Power Plants Storing Energy in Massive Flywheels

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Homework Help Overview

The discussion revolves around the concept of energy storage in power plants using flywheels. The original poster presents a problem involving the calculation of the diameter of a flywheel designed to store a specific amount of kinetic energy when spinning at a given angular velocity. The subject area includes mechanics and energy storage systems.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the diameter of the flywheel using given parameters and equations related to kinetic energy and moment of inertia. Some participants question the conversion of angular velocity from revolutions per minute to radians per second, while others confirm the conversion process.

Discussion Status

The discussion is ongoing, with participants exploring the calculations and questioning the original poster's approach. There is no explicit consensus on the correctness of the calculations, and some guidance has been offered regarding unit conversions.

Contextual Notes

Participants note that the online homework system provides specific units for answers, which may be relevant to the problem's resolution. There is an indication that the original poster's calculations may not align with the expected format or units required by the homework system.

Treefolk
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Homework Statement


It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 10.6 cm.

What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.4 MJ when spinning at an angular velocity of 90.0 rpm about an axis perpendicular to the disk at its center?

Homework Equations


V = h\pir^2 <--Volume of a cylinder
I = .5mr^2 <-- Moment of an inertia of a solid cylinder
KE = .5I\omega^2 <-- Kinetic Energy of a Cylinder


The Attempt at a Solution


\delta = 7800
h = .106m
m = \delta*V = 7800*.106*\pi*r^2 = 2597.5r^2
I = .5(2597.5r^2)r^2 = 1298.7r^4
KE = .5(1298.7r^4)\omega^2 = 649.4r^4\omega^2
\omega = 90rpm = 1.5rps = 3rad/s
KE = 649.4r^4(3)^2
KE = 14.4x10^6
14.4x10^6 = 649.4r^4*(3)^2
r^4 = 2463.9
r=7.05
diameter = 14.1

That sorts through my logic, but the online homework spits back the answer as wrong, so I'm rather at a loss for what I did wrong. Any advice would be quite welcome.
 
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Hi Treefolk! :smile:

(try using the X2 button just above the Reply box :wink:)
Treefolk said:
\omega = 90rpm = 1.5rps = 3rad/s

erm … :biggrin:
 
What about the angular velocity?
90 revolutions/minute => 1.5 Revolutions/Second => 3 Rad/s

Or did I muck this up?
 
yup! :smile:

one revolution = 2π radians :wink:
 
Treefolk said:
diameter = 14.1

Furlongs?
 
tiny-tim said:
yup! :smile:

one revolution = 2π radians :wink:

That would do it...lemme check that.

Borek said:
Furlongs?

Meters, the units weren't the issue here (online homework provides the unit next to the blank or asks for the units in the problem). Thank you anyways.
 
Treefolk said:
Meters, the units weren't the issue here (online homework provides the unit next to the blank or asks for the units in the problem). Thank you anyways.

We don't see the units you were shown, so we can't rule them out as a problem. Good practice to include them always :smile:
 

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