Governors type Hartnell type has equal balls of mass 3kg. set initially at a radius of 200 mm. The arms of bell crank lever are 110 mm vertically and 150 mm horizontally. a) Find the initial compressive force on the spring. If the speed for an initial ball radius of 200mm is 240 rpm; and b) the stiffness of the spring required to permit a sleeve movement of 4mm on a fluctuation of 7.5% in engine speed.

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

Marks: 8 Marks

Year: May 2016

M = 4 kg

h = 40 mm = 40 * 10 -3 = 0.04 m

$N_1= 200rpm \\ \therefore ω_1 = 2П N_1 / 60 \\ = 2 * П *200 / 60 = 20.94 rad / sec \\ r_1 = 90 mm = 0.09m \\ a = 100 mm = 0.1 m \\ b= 80 mm = 0.08 m \\ r = 115 mm = 0.115 m$

Mean speed is 16 time the range of the speed

$\dfrac{[ N_1 + N_2 ]}{2} = 16 [ N_2 - N_1 ] \\ \dfrac{[200 +N 2 ]}{2} = 16 [N_2 - 200] \\ \therefore N_2 = ? \\ [200 +N_2] = 32 [N_2 - 200] \\ 6600= 31 N_2 \\ \therefore N_2 = 212. 90 rpm \\ \therefore ω_2 = 2П N 2 / 60 = 22.29 rad /sec$

Case 1:

Neglecting the friction calculating the initial

$\therefore h = b \dfrac{(r_2 - r_1 )}{ a} \\ 0.09 = \dfrac{0.08}{0.1 [r_2 – 0.09]} \\ \dfrac{0.04 * 0.1}{0.08} =[r_2 – 0.09] \\ 0.05 = [r2 – 0.09] \\ r_2 =0.14 \\ \therefore h = b \dfrac{r2 – 0.09}{a} \\ h_1= \dfrac{0.08}{0.1 [0.115– 0.09]} \\ h_1 =0.02 m$

But , $h = h_1 + h_2 \\ 0.04 = 0.02 + h_2 \\ h_2 = 0.02 m$

$F C_1 = m * r_1 * ω_1^2 = 4 * 0.09 * (20.14)2 = 146.02 N \\ F C_2 = m * r_1 * ω_2^2 = 4 * 0.14 * (22.29)2 = 278.23 N$

Step 2 – Calculating the spring force exerted on the sleeve.

$a_1 = \text{from fig a} , \\ a_1 = \sqrt{a^2+ßb^2} \\ = \sqrt{a^2+(r_1-r_2)^2} \\ =\sqrt{a^2+[0.115– 0.09]^2} \\ =0.096 km$

Similarly from fig (a)

$b_1 = \sqrt{h_1^2+b^2} \\ =\sqrt{0.02^2+0.08^2} \\ b_1 = 0.077 m$

consider fig b,

$a_2 = \sqrt{a^2-(r_1-r_2)^2} \\ =\sqrt{0.1^2-(0.14-0.115)^2} \\ a_2 = 0.0968 m \\ b_2 = \sqrt{b^2-h_2^2} \\ =\sqrt{0.08^2-0.02^2} \\ b_2 = 0.077$

For minimum radins position

$F_{c1} * a_1 = m*g * (r-r1) + [Mg + S_1]*b_1 / 2 \\ 146.02 * 0.096 = 4 * 9.81 * [0.115– 0.09]+[0 * S1]/2 \\ 146.02 * 0.096 = 0.981 + (S_1 /2 )*0.077 \\ S_1 = 338.62 N$

$F_{c2} * a_2 = m*g * (r2-r) = [Mg + S_2]*b_2 / 2 \\ 278.23 *0.0968 +9 * 9.81 [0.14 – 0.115]= (0+S_1 /2 )*0.077 \\ 27.912 = (S_1 /2 )*0.077 \\ \text{Calculate the initial compression of the spring} \\ ks = [S_2 - S_1]/N = [725.03 – 338.62]/0.04 \\ ks =96600 N / m \\ Si =S_1 / kS \\ =338.62/9660.25 \\ =0.033m$