0
911views
The TMD for a petrol engine is drawn to the following scales TM, $1mm = 5N m$ crank angle $1 m = 1 degree$ .
1 Answer
0
10views

The TMD repeats itself at every half revolution of the engine and the areas above and below the mean TM line taken in order are: $295, 685, 40, 340, 960, 270\ mm^2$. The rotating parts are equivalent to a mass of $36\ kg$ at a radius of gyration of $150\ mm$. Determine the coefficient of fluctuation of speed when the engine runs at $1800\ rpm$.

enter image description here

Scale: TM $\rightarrow$ 1 mm = 5N – m

Crank angle $\rightarrow$ $1 \ mm = 1 \times \frac{\pi }{180}$ {Given}

$\therefore$ $Area = 5 \times \frac{\pi}{180} = \frac{\pi}{36} \ mm^2$

Let,

Energy @ $A = E$

$B = E + 295$ - - - - Max energy = $E_1$

$C = (E + 295) – 685 = E – 390$

$D = (E – 390) + 40 = E – 350$

$E = (E – 350) – 340 = E – 690$ - - - Min energy = $E_2$

$F = (E – 690) + 960 = E + 270$

$G = (E + 270) – 270 = E$

$\rightarrow \triangle E = E_1 – E_2$

$= (E + 295) – [E – 690]$

$= 985 \ mm^2$

$= 985 \times \frac{\pi}{36}$

$\triangle E = 85.95 \ N - m$

$\triangle E = mk^2 (\frac{\pi^2}{900}) N^2 \ C_s$

$\therefore$ $85.95 = 36 (0.150)^2 (\frac{\pi^2}{900}) (1800)^2$

$\therefore$ $C_s = 0.003 = 0.3$%

Please log in to add an answer.