Dynamics of Machine (ME-503 (B)) Unit 5 Notes

UNIT 5: Dynamics of Machines - Short Notes

1. Flywheels and Energy Fluctuation

Function of Flywheel vs. Governor:

Flywheel: Stores kinetic energy during power stroke and releases it during other strokes to reduce speed fluctuation caused by cyclic torque variation. It does not control the mean speed.
Governor: Controls the mean speed of an engine by automatically regulating the fuel/steam supply in response to load changes.

Turning Moment Diagram for Four-Stroke IC Engine:

Represents the turning moment (torque) on the crank vs. crank angle for one complete cycle (2 revolutions).
Key Feature: Power stroke (combustion) produces a large positive torque. The other three strokes (intake, compression, exhaust) consume power, resulting in negative or low torque.
The mean torque line is drawn such that the area above it equals the area below it over the cycle. The net area represents the work output per cycle.

[!TIP] Exam Focus: You will be given scales (e.g., 1 mm = X N-m vertically, 1 mm = Y° horizontally) and areas (in mm²) above/below the mean line. You must convert these areas to energy (Nm) using the vertical scale.

Fluctuation of Energy (ΔE):

The maximum excess or deficiency of energy stored in the flywheel relative to the mean energy level. It is the maximum deviation of the energy curve from the mean energy line.

\boxed{\Delta E_{\text{max}} = \text{Maximum area of the turning moment diagram above or below the mean torque line}}

Fluctuation of Speed (ΔN):

The difference between the maximum and minimum speeds of the flywheel.

\boxed{\Delta N = N_{\text{max}} - N_{\text{min}}}

Coefficients:

Coefficient of Fluctuation of Energy (K_E):

\boxed{K_E = \frac{\Delta E_{\text{max}}}{\text{Work done per cycle}}}
Coefficient of Fluctuation of Speed (K_N):

\boxed{K_N = \frac{\Delta N}{N_{\text{mean}}}}

Calculations from Turning Moment Diagram:

Given: Scales, areas (A₁, A₂, ...), mean speed (N), mass (m) or radius of gyration (k).

Convert Areas to Energy: ΔE_max (in Nm) = (Largest area in mm²) × (Vertical scale in Nm/mm).
Energy Stored in Flywheel: The maximum kinetic energy change equals ΔE_max.

\boxed{\Delta E_{\text{max}} = \frac{1}{2} I (\omega_{\text{max}}^2 - \omega_{\text{min}}^2) = I \omega_{\text{mean}} \Delta \omega}

where \( I = m k^2 \), \( \omega = \frac{2\pi N}{60} \).
Relate Δω to ΔN: \( \Delta \omega = \frac{2\pi \Delta N}{60} \).
For small fluctuations: \( \omega_{\text{max}}^2 - \omega_{\text{min}}^2 \approx 2 \omega_{\text{mean}} \Delta \omega \).

\boxed{\Delta E_{\text{max}} = I \omega_{\text{mean}} \Delta \omega = m k^2 \cdot \frac{2\pi N_{\text{mean}}}{60} \cdot \frac{2\pi \Delta N}{60}}

\boxed{\Delta E_{\text{max}} = \frac{m k^2 \pi^2 N_{\text{mean}} \Delta N}{900}}
Find Unknown (m or k): Substitute known values and solve.
Find N_max & N_min: Given ΔN and N_mean:

\boxed{N_{\text{max}} = N_{\text{mean}} + \frac{\Delta N}{2}}, \quad \boxed{N_{\text{min}} = N_{\text{mean}} - \frac{\Delta N}{2}}

Energy Stored in Flywheel:

\boxed{E = \frac{1}{2} I \omega^2 = \frac{1}{2} m k^2 \omega^2}

2. Centrifugal Governors

Function: To maintain the mean speed of an engine constant by automatically adjusting the fuel/steam supply as the load varies.

Classification:

Centrifugal Governors: (Watt, Porter, Proell) - Use centrifugal force of rotating balls.
Inertia Governors: Use inertia forces of a rotating mass (e.g., pendulum type).

Watt Governor

Construction: Two arms (length l) with balls (mass m) at ends, hinged to a rotating spindle. Arms are connected to a sleeve on the spindle via links.
Working: As speed increases, centrifugal force (mω²r) lifts the balls and sleeve. The sleeve is linked to the throttle valve to reduce fuel supply.
Derivation of Height (h):

For equilibrium at radius r:

Centrifugal force = Controlling force

\( m \omega^2 r = \frac{m g}{\cos \theta} \cdot \sin \theta = m g \tan \theta \)

But \( r = h \tan \theta \). Substituting:

\( m \omega^2 (h \tan \theta) = m g \tan \theta \)

\boxed{h = \frac{g}{\omega^2} = \frac{g}{4\pi^2 N^2}} \quad \text{(where } \omega = \frac{2\pi N}{60}\text{)}
Proof h ∝ 1/N²: Directly from formula \( h \propto \frac{1}{\omega^2} \propto \frac{1}{N^2} \).

Porter Governor

Construction: Similar to Watt, but has an additional central load (M) on the sleeve. The ball arms are connected to a lower hinged link, which is connected to the sleeve.
Effect of Friction: Friction in the sleeve and linkages causes dead band. The governor does not respond until the speed change overcomes friction, leading to hunting.
Range of Speed (with friction F):

Let:
- m = mass of each ball
- M = mass of central load + sleeve
- l = length of each arm (upper & lower, usually equal)
- r = radius of rotation of balls
- h = height of governor (distance from pivot to sleeve axis)
- F = constant friction force (acting downwards on sleeve)
Without Friction:

\boxed{\frac{m g}{M + m} = \frac{g}{4\pi^2 N^2} \cdot \frac{h}{r}} \quad \text{or} \quad N^2 \propto \frac{h}{r}

With Friction (F):

The friction adds/subtracts to the controlling force (M+m)g.

For maximum speed (sleeve just about to rise): Effective load = (M+m)g - F.

For minimum speed (sleeve just about to fall): Effective load = (M+m)g + F.

\boxed{N_{\text{max}}^2 = \frac{(M+m)g - F}{4\pi^2 m r} \cdot \frac{h}{r}}, \quad \boxed{N_{\text{min}}^2 = \frac{(M+m)g + F}{4\pi^2 m r} \cdot \frac{h}{r}}

Using geometry: \( h = \sqrt{l^2 - r^2} \) (if pivots on axis). If pivots at distance d from axis, \( h = \sqrt{l^2 - (r-d)^2} \).

Proell Governor

Construction: The lower arms are extended beyond their pivots (at distance d from axis). The balls are attached to these extensions (length b). The upper arms connect to the spindle axis.
Minimum Speed Calculation (extensions parallel to axis):

At minimum radius r, the lower extensions are vertical. The centrifugal force acts horizontally at the ball.

Taking moments about lower pivot O:

\( m \omega^2 r \cdot b = (M + m) g \cdot h \)

where h is the vertical distance from O to the line of action of (M+m)g (usually h = l cos θ, but at min speed θ small, h ≈ l).

\boxed{\omega_{\text{min}}^2 = \frac{(M + m) g \cdot h}{m r b}} \quad \Rightarrow \quad N_{\text{min}} \propto \sqrt{\frac{h}{r b}}
Maximum Speed: When arms are inclined, geometry gives r_max and h_max. Use general force/moment equilibrium.

Comparison: Porter vs. Proell

Feature	Porter Governor	Proell Governor
Ball Arm Connection	Lower arms pivot on sleeve axis.	Lower arms pivot away from axis (d > 0).
Sensitivity	Less sensitive.	More sensitive (higher ΔN for same Δh).
Height-Speed Relation	h ∝ 1/N² only if d=0.	More complex, but generally higher sensitivity.
Friction Effect	Significant dead band.	Less affected by friction due to longer lever arm (b).

Governor Performance Terms

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

\boxed{\text{Sensitiveness} = \frac{\text{Range of speed } (N_1 - N_2)}{\text{Mean speed } N_{\text{mean}}}}

Higher sensitiveness ⇒ larger ΔN for same Δh.
Isochronism: Ideal condition where the governor maintains constant speed (N_max = N_min) for all positions of the sleeve within its range. Requires infinite sensitiveness. Unstable in practice.
Hunting: Oscillations of the governor sleeve about its mean position due to over-sensitivity. Causes continuous speed fluctuation.
Stability: Governor returns to its mean position for a given speed without oscillation. Stable governors have a positive slope in controlling force diagram (F_c vs r).

Controlling Force Diagram & Stability

Plots controlling force (F_c) (vertical) vs. radius (r) (horizontal).
Stable Governor: F_c increases with r. Slope > 0. For a given Δr, ΔF_c is positive.
Unstable Governor: F_c decreases with r. Slope < 0.
Isochronous Governor: F_c ∝ r² (parabolic through origin). Slope increases with r.
Stability Condition: For a stable governor, the steepness of the F_c-r curve must be greater than that of the centrifugal force curve (mω²r vs r).

\boxed{\frac{dF_c}{dr} > m \omega^2} \quad \text{(for stability)}

Coefficient of Insensitiveness (K_i)

Quantifies the effect of friction. It is the fraction of the total force that is due to friction.

\boxed{K_i = \frac{F}{(M+m)g}} \quad \text{where F is friction force.}

The range of speed is increased by \( 2K_i \times \text{isochronous speed} \).

[!TIP] Proof: Sensitiveness of Proell > Porter: For same (M+m), m, r, h, the Proell governor has an additional lever arm (b) in the moment equation, making \( \omega^2 \propto \frac{1}{b} \). Since b > effective lever arm in Porter, the change in ω² for a given change in r is larger in Proell ⇒ higher sensitiveness.

3. Balancing of Engines

Primary and Secondary Balancing:

Primary Balance: Balancing of forces due to reciprocating masses considering only first-order (cos θ) terms. Assumes connecting rod length l → ∞ (i.e., piston motion is simple harmonic).

Primary Unbalanced Force per cylinder: \( F_P = m r \omega^2 \cos \theta \)
Secondary Balance: Balancing of second-order (cos 2θ) forces due to the finite length of the connecting rod. The reciprocating mass has an acceleration component \( \frac{r}{l} \omega^2 \cos 2\theta \).

Secondary Unbalanced Force: \( F_S = m r \omega^2 \frac{r}{l} \cos 2\theta \)

Partial Balancing of Reciprocating Masses:

Only a fraction (c) of the reciprocating mass is balanced by an equivalent revolving mass (placed opposite to crank). This reduces primary unbalanced force but introduces a vertical unbalanced force (hammer blow) and a swaying couple in locomotives.

Balanced Primary Force = \( c \cdot m r \omega^2 \cos \theta \)

Residual Primary Force = \( (1-c) \cdot m r \omega^2 \cos \theta \)

Key Terms (Locomotive Engines):

Hammer Blow: The vertical unbalanced force transmitted to the rails due to the unbalanced primary reciprocating masses. It varies with crank angle and causes dynamic loading of rails.

\boxed{Hammer Blow = (1-c) \cdot m r \omega^2 \cos \theta} \quad (for one cylinder)
Swaying Couple: The couple in the horizontal plane (perpendicular to the line of cylinders) due to unbalanced primary forces of two or more cylinders with crank angles not 180° apart. Causes the locomotive to sway.

For two cylinders at angle α: \( S = (1-c) m r \omega^2 d \sin(\theta + \alpha/2) \cdot \text{sign factor} \)

where d = distance between cylinder center lines.
Variation in Tractive Effort: The fluctuation in the horizontal force available for pulling the train due to the horizontal component of the unbalanced reciprocating forces.

Tractive Effort Variation = \( (1-c) m r \omega^2 \cos \theta \cdot \cos \phi \) (φ = angle of connecting rod)

In-line Engines & Complete Balancing:

In-line engines: Cylinders in a single row (e.g., 4, 6, 8 cylinders).
Complete Primary Balance: Possible for even-numbered in-line engines with cranks at 180° (e.g., 4-stroke 4-cylinder with cranks at 0°, 180°, 180°, 0°? Actually typical is 0°, 180°, 0°, 180° for even firing). The primary forces cancel in pairs.
Complete Secondary Balance: Possible for even-numbered in-line engines with cranks at 180° and connecting rods of equal length. Secondary forces also cancel.
Odd-numbered cylinders (e.g., 3, 5) cannot be completely balanced for both primary and secondary forces simultaneously. Residual forces remain.

Multi-cylinder Engine Balancing Examples

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

Crank angles: 0°, 120°, 240°.
Primary Force:

Vector sum of \( m r \omega^2 \cos(\theta + \phi_i) \) where φ_i are crank angles.

For 3 cylinders at 120°: \( \sum \cos(\theta + 0, 120, 240) = 0 \). Primary forces completely balance.
Secondary Force:

Sum of \( m r \omega^2 \frac{r}{l} \cos 2(\theta + \phi_i) \).

For 120° separation: \( \sum \cos(2\theta + 0, 240, 480=120) = \cos 2\theta + \cos(2\theta+240) + \cos(2\theta+120) = 0 \).

Secondary forces also completely balance.

Result: A 3-cylinder radial engine with 120° cranks is completely balanced for both primary and secondary forces.

2. Locomotive Engines (Two-cylinder, 90° crank):

Crank angles: 0° and 90°.
Primary Unbalanced Forces:

\( F_{P1} = m r \omega^2 \cos \theta \), \( F_{P2} = m r \omega^2 \cos(\theta - 90^\circ) = m r \omega^2 \sin \theta \)

Resultant magnitude: \( F_P = m r \omega^2 \sqrt{\cos^2\theta + \sin^2\theta} = m r \omega^2 \) (constant!).

This constant horizontal force is balanced by a balancing mass on the driving wheel.
Balancing Fraction (c): To limit hammer blow, only a fraction (c) of reciprocating mass (m) is balanced by a revolving mass (m_b) on the wheel.

\boxed{m_b r_b = c \cdot m r} \quad (where r_b is radius of balancing mass)

The unbalanced reciprocating mass = (1-c)m.
Hammer Blow (Vertical): The vertical component of the unbalanced primary force.

For 90° crank, the resultant unbalanced force \( F_U = (1-c) m r \omega^2 \) acts at an angle (θ - 45°). Its vertical component:

\boxed{H = (1-c) m r \omega^2 \sin(\theta - 45^\circ)}

Maximum Hammer Blow = \( (1-c) m r \omega^2 \).
Swaying Couple: The couple about the centerline between cylinders.

\boxed{S = (1-c) m r \omega^2 d \cos \theta} \quad (for 90° crank, d = cylinder spacing)

Maximum Swaying Couple = \( (1-c) m r \omega^2 d \).
Tractive Effort Variation: The horizontal component of the unbalanced force available for pulling.

\boxed{\Delta T = (1-c) m r \omega^2 \cos(\theta - 45^\circ) \cos 45^\circ = \frac{(1-c) m r \omega^2}{\sqrt{2}} \cos(\theta - 45^\circ)}

Maximum Variation = \( \frac{(1-c) m r \omega^2}{\sqrt{2}} \).

3. Single Cylinder with Revolving & Reciprocating Masses:

Given: Mass of reciprocating parts (m_r), mass of revolving parts (m_re) at crank radius r.
Balancing Mass (m_b): Placed opposite to crank at radius R.

To balance all revolving mass and fraction (c) of reciprocating mass:

\boxed{m_b R = m_{re} r + c \cdot m_r r}
Residual Force: Due to unbalanced (1-c) of reciprocating mass.

\boxed{F_{\text{res}} = (1-c) m_r r \omega^2 \cos \theta}

At crank angle θ from IDC, substitute θ to find magnitude.

Dynamically Equivalent System:

A two-mass system (mass m₁ at distance l₁, mass m₂ at distance l₂ from a reference point) is dynamically equivalent to a distributed mass if:

\( m_1 + m_2 = m \) (total mass equal)
\( m_1 l_1 + m_2 l_2 = m r \) (first moment equal, where r is C.G. location)
\( m_1 l_1^2 + m_2 l_2^2 = m k^2 \) (second moment equal, where k is radius of gyration).

For a connecting rod, the two masses are placed at the crank pin and at the center of gravity.

4. Friction Clutches and Brakes

Friction Circle:

Definition: In journal bearings, the resultant reaction (R) passes through a point on the circumference of a circle of radius \( r_f \), called the friction circle.
Radius Expression:

\boxed{r_f = r \sin \phi}

where r = journal radius, φ = angle of friction (tan φ = μ).
Application: Used in analysis of friction in bearings and clutches to find frictional torque.

Single Plate Clutch

Working: Friction lining on one or both sides of a rotating plate. Axial pressure (P) applied by springs. Torque transmitted by friction.
Pressure Distribution Assumptions:
1. Uniform Pressure (p = constant): New clutch. \( p = \frac{P}{\pi (R^2 - r^2)} \)
2. Uniform Wear (p r = constant): Worn clutch. \( p = \frac{P}{2\pi R r} \)
Torque Transmission (for one side):
- Uniform Pressure: \( T = \mu p \int_{r}^{R} 2\pi r^2 dr = \frac{2}{3} \mu P \frac{R^3 - r^3}{R^2 - r^2} \)
- Uniform Wear: \( T = \mu \cdot \frac{P}{2\pi R r} \int_{r}^{R} 2\pi r^2 dr = \mu P R \)
Mean Radius (R_m): For uniform wear, \( T = \mu P R_m \), so \( R_m = R \) (for single plate). For uniform pressure, \( R_m = \frac{2}{3} \frac{R^3 - r^3}{R^2 - r^2} \).
Calculations: Given Power (P_w), Speed (N), Pressure (p_max), Coefficient (μ), find R, r, face width (b).
1. Torque: \( T = \frac{P_w}{\omega} = \frac{60 P_w}{2\pi N} \)
2. Use appropriate torque formula.
3. Pressure condition: \( p = \frac{P}{\pi (R^2 - r^2)} \leq p_{\text{max}} \) (uniform pressure) or \( p_{\text{max}} = \frac{P}{2\pi R r} \) (uniform wear, max p at inner radius).
4. Often given \( R/r \) ratio or \( R_m/b \) ratio. Use \( b = \frac{R - r}{\text{number of surfaces}} \).

Conical Clutch

Working: Friction surface is conical. Axial force (P) produces normal force \( N = \frac{P}{\sin \alpha} \) (α = cone angle). Friction force \( F = \mu N \).
Torque: \( T = \mu N \cdot \text{mean radius} \).

For uniform pressure/wear, mean radius \( R_m = \frac{R_2^2 - R_1^2}{2(R_2 - R_1)} \) or \( \frac{R_2 + R_1}{2} \).

\boxed{T = \mu \frac{P}{\sin \alpha} R_m}

Band Brakes

Simple Band Brake with Lever: One end of band fixed to fulcrum, other to lever. Tension \( T_1 \) (tight side) and \( T_2 \) (slack side).
Braking Torque: \( T_b = (T_1 - T_2) r_d \), where \( r_d \) = drum radius.
Friction Equation: \( \frac{T_1}{T_2} = e^{\mu \theta} \) (θ in radians, μ = coefficient).
Lever Analysis: Taking moments about fulcrum: \( F \cdot L = T_1 \cdot l_1 - T_2 \cdot l_2 \) (L = lever effort arm, l₁, l₂ = distances from fulcrum to band attachments).

Solve for \( T_1, T_2 \), then \( T_b \).

Internal Expanding Brake

Working: Two shoes (with friction lining) inside a rotating drum. A spring pulls shoes together (releasing brake). A hydraulic or mechanical linkage pushes a cam/lever to expand shoes against drum.
Self-energizing: The rotation of drum helps to increase the force on the leading shoe.
Torque Calculation: For each shoe, find normal force (N) from linkage geometry. Then \( T = \mu N R_m \cdot \text{effective radius} \). Consider leading/trailing shoe effects.

5. Bearings and Friction Loss

Conical Pivot Bearing

Supports vertical shaft with conical surface (angle 2α). Load W vertical.

Assumption 1: Uniform Pressure (p = constant)

Normal force on area: \( dN = p \cdot 2\pi r \cdot \frac{dr}{\sin \alpha} \)

Frictional force: \( dF = \mu dN \)

Torque: \( dT = r dF = \mu p \cdot 2\pi r^2 \frac{dr}{\sin \alpha} \)

Integrate r from r₁ to r₂:

\boxed{T = \frac{2}{3} \mu p \pi \frac{(r_2^3 - r_1^3)}{\sin \alpha}}

Also, \( W = \int dN \cos \alpha = p \pi \frac{(r_2^2 - r_1^2)}{\sin \alpha} \). Eliminate p.

\boxed{T = \frac{2}{3} \mu W \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2}}
Assumption 2: Uniform Wear (p r = constant)

Let \( p = \frac{c}{r} \). Then:

\( W = \int \frac{c}{r} \cdot 2\pi r \frac{dr}{\sin \alpha} \cos \alpha = \frac{2\pi c}{\sin \alpha} \cos \alpha \int_{r_1}^{r_2} dr = \frac{2\pi c}{\sin \alpha} \cos \alpha (r_2 - r_1) \)

\( T = \int \mu \frac{c}{r} \cdot 2\pi r^2 \frac{dr}{\sin \alpha} = \frac{2\pi \mu c}{\sin \alpha} \int_{r_1}^{r_2} r dr = \frac{\pi \mu c}{\sin \alpha} (r_2^2 - r_1^2) \)

Eliminate c:

\boxed{T = \frac{1}{2} \mu W \frac{r_2^2 - r_1^2}{r_2 - r_1}}
Power Loss: \( P = T \omega \), where \( \omega = \frac{2\pi N}{60} \).

Collar Bearings

Construction: Shaft with multiple collars (annular rings) on which load is borne.
Pressure Distribution: Usually assumed uniform over the projected area (for new bearings) or uniform wear.
Power Absorbed (Uniform Pressure):

Similar to conical pivot but with α = 90° (sin α = 1). For one collar with inner radius r₁, outer r₂:

\boxed{T = \frac{2}{3} \mu W \frac{r_2^3 - r_1^3}{r_2^2 - r_1^2}}

Total torque for n collars: \( T_{\text{total}} = n \times T \).

Power: \( P = T_{\text{total}} \omega \).
Number of Collars Required:

Given total load W_total, pressure limit p_max, find n.

For uniform pressure: Area per collar \( A = \pi (r_2^2 - r_1^2) \), load per collar \( W = p_{\text{max}} A \).

\boxed{n = \frac{W_{\text{total}}}{p_{\text{max}} \pi (r_2^2 - r_1^2)}}

Check power loss after finding n.

6. Dynamometers

Function: To measure the power and torque output of an engine or motor.