{\text{max}} $$, find $$\displaystyle r_1, r_2 $$ from load equation.

5.3 Collar Bearing (Thrust Bearing)

• Uniform Pressure:

  • Load: $$\displaystyle W = \frac{\pi p (r_2^2 - r_1^2)}{2} $$

  • Friction Torque: $$\displaystyle T_f = \frac{2}{3} \mu W \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2} $$

• Uniform Wear:

  • Load: $$\displaystyle W = \pi \mu p r_m (r_2 - r_1) $$

  • Friction Torque: $$\displaystyle T_f = \frac{1}{2} \mu W (r_2 + r_1) $$

• Number of Collars: If total load $W$ is shared by $n$ collars, load per collar $W/n$ → use above formulas.

6.0 Dynamometers (Power Measurement)

6.1 Classification

6.2 Torsion Dynamometers

• Working Principle: Measures torsional deflection or strain in a shaft between engine and load.

• Types:

  • Shaft with Strain Gauges: Torque $$\displaystyle T = \frac{\pi d^3}{16} \tau_{\text{max}} $$ from measured strain.

  • Torsion Bar with Lever: Deflection $$\displaystyle \phi = \frac{T L}{G J} $$ measured by lever system.

• Power Calculation:

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

7.0 Kinematic Analysis of Mechanisms

7.1 Four-Bar Chain - Velocity & Acceleration

• Instantaneous Center (IC) Method:

  • Velocity: $$\displaystyle v = \omega \times r $$. For any point, $$\displaystyle v = \omega_{\text{link}} \times \text{distance to IC} $$.

  • Steps:

    1. Locate ICs ($$\displaystyle I_{12}, I_{13}, I_{14}, I_{23}, I_{34} $$).

    2. $$\displaystyle v_B = \omega_{AB} \cdot AB $$ (perpendicular to AB).

    3. $$\displaystyle \omega_{BC} = \frac{v_B}{I_{13}B} = \frac{v_C}{I_{13}C} $$.

    4. $$\displaystyle \omega_{CD} = \frac{v_C}{I_{14}C} $$.

  • Acceleration: Use relative acceleration equation:

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

    Where $$\displaystyle \vec{a}_{C/B} = \vec{\alpha}_{BC} \times \vec{r}_{C/B} - \omega_{BC}^2 \vec{r}_{C/B} $$.

    Resolve into tangential & radial components.

• Velocity/Acceleration of Mid-point: Same as any point on link. Use $$\displaystyle v = \omega \cdot r_{\text{mid to IC}} $$.

7.2 Turning Moment Diagram for Four-Stroke IC Engine

• Cycle: 720° (2 revolutions).

• Diagram Shape:

  • Power Stroke (Combustion): High positive torque.

  • Compression Stroke: High negative torque (resistance).

  • Exhaust & Intake: Low positive/negative torques.

• Significance: Shows cyclic energy fluctuation → determines flywheel size. Area under curve = work done per cycle.

8.0 Special Topics & Problem-Solving Applications

8.1 Dynamically Equivalent System

• Concept: Replace a distributed mass system with a lumped mass system having:

  1. Same total mass $m$.

  2. Same center of gravity (CG) position.

  3. Same mass moment of inertia about CG.

• Two-Mass System: Place masses $$\displaystyle m_1, m_2 $$ at distances $$\displaystyle l_1, l_2 $$ from CG such that:

  $$\displaystyle m_1 l_1 = m_2 l_2 $$ and $$\displaystyle I_{\text{CG}} = m_1 l_1^2 + m_2 l_2^2 $$.

• Importance: Simplifies dynamic analysis of connecting rods, etc.

8.2 Cams: Radial Cams with Offset (Circular Disc Cam)

• Follower Acceleration Expression:

  For a flat-faced follower in contact with a circular disc cam of radius $R$, center offset $e$ from camshaft axis.

  Let $\theta$ = cam rotation angle from start of lift.

  Follower lift $$\displaystyle s = R - \sqrt{R^2 - e^2 \sin^2\theta} - e \cos\theta $$.

  Acceleration: $$\displaystyle a = \frac{d^2s}{dt^2} = e \omega^2 \left[ \frac{\cos\theta}{(1 - k^2 \sin^2\theta)^{3/2}} + \cos\theta \right] $$, where $$\displaystyle k = e/R $$.

• Critical Speed Condition: Follower loses contact when normal force becomes zero. This occurs when radial acceleration of cam center towards follower exceeds spring force/mass.

$$ \omega_{\text{crit}} = \sqrt{\frac{k_s}{m} \cdot \frac{1}{\text{max}\left( \frac{d^2s}{d\theta^2} \right)}} $$

Where $$\displaystyle k_s $$ = spring stiffness, $m$ = follower mass.

8.3 Integrated Problem Solving

• Locomotive Example: Combine:

  • Balancing: Determine balancing fraction $c$ from hammer blow limit.

  • Flywheel: Use turning moment diagram (including unbalanced forces) to find $\Delta E$ and required $I$.

  • Governor: Ensure speed regulation matches engine characteristics.

• Engine Example: For inline engine, use partial balancing to find residual forces → include in crank effort diagram → design flywheel accordingly.

[!TIP] Exam Strategy: For integrated problems, first draw clear diagrams, identify knowns/unknowns, and apply concepts stepwise (e.g., balance → forces → energy → flywheel).