Hartnell Governor

The Hartnell governor is shown in Figure 14.7(a). The two bell crank levers are used which can rotate about fulcrums O and O′. One end of both bell crank lever carries a ball and a roller at the other end of the arm. The rollers make contact with the sleeve. A helical spring is mounted on the spindle between frame and sleeve. With the rotation of the spindle, all these parts rotate.

Figure 14.7 (a) Hartnell Governor and (b) Force Analysis of Hartnell Governor

With the increase in speed, the radius of rotation of the balls increases and the rollers lift the sleeve against the spring force. With the decrease in speed, the sleeve moves downwards. The movement of the sleeve is transferred to the throttle of the engine through linkages.

Let

r1	=	Minimum radius of rotation of ball centre from spindle axis, in m
r2	=	Maximum radius of rotation of ball centre from spindle axis, in m
S1	=	Spring force exerted on sleeve at minimum radius, in N
S2	=	Spring force exerted on sleeve at maximum radius, in N
m	=	Mass of each ball, in kg
M	=	Mass of sleeve, in kg
N1	=	Minimum speed of governor at minimum radius, in rpm
N2	=	Maximum speed of governor at maximum radius, in rpm
ω1 and ω2	=	Corresponding minimum and maximum angular velocities, in rad/s
(FC)1	=	Centrifugal force corresponding to minimum speed
(FC)2	=	Centrifugal force corresponding to maximum speed
s	=	Stiffness of spring or the force required to compress the spring by 1 m
r	=	Distance of fulcrum O from the governor axis or radius of rotation
a	=	Length of ball arm of bell crank lever, i.e., distance OA
b	=	Length of sleeve arm of bell crank lever, i.e., distance OC

 

Considering the position of the ball at radius ‘r₂’ as shown in Figure 14.7(b) and taking the moments of all the forces about O′.

If θ₁ and θ₂ are small. Weight w is small in comparision to W and spring force S, w tan θ₁ and w tan θ₂ can be neglected.

Example 14.5: A Hartnell governor having a central sleeve spring and two right angled bell crank levers moves between 280 and 320 rpm for a sleeve lift of 15 mm. The sleeve arms and the ball arms are 100 and 120 mm, respectively. The levers are pivoted at 120 mm from the governor axis and mass of each ball is 4 kg. The ball arms are parallel to the governor axis at the lowest equilibrium speed. Determine load on spring at lowest and highest speeds and stiffness of the spring.

Solution:

Let

S₁ = Load on spring at lowest speed

S₂ = Load on spring at highest speed

 

Since the ball arms are parallel to governor axis at the lowest speed, therefore, r = r₁ = 120 mm

Let r₂ is radius of rotation at N₂ = 320 rpm.