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Assuming that the power of the governor is sufficient to overcome the friction by 1% change of speed on each side of mean position.

The arms of Hartnell governor are of equal length. When the sleeve is in the mid position, the masses rotate in a circle with diameter of 150mm (the arms are vertical in the mid position). Neglecting friction, the equilibrium speed for this position is 360rpm. Maximum variation/ of speed taking friction into account, is to be 6% of the midposition speed for a maximum sleeve movement of 30mm. The sleeve mass is 5kg and the friction of sleeve is 35N. -
Assuming that the power of the governor is sufficient to overcome the friction by 1% change of speed on each side of mean position. Find (neglecting obliquity effects of the arms), the
- (i) mass of each rotating of ball. - (ii) spring stiffness. - (iii) initial compression of the spring. -

Mumbai University > MECH > Sem 5 > Theory Of Machines 2

Marks: 12 M

Year: May 2015

1 Answer
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Given:

w= 360 rpm

m = 5kg

F = 35N

X= 30mm

To find:

M, K and x1

Fig: Hartell governer considering only the left side.

$$\text{Fig: Hartell governer considering only the left side.}$$


    1. Mass

      It is said that 1% change in speed can overcome friction of 35N. In this case only friction will be balanced and not mg and S. hence mg and S will be 0

      That is, at the rpm of 1.01 x 360 = 363.6 rpm

      Taking moment about hinge I1

      2 Fc x A = F x A

      2Fc = F

      $2M r w^2 = 35 \\ 2M \times 0.075 \times (2π \times 363.6/60)^2 = 35 \\ M = 0.16 kg$

    2. Spring stiffness

      When the speed has varied by 6%, the sleeve has risen 30mm.

      So N2 = 1.06 x N = 381.6rpm

      $W2 = 2πN_2/60 = 40$

      The force on sleeve S can again be found by taking moment about I1

      $2Fc \times A = -(F + mg + S) \times A \\ 2 \times M \times r \times w_2^2 = -(35 + 5 \times 9.8 + S) \\ 2 \times 0.16 \times 0.075 \times 402 = -(35 + 5 \times 9.8 + S) \\ S = 122.4 N$

      Since the sleeve rises by 30 mm or 0.03m

      K = S/x = 122.4/0.03 = 4080 N/m

    3. Initial compression

      Since equilibrium speed is 300 rpm, at this speed the sleeve collar will not rise and only active forces will be F and S

      W = 2πN/60 = 31.42 rad/s

      $2Fc \times A = -(F + S1) \times A \\ 2 \times M \times r \times w2 = -(35 + S1) \\ 2 \times 0.16 \times 0.075\times 31.422 = -(35 + S1) \\ =58.69 N$

      So, initial compression required will be

      K = S1 / x1

      X1 = F / K = 58.69 / 4080 = 0.014 m = 14cm

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