Radius of gyration/power/rev per minute

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SUMMARY

The discussion centers on calculating the mass of water impacting a water wheel and the rotational speed of the generator. The user has determined the velocity of water hitting the wheel to be 44.3 m/s and calculated the mass flow rate required to produce 20 kW of power to be approximately 10.2 kg/s. The radius of gyration for the wheel and generator is 3 m, and the mass is 120 kg. The user also explored the relationship between kinetic energy and power, ultimately deriving an angular velocity of 6.086 rad/s using the equation E = 0.5 mk²ω².

PREREQUISITES
  • Understanding of kinetic energy and potential energy equations (K.E = 0.5mv², P.E = mgh)
  • Familiarity with the concept of radius of gyration (J = mk²)
  • Knowledge of angular velocity and its relation to power (P = Mk²ω²)
  • Basic principles of fluid dynamics and energy conversion in mechanical systems
NEXT STEPS
  • Study the derivation and application of the kinetic energy equation for rotating bodies (E = 0.5 mk²ω²)
  • Learn about the conservation of energy in mechanical systems, particularly in water wheels
  • Research the calculation of angular velocity from power and mass flow rate
  • Explore the impact of varying mass flow rates on the efficiency of water wheels and generators
USEFUL FOR

Engineering students, mechanical engineers, and anyone involved in renewable energy systems, particularly those focusing on hydropower and rotational dynamics.

tommy56
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Homework Statement


A water wheel rotates a generator producing power from vertically flowing water onto its blades. Height of water is 100m above blades. init vert velocity is 0.
I have calc velocity at hitting wheel as 44.3m/s
calc the mass of water hitting the wheel per second to create 20kW power
Calc rev/min of gene (wheel and gene combined radius of gyration 3m and mass 120kg)

Homework Equations


v²=u²+2as
K.E=0.5mv²
P.E=mgh
J=mk²
g=9.8m/s²

The Attempt at a Solution


For mass I have calculated P.E and K.E in terms of m and made equal to 2000 giving answer 10.2Kg? not sure if this is right.
For rev/min I have worked out J=120x3² = 1080kg m², not sure where to go from here?
 
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tommy56 said:

The Attempt at a Solution


For mass I have calculated P.E and K.E in terms of m and made equal to 2000 giving answer 10.2Kg? not sure if this is right.
For rev/min I have worked out J=120x3² = 1080kg m², not sure where to go from here?

10.19 or 10.2 kg/s is correct. Remember to put the correct units of mass per second.

For the second part, remember that you can the kinetic energy of a rotating component as E = Iω2 = mk2ω2.

or in terms of power P = Mk2ω2 where M is the mass per second and ω is in rad/s.
 
rock.freak667 said:
10.19 or 10.2 kg/s is correct. Remember to put the correct units of mass per second.

For the second part, remember that you can the kinetic energy of a rotating component as E = Iω2 = mk2ω2.

or in terms of power P = Mk2ω2 where M is the mass per second and ω is in rad/s.

Thanks for the help, but I cannot find that equation, I can however find the equation E = 0.5 mk2ω2.
If I use this then I get 20000=0.5x120x32ω2
which gives me 6.086rad/s.
Am I right to use the mass of 120kg, as this is given in the question, and I'm not sure if i need the 0.5 in the equation? Thanks.
 
tommy56 said:
Thanks for the help, but I cannot find that equation, I can however find the equation E = 0.5 mk2ω2.
If I use this then I get 20000=0.5x120x32ω2
which gives me 6.086rad/s.
Am I right to use the mass of 120kg, as this is given in the question, and I'm not sure if i need the 0.5 in the equation? Thanks.

That's equating energy with power,

you need to find E which would normally be the KE+PE but you aren't given a time element so I am not sure if you can get it.
 

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