9.3 Flywheels, Centrifugal/Inertia Governors & Rotor Balancing

Flywheels, Centrifugal/Inertia Governors & Rotor Balancing

Quick Reference: Flywheels control cyclic speed fluctuation within a single cycle ($I = \Delta E / [C_s \omega^2]$). Governors control mean engine speed under changing external load by throttling fuel. Static balance requires $\sum \vec{F} = 0$; dynamic balance requires $\sum \vec{F} = 0$ and $\sum \vec{C} = 0$. In reciprocating balancing, balancing fraction $c$ creates hammer blow $F_{\text{hb}} = c m r \omega^2$.

In mining power plants, mine ventilation fan drives, diesel-electric haul trucks, and crushing/pulverizing equipment, mechanical control and balancing are paramount. Massive cyclical torque variations and high-speed rotor unbalance induce destructive structural vibrations, bearing fatigue, and shaft failure if not precisely controlled.

---

1. Turning Moment Diagrams & Flywheel Dynamics

Turning Moment Diagram (TMD)

A turning moment diagram (crank effort diagram) plots output driving torque ($T$) against crank angle ($\theta$). The area under the TMD curve represents the total work done during the cycle:

$W_{\text{cycle}} = \int_0^{\theta_{\text{cycle}}} T \, d\theta = T_{\text{mean}} \cdot \theta_{\text{cycle}}$

$\text{Power } P = T_{\text{mean}} \cdot \omega = \frac{2\pi N T_{\text{mean}}}{60}$

  Torque T
     ^
     |        Area A1 (+)
     |       /-----------\              Area A3 (+)
     |      /             \            /-----------\
  T_m|=====+---------------+----------+-------------+==== Mean Torque Line
     |      \             /            \           /
     |       \-----------/              \---------/
     |         Area A2 (-)               Area A4 (-)
     +----------------------------------------------------> Crank Angle theta
             0                   pi                 2pi

Energy Fluctuation & Flywheel Equations

1. Maximum Fluctuation of Energy ($\Delta E$): The maximum difference between the maximum kinetic energy and minimum kinetic energy of the flywheel during the cycle: $\Delta E = E_{\max} - E_{\min}$
2. Coefficient of Fluctuation of Energy ($C_e$): $C_e = \frac{\Delta E}{W_{\text{cycle}}} = \frac{\Delta E}{T_{\text{mean}} \cdot \theta_{\text{cycle}}}$
3. Coefficient of Fluctuation of Speed ($C_s$): $C_s = \frac{\omega_1 - \omega_2}{\omega_{\text{mean}}} = \frac{N_1 - N_2}{N_{\text{mean}}} = \frac{2(N_1 - N_2)}{N_1 + N_2}$
4. Flywheel Sizing Formula: $\Delta E = \frac{1}{2} I (\omega_1^2 - \omega_2^2) = I \omega_{\text{mean}} (\omega_1 - \omega_2) = I \omega_{\text{mean}}^2 C_s = m k^2 \omega_{\text{mean}}^2 C_s$ Where $I = m k^2$ is the mass moment of inertia, $m$ is flywheel mass, and $k$ is radius of gyration.

Flywheel vs. Governor Comparison

---

2. Centrifugal & Inertia Governors

Governor Classifications & Equilibrium Equations

+-------------------------------------------------------------------------+
|                        GOVERNOR CLASSIFICATION                          |
+-------------------------------------------------------------------------+
                                     |
     +-------------------------------+-------------------------------+
     |                                                               |
Centrifugal Governors                                         Inertia Governors
(Respond to Angular Speed w)                               (Respond to Angular Accel alpha)
     |
     +-------------------------------+-------------------------------+
     |                               |                               |
Gravity-Controlled              Spring-Controlled (Hartnell, Wilson-Hartnell, Pickering)
- Watt Governor (Simple)        - High sensitivity & operating speed
- Porter Governor (Central Load)- Operates in any physical orientation
- Proell Governor (Extended Arms)

Governing Equations for Standard Types:

1. Watt Governor (Gravity-Controlled): $h = \frac{g}{\omega^2} = \frac{895}{N^2} \text{ metres}$ Limitation: Highly insensitive at high speeds ($N > 80\text{ rpm}$) because $h$ becomes minuscule ($h \propto 1/N^2$).

2. Porter Governor (Dead-Weight Loaded): Incorporates a central sleeve mass $M$ in addition to fly-ball mass $m$: $N^2 = \frac{895}{h} \left[ \frac{m + \frac{M}{2}(1 + q)}{m} \right]$ Where $q = \frac{\tan\beta}{\tan\theta}$. For equal upper and lower arm lengths ($q = 1$): $N^2 = \frac{895}{h} \left( \frac{m + M}{m} \right)$ With sleeve friction force $f$: $N^2 = \frac{895}{h} \left[ \frac{m \cdot g + (M g \pm f)}{m \cdot g} \right]$

3. Proell Governor: Fly-balls are mounted on extensions of the lower links. Provides greater sleeve lift for a given speed variation than the Porter governor.

4. Hartnell Governor (Spring-Loaded): Fly-balls are mounted on vertical arms of bell-crank levers of length $a$; horizontal arms of length $b$ actuate the sleeve. $F_{c1} = \frac{b}{a}\left( \frac{M g + S_1 \pm f}{2} \right), \quad F_{c2} = \frac{b}{a}\left( \frac{M g + S_2 \pm f}{2} \right)$ Spring Stiffness ($s$): $s = \frac{S_2 - S_1}{h_s} = 2 \left(\frac{a}{b}\right)^2 \left( \frac{F_{c2} - F_{c1}}{r_2 - r_1} \right)$ Where sleeve lift is $h_s = \frac{b}{a}(r_2 - r_1)$.

---

3. Governor Characteristics & Stability Criteria

• Sensitiveness: Ratio of mean speed to the range of speed: $\text{Sensitiveness} = \frac{N_1 - N_2}{N_{\text{mean}}} = \frac{2(N_1 - N_2)}{N_1 + N_2}$
• Stability: A governor is stable if for every speed within the working range there is only one unique equilibrium radius $r$, and radius $r$ increases as speed $N$ increases ($dN/dr > 0$).
• Isochronism: Extreme condition of infinite sensitivity where equilibrium speed is identical at all radii ($N_1 = N_2 = N_{\text{const}}$). A gravity governor cannot be isochronous; a Hartnell governor is isochronous when initial spring tension satisfies $F_c = a r$ (intercept $b = 0$).
• Hunting: Severe dynamic instability where the governor continuously overshoots and undershoots the target speed due to excessive sensitiveness or insufficient dashpot damping.
• Controlling Force ($F_c = m \omega^2 r$): For spring-controlled governors: $F_c = a r + b$.
  • Stable: $b > 0$ (positive intercept on $F_c$ axis).
  • Isochronous: $b = 0$ (line passes through origin).
  • Unstable: $b < 0$ (negative intercept on $F_c$ axis).

4. Static & Dynamic Balancing of Rotors

Static vs. Dynamic Balancing

1. Static Balancing (Force Equilibrium):

   • The center of gravity of the rotating assembly lies precisely on the axis of rotation.
   • Net centrifugal force is zero: $\sum \vec{F} = 0 \implies \sum m_i r_i \omega^2 = 0 \implies \sum m_i r_i = 0$
   • A single revolving mass can be statically balanced by adding a single counterweight in the same plane of rotation.

2. Dynamic Balancing (Couple Equilibrium):

   • When rotating, no resultant centrifugal couple acts along the shaft.
   • Both resultant force and resultant moment about any reference plane must vanish: $\sum \vec{F} = 0 \quad \text{and} \quad \sum \vec{C} = \sum m_i r_i l_i = 0$
   • Dynamic balancing of any multi-mass rotor requires a minimum of two balancing masses in two distinct balancing planes.

  DYNAMIC COUPLE UNBALANCE IN ROTATING SHAFTS:
  
        Plane 1                      Plane 2
          m1 (Up)                      m2 (Down)
           o                            |
           |   Centrifugal Couple       |
  =========+============================+======== Axis of Rotation
           |    <--- Distance L --->    |
           |                            o
  
  Net Force = 0 (Static Balanced), but Net Couple = m*r*L*w^2 != 0 (Dynamic Unbalanced)!

---

5. Primary & Secondary Balancing of Reciprocating Engines

Reciprocating Inertia Force Expansion

In an internal combustion engine or reciprocating compressor with piston mass $m$, crank radius $r$, connecting rod length $l$, and obliquity ratio $n = l/r$:

$F_I \approx m r \omega^2 \left( \cos\theta + \frac{\cos 2\theta}{n} \right)$

• Primary Unbalanced Force ($F_p$): $F_p = m r \omega^2 \cos\theta$ (Acts at engine rotational frequency $\omega$).
• Secondary Unbalanced Force ($F_s$): $F_s = m r \omega^2 \frac{\cos 2\theta}{n}$ (Acts at twice engine frequency $2\omega$; maximum at TDC and BDC, magnitude is $1/n$ of primary).

Partial Primary Balancing

Balancing revolving counterweights cannot completely balance a reciprocating mass. If a balance mass $m_b$ is placed opposite the crank at radius $r_b$ such that $m_b r_b = c \cdot m r$ (where $c$ is the fraction balanced, typically $0.50$ to $0.75$):

1. Unbalanced Primary Force along line of stroke: $F_x = (1 - c) m r \omega^2 \cos\theta$
2. Unbalanced Primary Force perpendicular to line of stroke: $F_y = c \cdot m r \omega^2 \sin\theta$
3. Resultant Unbalanced Primary Force: $F_{\text{res}} = m r \omega^2 \sqrt{(1 - c)^2 \cos^2\theta + c^2 \sin^2\theta}$

Effects of Partial Balancing in Locomotives:

• Hammer Blow: The maximum vertical unbalanced force exerted on the rails by the revolving balance mass: $F_{\text{hb}} = c m r \omega^2$. If hammer blow exceeds static wheel load ($W$), wheel lifting (derailment hazard) occurs.
• Variation of Tractive Force: Along the line of stroke: $\Delta F_T = \pm \sqrt{2}(1 - c) m r \omega^2$.
• Swaying Couple: Creates lateral yawing moments about the vertical locomotive center: $C_{\text{sway}} = \pm \frac{a}{\sqrt{2}}(1 - c) m r \omega^2$ (where $a$ is distance between cylinder axes).

---

6. Worked Numerical Examples

Example 1: Flywheel Sizing for a Punching Press

Problem: A coal briquette stamping press executes 30 punches per minute. Each punching operation requires $1500\text{ J}$ of energy and takes $1/10\text{th}$ of the cycle time. A continuous electric motor drives the press via reduction gears at a mean flywheel speed of $\omega = 25\text{ rad/s}$. If the coefficient of fluctuation of speed is limited to $C_s = 0.04$ and the radius of gyration is $k = 0.6\text{ m}$, determine:

1. The power rating of the driving electric motor.
2. The required flywheel mass moment of inertia $I$ and flywheel mass $m$.

Solution:

1. Cycle Duration & Motor Power:

   • Total cycle time: $t_{\text{cycle}} = \frac{60}{30} = 2.0\text{ seconds}$.
   • Punching duration: $t_{\text{punch}} = \frac{2.0}{10} = 0.2\text{ seconds}$.
   • Motor Power ($P$): Motor supplies energy uniformly over entire $2.0\text{ s}$: $P = \frac{W_{\text{cycle}}}{t_{\text{cycle}}} = \frac{1500\text{ J}}{2.0\text{ s}} = 750\text{ W} = 0.75\text{ kW}$

2. Energy Supplied by Motor During Punching: $E_{\text{motor}} = P \times t_{\text{punch}} = 750\text{ W} \times 0.2\text{ s} = 150\text{ J}$

3. Maximum Fluctuation of Energy ($\Delta E$) Absorbed from Flywheel: $\Delta E = E_{\text{punch}} - E_{\text{motor}} = 1500 - 150 = 1350\text{ J}$

4. Flywheel Sizing: $\Delta E = I \omega^2 C_s \implies 1350 = I \times (25)^2 \times 0.04$ $1350 = I \times 625 \times 0.04 = I \times 25$ $I = \frac{1350}{25} = 54.0\text{ kg}\cdot\text{m}^2$

   Flywheel mass ($m$): $m = \frac{I}{k^2} = \frac{54.0}{(0.6)^2} = \frac{54.0}{0.36} = 150.0\text{ kg}$

---

Example 2: Hartnell Spring-Loaded Governor Design

Problem: In a Hartnell governor, the ball arm is $a = 120\text{ mm}$ and the sleeve arm is $b = 90\text{ mm}$. The mass of each rotating ball is $m = 2.5\text{ kg}$. At the minimum equilibrium radius $r_1 = 80\text{ mm}$, the speed is $N_1 = 300\text{ rpm}$. At the maximum radius $r_2 = 120\text{ mm}$, the speed is $N_2 = 330\text{ rpm}$. Neglecting friction and sleeve mass, determine:

1. The centrifugal forces $F_{c1}$ and $F_{c2}$.
2. The required spring stiffness $s$.

Solution:

1. Angular Velocities: $\omega_1 = \frac{2\pi(300)}{60} = 31.416\text{ rad/s}$ $\omega_2 = \frac{2\pi(330)}{60} = 34.558\text{ rad/s}$

2. Centrifugal Forces: $F_{c1} = m \omega_1^2 r_1 = 2.5 \times (31.416)^2 \times 0.080 = 2.5 \times 986.96 \times 0.080 = 197.39\text{ N}$ $F_{c2} = m \omega_2^2 r_2 = 2.5 \times (34.558)^2 \times 0.120 = 2.5 \times 1194.25 \times 0.120 = 358.28\text{ N}$

3. Spring Stiffness ($s$): $s = 2 \left(\frac{a}{b}\right)^2 \left( \frac{F_{c2} - F_{c1}}{r_2 - r_1} \right)$ $\frac{a}{b} = \frac{120}{90} = \frac{4}{3} \implies \left(\frac{a}{b}\right)^2 = \frac{16}{9} \approx 1.778$ $r_2 - r_1 = 0.120 - 0.080 = 0.040\text{ m}$ $s = 2 \times \left(\frac{16}{9}\right) \times \left( \frac{358.28 - 197.39}{0.040} \right) = 3.556 \times \frac{160.89}{0.040} = 3.556 \times 4022.25 = 14303\text{ N/m} = 14.30\text{ N/mm}$