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QUESTION 10.13

A dc shunt generator rated at 85 kW produces a voltage of 280 V. The brush voltage drop is 2.5 V, and the armature and field resistances are 0.09 ? and 115 ?, respectively. The generator delivers rated current at rated speed and rated voltage.

(a) Calculate the field, armature, and load currents versus load. Test the equations with the rated load. 

(b) Determine the terminal voltage at no-load and at rated load conditions.

 (c) Calculate the voltage regulation of the generator. Use the no-load voltage as the base value. 

(d) Plot the terminal voltage as a function of the load. 

Determine the load that corresponds to a 5% voltage drop using the no-load voltage as the base.

Answers:

  

10.6. nmot 649 rpm 86 rpm = 10.7. nmot = 10.8. V 258 V = 10.9. Vs = sup = 108 V, Tstart 1126 N-m; motor can start the pump 10.10. (a) Km = 0.837 V-s/A; (b) n = 1367 rpm; (c) 199% = 24.7 A 10.11. (a) Km = 0.06 V-s/A; (b) n = 858 rpm, T = 96.5 N-m; (c) I = 18.2 A, n = 1938 rpm 10.12. (a) diagram; (b) Km = 0.651 V-s/A; (c) Tstart=3175 rpm, n = 1241 rpm - 10.13. (a) If = 2.14 A, I = 347 A, Iload = 345 A; (b) V(0) = 277 V, V(P) = 246 V; (c) Reg = 12.6%; (d) P5% = 40.6 kW 10.14. (a) Km = 7.86 V-s/A; (b) Iload = 149 A, E = 393 V, n = 672 rpm, (c) Pin = 56.8 kW, Pout= 75.8 hp, = 99.6% 10.15. derivation CHAPTER 11: INTRODUCTION TO POWER ELECTRONICS AND MOTOR CONTROL 11.1. (a) and (b) a = VM, otherwise a = 0 b; (c) a, (cosine) = b, (sine) 11.2. If(0) = 21.6 A, Ifw(1) = 0, Ifw(2) = 14.4 A, Ifw(3) = 0, Ifw (4) = 2.88 A, Ifw (5) = 0. iO 1440