Design Problem EE517 Fall 2000 Revision 6

Parts 1) through 12) are due November 21, 2000 and

parts 13) through 24) are due November 28, 2000.

You are to design a 2-pole, 3-phase, synchronous motor that will be supplied power from a 440Vrms (line to line) 60Hz source. The armature windings are Y connected. The rating of the motor is 2.5HP.The maximum allowed magnetic flux density at any point in the iron and at any time is Bmax = 1.65Tesla. The maximum rms current density in the copper at rated conditions is Jrmsrated = 350A/cm2. You are to design the machine so that the DC field current is 0.1 times the peak armature current. The minimum allowed air gap is 0.02 inches to insure it is not to difficult to build the machine. You may assume the machine is sinusoidally wound. You may use the design factors and the values given in table 1. To complete your design you must compute the quantities in table 2. All dimensions are to be given in inches though it is recommended that all calculations be done using MKS units.

 Design Factor Definition Value Kw = Winding factor, fraction of the rotor and stator slot area occupied by copper. 0.25 adslot = depth of slot / radius to the air gap (rotor radius) 0.333 alength = stack length / radius to the air gap 1.167 a%slot = percent (fraction) of the rotor or stator circumference occupied by slots 0.5 (50%) NaIa = NfIf

Table 1 Design factors and their values

1. Draw the single-phase equivalent circuit for the machine for balanced three-phase excitation. What is the difference in the schematic between constant voltage input and constant current input?
2. What power in watts must the motor produce and how much power is supplied by each phase? What is the machine's torque in N-m?
3. One approach to designing the machine is to design it so that the maximum power it can produce is the rated power. What is g when the motor is producing its maximum (rated) power/torque driven from a current source? Using the fact that the input phase voltage is fixed and equal to the vector sum of the inductive voltage and the BEMF, what is the phase angle between the armature voltage and current. What is the motor's power factor, what is d, and is the phase current leading or lagging the phase voltage at the maximum power condition?
4. At the maximum power condition, what must the BEMF () be and what is the armature current at this condition?
5.

 Design Results Value Rg = radius to the air gap = rotor radius 2.572 in Dg = rotor diameter 5.145 in g = air gap 0.0499 in Awa = 2 AW3f = total area required for the stator (armature) windings 6.922 in2 Ds = diameter to the bottom of the stator slot 6.858 in Awf = total area required for the rotor (field) windings 2.569 in2 Dr = Diameter to the bottom of the rotor slot 3.873 Dshaft = largest allowed shaft diameter 1.301 Dout = outside diameter of the stator 9.43 in lstk = stack length = axial length of iron. 3.002 in aw = area of the wire 1.954 10-3 in2 Wire gauge # 16 Na = number of stator (armature) turns per phase 148 Nf = number of rotor (field turns) 1,480 Rph = resistance of one stator phase winding 1.275 W

Table 2 Design results to be computed.

6. Starting with the machine's output power in terms of Ia, If, and g, show that the area product for the machine must obey
7. where Bfgmax is the peak B field due to the field winding in time and space neglecting the effect of the slots.

8. Show that the area occupied by the three windings (assuming uniform slot size around the machine) of a three-phase machine is roughly 2 / p as large as the area of the one winding of a similarly rated single-phase machine.
9. What is the three phase machine's area product in terms of the torque?
10. What is the winding area Aw3f in terms of the machine's air gap radius Rg and the appropriate design factors. Be careful to not over count or under count the area.
11. What is the peak B field Bth between the slots (in the tooth) in terms of Bfgmax. Show how you obtained your result.
12. Given and your result for Aw3f in part 8, write the area product in terms of lstk and Rg. Using the design factors write the area product in terms of Rg only.
13. What are Rg, lstk, and the slot depth for the motor being designed.
14. For an arbitrary position of the rotor q, what is the field in the stator yoke at any instant of time t and position in the yoke f? Assume balanced three-phase excitation and be sure to account for all three phases and the field current. Where in the yoke is the B field a maximum? How thick should the yoke be?
15. Repeat part 12 for the rotor yoke.
16. Make a sketch of the machine's cross section (r - q plane) assuming each slot is 20 degrees wide. How many slots are on the rotor, on the stator?
17. Write the line to neutral voltage in terms of the BEMF at the maximum power condition. Write the mutual inductance in terms of the geometry and the rotor current generated B field. What is the required number of stator/armature turns for each phase? How would the number of stator turns change if the motor was to operate from 220Vrms (line to line)? Comment on your result.
18. What is the required wire size for the armature windings for 440V and 220V excitation? Comment on your result.
19. Estimate the number of rotor turns. Should the air gap be small or large?
20. What is the number of turns per degree for phase A? How many phase A turns should there be in each stator slot? Repeat for phase B and C. Number the slots and make a table.
21. What is the total number of turns in each stator slot? Add a column to your table in 18.
22. Make a drawing of a representative stator slot and draw the turns in the slot to scale indicating the A, B, and C phase turns.
23. What is the number of turns per degree for the field winding? How many field turns should there be in each rotor slot? Number the slots and make a table.
24. Estimate the length of an individual end turn. Estimate the length of the phase A winding. Estimate the resistance of the phase A winding.
25. Estimate the power dissipated in the phase A winding using your resistance from part 22 and using the rms current density.
26. What is the efficiency of the motor neglecting iron and windage losses.

 Slot No. Phase A turns Phase B turns Phase C turns Sum 1 17+ 50- 32+ 99 2 44+ 44- 5+ & 5- (10 wires) 98 3 50+ 17- 32- 99 4 32+ 17+ 50- 99 5 5+ & 5- (10 wires) 44+ 44- 98 6 32- 50+ 17- 99 7 50- 32+ 17+ 99 8 44- 5+ & 5- (10 wires) 44+ 98 9 17- 32- 50+ 99 Total (half machine) 148 148 148 444 / 2

Summary of stator turns

 Slot No. Field turns 1 173+ 2 438+ 3 499+ 4 325+ 5 45+ & 45- (90 wires) 6 325- 7 499- 8 438- 9 173- Total (half machine) 1480

Summary of rotor turns