Motor Efficiency

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Anonymous Poster	Motor Efficiency 11/04/2007 10:35 PM
	I am new to induction motors, I am trying to solve this problem but I am not quiet sure of the answer of n=51.123. here is the problem. A single phase capacitor start induction motor is rated at 1750 RPM, 120 VAC, 1/2hp, and 6.08 A line current at 0.7 lagging power factor. Under rated operating conditions, what is the motor efficiency?
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Paulusgnome Power-User Join Date: Oct 2007 Location: Christchurch, New Zealand Posts: 328 Good Answers: 29	#1 Re: MOTOR EFFICIENCY 11/05/2007 12:20 AM
	This is how I would calculate the motor efficiency : 1) Define efficiency n = Pout/Pin. Pout is the mechanical power delivered to the load, here given as 1/2hp (=373W assuming your horsepower is 746W). 2) Pin is the electrical power consumed by the motor. Pin = V x I x PF = 120 x 6.08 x 0.7 = 510.72W 3) Efficiency n = 373/510.72 = 0.73 or 73%. __________________ paulusgnome
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drcmuthu Participant Join Date: Oct 2007 Posts: 2	#2 Re: Motor Efficiency 11/06/2007 9:02 AM
	the method of calcultion provided in reply #1 is ok & correct for the details provided
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nesubra Guru Join Date: Nov 2006 Posts: 729 Good Answers: 2	#3 Re: Motor Efficiency 11/07/2007 12:28 PM
	The Basic assumption in the answer is motor is running at full load .However it is a practice to have margin while selecting the motor normally 15% in such small ratings and 10% in higher ratings. The efficiency of the motor varies with load.It may be prudent to take atleast 90%of 1/2 hp and workout the efficiency.It is however best to refer the name plate where full load efficiency is normally given and refer test cert.for other part load eff. __________________ To avoid crticism do nothing,say nothing,be nothing
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