ee341_h05_Wi2012_Solution

ee341_h05_Wi2012_Solution - ECE 341 Homework #5 Solution...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 341 Homework #5 Solution Due: Mar. 2, Feb. 12 Problem 1 : An eight-pole, 60 Hz, 3-phase, wye-connected induction motor has mechanical losses of 620 W at s=0.04. At the same time, the motor delivers 30 HP at its shaft. Calculate (a) the input power to the rotor (P AG ) (2 points) (b) the load torque and total torque (2 points) Solution: ns=900 rpm, Pout=30 HP8760=22.38 kW a. Pconv=Pout+Pmech_loss=22.38+0.62=23 kW P AG = Pconv/(1-s)=23/0.96=23.96 kW b. 35 . 247 900 96 . 2 22380 m out load P Nm 21 . 254 900 96 . 2 23000 m conv total P Nm Problem 2: A six-pole, 480 V, 3-phase, 60 Hz induction motor takes 200 A of current in each phase at starting, and 30 A/Phase when it runs with full load. The starting torque is 1.8 times the torque at the full load at 480 V. If it is desired that the starting torque be the same as the full load torque, determine (a) The applied line-to-line voltage (2 points) (b) The corresponding line current (phase) of the motor...
View Full Document

Page1 / 3

ee341_h05_Wi2012_Solution - ECE 341 Homework #5 Solution...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online