written 5.6 years ago by |
Numerical:
$m = 5\ kg$
$M = 25\ kg$
$F = 10\ N$
$q = 1$
[A] When, $r_1 = 150\ mm$, $N_1 = ?$
$h_1 = \sqrt{250^2 – 150^2} = 0.2\ m$
$0.2 = h_1 = \frac{895}{N_1^2} [ 1 + \frac{25}{2 \times 5} (1 + 1)]$
$N_1 = 164 \ rpm.$
For $r_2 = 200\ mm$, $N_2 = ?$
$h_2 = \sqrt{ 250^2 – 200^2}$
$h_2 = 0.15\ m$
$h_2 = \frac{895}{N_2^2} ( 1+ \frac{M}{2m} (1 + q))$
$0.15 = \frac{895}{N_2^2} (1 + \frac{25}{2 \times 5} (1 + 1))$
$N_2 = 189.208\ rpm.$
Range = $N_2 – N_1$
$= 25.208\ rpm.$
Sleeve lift = $x = 2 (h_1 – h_2)$
$= 2(0.2\ – \ 0.15)$
$= 100\ mm = 0.1\ m$
Case 1: Neglect friction force:
$C = \frac{N_2 - N_1}{N_1}$ $= \frac{Range \ of \ speed}{Minimum \ speed \ of \ rotation}$
$C = 0.152$
$p = C[mg + Mg]$
$= 0.192 (5 \times 9.81 + 25 \times 9.81)$
$P = 44.7336\ N$
$Power = p \times x$
$= 44.7336 \times 0.1$
$Power = 4.47\ N.m$
Case 2: Friction force of 10N:
1] for $r_1 = 150\ mm, \ N_1 = ?$
$h_1 = 0.2\ m$
$0.2 = \frac{895}{N_1^2} [ 1+ \frac{25 \times 9.81 – 10}{2 \times 5 \times 9.81} (1+1)]$
$N_1 = 161.05\ rpm.$
2] for $r_2 = 200\ mm, \ N_2 = ?$
$h_2 = 0.15\ m$
$\therefore$ $0.15 = \frac{895}{N_2^2} [ 1+ \frac{25 \times 9.81 + 10}{2 \times 5 \times 9.81} \times (1+1)]$
$N_2 = 192.3965 \ rpm$
Range = $N_2 – N_1 = 31.3465 \ rpm.$
$\rightarrow$ sleeve lift $= x = 0.1\ m$
$\rightarrow$ $C = \frac{N_2 – N_1}{N_1} = 0.195$
$\rightarrow P = c (mg + Mg + F)$
$= 0.195 [ 5 \times 9.81 + 25 \times 9.81 + 10]$
$P = 55.3385\ N$
$\rightarrow power = p \times x$
$= 55.3385 \times 0.1$
$Power = 5.53\ N.m$