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The TMD for a petrol engine is drawn to the following scales TM, 1mm=5Nm crank angle 1m=1degree .
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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.

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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%

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