Physics Help Forum Wind generator

 Energy and Work Energy and Work Physics Help Forum

 Apr 9th 2012, 02:58 AM #1 Junior Member   Join Date: Aug 2011 Posts: 7 Wind generator I am designing a small wind generator. The design speed is to achieve an output of 1KW in a wind of 10 mph. If I can achieve an efficiency of 40% at this design speed, what diameter do I need for the rotor? Givens Output 1KW Wind 10mph Efficiency 40% Unknowns Rotor Diameter ? Answer 3.7m Would like some help getting to this answer. Thanks.
 Apr 17th 2012, 12:16 AM #2 Member   Join Date: Oct 2010 Location: Sri Lanka Posts: 63 Since you need an output of 1kW, you need 2.5KW power (1kW/.4) from the wind. lets assume that all the kinetic energy of the wind is used in generating power to the the generator. so the Power achived from the wind = Kinetic energy of the mass of the wind that passes through the rotor per second mass of the wind that passes throught the rotor per sec.=p.V where p= density of the wind V=volume of the wind that passes through the rotor per sec V= (pi.(d^2)/4)*v where d=diameter of the rotor and v= velocity of the wind. So going back through the path you'll get, power achived from the wind = 1/2(pV)v^2 = (1/2)p(pi(d^2)/4)v.v^2 =(1/8)pi.p.d^2.v^3 this can be equated to 2.5kW now find d
 Apr 19th 2012, 04:10 AM #3 Junior Member   Join Date: Aug 2011 Posts: 7 Thank you very much that 100% helped I fully understand now.
 Apr 19th 2012, 07:40 AM #4 Member   Join Date: Oct 2010 Location: Sri Lanka Posts: 63 It's my pleasure
Apr 28th 2012, 06:21 AM   #5
Physics Team

Join Date: Apr 2009
Location: Boston's North Shore
Posts: 1,576
 Originally Posted by BAdhi Since you need an output of 1kW, you need 2.5KW power (1kW/.4) from the wind. lets assume that all the kinetic energy of the wind is used in generating power to the the generator. so the Power achived from the wind = Kinetic energy of the mass of the wind that passes through the rotor per second mass of the wind that passes throught the rotor per sec.=p.V where p= density of the wind V=volume of the wind that passes through the rotor per sec V= (pi.(d^2)/4)*v where d=diameter of the rotor and v= velocity of the wind. So going back through the path you'll get, power achived from the wind = 1/2(pV)v^2 = (1/2)p(pi(d^2)/4)v.v^2 =(1/8)pi.p.d^2.v^3 this can be equated to 2.5kW now find d
Where do show the number of blades in this wind generator?

 Tags generator, wind

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post OldTrafford Thermodynamics and Fluid Mechanics 0 Mar 9th 2014 12:56 PM whatever Electricity and Magnetism 1 Jun 10th 2013 07:16 AM YellowPeril Thermodynamics and Fluid Mechanics 8 Nov 29th 2010 01:52 PM LeopardGecko Kinematics and Dynamics 1 Sep 19th 2009 11:44 PM