Question 23.1: The turning moment diagram for a petrol engine is drawn to t......
The turning moment diagram for a petrol engine is drawn to the following scale:
Along x-axis crank angle is taken with a scale as 1 mm = 1 ° and along y-axis, turning moment is taken with a scale 1 mm = 10 N·m
The areas above and below the mean turning moment line, taken in order, are – 400, +800, -340, +900, and -650 mm²
The nominal speed of the engine is 300 rad/s. Determine (a) the mass of the flywheel
(b) Dimension of rim if width is twice of the thickness. Assume diameter of flywheel rim is 300 mm, density of wheel rim material =7000 \mathrm{~kg} / \mathrm{m}^3, C_s=0.3 \%.
The turning moment diagram is shown in Figure 23.3.
Conversion from given scale:
1 \mathrm{~mm}=10 \mathrm{~N} \cdot \mathrm{m} \text { and } 1 \mathrm{~mm}=1^{\circ}=\frac{\pi}{180} \text { radian }∴ 1 mm2 on turning moment diagram= 10 \times \frac{\pi}{180}=0.1745 \mathrm{~N} \cdot \mathrm{m}
Let at A, energy be = E
From Figure 23.2
Energy at B=E-400
Energy at C=E-400+800=E+400
Energy at D=E+400-340=E+60
Energy at E=E+60+900=E+960
Energy at F=E+960-650=E+310
Hence, the maximum energy =E+960
the minimum energy =E-400
∴ Maximum fluctuation of energy
\Delta E=E_2-E_1=(E+960)-(E-400)
=1360 \mathrm{~mm}^2
=1360 \times 0.1745=237.32 \mathrm{~N} \cdot \mathrm{m}
The maximum fluctuation of energy is also obtained from,
\Delta E=m R^2 \omega^2 C_swhere m = mass, and R = radius of flywheel
237.32=m \times\left(\frac{150}{1000}\right)^2 \times(300)^2 \times\left(\frac{0.3}{100}\right)Solving, we get
m=39.06 \mathrm{~kg}Cross-section of rim
Let t be the thickness, and b be the width = 2t,
Cross-sectional area A=b \times t=2 t^2
we know mass of rim
m=A \times(2 \pi R) \times p
39.06=2 t^2 \times 2 \pi \times\left(\frac{150}{1000}\right) \times 7000
Solving,
t=0.0544 \mathrm{~m}=54.4 \mathrm{~mm}
t=55 \mathrm{~mm} \text { (say) }
b=55 \times 2=110 \mathrm{~mm}
