Question

The radius of rotation of the balls of a Hartnell governor is 80 mm at the minimum speed of 300 r.p.m. Neglecting gravity effect, determine the speed after the sleeve has lifted by 60 mm. also determine the initial compression of the spring, the governor effort and the power. The particulars of the governor are given below: Length of ball arm = 150 mm, length of sleeve arm = 100 mm, mass of each ball = 4 kg and stiffness of the spring = 25 N/mm.

Answer 1

Given r₁ =80 mm = 0.08 m; N₁ = 300 r.p..m. or ω₁ = 2π x 300/60 = 31.42 rad/s ; h = 60 mm; x = 150 mm = 0.15 m; y = 100 mm = 4 kg ; s = 25 N/mm

The minimum and maximum position of the governor is fig. (a) and (b) respectively. First of all, let us find the maximum radius of rotation(r₂). We known that lift of the sleeve,

Maximum speed of rotation

Let N₂ = Maximum speed of rotation, and

S₁ and S₂ = Spring force at the minimum and speed respectively, in newtons.

We know centrifugal force at the minimum speed,

Now taking moments about the fulcrum O of the bell crank lever when in minimum position as shown in Fig (a). The gravity effect is neglected, i.e. the moment due the weight of balls, sleeve and the bell crank crank lever arms is neglected.

 We know that S₂ - S₁ = h.s or S₂ = S₁ + h.s 

= 948 + 60 x 25 = 2448 N

We know that centrifugal at the maximum speed,

Now taking moments about the fulcrum O when in maximum, as shown in fig (b),

Initial compression of the spring

We know that initial compression of the spring

= S₁/s = 948/25 = 37.92 mm

Governor effort

We know the governor effort,

P = S₂ - S₁/2 = 2448 - 948/2 = 750 N

Governor Power 

We know that the governor power

= P x h= 750 x 0.06 = 45 N-m