We aren't endorsed by this school 
ECE 321  Electromechanical Motion Devices  Purdue Study Resources

hw 4 solutions
School: Purdue
Course: Electromechanical Motion Devices
EE321, Spring 2013 Homework 4 Problem 1 1 2 Wc = 5 + 2 sin 4 rm i 2 ( ( ( ) Te = 4 cos 4 rm i ) 2 Problem 2 We may express the system as 1 2 7 5 2 5 + x i1 = 7 i 2 2 2 5+x which is of the form i 1 = L 1 2 i2 where L is independent of both c

HW1 Solution
School: Purdue
Homework 1, Problem 1 Let the elements of the vector correspond to the x, y, and z components 1 H := 0 0 Now 1 Point1 := 1 1 2 Point2 := 5 1 Since the Hfield is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

hw9
School: Purdue
EE321 Spring 2008 / Homework 9 Problem 37 Discrete winding function The number of conductors in each slot of the aphase of the stator of the machine are as follows: N as = [10 20 20 10 10 20 20 10 10 20 20 10 10 20 20 10]T Compute and graph the winding f

hw3[1]
School: Purdue
Course: Electromechanical Motion Devices
Fall 2010 ECE 321 Homework Set 3 Due Wed. Sept. 29 Work on separate sheets of paper. Must be turned in at beginning of class. First page blank with only your name and should be stapled. Homework will be collected promptly at 2:30. If not submitted in time

hw1
School: Purdue
EE321 Spring 10 Homework 1 Problem 1 Review of lineintegral and application to MMF drop. Consider a Cartesian coordinate system (x,y,z). Suppose a uniform Hfield of 1 A/m exists in the direction of the xaxis. Calculate the MMF drop from the point (1,1

hw 1 solutions
School: Purdue
Course: Electromechanical Motion Devices
Homework 1, Problem 1 Let the elements of the vector correspond to the x, y, and z components 1 H := 0 0 Now 1 Point1 := 1 1 5 Point2 := 2 1 Since the Hfield is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

HW1
School: Purdue
EE321 Spring 2012 Homework 1 Problem 1 Review of lineintegral and application to MMF drop. Consider a Cartesian coordinate system (x,y,z). Suppose a uniform Hfield of 1 A/m exists in the direction of the xaxis. Calculate the MMF drop from the point (1

HW2 Solution
School: Purdue
Problem 1  Simple UI Core Analyiss Dimensions, etc 2 3 cm := 1 10 mm := 1.0 10 w := 1 cm d s := 2 cm g := 1.5 mm ws := 5 cm d := 5 cm N := 100 B sat := 1.3 Point where saturation occurs 7 u 0 := 4 10 Now let's compute some reluctances. For flux densities

hw 4
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 4 Problem 1 Calculation of Torque The fluxlinkage of a certain rotational electromechanical device may be expressed = (5 + 2sin 4 rm )i where rm is the rotor position and i is the current. What is the electromagnetic torque ?

hw8
School: Purdue
EE321 Spring 2008 / Homework 8 Problem 33 Buck converter operation Consider the example on page 11 of the lecture notes. Suppose the dc voltage is changed to 150 V and the speed to 450 rad/s. Find the average armature current, the average switch current,

hw4
School: Purdue
EE321 Spring 2010 Homework 4 Problem 1 Calculation of Torque The fluxlinkage of a certain rotational electromechanical device may be expressed = (5 + 2 sin 4 rm )i where rm is the rotor position and i is the current. What is the electromagnetic torque ?

HW6
School: Purdue
ECE321/ECE595 Spring 2012 Homework 6 Problem 1 Permanent Magnet DC Machine A permanent magnet dc machine has ra = 8 and kv = 0.01 Vs/rad. The shaft load torque is approximated as TL = Kr, where K = 5106 Nms. The applied voltage is 6 V and Bm = 0. Calcula

abet
School: Purdue
EE321. ABET Exam Spring 2004 Name: Student ID: Instructions: Work ALL Problems. When you have completed exam, turn in to Professor Sudhoff, Brandon Cassimere, or Brant Cassimere, any of whom will check it on the spot, and let you know which ones are wrong

hw3
School: Purdue
EE321 Spring 2010 Homework 3 Problem 1 UI Inductor Analysis Consider the UI inductor design we did in class. Recall we had N = 260 Turns d = 8.4857 cm g = 13.069 mm w = 1.813 cm aw = 21.5181 mm2 ds = 8.94 cm ws = 8.94 cm In our design, we assumed that the

Lecture Set 0
School: Purdue
Lecture Set 0 ECE321/ECE595 S.D. Sudhoff Electromechanical Motion Devices Spring 2012 Courses Meeting Together Courses ECE321 Live (57) cfw_321L ECE321 Video (9) cfw_321V ECE595 On Campus (3) cfw_595C ECE595 Off Campus Pro Ed (11) cfw_595P Differences 321

hw12 solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework #12 Problem 1 500 e Xphasor := 5+ 1 20 ( 100j) + Xphasor = 17.379 ( j 1 300 ( 100j) 2 Xphasor 2 = 24.577 ) arg Xphasor = 1.967 x=17.4*sqrt(2)*cos(100t1.97) If the cosine term were a sine, I simply would have adjusted the phase

hw 2
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework 2 Problems 1 UI Inductor Analysis Consider the UI core below. Consider the following parameters: w = 1 cm; ws = 5 cm; d s = 2 cm; d = 5 cm; g = 1.5 mm; N = 100 . Suppose the material used is such that for a flux density less tha

hw 2 solutions
School: Purdue
Course: Electromechanical Motion Devices
Problem 1  Simple UI Core Analyiss Dimensions, etc 2 3 cm := 1 10 mm := 1.0 10 w := 1 cm d s := 2 cm g := 1.5 mm ws := 5 cm d := 5 cm N := 100 B sat := 1.5 Point where saturation occurs 7 u 0 := 4 10 Now let's compute some reluctances. For flux densities

hw14 solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Spring 2013 Homework #14 Problem 1 Machine Parameters 3 3 rs := 72.5 10 3 Lls := 1.32 10 3 rr := 41.3 10 P := 4 Llr := 1.32 10 N Lrr := Llr + Lms 2 N := 2 Source vs := 460 e := 2 60 3 Mechanical Load 50 746 = 1800 2 := 3 2 1800 60 60 50 746 1

exam 2 solutions
School: Purdue
Course: Electromechanical Motion Devices
ECE3211ECE595 Exam 2 Spring 2013 Notes: You must show work for credit. This exam has 5 problems and 12 pages. Note that problems 2, 4, and 5 have different specications depending on if you are in ECE321 or ECE595 Good luck! i) 20 pts. Consider a 4pha

exam 1 solutions
School: Purdue
Course: Electromechanical Motion Devices
SCA Whom 32% VS 915 50 ECE321IECE595 Exam 1 Spring 2013 Notes: You must show work for credit. This exam has 5 problems and 12 pages. Note that problems 1, 4, and 5 have different specications depending on if you are in ECE321 or ECE595 Good luck!

practice exam5
School: Purdue
Course: Electromechanical Motion Devices
EE321 Exam 5 Spring 2008 Write your name and student ID on the bluebook. Only turn in the bluebook. Notes: You must show work for credit. Getting 70% on problems 1 or 2 satisfies objective 1 and 2. Getting 70% on problems 4 or 5 satisfies objective 4. Get

practice exam5_solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Exam 5 Solution Outline Problem 1 cm := 0.01 mm := 0.001 w := 1 cm d := 5 cm ws := 5 cm d s := 2 cm g := 1 mm N := 100 r := 1000 7 0 := 4 10 Fe := 25 A := w d w ds + 2 R 45 := A 0r R 56 := ws + w 0 r A R 45 = 39.789 10 3 R 56 = 95.493 10 3 w R 81

practice exam4
School: Purdue
Course: Electromechanical Motion Devices
EE321 Exam 4 Spring 2008 Write your name and student ID on the bluebook. Only turn in the bluebook. Notes: You must show work for credit. Getting 70% of problems 25 satisfies objective 4. Getting 70% of problem 3 or 4 satisfies objective 2. Getting 70% o

practice exam4_solution
School: Purdue
Course: Electromechanical Motion Devices
EE321 Exam 4 Solution Outline Problem 1 By inspection, the secondary to primary turns ratio is 10. Nsp := 10 Lm := 500 Referred to secondary rs := 2 From secondary rp := 1 Nsp 2 0 rp = 100 10 From secondary The leakage inductances are zero. Problem 2 Lls

practice exam3
School: Purdue
Course: Electromechanical Motion Devices
EE321 Exam 3 Spring 2008 Write your name and student ID on the bluebook. Only turn in the bluebook. Notes: You must show work for credit. Getting 75% of problems 13 satisfies ABET objective 2. Getting 75% of problem 5 satisfies ABET objective 3. Getting

practice exam3_solution
School: Purdue
Course: Electromechanical Motion Devices
ECE321 Spring 2008 Exam 3 Solution Outline Problem 1 T Nas := ( 3 3 6 3 3 6 3 3 6 3 3 6 ) Nslts := 12 P := 4 Nslts P =3 1 Was := [ 3 + ( 3 ) + ( 6 ) ] 1 2 j := 2 . 12 k := 1 . 11 Was := Was Nas j j 1 j 1 T 1 Was = 2 3 1 6 3 3 Problem 2 () was = 100 co

hw 3 solutions
School: Purdue
Course: Electromechanical Motion Devices
EE321, Homework 3 Problem 1 From minimum cost solution in class N := 260 2 d := 8.4857 10 2 d s := 8.94 10 ws := 8.94 10 3 i := 40 3 g := 13.069 10 Ldes := 5 10 2 2 w := 1.813 10 7 0 := 4 10 r := 2000 Recomputing the reluctance R := 2 ( ws + 2w) + 2d s 2