Governors: Watt, Porter, Proell and Hartnell

How Watt, Porter, Proell and Hartnell centrifugal governors find equilibrium, their speed and spring formulas, and the meaning of sensitiveness, stability, isochronism, hunting and the controlling force diagram.

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Why it matters

When the load on an engine or turbine changes, its mean speed drifts unless the fuel or steam supply is adjusted. A centrifugal governor senses speed with rotating masses and moves a sleeve linked to the throttle. Electronic governors have replaced most of them, but the same ideas, equilibrium, sensitivity, stability, isochronism and hunting, are the language of any speed-control loop, and the mechanical governors remain classic exam and interview material.

Key ideas

Governor versus flywheel. A flywheel limits speed variation within a cycle at constant load; a governor maintains the mean speed over many cycles when the load changes, by changing the energy supplied.

Centrifugal governors. Balls (mass m each) rotate with the spindle at radius r. At equilibrium the controlling force (from gravity, a dead weight or a spring, acting through the linkage) supplies the centripetal force m·ω²·r. If speed rises, the balls fly out, the sleeve lifts and the throttle closes, and vice versa.

  • Gravity (dead-weight) controlled: Watt, Porter, Proell.
  • Spring controlled: Hartnell, Hartung, Wilson-Hartnell.

Watt governor. Two balls on arms pivoted on (or near) the spindle axis, no central load. The height h (from the plane of the balls to where the arms, extended, meet the axis) satisfies h = g/ω²; it depends only on speed, so the Watt governor is suited only to slow engines (at higher speeds h becomes tiny and the sleeve hardly moves).

Porter governor. A Watt governor with a heavy central load M on the sleeve. For equal arm and link lengths, pivoted on the axis and without friction, the speed for a given h is increased by the factor (m + M)/m, so it can work at higher speeds with a usable sleeve lift and a larger effort.

Proell governor. A Porter governor whose balls are fixed to extensions of the lower links (the links joined to the sleeve), not at the arm–link joint. For the same speed it needs smaller balls and gives a more powerful, more sensitive governor.

Hartnell governor. Balls on the vertical arms of bell-crank levers pivoted on a frame that rotates with the spindle; their horizontal arms press on the sleeve against a compressed helical spring. Spring force rather than weight provides the controlling force, so it can run fast and in any orientation, and the spring setting adjusts the speed.

Performance terms.

  • Sensitiveness: sleeve movement for a given speed change; often expressed as (N₂ − N₁)/N_mean (smaller is more sensitive).
  • Stability: for each speed in the working range there is exactly one equilibrium radius, and the radius rises as speed rises.
  • Isochronism: the equilibrium speed is the same at all radii (infinitely sensitive). An isochronous governor is not stable; in practice it hunts.
  • Hunting: the governor overshoots, the sleeve oscillating between extremes. It results from too sensitive a governor (or too little friction or damping).
  • Effort: mean force on the sleeve for a given fractional change of speed; power: effort × sleeve lift.
  • Controlling force diagram: plot of controlling force F against r. A spring governor with F = a·r − b (line cutting the force axis below the origin) is stable; F = a·r (through the origin) is isochronous; F = a·r + b is unstable.

Formulas

N² = 895 / h (Watt; N in rpm, h in m; 895 ≈ (60/2π)²·g with g = 9.81 m/s²) h = g / ω² (Watt, h in m, ω in rad/s)

N² = ((m + M) / m) × 895 / h (Porter: equal arms and links, pivots on the axis, friction neglected; m = mass of each ball, M = central load, kg) With link/arm angle ratio q = tan β / tan α: N² = ((m + M(1 + q)/2) / m) × 895 / h

N² = (FM / BM) × ((m + M) / m) × 895 / h (Proell, equal arms and links; FM/BM is the ratio set by where the ball sits on the extended link, from the sketch)

Hartnell (ball weight and arm obliquity neglected; x = ball arm, y = sleeve arm, M = sleeve mass):

  • F_c = m·ω²·r
  • F_c·x = ((S + M·g) / 2)·y → S = 2·F_c·x / y − M·g
  • sleeve lift h = (r₂ − r₁)·y / x
  • spring stiffness s = (S₂ − S₁) / h (N/m)

Sensitiveness = (N₂ − N₁) / N_mean

Worked examples

Example 1 (standard): Porter governor speed range. Given: each ball 5 kg, central load 30 kg, arms and links 250 mm, all pivoted on the axis. Ball radius ranges from 150 mm to 200 mm.

  1. At r = 0.15 m: h₁ = √(0.25² − 0.15²) = 0.200 m.
  2. N₁² = ((5 + 30)/5) × 894.6 / 0.200 = 7 × 4473 = 31 310 → N₁ = 176.9 rpm.
  3. At r = 0.20 m: h₂ = √(0.25² − 0.20²) = 0.150 m → N₂² = 7 × 894.6 / 0.150 = 41 750 → N₂ = 204.3 rpm.
  4. Speed range = 204.3 − 176.9 = 27.4 rpm; sensitiveness = 27.4 / 190.6 = 0.144. Answer: about 177 rpm to 204 rpm. (A Watt governor with h = 0.2 m would run at only 66.9 rpm.)

Example 2 (GATE level): Hartnell spring design. Given: each ball 2 kg; ball arm x = 100 mm, sleeve arm y = 80 mm. At the lowest equilibrium speed, 300 rpm, the ball radius is 120 mm; at the highest, 320 rpm, it is 150 mm. Neglect sleeve mass, ball weight and arm obliquity. Find the spring stiffness and initial compression.

  1. ω₁ = 31.42 rad/s → F_c1 = 2 × 31.42² × 0.12 = 236.9 N
  2. ω₂ = 33.51 rad/s → F_c2 = 2 × 33.51² × 0.15 = 336.9 N
  3. S₁ = 2 × 236.9 × (100/80) = 592.2 N ; S₂ = 2 × 336.9 × 1.25 = 842.2 N
  4. Sleeve lift h = (0.15 − 0.12) × 80/100 = 0.024 m
  5. s = (842.2 − 592.2) / 0.024 = 10 420 N/m ≈ 10.4 N/mm
  6. Initial compression = S₁ / s = 592.2 / 10 420 = 0.0568 m ≈ 56.8 mm Answer: spring stiffness ≈ 10.4 N/mm, initial compression ≈ 57 mm.

Common mistakes

  • Taking h as arm length minus radius; h = √(L² − r²) for arms pivoted on the axis.
  • Using N in rad/s with the 895 constant; 895 gives N in rpm with h in metres.
  • Forgetting the factor 2 in S = 2·F_c·x/y (the sleeve force is shared by two bell cranks).
  • Calling an isochronous governor ideal; it hunts.
  • Saying the central load makes a Porter governor "more stable at high speed"; it raises the operating speed and the effort.
  • Confusing Proell (balls on link extensions) with Hartnell (balls on bell cranks).

For GATE ME

Typical questions: equilibrium speed of Watt or Porter governors for given geometry; Hartnell spring stiffness and initial compression; definitions and conditions for stability, isochronism and hunting; controlling force diagrams. Practise the Porter formula with h from geometry and the Hartnell moment balance.

Quick check

  1. Height of a Watt governor running at 60 rpm?
  2. What is the speed factor for a Porter governor with 4 kg balls and a 20 kg central load?
  3. A spring governor has controlling force F = 800r + 20 (N, r in m). Is it stable?
  4. What is hunting?
  5. Which governor carries its balls on bell-crank levers?

Answers: 1. h = 895/3600 = 0.249 m. 2. (4 + 20)/4 = 6 (speed scales by √6). 3. No; F = ar + b is unstable. 4. Continuous oscillation of the sleeve and speed above and below the mean, caused by an over-sensitive governor. 5. Hartnell.

Interview questions

All Theory of Machines and Vibrations interview questions

Try answering each one aloud before you open it.

  1. 1.What is a governor in the context of mechanical engineering?Concept

    A governor is a device used in engines to regulate the speed by adjusting the fuel supply. It maintains a constant speed despite load variations by controlling the engine's input. Governors are crucial in applications where speed stability is essential.

  2. 2.Explain the working principle of a Watt governor.Concept

    A Watt governor is a simple type of centrifugal governor that uses two rotating balls attached to arms. As the engine speed increases, the balls move outward due to centrifugal force, lifting a sleeve that adjusts the fuel supply to reduce speed. Conversely, when the speed decreases, the balls move inward, lowering the sleeve and increasing fuel supply.

  3. 3.How does a Porter governor differ from a Watt governor?Concept

    A Porter governor is a Watt governor with a heavy central load M on the sleeve. For equal arms and links pivoted on the axis, N² = ((m + M)/m) × 895/h, so for the same height it runs at a speed √((m + M)/m) times higher. That lets it work at engine speeds where a Watt governor's height would be uselessly small, and the dead weight gives a much larger effort and power to move the throttle linkage against friction.

  4. 4.Describe the construction and working of a Proell governor.Concept

    A Proell governor is a Porter governor in which the balls are not at the joint of the upper arm and lower link but are fixed to an upward extension of each lower link, the link joined to the sleeve, with a central dead load on the sleeve. Because the ball's moment arm about the instantaneous centre of the link is larger, the equilibrium speed is N² = (FM/BM)((m + M)/m) × 895/h for equal arms. For the same speed it needs smaller balls than a Porter governor and is more sensitive.

  5. 5.What is a Hartnell governor and how does it function?Concept

    A Hartnell governor is spring controlled: the balls sit on the vertical arms of bell-crank levers pivoted on a frame that rotates with the spindle, and the horizontal arms bear on the sleeve, which is pushed down by a compressed helical spring. As speed rises, the centrifugal force m·ω²·r on each ball turns its bell crank and lifts the sleeve against the spring; moment balance gives m·ω²·r·x = ((S + Mg)/2)·y. Because a spring, not gravity, provides the controlling force, it works at high speed and in any orientation, and its speed is set by the spring stiffness and initial compression.

  6. 6.Why is a governor used in steam engines?Application

    Governors are used in steam engines to maintain a constant speed despite load variations. This is crucial for applications where consistent speed is necessary, such as in power generation or industrial machinery. Without a governor, the engine speed would fluctuate with changes in load, leading to inefficiency and potential damage.

  7. 7.What happens if the spring in a Hartnell governor is made stiffer?Application

    A stiffer spring needs a larger increase in centrifugal force, so a larger speed change, to move the sleeve through its lift, so the governor becomes less sensitive and the speed band between the lowest and highest equilibrium speeds widens. In return it is more stable and less prone to hunting. Increasing the initial compression without changing stiffness instead raises the whole speed range.

  8. 8.A Watt governor has arms 0.3 m long pivoted on the spindle axis, and the balls are 0.2 m apart. Find its equilibrium speed.Numerical

    The ball radius is r = 0.1 m, so the governor height is h = √(0.3² − 0.1²) = 0.283 m. For a Watt governor N² = 895/h with N in rpm and h in metres, so N = √(895 / 0.283) = 56.2 rpm. The ball mass does not appear, because both gravity and the centrifugal force are proportional to it.

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