In the porter governor links and arms are each $300 \ mm$ long. Each ball have mass $2.5 \ kg$ and central load is $250 \ N$ for the highest and lowest position of sleeve the arms are inclined @ 40° and 30° to vertical. The friction at the governor and the mechanism connected to sleeve is equivalent in force of $25 \ N$. Assuming that links and arms are intersecting @ the axis. Find the minimum and maximum speed and range of governor.

Solution:

$m = 2.5 \ kg = 24.525 \ N$

$W = Mg = 250 \ N$

$F = 25 \ N$

A] When $\alpha$ = 30° , $N_1 = ?$

$cos \ 30° = \frac{h_1}{0.3}$ $h_1 = 0.2598 \ m$

$\therefore$ $h_1 = \frac{895}{N_1^2} [ 1 + \frac{Mg – f}{2 mg} (1 + q)]$

$0.2598 = \frac{895}{N_1^2} [ 1+ \frac{250 – 25}{2 \times 24.525} (1 + 1)]$

$N_1 = 187.45 \ rpm.$

B] When $\alpha$ = 40° then $N_2$ = ?

$cos \ 40° = \frac{h_2}{0.3}$

$h_2 = 0.2298 \ m$

$\therefore$ $h_2 = \frac{895}{N_2^2} [ 1+ \frac{Mg + f}{2 mg} (1 + q)]$

$\therefore$ $0.2298 = \frac{895}{N_2^2} [ 1+ \frac{250 + 25}{2 \times 24.525} (1 + 1)]$

$N_2 = 218.09 \ rpm$

Range $= 218.09 \ – \ 187.45$

$= 30.646 \ rpm$.