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Project 3 The Watt Governor, Feedback Control, and Stability
In the latter part of the 18th century, James Watt designed and built a steam engine with a
rotary output motion (see Figure 5.P.3).
It was highly desirable to maintain a uniform rotational speed for powering various types
of machinery, but fluctuations in steam pressure and work load on the engine caused the
rotational speed to vary. At first, the speed was controlled manually by using a throttle valve
to vary the flow of steam to the engine inlet. Then, using principles observed in a device
for controlling the speed of the grinding stone in a wind-driven flour mill, Watt designed a
flyball or centrifugal governor, based on the motion of a pair of centrifugal pendulums,
to regulate the angular velocity of the steam engine’s flywheel. A sketch of the essential
components of the governor and its mechanical linkage to the throttle valve is shown in
Figure 5.P.4.
FIGURE 5.P.3
The Watt steam engine, 1781.
To understand how the mechanism automatically controls the angular velocity of the
steam engine’s flywheel and drive shaft assembly, assume that the engine is operating under
a constant load at a desired equilibrium speed or operating point. The engine, via a belt
and pulley assembly, also rotates a straight vertical shaft to which a pair of flyballs are
connected. If the load on the engine is decreased or the steam pressure driving the engine
increases, the engine speed increases. A corresponding increase in the rotational speed of
the flyball shaft simultaneously causes the flyballs to swing outward due to an increase in
centrifugal force. This motion, in turn, causes a sliding collar on the vertical shaft to move
downward. A lever arm connected to the sliding collar on one end and a throttle control
rod on the other end then partially closes the throttle valve, the steam flow to the engine is
reduced, and the engine returns to its operating point. Adjustable elements in the linkage
allow a desirable engine speed to be set during the startup phase of operation.
A Problem with Instability. Although some governors on some steam engines were
successful in maintaining constant rotational speed, others would exhibit an anomalous
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Chapter 5
October 21, 2010
The Laplace Transform
Sliding collar
Steam flow
to engine
Throttle valve
φ
θ0
FIGURE 5.P.4
The centrifugal governor.
rhythmic oscillation in which the system appeared to “hunt” unsuccessfully about its equilibrium value for a constant speed. The amplitude of oscillations would increase until
limited by mechanical constraints on the motion of the flyballs or the throttle valve.
The purpose of this project is to develop a relatively simple mathematical model of the
feedback control system and to gain an understanding of the underlying cause of the unstable
behavior. We will need a differential equation for the angular velocity of the steam engine
flywheel and drive shaft and a differential equation for the angle of deflection between the
flyball connecting arms and the vertical shaft about which the flyballs revolve. Furthermore,
the equations need to be coupled to account for the mechanical linkage between the governor
and the throttle valve.
The Flyball Motion. The model for the flyball governor follows from taking into account
all of the forces acting on the flyball and applying Newton’s law, ma = F. The angle between
the flyball connecting arm and the vertical shaft about which the flyballs revolve will be
denoted by θ. Assuming that the angular velocity of the vertical shaft and the rotational
speed of the engine have the same value, , there is a centrifugal acceleration acting on the
flyballs in the outward direction of magnitude 2 L sin θ (see Figure 5.P.5).
Ω
θ
mΩ 2L sin θ cos θ
L
m
–mg sin θ
mΩ 2L sin θ
–mg
FIGURE 5.P.5
The angle of deflection of the centrifugal pendulum is determined by the
opposing components of gravitational force and centrifugal force.
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Recall from calculus that the magnitude of this acceleration is the curvature of the motion,
1/L sin θ, times the square of the tangential velocity, 2 L 2 sin2 θ. Taking into account the
force due to gravitational acceleration and assuming a damping force of magnitude γ θ , we
obtain the following equation for θ:
m Lθ = −γ θ − mg sin θ + m2 L sin θ cos θ.
(1)
Equation (1) results from equating components of inertial forces to impressed forces parallel
to the line tangent to the circular arc along which the flyball moves in the vertical plane
determined by the pendulum arm and the vertical shaft (Problem 1).
Angular Velocity of the Steam Engine Flywheel and Drive Shaft. The equation for
the rotational speed of the flywheel and drive shaft assembly of the steam engine is assumed
to be
d
J
= −β + τ,
(2)
dt
where J is the moment of inertia of the flywheel, the first term on the right is the torque due
to the load, and the second term on the right is the steam generated torque referred to the
drive shaft.
Linearization About the Operating Point. In order to use the feedback control concepts
of Section 5.9, it is necessary to determine the linear equations that are good approximations
to Eqs. (1) and (2) when θ and are near their equilibrium operating points, a mathematical
technique known as linearization. The equilibrium operating point of the steam engine will
be denoted by 0 , the equilibrium angle that the flyball connecting arm makes with the
vertical will be denoted by θ 0 , and the equilibrium torque delivered to the engine drive shaft
will be denoted by τ 0 . Note that in the equilibrium state, corresponding to θ = 0, θ = 0,
and = 0, Eqs. (1) and (2) imply that
g = L20 cos θ0 .
(3)
τ0 = β0 .
(4)
and
To linearize Eqs. (1) and (2) about θ 0 , 0 , and τ 0 , we assume that θ = θ0 + φ, = 0 + y,
and τ = τ0 + u, where φ, y, and u are perturbations that are small relative to θ 0 , 0 , and
τ 0 , respectively. Note that φ represents the error in deflection of the flyball connecting arm
from its desired value θ 0 , in effect, measuring or sensing the error in the rotational speed of
the engine. If φ > 0, the engine is rotating too rapidly and must be slowed down; if φ < 0,
the engine is rotating too slowly and must be sped up. Substituting the expressions for θ,
, and τ into Eqs. (1) and (2), retaining only linear terms in φ, y, and u, and making use of
Eqs. (3) and (4) yield
φ + 2δφ + ω02 φ = α0 y
(5)
J y = −βy + u,
(6)
and
where δ = γ /2m L, ω02 = 20 sin2 θ0 , and α0 = 0 sin 2θ0 (Problem 3).
The Closed-Loop System. Regarding the error in rotational speed, y, as the input, the
transfer function associated with Eq. (5) is easily seen to be
α0
G(s) = 2
.
s + 2δs + ω02
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Chapter 5
October 21, 2010
The Laplace Transform
Thus (s) = G(s)Y (s), where (s) = L{φ(t)} and Y (s) = L{y(t)}. Regarding u as the
input, the transfer function associated with Eq. (6) is
H (s) =
1
Js + β
so that Y (s) = H(s)U(s), where U (s) = L {u(t)}. The closed-loop system is synthesized by
subtracting (s) at the summation point, as shown in Figure 5.P.6.
F +
Σ
K
FIGURE 5.P.6
Y
H
–
G
Block diagram of the feedback control system corresponding to the Watt
governor linearized about the operating point.
Thus a positive error in deflection of the flyball connecting arm causes a decrease in steam
generated torque and vice versa. A proportional gain constant K has been inserted into
the feedback loop to model the sensitivity of the change in steam generated torque to the
error (s) in the deflection of the flyball connecting arm. Physically, K can be altered by
changing the location of the pivot point of the lever arm that connects the sliding collar on
the governor to the vertical rod attached to the throttle valve. Note that any external input
affecting engine speed is represented by F(s).
Project 3 PROBLEMS
In the following problems, we ask the reader to supply
some of the details left out of the above discussion, to
analyze the closed-loop system for stability properties,
and to conduct a numerical simulation of the nonlinear
system.
1. Work out the details leading to Eq. (1).
2. Give physical explanations for the meaning of
Eqs. (3) and (4).
3. Derive the linearized system (5) and (6).
4. Show that the transfer function of the closed-loop
system linearized about the operating point is
HK (s) =
s 2 + 2δs + ω02
.
(J s + β)(s 2 + 2δs + ω02 ) + K
5. Use the Routh criterion to show that if the gain
factor K is sufficiently large, H K (s) will have two
poles with positive real parts and the corresponding
closed-loop system is therefore unstable. Derive an
expression for K c , that value of K at which the real
parts of a pair of conjugate complex poles of H K (s)
are equal to 0.
6. The Nonlinear Feedback Control System. Using
the relations (3) and (4), show that Eqs. (1) and (2)
can be expressed in terms of φ and y as
m Lφ = −γ φ − m20 L cos(θ0 ) sin(θ0 + φ)
+ m(0 + y)2 L sin(θ0 + φ) cos(θ0 + φ)
(i)
and
J y = −βy − K φ,
(ii)
where the negative feedback loop has been incorporated into the nonlinear system.
7. Simulations. Consider the following parameter
values expressed in SI units:
m = 12,
J = 400,
L = 12 ,
β = 20,
g = 9.8,
θ0 = π/6,
γ = 0.01,
with 0 determined by the relation (3).
(a) Using the above parameter values, construct
a root locus plot of the poles of H K (s) as K
varies over an interval containing K c . Verify that
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a pair of poles cross from the left half plane to
the right half plane as K increases through K c .
(b) Conduct computer simulations of the system
(i), (ii) in Problem 6 using the above parameter values. Do this for various values of K less
than K c and greater than K c while experimenting with different values of y(0) = 0 to represent
departures from the equilibrium operating
point. For simplicity, you may always assume that φ(0) = φ (0) = 0. Plot graphs of the
393
functions φ(t) and y(t) generated from the simulations. Note that if φ(t) wanders outside the
interval [−θ0 , π − θ0 ], the results are nonphysical. (Why?) Are the results of your simulations
consistent with your theoretical predictions?
Discuss the results of your computer experiments addressing such issues as whether the
feedback control strategy actually works, good
choices for K, stability and instability, and the
“hunting” phenomenon discussed above.
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