# Mathcad - 11-04 - f 59.167Hz = f n rated = since it is a 2...

This preview shows page 1. Sign up to view the full content.

E anew 266.76V = E anew 0.95 E aspeed = but reducing the flux by 5% reduces the voltage by 5%, so E aspeed 280.8V = E aspeed 1.17 E a = Increasing the speed would increase the voltage by 17% The generated voltage is affected by the speed change and by the flux decrease. f new 69.225 Hz = f new 1.17 f = b. if the speed is increased by 17%, so will the frequency
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f 59.167Hz = f n rated = since it is a 2 pole machine, the frequency is equal to the rev/sec n rated 59.167 rev sec = a. Dividing rpm by 60 yields: E a 240 volt ⋅ = P rated 100 kW ⋅ = n rated 3550 rev min ⋅ = P 2 poles ⋅ = Problem 11-4:...
View Full Document

## This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue.

Ask a homework question - tutors are online