Power Plants Storing Energy in Massive Flywheels

AI Thread Summary
Power plants can optimize energy storage by generating mechanical energy during off-peak hours and storing it in flywheels. A specific problem involves calculating the diameter of an iron flywheel needed to store 14.4 MJ of kinetic energy at 90 rpm. The calculations involve determining the volume, mass, moment of inertia, and kinetic energy equations. The initial attempt yielded a diameter of 14.1 meters, but the solution was marked incorrect, prompting a request for advice. Participants emphasized the importance of checking angular velocity conversions and ensuring unit consistency in calculations.
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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