Mathcad - 11-08 - P rated 29.828kW = we can find the...

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P rot 6.577kW = P rot P in P dev - = the rotational losses are the difference between the input power and the developed power P in 34.285kW = P in P rated η = the input power is: and the developed power is P dev 27.708kW = P dev I a E a = E a 483.969V = therefore the CEMF is: E a V t I a R a - = I a 57.252A = I a I t I f - = so the armature current is: I f 2.404A = Problem 11-8: P rated 40 hp = V t 500 volt = R f 98 = R rh 110 = R a 0.28 = η 0.87 = n rated 1750 rev min = R f R rh R a converting the output to watts:
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Unformatted text preview: P rated 29.828kW = we can find the terminal current I t P rated V t = I t 59.656A = The field current is: I f V t R f R rh + = assuming rated armature current is Ia above, then the limit for the starting current is I st 2.I a = I st 114.504A = I st V t R a R x + = so R x V t I st R a-= R x 4.087 =...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.

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Mathcad - 11-08 - P rated 29.828kW = we can find the...

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