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The turning moment diagram for a four stroke gas engine may be assumed for simplicity to be represented by four triangles, the areas of which from the line of zero pressure are as follows:

The turning moment diagram for a four stroke gas engine may be assumed for simplicity to be represented by four triangles, the areas of which from the line of zero pressure are as follows: - Suction stroke $=0.45 \times 10^{-3} m^2$ , compression stroke$= 1.7X 10^{-3} m^2$ , Expansion stroke $= 6.8 X 10^{-3}m^2$ ,v Exhaust stroke $= 0.65 X 10^{-3} m^2 Each $m^2$ of area represents 3 MN-m of energy. All the areas except expansion stroke are negative. Assuming the resisting torque to be uniform, Find the mass of the rim of a flywheel required to keep the speed between 202 and 198rpm. The mean radius of the rim is 1.2m. - Mumbai University > MECH > Sem 5 > Theory Of Machines 2

Marks: 10 M

Year: Dec 2014

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Given:

Maximum and minimum speed = 202 rpm and 198rpm

R= 1.2m

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Solution:

Mean speed = (202 + 198) / 2 = 200

Therefore, fluctuation of speed = +/- 1% …i

But fluctuation of speed is given by

$\pm \bigg(\dfrac{\Delta E}{2I \omega 0^2}\times 100 \bigg) \%$…ii

Where

$∆E$ is energy above mean level.

I = moment of inertia = $MR^2 = M x 1.2^2 = M x 1.44$

Wo = 0.105 No = 0.105 x 200 = 21 rad/s

To find mean torque:

$A_{mean}$ $= ∑A / 4 \\ = (-0.45 – 1.7 + 6.8 – 0.65) x 10^{-3} / 4 = 10^{-3} m^2$

Therefore $T_{mean} = A_{mean} \times 3 MNm/m^2 = 3000 Nm$

So, $∆E = E_{\max} – E_{mean}$

But $E_{\max} = 6.8 m^2 = 6800 Nm$

Hence, ∆E = 3800 Nm

Using I and ii

$1 = ∆E/(2Iw^2) \times 100$

$1 = 3800 / (2 \times M \times 1.44 \times 21^2) x 100$

M = 299 kg

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