A flyball governor is an ingenious mechanical device designed to stabilize the r

Question

A flyball governor is an ingenious mechanical device designed to stabilize the rotational speed of an

engine. It is connected, directly or via a gear box, to the rotating shaft, and spins at a uniform

angular speed ? . The governor consists of two equal

weights of mass m, and a third heavier weight mass M. They

are linked by hinged bars, length l, and of negligible mass.

The weight M can slide up and down the central rod without

friction. As mechanical engineers, we wish to establish the

relationship between the angle ? and the angular speed ?

in terms of the various masses and lengths.

a) Explain conceptually how the governor works and

how it helps stabilize the rotational speed of the

engine

b) Using Newton

A flyball governor is an ingenious mechanical device designed to stabilize the rotational speed of an engine. It is connected, directly or via a gear box, to the rotating shaft, and spins at a uniform angular speed ? . The governor consists of two equal weights of mass m, and a third heavier weight mass M. They are linked by hinged bars, length l, and of negligible mass. The weight M can slide up and down the central rod without friction. As mechanical engineers, we wish to establish the relationship between the angle ? and the angular speed ? in terms of the various masses and lengths. Explain conceptually how the governor works and how it helps stabilize the rotational speed of the engine Using Newton's Laws, find the relationship between the variables (tensions T1and T2 should not be part of your answer) Show that this relationship can be written in the following dimensionless form (i.e. both sides of the equations are pure numbers) (m/(m + M))(l/g)?^2 = tan?/(sin? + a/l) Assess this relationship in terms of the effect of masses and rotational speed Which condition(s) on ? would allow this equation to be solved algebraically? Find ? One elegant way to solve this equation for any values of ? is to use a graphical method PRECISELY graph the function f (? ) = tan?/(sin? + (a/l)) for the value a/l = 0.10 . To help out, use an angular domain ? ?[0,80deg] and a range of [0,4]for your plot. (You are encouraged to use an Excel spreadsheet, or the likes) If ? = 300RPM , m=1kg, M=10kg, l=20cm, find the equilibrium angle using your plot

Explanation / Answer

a)

When the load on the engine increases, the engine

and the governor speed decreases. This results in

the decrease of centrifugal force on the balls. Hence

the balls move inwards and the sleeve moves downwards.

The downward movement of the sleeve operates

a throttle valve at the other end of the bell crank

lever to increase the supply of working fluid and thus

the engine speed is increased. In this case, the extra

power output is provided to balance the increased load.

When the load on the engine decreases, the engine and

the governor speed increases, which results in the increase

of centrifugal force on the balls. Thus the balls

move outwards and the sleeve rises upwards. This upward

movement of the sleeve reduces the supply of the

working fluid and hence the speed is decreased. In this

case, the power output is reduced.

b)

Centrifugal force on ball = mrw^2........where r = distance of ball from governor axis.

So r = a + L*sin alpha

Instantaneous centre I can be found by extending rod T1 until it touches the horizontal line. Takng moments about I,

mrw^2 *(L*cos alpha) = mg*(L*sin alpha) + (Mg / 2)*(2*L*sin alpha)

By simplifying, mrw^2 = (m + M)*g*(Tan alpha)

c)

m*(a + L*sin alpha)*w^2 = (m + M)*g*(Tan alpha)

(m / (m + M))*(1 / g)*w^2 = Tan alpha / (a + L*sin alpha)

(m / (m + M))*(L / g)*w^2 = Tan alpha / (a/L + sin alpha)

d)

As m / (m+M) increases , w will decrease.

e)

When a/L = 0, we'll get (m / (m + M))*(L / g)*w^2 = Tan alpha / (sin alpha)

(m / (m + M))*(L / g)*w^2 = Sec alpha

f)

i)

ii)

w = 2*pi*N/60 = 2*3.14*300/60 = 31.4 rad/s

(m / (m + M))*(L / g)*w^2 = Tan alpha / (a/L + sin alpha)

m = 1 kg, M = 10 kg, L = 20 cm = 0.2 m, a / L = 0.1

(1 / (1 + 10))*(0.2 / 9.81)*31.4^2 = Tan alpha / (0.1 + sin alpha)

Solving this alpha = 60.6 deg

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