Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 5 Marks

Year: May 2016

1. Sensitivity : A governor is said to be sensitive if it readily responds to any change in the speed. Further , it maybe defined as the ratio of the difference between max and min equilibrium speeds to the mean speeds

   Sensitivity λ = the difference between max and min equilibrium speeds/ the mean speed

   λ = N2- N1/N

   λ = N2- N1/[( N2+ N1)/2]

   where N1 is the minimum equilibrium speed in rpm

   and N2 is the maximum equilibrium speed in rpm

   N mean speed in rpm.

2. Stability : A governor is said to be stable when for every speed within working range the flyballs occupy the defined specified position , it will have only one radins of rotation which the governor in equilibrium.

3. Isochronous governor : When the equilibrium speed of the governor for all radii of rotation of the balls within the working range by neglecting the friction then the governor is called as Isochronous governor.

4. Coefficient of Insensitiveness: the frictional force is always present in all mechanism as well as in all governors , it may not be excluded . This frictional force acts in the opposite direction of the motion when speed of rotation decreases. The friction prevents the downward movement of the sleeve and radially inward movement of the ball.

   Coefficient of Insensitiveness = ( N´´-N´ )/N

   Where ,

   N = equilibrium speed at some radians

   N´´=maximum speed at some radians

   N´=minimum speed at some radians

5. The sleeve distance from axis= 3.76 cm = 0.0376 m

6. Length of the link = 30 cm =0.3m

7. The weight is 60 N

   W = m *g

   m = W /g = 60/ 9.81

   m = 1.01 kg

8. Load on the sleeve = 480 N = 480 / 9.81 = 48.92 kg

9. The extreme radia of rotation of the governor are 20 cm and 25 cm

Case 1) when the radius of rotation is 20 cm =0 .2 m

From triangle BPE,

$PB^2 = BE^2 + PE^2 \\ PE = \sqrt{PB^2-BE^2} \\ h = \sqrt{0.3^2+0.25^2} \\ h= 0.1658 \\ \tan α = \dfrac{BE}{PE} =\dfrac{r}{h} =\dfrac{0.25}{0.1658} = 1.507$

$BD$ $= BE –DE \\ = 0.25 – 0.0376 \\ = 0.2124$

Now consider, triangle BDC,

$DC = \sqrt{BC^2-BD^2} \\ DC = \sqrt{0.3^2-0.2124^2} \\ DC = 0.2118 \\ \tan ß = \dfrac{BD}{DC} = \dfrac{0.2124}{0.2118} = 1.0028 \\ q = \dfrac{\tan ß}{\tan α} =\dfrac{1.0028}{1.507} = 0.6654 \\ N^2_2 = (m + M(1+q)*g*91.2/m*n*2 \\ N^2_2 =(1.01+48.92[1+0.6654]/m*2)*9.81*91.20.1658 \\ N^2_2 = 136077.67 \\ N_1 = 368.88 rpm$

Therefore, equilibrium speeds are 402.07 and 368.88 rpm for 20 and 25 cm radii respectively