Alternate Automotive Fuels & Emissions (ME-503 (C)) - Unit 3 Short Notes

UNIT 3: Dynamics of Engines and Frictional Systems

I. Governors

Function: Automatically control the fuel supply to maintain constant engine speed under varying load.

Classification:

• Centrifugal Governors: Use centrifugal force of rotating masses (balls).

• Inertia Governors: Use inertia forces of rotating masses.

Watt Governor

• Construction: Two arms, each of length l, hinged at the top to a spindle. Balls at the lower ends, sleeve on the spindle.

• Working: As speed increases, balls fly out, sleeve rises, closing fuel supply.

• Derivation of Height (h):

  Consider forces on one ball:

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

  Weight: $$\displaystyle W = mg $$

  For equilibrium: $$\displaystyle \tan \theta = \frac{F_c}{W} = \frac{m \omega^2 r}{mg} = \frac{\omega^2 r}{g} $$

  But $$\displaystyle \sin \theta = \frac{r}{l} $$ and for small $\theta$, $\tan \theta \approx \sin \theta \approx \theta$.

  So, $$\displaystyle \frac{r}{l} = \frac{\omega^2 r}{g} \implies h = l \cos \theta \approx l = \frac{g}{\omega^2} $$

  Proof: $$\displaystyle h \propto \frac{1}{N^2} $$ (since $\omega \propto N$).

$$\boxed{h = \frac{g}{\omega^2} \approx \frac{895.5}{N^2} \ \text{(if h in m, N in rpm)}}$$

Porter Governor

• Construction: Similar to Watt, but with an additional lower link (each of length l') pivoted on the sleeve. Balls are attached to the lower link extensions.

• Expression for Equilibrium Speed (N):

$$\boxed{N = \frac{1}{2\pi} \sqrt{\frac{g(m + M)}{m \left( \frac{l'}{l} + 1 \right) r} \left( \frac{l + l'}{l} \right)}}$$

where *m* = mass of each ball, *M* = mass of sleeve, *l* = length of upper arm, *l'* = length of lower arm, *r* = radius of rotation of balls.

• Effect of Friction: Friction at sleeve/pivot causes dead band. Governor responds only after speed change exceeds friction resistance, reducing sensitiveness.

• Speed Range with Friction (Numerical):

  Given limiting inclinations $$\displaystyle \theta_1, \theta_2 $$ and friction force F equivalent at sleeve.

  1. Calculate equilibrium speeds $$\displaystyle N_1 $$ (no friction, $$\displaystyle \theta_1 $$) and $$\displaystyle N_2 $$ (no friction, $$\displaystyle \theta_2 $$).

  2. Speed range without friction: $$\displaystyle N_2 - N_1 $$.

  3. With constant friction F, the sleeve force changes by $\pm F$.

  4. New max speed $$\displaystyle N_{max} $$ occurs when friction opposes rise (sleeve force = Mg - F).

  5. New min speed $$\displaystyle N_{min} $$ occurs when friction opposes fall (sleeve force = Mg + F).

  6. Solve Porter equation for N with modified sleeve force.

Proell Governor

• Construction: Upper arms pivoted on spindle. Lower arms are extended backwards and pivoted on the sleeve. Balls attached to the ends of these extensions. When at minimum radius, lower arms are parallel to the axis.

• Expression for Minimum Speed (N_min): (When extensions are parallel)

$$\boxed{N_{min} = \frac{1}{2\pi} \sqrt{\frac{g(m+M)}{m \cdot r_{min}} \left(1 + \frac{l'}{l}\right)}}$$

*r_min* = minimum radius of ball path.

• Expression for Maximum Speed (N_max):

$$\boxed{N_{max} = \frac{1}{2\pi} \sqrt{\frac{g(m+M)}{m \cdot r_{max}} \left(1 + \frac{l'}{l}\right)}}$$

*r_max* = maximum radius.

• Comparison with Porter Governor (Sensitiveness):

  Proell governor is more sensitive than Porter governor for the same dimensions because the centrifugal force on balls acts directly on the sleeve via the lower link pivot, giving a greater lifting force for the same speed change.

Governor Characteristics

• Sensitiveness: $$\displaystyle \frac{\text{Change in speed range}}{\text{Mean speed}} $$. A sensitive governor responds quickly to small load changes.

$$\text{Sensitiveness} = \frac{N_2 - N_1}{N_{mean}}$$

• Isochronism: Condition where the governor maintains constant speed for all load positions within its range ($$\displaystyle N_1 = N_2 $$). Theoretical ideal, not practical due to friction.

• Hunting: Oscillations of the governor sleeve about its mean position due to over-sensitivity. Causes speed fluctuations.

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

• Controlling Force Diagram & Stability Condition:

  • Plot $$\displaystyle F_C $$ (controlling force) vs. r (radius).

  • Stable: Slope of $$\displaystyle F_C $$ curve > Slope of centrifugal force line ($$\displaystyle m\omega^2 r $$).

  • Unstable: Slope of $$\displaystyle F_C $$ curve < Slope of $$\displaystyle m\omega^2 r $$.

  • Isochronous: $$\displaystyle F_C $$ curve is a horizontal line (slope = 0).

Friction in Governors

• Effect: Reduces sensitiveness, creates dead band, causes hunting if excessive.

• Coefficient of Insensitiveness ($k$):

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

Quantifies the relative effect of friction.

II. Flywheels

Function: Store kinetic energy during power stroke and release during other strokes to smoothen speed fluctuations. (Governor controls mean speed, flywheel controls fluctuation).

Turning Moment Diagram for 4-Stroke Engine:

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

• Other strokes (compression, suction, exhaust): Torque < Mean torque (negative area).

• Net area over cycle = 0 (mean torque line divides areas equally).

Diagram:

DiagramSEARCH: "four stroke engine turning moment diagram"

Fluctuation of Energy (ΔE):

• Definition: Maximum excess or deficiency of energy supplied/required relative to the mean over a cycle.

• Coefficient of Fluctuation of Energy (δE):

$$\delta_E = \frac{\Delta E}{\text{Work done per cycle}}$$

Fluctuation of Speed (ΔN):

• Definition: Difference between maximum and minimum speeds in a cycle.

• Coefficient of Fluctuation of Speed (δN):

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

Energy Stored in Flywheel:

• Kinetic energy: $$\displaystyle E = \frac{1}{2} I \omega^2 $$

• Change in energy: $$\displaystyle \Delta E = I \omega_{mean} \Delta \omega $$ (for small Δω)

• Relationship:

$$\boxed{\Delta E = I \cdot \omega_{mean} \cdot \Delta \omega = I \cdot \frac{2\pi N_{mean}}{60} \cdot \frac{2\pi \Delta N}{60}}$$

$$\implies \Delta E = \frac{2\pi^2 I N_{mean} \Delta N}{900}$$

> **For Design:** $\Delta E$ is found from turning moment diagram areas. $\Delta N$ is specified (usually as % of $$\displaystyle N_{mean} $$). Solve for *I* (moment of inertia).

$$I = \frac{900 \Delta E}{2\pi^2 N_{mean} \Delta N}$$

Then mass *m* and radius of gyration *k*: $$\displaystyle I = m k^2 $$.

Numerical Approach from Turning Moment Diagram:

1. Determine scale: 1 mm = S_T N-m (torque), 1 mm = S_θ degrees.

2. Calculate areas above/below mean torque line: $$\displaystyle A_1, A_2, ... $$ (in mm²).

3. Convert to actual energy: $$\displaystyle \text{Actual Area} = A \times S_T \times S_θ \times \frac{\pi}{180} $$ (since S_θ in degrees).

4. Find maximum cumulative energy deviation from zero: This is $\Delta E$ (max energy fluctuation).

5. Use formula above to find I or m or k.

III. Forces in Reciprocating Engines and Balancing

Forces in Reciprocating Engines

• Inertia Force (F_I): $$\displaystyle F_I = m_R \cdot a = m_R \cdot \omega^2 r \left( \cos \theta + \frac{r}{l} \cos 2\theta \right) $$

  • $$\displaystyle m_R $$ = mass of reciprocating parts.

  • Primary: $$\displaystyle m_R \omega^2 r \cos \theta $$

  • Secondary: $$\displaystyle m_R \omega^2 r \frac{r}{l} \cos 2\theta $$

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

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

*P* = pressure, *A* = area, $$\displaystyle F_f $$ = friction (often given as $\mu \cdot$ piston weight or separate).

• Thrust on Cylinder Walls (F_T): $$\displaystyle F_T = \frac{F_P}{\tan \phi} \approx F_P \cdot \frac{r}{l} \sin \theta $$ (for small $\phi$).

• Thrust in Connecting Rod (F_CR): $$\displaystyle F_{CR} = \frac{F_P}{\cos \phi} $$.

• Crank Effort / Torque (T): $$\displaystyle T = F_T \cdot r = F_P \cdot r \cdot \sin(\theta + \phi) \approx F_P \cdot r \sin \theta $$ (for small $\phi$).

• Dynamically Equivalent System: A system of two masses (one at crank pin, one on connecting rod) that produces the same inertia forces and torques as the original connecting rod.

Fundamentals of Balancing

• Static Balancing: Center of mass lies on axis of rotation. (For rotors).

• Dynamic Balancing: Both static balance and couple balance (no unbalanced couples). Necessary for long, narrow rotors.

• Primary Balancing: Balancing of forces due to first-order inertia terms ($\cos \theta$).

• Secondary Balancing: Balancing of forces due to second-order inertia terms ($\cos 2\theta$).

Balancing of Single Cylinder Engines

• Revolving Masses: Balance completely with a mass m_b at radius r_b opposite crank: $$\displaystyle m_b r_b = m_R r $$.

• Reciprocating Masses: Only partial balancing is possible/desirable.

  • Balance fraction α (usually 1/2 to 2/3) of $$\displaystyle m_R $$ as revolving mass.

  • Unbalanced primary force remains: $$\displaystyle F_{UP} = (1-\alpha) m_R \omega^2 r \cos \theta $$.

  • Secondary force is always unbalanced (unless α=1, which causes large vertical forces).

Balancing of Multi-Cylinder In-Line Engines

• Primary Balance: Achieved by arranging crank angles so primary forces cancel.

  • For n cylinders: $$\displaystyle \sum m_R r \cos \theta_i = 0 $$ and $$\displaystyle \sum m_R r \sin \theta_i = 0 $$.

• Secondary Balance: Requires $$\displaystyle \sum m_R r \frac{r}{l} \cos 2\theta_i = 0 $$ and $$\displaystyle \sum m_R r \frac{r}{l} \sin 2\theta_i = 0 $$.

  • Possible only for even n (e.g., 4, 6) with specific crank intervals (e.g., 180° for 4-cyl).

• Fraction to Balance: Typically balance 100% of revolving and 20-40% of reciprocating mass to limit unbalanced vertical forces.

• Complete Balancing? No. Complete balance of both primary and secondary forces for an in-line engine with reciprocating masses is impossible because the secondary force direction is fixed (vertical) and cannot be cancelled by angular arrangement alone without creating unbalanced couples.

Balancing of Radial Engines

• Cylinders radiate from center. Crank pins on same crank throw or multiple throws.

• Example: 3-cylinder at 120°:

  • Primary forces: $$\displaystyle F_{P1} \angle 0°, 120°, 240° $$ → Resultant = 0 (vector sum zero).

  • Secondary forces: $$\displaystyle F_{S1} \angle 0°, 240°, 120° $$ → Resultant = 0.

  • ⇒ Complete primary and secondary balance possible for symmetric radial engines.

Locomotive Balancing (Uncoupled Two-Cylinder)

• Assumptions: Two cylinders (left/right), cranks at 90° (typical), driving wheels coupled.

• Reciprocating Mass per cylinder: $$\displaystyle m_R $$.

• Crank radius: r.

• Wheel diameter: D.

• Distance between cylinder center lines: d.

• Distance between wheel centers (wheelbase): L.

1. Hammer Blow (Vertical Force on Rail):

• Definition: Unbalanced vertical force due to revolving masses (including partially balanced reciprocating masses) acting at the wheel.

• Derivation:

  • Balanced fraction α of reciprocating mass is added to revolving mass.

  • Total revolving mass per wheel: $$\displaystyle M_{eff} = m_R \alpha \cdot \frac{r}{R} + m_R $$? Wait, careful.

  • Actually, for two-cylinder 90°:

    Primary unbalanced force magnitude: $$\displaystyle F_{UP} = m_R \omega^2 r \sqrt{1 + \alpha^2 - 2\alpha \cos 90°} = m_R \omega^2 r \sqrt{1+\alpha^2} $$.

    But Hammer Blow is from revolving component.

  • Better: The unbalanced force at wheel due to rotating masses (including the αm_R treated as rotating) is:

$$H = (m_R + \alpha m_R) \omega^2 r \cdot \text{(factor depending on crank angle)}$$

    At speed where $$\displaystyle H_{max} $$ is vertical.

*   **Maximum Hammer Blow (when vertical):**

$$\boxed{H_{max} = (m_R + \alpha m_R) \omega^2 r \sqrt{2(1-\cos 90°)}?}$$

Let's derive properly.

    For two cranks at 90°, the resultant of two equal forces at 90° is $\sqrt{2} \times$ one force.

    But the rotating mass force from each cylinder is $$\displaystyle (m_R + \alpha m_R) \omega^2 r $$ at angle $$\displaystyle \theta_i $$.

    Resultant magnitude: $$\displaystyle H = (m_R + \alpha m_R) \omega^2 r \sqrt{2 + 2\cos(\theta_1-\theta_2)} $$? No.

    Actually, for two cylinders with cranks at 90°, the **unbalanced rotating force** (from *m_R + αm_R* treated as rotating) is:

$$H = (m_R + \alpha m_R) \omega^2 r \sqrt{2(1 + \cos 90°)}?$$

This is messy.

*   **Standard Result (for 90° crank):**

    Primary unbalanced force magnitude (from reciprocating part) is $$\displaystyle m_R \omega^2 r \sqrt{1+\alpha^2} $$.

    But Hammer Blow is the **vertical component** of the total unbalanced force from rotating masses.

    Total unbalanced force at wheel = $$\displaystyle \sqrt{ (m_R \omega^2 r \cos\theta_1 + ... )^2 + ... } $$.

    For two cylinders at 90°, the maximum vertical hammer blow occurs when one crank is at 45° to vertical? Actually, the **maximum vertical force** is:

$$H_{max} = (m_R + \alpha m_R) \omega^2 r \sqrt{2}$$

    because the two rotating mass forces are perpendicular at some instant? Let's think: if cranks are at 90°, the angle between their force vectors is 90°. The maximum vertical component of their resultant is when the resultant is vertical, which gives $$\displaystyle H_{max} = \sqrt{ (F_1 \cos\theta_1 + F_2 \cos\theta_2)^2 + ... } $$ max.

    Actually, simpler: The unbalanced force on each wheel due to its cylinder's rotating mass is $$\displaystyle (m_R + \alpha m_R) \omega^2 r $$ at angle $\theta$. For two wheels, the forces are at 90° relative to each other? Not exactly, because cranks are 90° apart, so the forces on the wheels are also 90° apart in phase.

    The **resultant vertical force on the rail** (sum of both wheels) varies. Hammer Blow is usually defined as the **maximum vertical force exerted by a wheel**.

    For a single wheel, the force from its cylinder is $$\displaystyle (m_R + \alpha m_R) \omega^2 r \sin(\theta + \phi) $$? No.

    Let's use standard formula from textbooks:

    For two-cylinder locomotive with cranks at 90°:

    Unbalanced primary force per cylinder: $$\displaystyle m_R \omega^2 r \cos\theta $$ (if we ignore α for hammer blow? No, hammer blow includes α).

    Actually, the **revolving mass** on each crank is $$\displaystyle m_R + \alpha m_R $$ (since we balance α fraction of reciprocating as revolving).

    So force from each crank: $$\displaystyle F_i = (m_R + \alpha m_R) \omega^2 r \sin(\theta + \delta_i) $$? Direction depends on crank orientation.

    Typically, if crank 1 is at angle θ from horizontal, its vertical force is $$\displaystyle (m_R + \alpha m_R) \omega^2 r \sin\theta $$.

    For crank 2 at θ+90°, vertical force = $$\displaystyle (m_R + \alpha m_R) \omega^2 r \sin(\theta+90°) = (m_R + \alpha m_R) \omega^2 r \cos\theta $$.

    So total vertical force on **both wheels** (assuming wheels are coupled and forces add) is:

$$H_{total} = (m_R + \alpha m_R) \omega^2 r (\sin\theta + \cos\theta)$$

    Maximum when $$\displaystyle \theta = 45° $$: $$\displaystyle H_{max,total} = (m_R + \alpha m_R) \omega^2 r \sqrt{2} $$.

    But Hammer Blow is often per wheel? Actually, the force on the rail is the sum from both wheels. So:

$$\boxed{H_{max} = \sqrt{2} \ (m_R + \alpha m_R) \ \omega^2 \ r}$$

    **Effect at Different Speeds:** $$\displaystyle H \propto \omega^2 \propto N^2 $$. Increases rapidly with speed.

2. Swaying Couple (Frictional Couple):

• Definition: Unbalanced couple about the center of gravity of the locomotive due to horizontal components of unbalanced primary forces.

• Derivation:

  Horizontal forces: $$\displaystyle F_{H1} = (m_R + \alpha m_R) \omega^2 r \cos\theta $$, $$\displaystyle F_{H2} = (m_R + \alpha m_R) \omega^2 r \cos(\theta+90°) = -(m_R + \alpha m_R) \omega^2 r \sin\theta $$.

  Distance between cylinders = d.

  Swaying couple magnitude:

$$S = F_{H1} \cdot \frac{d}{2} + F_{H2} \cdot \frac{d}{2} = \frac{d}{2} (F_{H1} + F_{H2}) = \frac{d}{2} (m_R + \alpha m_R) \omega^2 r (\cos\theta - \sin\theta)$$

Maximum when $$\displaystyle \cos\theta - \sin\theta = \sqrt{2} $$ (at $$\displaystyle \theta = -45° $$ or 315°):

$$\boxed{S_{max} = \frac{\sqrt{2}}{2} \ d \ (m_R + \alpha m_R) \ \omega^2 \ r = \frac{d}{\sqrt{2}} \ (m_R + \alpha m_R) \ \omega^2 \ r}$$

3. Variation in Tractive Effort:

• Definition: Fluctuation in the net force available for pulling the train due to unbalanced horizontal forces.

• Net Tractive Effort (F_T):

$$F_T = F_{Piston} \cdot \frac{r}{l} \ \text{(approx)} - \text{Unbalanced Horizontal Force}$$

Unbalanced horizontal force = $$\displaystyle (m_R + \alpha m_R) \omega^2 r (\cos\theta - \sin\theta) $$ (from above).

So $$\displaystyle F_T $$ varies with crank angle $\theta$.

4. Balancing Criteria: Limiting Hammer Blow

• Hammer Blow must not exceed a safe value $$\displaystyle H_{allow} $$ (to avoid rail damage).

• At maximum speed $$\displaystyle N_{max} $$:

$$H_{max} = \sqrt{2} \ (m_R + \alpha m_R) \ \left( \frac{2\pi N_{max}}{60} \right)^2 \ r \le H_{allow}$$

• Solve for maximum allowable α:

$$\alpha \le \frac{H_{allow}}{\sqrt{2} \ m_R \ \omega_{max}^2 \ r} - 1$$

This gives the fraction of reciprocating mass that can be balanced without exceeding hammer blow limit.

IV. Friction Clutches

Function: To engage/disengage power transmission from engine to gearbox.

Single Plate Clutch

• Construction: Two friction discs (one splined to engine shaft, one to gearbox shaft), pressed by spring force F.

• Torque Transmission: $$\displaystyle T = \mu F r_f $$, where $$\displaystyle r_f $$ = effective radius.

• Pressure Distribution Assumptions:

  1. Uniform Pressure (p = constant):

     • Pressure: $$\displaystyle p = \frac{F}{\pi (R^2 - r^2)} $$

     • Torque: $$\displaystyle dT = \mu p \cdot 2\pi r \cdot dr \cdot r = 2\mu \pi p r^2 dr $$

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

    For $R/r \approx 1.2-1.5$, approximate: $$\displaystyle T \approx \mu F R_{mean} $$, where $$\displaystyle R_{mean} = \frac{2}{3} \frac{R^3 - r^3}{R^2 - r^2} $$.

2.  **Uniform Wear (p r = constant):**

    *   Pressure: $$\displaystyle p = \frac{F}{2\pi (R - r) r} $$ (at inner radius *r* max pressure).

    *   Torque:

$$\boxed{T = \frac{1}{2} \mu F (R + r)}$$

    *   **Maximum pressure** at inner radius: $$\displaystyle p_{max} = \frac{F}{2\pi r (R - r)} $$

    *   **Minimum pressure** at outer radius: $$\displaystyle p_{min} = \frac{F}{2\pi R (R - r)} $$

    *   **Average pressure:** $$\displaystyle p_{avg} = \frac{F}{\pi (R^2 - r^2)} $$ (same as uniform pressure formula for *p*).

Conical Clutch

• Construction: Friction lining on conical surface. Axial force F.

• Torque Expression:

  Consider elemental ring at radius r, width dr (along slant height).

  Normal force: $$\displaystyle dN = p \cdot 2\pi r \cdot \frac{dr}{\sin\alpha} $$ ($\alpha$ = cone angle).

  Friction: $$\displaystyle dT = \mu dN \cdot r = \mu p \cdot 2\pi r^2 \frac{dr}{\sin\alpha} $$.

  For uniform wear: $$\displaystyle p r = c \implies p = c/r $$.

$$T = \int_r^R \mu \frac{c}{r} \cdot 2\pi r^2 \frac{dr}{\sin\alpha} = \frac{2\pi \mu c}{\sin\alpha} \int_r^R r dr = \frac{\pi \mu c (R^2 - r^2)}{\sin\alpha}$$

But $$\displaystyle c = p_{max} \cdot r $$, and axial force $$\displaystyle F = \int dN \cos\alpha = \frac{\pi c (R^2 - r^2)}{\sin\alpha} $$.

Hence:

$$\boxed{T = \mu F}$$

Friction Circle

• Definition: In a journal bearing or when considering friction in a pin joint, the friction circle is an imaginary circle of radius $$\displaystyle r_f $$ such that the frictional force vector is always tangent to it.

• Derivation of Radius:

  For a journal of radius r and coefficient of friction $$\displaystyle \mu = \tan\phi $$ ($\phi$ = angle of friction):

  Frictional force $$\displaystyle F_f = \mu N = N \tan\phi $$.

  The frictional torque is $$\displaystyle T_f = F_f \cdot r = N \tan\phi \cdot r $$.

  This torque can be represented by a force $N$ acting at radius $$\displaystyle r_f $$: $$\displaystyle T_f = N \cdot r_f $$.

  Equating: $$\displaystyle N r_f = N r \tan\phi \implies \boxed{r_f = r \tan\phi \approx r \sin\phi} $$ (for small $\phi$).

  Use: In analysis of bearings and clutches with uniform wear, the effective radius for torque is often $$\displaystyle r_f $$.

V. Brakes

Function: To reduce speed or stop a moving machine by absorbing kinetic energy.

Band Brake

• Simple Band Brake: One end of band fixed, other end pulled by force P. Tension $$\displaystyle T_1 $$ (tight side), $$\displaystyle T_2 $$ (slack side).

$$\frac{T_1}{T_2} = e^{\mu \theta}$$

($\theta$ in radians, angle of wrap).

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

Force *P* related to $$\displaystyle T_1, T_2 $$ by lever principle.

• Differential Band Brake: Both ends of band attached to lever at different distances from fulcrum. Allows self-energizing (braking force assists itself). Condition: $$\displaystyle P \cdot l_1 > T_2 \cdot l_2 $$ for clockwise drum rotation (typical).

Internal Expanding Brake (Shoe Brake)

• Construction: Two friction shoes inside a rotating drum. Cam mechanism (or hydraulic) pushes shoes outward. Return springs.

• Force Analysis:

  For each shoe:

  Normal force: $$\displaystyle N = \frac{F_{cam}}{2 \sin(\theta/2)} $$? Actually, cam force F is resolved.

  Better: For a given shoe with contact angle $2\theta$, the normal force N produces:

  • Tangential friction force: $$\displaystyle F_t = \mu N $$

  • Radial force on drum: $N$ (distributed).

  Braking torque from one shoe: $$\displaystyle T_{shoe} = \mu N \cdot r_d \cdot \text{(effective radius factor)} $$.

  For uniform pressure, effective radius = $$\displaystyle \frac{2}{3} \frac{R^3 - r^3}{R^2 - r^2} $$.

  For uniform wear, effective radius = $$\displaystyle \frac{R+r}{2} $$.

  Total torque = sum from both shoes.

Double Shoe Brake

• Construction: Two shoes on opposite sides of drum, each with its own actuator (or linked). More balanced braking.

• Force & Pressure: Similar to single shoe, but forces from both shoes act on drum. Spring force S determines normal force.

VI. Bearings (Friction in Bearings)

Conical Pivot Bearing

• Construction: Shaft with conical end, mating conical housing.

• Uniform Pressure Assumption:

  • Pressure: $$\displaystyle p = \frac{W}{\pi (R^2 - r^2)} $$ (constant).

  • Frictional Torque:

$$T_f = \mu W \cdot \frac{R^3 - r^3}{3(R^2 - r^2)} \cdot \csc\alpha$$

    ($\alpha$ = cone angle).

*   Power Loss: $$\displaystyle P = T_f \cdot \omega $$.

• Uniform Wear Assumption:

  • Pressure: $$\displaystyle p = \frac{W}{2\pi (R - r) r} $$ (max at inner radius r).

  • Frictional Torque:

$$\boxed{T_f = \frac{1}{2} \mu W (R + r) \csc\alpha}$$

*   Power Loss: $$\displaystyle P = T_f \cdot \omega $$.

Collar Bearing

• Construction: Shaft with one or more collars (discs) bearing against a surface.

• Uniform Pressure Assumption:

  • Pressure: $$\displaystyle p = \frac{W}{\pi (R^2 - r^2)} $$.

  • Frictional Torque:

$$T_f = \mu W \cdot \frac{R^3 - r^3}{3(R^2 - r^2)}$$

*   **Number of Collars:** To limit pressure *p* ≤ allowable, choose number *n* such that:

$$p = \frac{W}{n \cdot \pi (R^2 - r^2)} \le p_{allow}$$

*   Power Loss: $$\displaystyle P = T_f \cdot \omega $$.

• Uniform Wear Assumption:

  • Pressure: $$\displaystyle p = \frac{W}{2\pi n (R - r) r} $$.

  • Frictional Torque:

$$\boxed{T_f = \frac{1}{2} \mu W (R + r)}$$

    (Independent of *n*? Actually, *W* is total load, so torque per collar is $$\displaystyle \frac{1}{2} \mu \frac{W}{n} (R+r) $$, total $$\displaystyle T_f = \frac{1}{2} \mu W (R+r) $$).

*   Power Loss: $$\displaystyle P = T_f \cdot \omega $$.

VII. Dynamometers

Purpose: Measure power, torque, and speed of engines.

Classification:

• Absorption Dynamometers: Absorb engine power as heat (e.g., Prony brake, hydraulic).

• Transmission Dynamometers: Transmit power while measuring (e.g., epicyclic, belt, torsion).

Torsion Dynamometers

• Epicyclic Torsion Dynamometer:

  • Construction: Sun gear (on engine shaft), planet gears (on torque arm), ring gear fixed. Torque arm with spring balance.

  • Working: Engine torque T causes planet gears to push against torque arm. Force F measured.

  • Power Calculation:

    Let R = pitch radius of planet gear, n = number of planet gears.

    Torque on one planet gear: $$\displaystyle T_{planet} = F \cdot R $$.

    Total torque on sun gear: $$\displaystyle T = n \cdot T_{planet} = n F R $$.

$$\boxed{T = n F R}$$

    Power: $$\displaystyle P = T \cdot \omega $$.

• Belt Transmission Dynamometer: Engine drives a pulley, belt tension difference measured by spring balance. $$\displaystyle T = (T_1 - T_2) \cdot r $$.

Other Types:

• Prony Brake: Absorption. Brake band on drum, lever with weights. $$\displaystyle T = W \cdot L $$ (ignoring friction on lever).

• Hydraulic Dynamometer: Absorption. Water flow resistance, torque from torque tube.

• Electrical Dynamometer: Absorption/Transmission. Generator/brake, measure electrical output.

VIII. Kinematics of Mechanisms

Four-Bar Chain

• Terminology: Fixed link (frame), crank (input), coupler, follower (output).

• Velocity Analysis:

  1. Instantaneous Center (IC) Method:

     • Find all ICs (Kennedy's theorem).

     • Velocity of any point = $$\displaystyle \omega_{link} \times $$ distance from IC.

     • Relative velocity: $$\displaystyle v_B = v_A + v_{BA} $$.

  2. Relative Velocity Method:

     • $$\displaystyle \vec{v}_B = \vec{v}_A + \vec{\omega}_{AB} \times \vec{r}_{AB} $$ (if AB is crank).

     • Or: $$\displaystyle \vec{v}_B = \vec{v}_C + \vec{\omega}_{BC} \times \vec{r}_{BC} $$.

     • Solve vectorially or by components.

• Acceleration Analysis:

  1. Instantaneous Center of Acceleration: Less common.

  2. Relative Acceleration Method:

$$\vec{a}_B = \vec{a}_A + \vec{\alpha}_{AB} \times \vec{r}_{AB} - \omega_{AB}^2 \vec{r}_{AB}$$

    For coupler (BC): $$\displaystyle \vec{a}_B = \vec{a}_C + \vec{\alpha}_{BC} \times \vec{r}_{BC} - \omega_{BC}^2 \vec{r}_{BC} $$.

    Solve for $$\displaystyle \omega_{BC}, \alpha_{BC} $$.

Cams and Followers

• Disc Cam with Reciprocating Flat-Faced Follower:

  • Geometry: Cam is a circular disc of radius R with center O offset by e from camshaft axis C. Follower moves vertically along line through C.

  • Lift (h): Maximum vertical displacement of follower.

  • Base Circle: Imaginary circle of radius R about O.

• Derivation of Acceleration Expression:

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

  Follower displacement s = distance from cam center O to cam profile along vertical line through C.

  From geometry:

$$s = e \cos\theta + \sqrt{R^2 - e^2 \sin^2\theta}$$

(Assuming follower always in contact).

Velocity: $$\displaystyle v = \frac{ds}{dt} = \frac{ds}{d\theta} \cdot \omega $$.

Acceleration: $$\displaystyle a = \frac{dv}{dt} = \frac{d^2s}{d\theta^2} \cdot \omega^2 $$.

Compute derivatives:

$$\displaystyle \frac{ds}{d\theta} = -e \sin\theta - \frac{e^2 \sin\theta \cos\theta}{\sqrt{R^2 - e^2 \sin^2\theta}} $$

$$\displaystyle \frac{d^2s}{d\theta^2} = -e \cos\theta - \frac{e^2 (\cos^2\theta - \sin^2\theta)}{\sqrt{...}} - \frac{e^4 \sin^2\theta \cos^2\theta}{(R^2 - e^2 \sin^2\theta)^{3/2}} $$

Then $$\displaystyle a = \omega^2 \frac{d^2s}{d\theta^2} $$.

• Critical Speed and Lift Condition:

  • Critical Speed: Speed at which follower loses contact (spring force becomes zero). Condition: Required acceleration $$\displaystyle a > g $$ (downward) for positive spring force.

  • Lift Condition: Cam profile must have positive radius of curvature to avoid cusp. Minimum radius of curvature occurs at maximum lift point. Check $$\displaystyle \rho_{min} > 0 $$.

[!TIP]

Exam Focus: Past papers heavily test:

1. Governors: Porter with friction (speed range), Proell min/max speed, comparison, characteristics.

2. Flywheels: Energy fluctuation from turning moment diagram areas, finding I or k.

3. Balancing: Locomotive terms (hammer blow, swaying couple derivations), fraction to balance, in-line engine balancing possibility.

4. Friction Devices: Clutch torque (uniform pressure/wear), brake torque (band, internal), bearing power loss (conical, collar).

5. Kinematics: Four-bar velocity/acceleration (IC method), cam acceleration derivation and critical speed.

6. Dynamometers: Epicyclic power calculation, types comparison.