a flywheel of mass 60 kg, radius 40cm is revolving 300 revolutions per min. what is its kinetic energy?
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Flywheel Kinetic Energy
Kinetic energy stored in a flywheel - moment of inertia
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A flywheel can be used to smooth energy fluctuations and make the energy flow intermittent operating machine more uniform. Flywheels are used in most combustion piston engines.
Energy is stored mechanically in a flywheel as kinetic energy.
Kinetic Energy
Kinetic energy in a flywheel can be expressed as
Ef = 1/2 I ω2 (1)
where
Ef = flywheel kinetic energy (Nm, Joule, ft lb)
I = moment of inertia (kg m2, lb ft2)
ω = angular velocity (rad/s)
Angular Velocity - Convert Units
1 rad = 360o / 2 π =~ 57.29578o1 rad/s = 9.55 rev/min (rpm) = 0.159 rev/s (rps)
Moment of Inertia
Moment of inertia quantifies the rotational inertia of a rigid body and can be expressed as
I = k m r2 (2)
where
k = inertial constant - depends on the shape of the flywheel
m = mass of flywheel (kg, lbm)
r = radius (m, ft)
Inertial constants of some common types of flywheels
wheel loaded at rim like a bicycle tire - k =1flat solid disk of uniform thickness - k = 0.606flat disk with center hole - k = ~0.3solid sphere - k = 2/5thin rim - k = 0.5radial rod - k = 1/3circular brush - k = 1/3thin-walled hollow sphere - k = 2/3thin rectangular rod - k = 1/2
Moment of Inertia - Convert Units
1 kg m2 = 10000 kg cm2 = 54675 ounce in2 = 3417.2 lb in2 = 23.73 lb ft2
Flywheel Rotor Materials
MaterialDensity
(kg/m3)
DesignStress
(MPa)SpecificEnergy
(kWh/kg)Aluminum alloy2700 Birch plywood70030 Composite carbon fiber - 40% epoxy15507500.052E-glass fiber - 40% epoxy19002500.014Kevlar fiber - 40% epoxy140010000.076Maraging steel80009000.024Titanium Alloy45006500.031"Super paper"1100 S-glass fiber/epoxy19003500.020
1 MPa = 106 Pa = 106 N/m2 = 145 psiMaraging steels are carbon free iron-nickel alloys with additions of cobalt, molybdenum, titanium and aluminum. The term maraging is derived from the strengthening mechanism, which is transforming the alloy to martensite with subsequent age hardening.
Example - Energy in a Rotating Bicycle Wheel
A typical 26-inch bicycle wheel rim has a diameter of 559 mm (22.0")and an outside tire diameter of about 26.2" (665 mm). For our calculation we approximate the radius - r - of the wheel to
r = ((665 mm) + (559 mm) / 2) / 2
= 306 mm
= 0.306 m
The weight of the wheel with the tire is 2.3 kg and the inertial constant is k = 1.
The Moment of Inertia for the wheel can be calculated
I = (1) (2.3 kg) (0.306 m)2
= 0.22 kg m2
The speed of the bicycle is 25 km/h (6.94 m/s). The wheel circular velocity (rps, revolutions/s) - nrps - can be calculated as
nrps = (6.94 m/s) / (2 π (0.665 m) / 2)
= 3.32 revolutions/s
The angular velocity of the wheel can be calculated as
ω = (3.32 revolutions/s) (2 π rad/revolution)
= 20.9 rad/s
The kinetic energy of the rotating bicycle wheel can then be calculated to
Ef = 0.5 (0.22 kg m2) (20.9 rad/s)2
= 47.9 J
Kinetic energy stored in a flywheel - moment of inertia
Sponsored Links
A flywheel can be used to smooth energy fluctuations and make the energy flow intermittent operating machine more uniform. Flywheels are used in most combustion piston engines.
Energy is stored mechanically in a flywheel as kinetic energy.
Kinetic Energy
Kinetic energy in a flywheel can be expressed as
Ef = 1/2 I ω2 (1)
where
Ef = flywheel kinetic energy (Nm, Joule, ft lb)
I = moment of inertia (kg m2, lb ft2)
ω = angular velocity (rad/s)
Angular Velocity - Convert Units
1 rad = 360o / 2 π =~ 57.29578o1 rad/s = 9.55 rev/min (rpm) = 0.159 rev/s (rps)
Moment of Inertia
Moment of inertia quantifies the rotational inertia of a rigid body and can be expressed as
I = k m r2 (2)
where
k = inertial constant - depends on the shape of the flywheel
m = mass of flywheel (kg, lbm)
r = radius (m, ft)
Inertial constants of some common types of flywheels
wheel loaded at rim like a bicycle tire - k =1flat solid disk of uniform thickness - k = 0.606flat disk with center hole - k = ~0.3solid sphere - k = 2/5thin rim - k = 0.5radial rod - k = 1/3circular brush - k = 1/3thin-walled hollow sphere - k = 2/3thin rectangular rod - k = 1/2
Moment of Inertia - Convert Units
1 kg m2 = 10000 kg cm2 = 54675 ounce in2 = 3417.2 lb in2 = 23.73 lb ft2
Flywheel Rotor Materials
MaterialDensity
(kg/m3)
DesignStress
(MPa)SpecificEnergy
(kWh/kg)Aluminum alloy2700 Birch plywood70030 Composite carbon fiber - 40% epoxy15507500.052E-glass fiber - 40% epoxy19002500.014Kevlar fiber - 40% epoxy140010000.076Maraging steel80009000.024Titanium Alloy45006500.031"Super paper"1100 S-glass fiber/epoxy19003500.020
1 MPa = 106 Pa = 106 N/m2 = 145 psiMaraging steels are carbon free iron-nickel alloys with additions of cobalt, molybdenum, titanium and aluminum. The term maraging is derived from the strengthening mechanism, which is transforming the alloy to martensite with subsequent age hardening.
Example - Energy in a Rotating Bicycle Wheel
A typical 26-inch bicycle wheel rim has a diameter of 559 mm (22.0")and an outside tire diameter of about 26.2" (665 mm). For our calculation we approximate the radius - r - of the wheel to
r = ((665 mm) + (559 mm) / 2) / 2
= 306 mm
= 0.306 m
The weight of the wheel with the tire is 2.3 kg and the inertial constant is k = 1.
The Moment of Inertia for the wheel can be calculated
I = (1) (2.3 kg) (0.306 m)2
= 0.22 kg m2
The speed of the bicycle is 25 km/h (6.94 m/s). The wheel circular velocity (rps, revolutions/s) - nrps - can be calculated as
nrps = (6.94 m/s) / (2 π (0.665 m) / 2)
= 3.32 revolutions/s
The angular velocity of the wheel can be calculated as
ω = (3.32 revolutions/s) (2 π rad/revolution)
= 20.9 rad/s
The kinetic energy of the rotating bicycle wheel can then be calculated to
Ef = 0.5 (0.22 kg m2) (20.9 rad/s)2
= 47.9 J
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❤ hey buddy ❤
⭐ EF = 0•5 (0•22 kg m²)(20•9Rad/S)²⭐
⭐ = 47•9 j ⭐
✌ Hope it's help you✌
✔✔ Mark me brainlist ✔✔
⭐ EF = 0•5 (0•22 kg m²)(20•9Rad/S)²⭐
⭐ = 47•9 j ⭐
✌ Hope it's help you✌
✔✔ Mark me brainlist ✔✔
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