ece643_e2_au09_ans - ECE 643 Exam#2 answers au/09 1(a(i...

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Unformatted text preview: ECE 643 Exam #2 answers, au/09 1.(a)(i) Torque α voltage, therefore if V @ ½ then T @ ¼; same nsync, (ii) nsync reduced to ½ of original value, T reduced also. (b) from Faraday’s law, v(t) = dλ(t) / dt & λ = turns*φ = turns*area*B thus volts per hertz represents the level of magnetic flux density, so if volts/hertz toooooo high then, will saturate. 2.(a) at any given instant of time, ia(t) + ib(t) + ic(t) =0 , and all phase currents are the sum of a decaying exponential (i.e., dc offset) plus a “symmetrical” component. Refer to the homework problem and the difference in the two “symmetrical” waveforms as 0 was varied. Since the generator was initially unloaded, i.e., stator current = 0, then the sum of the dc offset and the “symmetrical” component at t = 0+ must also be zero. Any one phase may have a zero dc offset which means that the dc offset of the other two phases must have equal magnitudes with opposite signs. (b) The ac component of field current is a result of the DC components of the stator currents; (c) the amortisseur windings (damper windings) on the rotor dominate the subtransient time regime. 3. Laa0 = 6, Lal = 1, Lg2 = 2, Laf = 8, Lff = 12 all values in millihenries Xd = 2fLd = 2f*[1 + (3/2)*(6 + 2)]*10‐3 = 2f*[13]*10‐3 = 4.9 ohms Xd ‘ = 2fLd ‘ = 2f*[13 ‐ (3/2)*(82/12)]*10‐3 = 2f*[5]*10‐3 = 1.885 ohms 4. (a) when fault (short‐circuit) occurs, all electrical power output goes to zero, whereas the mechanical power in stays the same, thus that mechanical power now goes to accelerating the machine. (b) The orientation of the magnetic axis of the stator and rotor of a DC machine are fixed in space by construction, thus does not change with load. However, for both synchrous and induction, angle between these two magnetic axis increase with load. (c) Initially machine operating at (P0,0), thus mechanical power input = Po, neglecting losses. Then when load removed, operating point “jumps” to (0,0) and machine starts to accelerate, thus increases. Then when load reconnected, operating point “jumps” to sin() curve and continues to increase for a bit. To be stable, energy used for acceleration, i.e., area A1 must be less than or equal to energy used to decelerate, i.e., area A2. 5. (a) frequency of rotor current = |slip|*frequency of stator current= 3 Hz; (b) speed of rotor produced flux relative to rotor = slip*nsync = ‐90 rpm, where nsync = 120*frequency/poles; (c) nsync = 1800 rpm; (d) nsync = 1800 rpm; (d) speed of rotor = (1 – slip)*nsync = 1890 rpm. ...
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