1. 高等電機機械 學號： MA020211 學生：周士平

2. 題目 5.12 Consider the motor of Problem 5.10. a. Compute the filed current required when the motor is operating at rated voltage, 4200kW input power factor leading. Account for saturation inder load by the method described in the paragraph relating to Eq. 5.29. b. In addition to the data given in Problem 5.10, additional points on the open-circuit characteristic are given below: If the circuit breaker supplying the motor of part (a) is tripped, leaving the motor suddenly open-circuited, estimate the value of the motor terminal voltage following the trip. Field current, A200250300350 Line voltage, V4250458048205000

3. 題目 5.12 Ans(a) The total power is and

4. 題目 5.12 Ans(b) If the machine speed remains constant and the field current is not reduced, the terminal voltage will increase to the value corresponding to 306 A of field current the open- circuit saturation characteristic. Interpolating the given data shows that this corresponds to a value of around 4850 V line-to-line.

5. 題目 5.14 Loss data for the motor of Problem 5.10 are as follows ： Open-circuit core loss at 4160V = 37kW friction and windage loss = 46kW field-winding resistance at 75 ℃ =0.279Ω Compute the output power and efficiency when the motor is operating at rated input power, unity power factor, and rated voltage. Assume the field-winding to be operating At a temperature of 125 ℃

6. 題目 5.14 Ans At rated power, unity power factor, the armature current will be Ia =5000 kW/(√3 4160 V) = 694 A. The power dissipated in the armature winding will then equal Parm = 3× 6942 × 0.011 = 15.9 kW. The field current can be found from |Eaf | = |Va − ZsIa| = |4160√3 − ZsIa| = 2394 V, line-to-neutral If = AFNL_(2394/4160/√3) = 238 A At 125◦C, the field-winding resistance will be Rf = 0.279(234.5 + 125/234.5+ 75)= 0.324 Ω and hence the field-winding power dissipation will be Pfield = I2f Rf = 18.35 kW.The total loss will then be Ptot = Pcore + Parm + Pfriction/windage + Pfield = 120 kW Hence the output power will equal 4882.75 kW and the efficiency will equal 4882.75/5000= 0.976 = 97.6%.