written 5.8 years ago by |
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 mm2. 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.
Scale: TM → 1 mm = 5N – m
Crank angle → 1 mm=1×π180 {Given}
∴ Area=5×π180=π36 mm2
Let,
Energy @ A=E
B=E+295 - - - - Max energy = E1
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%