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ECE 321 - Electromechanical Motion Devices - Purdue Study Resources
  • 6 Pages hw 4 solutions
    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

  • 6 Pages HW1 Solution
    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 H-field is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

  • 1 Page hw9
    hw9

    School: Purdue

    EE321 Spring 2008 / Homework 9 Problem 37 Discrete winding function The number of conductors in each slot of the a-phase 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

  • 2 Pages hw3[1]
    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

  • 1 Page hw1
    hw1

    School: Purdue

    EE321 Spring 10 Homework 1 Problem 1 Review of line-integral and application to MMF drop. Consider a Cartesian co-ordinate system (x,y,z). Suppose a uniform H-field of 1 A/m exists in the direction of the x-axis. Calculate the MMF drop from the point (1,1

  • 6 Pages hw 1 solutions
    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 H-field is constant Point2 H dl = H ( Point2 Point1) Point1 Thus the MMF drop is given by T H (

  • 1 Page HW1
    HW1

    School: Purdue

    EE321 Spring 2012 Homework 1 Problem 1 Review of line-integral and application to MMF drop. Consider a Cartesian co-ordinate system (x,y,z). Suppose a uniform H-field of 1 A/m exists in the direction of the x-axis. Calculate the MMF drop from the point (1

  • 11 Pages HW2 Solution
    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

  • 2 Pages hw 4
    hw 4

    School: Purdue

    Course: Electromechanical Motion Devices

    EE321 Spring 2013 Homework 4 Problem 1 Calculation of Torque The flux-linkage 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 ?

  • 1 Page hw8
    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,

  • 2 Pages hw4
    hw4

    School: Purdue

    EE321 Spring 2010 Homework 4 Problem 1 Calculation of Torque The flux-linkage 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 ?

  • 2 Pages HW6
    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 = 510-6 Nms. The applied voltage is 6 V and Bm = 0. Calcula

  • 4 Pages abet
    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

  • 3 Pages HW5 Solution
    HW5 Solution

    School: Purdue

    EE321 HW #5 Problem 1 Problem 2 Problem 3

  • 2 Pages hw3
    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

  • 27 Pages Lecture Set 0
    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

  • 5 Pages hw12 solution
    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(100t-1.97) If the cosine term were a sine, I simply would have adjusted the phase

  • 2 Pages hw 2
    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

  • 11 Pages hw 2 solutions
    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

  • 8 Pages hw14 solution
    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

  • 13 Pages exam 3 solutions
    exam 3 solutions

    School: Purdue

    Course: Electromechanical Motion Devices

  • 12 Pages exam 2 solutions
    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 4-pha

  • 8 Pages exam 1 solutions
    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!

  • 2 Pages practice exam5
    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

  • 5 Pages practice exam5_solution
    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

  • 2 Pages practice exam4
    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 2-5 satisfies objective 4. Getting 70% of problem 3 or 4 satisfies objective 2. Getting 70% o

  • 4 Pages practice exam4_solution
    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

  • 2 Pages practice exam3
    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 1-3 satisfies ABET objective 2. Getting 75% of problem 5 satisfies ABET objective 3. Getting

  • 6 Pages practice exam3_solution
    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

  • 6 Pages hw 3 solutions
    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

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