0
3.4kviews
In an engine governor of the porter type, the upper and lower arms are $200mm$ and $250mm$ respectively and pivoted on the axis of rotation. The mass of the central load is $15kg.$
1 Answer
0
443views

The mass of each ball is $2 \ kg$ and friction of the sleeve together with the resistance of the operating gear is equal to a load of $25 \ N$ and sleeve. If the limitations inclinations of the upper arms to the vertical are 30° and 40° Find, taking friction into account, range of speed of the governor.

enter image description here

$m = 2kg$

$M = 15 \ kg$

$f = 25 \ N$

A] $\alpha = 30°$

$cos \ 30° = \frac{h_1}{0.2}$

$h_1 = 0.173$

$sin \beta = \frac{r_1}{0.25}$

$= \frac{0.1}{0.25}$

$\beta = 23.58°$

$sin 30° = \frac{r_1}{0.2}$

$r_1 = 0.1m$

$q = \frac{tan \ 23.58}{tan \ 30} = 0.7559$

$0.173 = \frac{895}{N_1^2} [ 1 + \frac{15 \times 9.81 - 25}{2 \times 9.81 \times 2} \times (1 + 0.7531)]$

$N_1 = 182.77 \ rpm$

$\alpha =$ 40°

$cos \ 40° = \frac{h_2}{0.2}$

$h_2 = 0.1532$

$sin \beta = \frac{r_2}{0.250} = \frac{0.1286}{0.250}$

$\beta = 30.946°$

$sin \ 40° = \frac{r_2}{0.2}$

$r_2 = 0.1286$

$q = \frac{tan \ 30.946}{tan \ 40} = 0.7101$

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

$0.1532 = \frac{895}{N_2^2} [ 1 + \frac{15 \times 9.81 + 25}{2 \times 2 \times 9.81} (1 + 0.7101)]$

$N_2 = 222.87 \ rpm$

Range = $N_2 – N_1 = 222.87 – 182.77$

$= 40.1 r.p.m$.

Please log in to add an answer.