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Find :

A governor of 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.

Find

1. The initial compressive force on the spring, If the speed for an initial ball radius of 200 mm is 240 rpm; and
2. 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: 10 Marks

Year: May 2016

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Type of governor: hartnell

$m= 30 kg \\ r_1 =200 mm = 0.2 m \\ M= 6kg \text{(Assumed not given)} \\ N_1 = 240 rpm \\ ω_1 = \dfrac{2П N}{60} = \dfrac{25.1327 rad}{sec} \\ \text{speed fluctuation} = 7 \%$

To find : the initial compress force:

Spring stiffness $F_{c1} = m * r_1 * ω_1^2 \\ = 3*0.2*25.1327 \\ F_{c1} = 378.9915 N$ $Mg + S_2 = 2a /b * F_{c2} \\ 6* 9.81 +S_2 = 2*0.11*444.366 \\ = 651.7368 \\ S_2 = 592.87N \\ \text{Spring stiffness} = S_2 – S_1 /h \\ k_S = (592.87 – 496.99 )/0.004 \\ k_S =23970 N/m \\ k_S = 23.970 * 103 N/m \\ k_S = 23.970 N/mm \\ h = b[r_2-r_1]/a \\ 0.004 = 0.15 [r_2-r_1]/0.11 \\ 2.933 = r_2 – 0.2 \\ r_2 =0.20293$

$FCR = m * r_2 * ω_2 \{ ω_2 =1+0.075=1.075 *240=258rpm \\ ω+2 =2П N / 60 = 27.017 \} \\ = 3*(0. 20293) * (27.017)2 \\ FCR =444.366 \\ \text{Calculating the initial spring force stiffness of the spring } \\ Mg + S_1 = 2a /b * F_{c1} \\ 6* 9.81 +S_1 = (2*0.11*378.99)/0.15 \\ S_1 = 496.99 N$

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