# Flywheel inertia

## Homework Statement

A flywheel is to be designed for a single cylinder double acting piston engine. The crankshaft has a mean speed of 1000rpm and a maximum and minimum rpm of 0.3% above and below this respectively.

The torque deliverd to the crankshaft is 100 Nm from top dead centre to 60°, then falls steadily from 100 Nm to 20 Nm over the next 60°, then fixed at 20 Nm for the final 60° to bottom dead centre. The pattern repeats as the crank returns to top dead centre.

As the energy stored in a flywheel is 0.5 * I * ω^2 determine the mass moment of inertia, I, that the flywheel must have.

## Homework Equations

ΔE = 0.5 * I * (ω(MAX)^2 - ω(MIN)^2)

Where, ΔE is the area of the torque time diagram above the mean torque.

ω(MAX) = 1003 rpm = 105.034 rad/s

ω(MIN) = 997 rpm = 104.406 rad/s

## The Attempt at a Solution

I determined the mean torque over 360° to be 60Nm by drawing a torque-time diagram with the triangular areas were the torque varies from 100 Nm to 20Nm drawn as rectangular areas and then sketching a line roughly through the middle and equated the areas above this line to the missing area below the line to determine the height of the line.

The diagrams were drawn using radians for the crank angle.

Then, I'm assuming that the flywheels moment of inertia should be large enough to absorb the energy when the torque is above 60Nm (the area on the diagram) and then supply this energy when the torque is below 60 Nm in order to keep the rpm within the range given.

So I calculated the area above 60 Nm to be ΔE = 146.64 and then using the equation given above, I get an inertia of 2.23 kg.m^2 but the answer is given as 0.793 kg.m^2

Using the answer given with that formula I get an area of ΔE = 521.77, which is bigger than the total area of the torque time graph.

Obviously, I'm doing something wrong or the answer is wrong.

Thanks for any help.

NascentOxygen
Staff Emeritus