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PROBLEM SOLUTIONS: Chapter 7
Problem 7. part (a): ωm ∝ Va. part (b): ωm ∝ (^) I^1 f part (c): ωm will be constant.
Problem 7. part (a): For constant terminal voltage, the product nIf (where n is the motor speed) is constant. Hence, since If ∝ 1 /Rf
Rf 1180
Rf + 5 1250
and hence Rf = 84.2 Ω. part (b): 1380 r/min
Problem 7. Check this part (a): ωm halved; Ia constant part (b): ωm halved; Ia doubled part (c): ωm halved; Ia halved part (d): ωm constant; Ia doubled part (e): ωm halved; Ia reduced by a factor of 4.
Problem 7. part (a): Rated armature current = 25 kW/250-V = 100 A. part (b): At 1200 r/min, Ea can be determined directly from the magneti- zation curve of Fig. 7.27. The armature voltage can be calculated as
Va = Ea + IaRa
and the power output as Pout = VaIa. With Ia = 100 A
If [A] Ea [V] Va [V] Pout [kW] 1.0 150 164 16. 2.0 240 254 25. 2.5 270 284 28.
part (c): The solution proceeds as in part (b) but with the generated voltage equal to 900/1200 = 0.75 times that of part (b)
If [A] Ea [V] Va [V] Pout [kW] 1.0 112 126 12. 2.0 180 194 19. 2.5 202 216 21.
Problem 7. part (a):
part (b): (i)
(ii)
Problem 7. The solution is similar to that of Problem 7.6 with the exception that the assumed straight-line magnetization characteristic is replaced by the non-linear characteristic of Fig. 7.27. MATLAB, with the ’spline()’ function used to represent the non-linear characteristic of Fig. 7.27, then produces the following plots. part (a):
part (b):
Problem 7. part (a): From the load data, the generated voltage is equal to 254 +62. 7 × 0 .175 = 265 A. From the magnetizing curve (using the ’spline()’ function of MATLAB), the corresponding field current is 1.54 A. Hence the demagnetizing effect of this armature current is equal to (1. 95 − 1 .54)500 = 204 A-turns/pole. part (b): At the desired operating point, the generator output power will be 250 V × 61 .5 A = 15.4 kW. Therefore, the motor speed will be
n = 1195 − 55
= 1139 r/min
Because the machine terminal voltage at no load must equal 230 V, from the magnetizing curve we see that the shunt field under this operating condition
must equal 1.05 A and hence the shunt field resistance must be 219 Ω. Hence, under this loading condition, with a terminal voltage of 250 V, the armature voltage will be 250 +61. 5 × 0 .065 = 250.8 V, the shunt field current will equal
- 8 /219 = 1.15 A and thus the armature current will equal 61.5 + 1.15 = 62 .7 A. The generated voltage can now be calculated to be 250.8 + 62.7(0.175) = 286 V. The corresponding voltage on the 1195 r/min mag curve will be Ea = 286(1195/1139) = 285 V and hence the required net field ampere-turns is (using the MATLAB ‘spline()’ function) 1042 A-turns. The shunt-field ampere-turns is 1. 15 × 500 = 575 A-turns, the demagnetizing armature amp-turns are 204 A- turns and hence the required series turns are
Ns =
= 10. 6 ≈ 11 turns
Problem 7. From the given data, the generated voltage at Ia = 90A and n(90) = 975 r/min is
Ea(90) = Va − Ia(Ra + Rs) = 230 − 90(0.11 +0 .08) = 212.9 V
Similarly, the generated voltage at Ia = 30 A is
Ea(30) = 230 − 30(0.11 +0 .08) = 224.3 V
Since Ea ∝ nΦ
Ea(30) Ea(90)
n(30) n(90)
Making use of the fact that Φ(30)/Φ(90) = 0.48, we can solve for n(30)
n(30) = n(90)
Ea(30) Ea(90)
= 2140 r/min
Problem 7.
constant armature current. Constant torque (motor B) will require that the armature current variation be proportional to the motor speed. Thus
motor A: Ia = 125 A motor B: Ia = 125(500/1800) = 34.7 A
part (b):
motor A: Ia = 125 A motor B: Ia = 125(1800/125) = 450 A
part (c): Under armature voltage control and with constant field current, the speed will be proportional to the armature voltage. The generated voltage will be proportional to the speed. Constant-power operation (motor A) will re- quire aramture current that increases inversely with speed while constant torque operation (motor B) will require constant armature current. For the conditions of part (a):
motor A: Ia = 125(1800/125) = 450 A motor B: Ia = 125 A
For the conditions of part (b):
motor A: Ia = 125(500/1800) = 34.7 A motor B: Ia = 125 A
Problem 7.
ωm = Ea KaΦd
Va − IaRa KaΦd
Ia =
T
KaΦd
Thus
ωm =^
KaΦd
Va −^
T Ra KaΦd
The desired result can be obtained by taking the derivative of ωm with Φd
dωm dΦd
KaΦ^2 d
2 T Ra KaΦd
− Va
KaΦ^2 d
(2IaRa − Va)
KaΦ^2 d
(Va − 2 Ea)
From this we see that for Ea > 0. 5 Va, dωa/dΦd < 0 and for Ea < 0. 5 Va, dωa/dΦd > 0. Q.E.D.
Problem 7. part (a): Synchronous machine:
Ia,ac =
30 × 103
= 37.7 A
Eaf = |Va,ac + jXsIa,ac| = | 460 /
3 + j 5. 13 × 37. 7 | = 328.4 V, l − n
DC machine:
P = EaIa,dc = 30 kW
Ea = Va,dc − Ia,dcRa
Thus,
Ea^2 =
Va,dc +
V (^) a^2 ,dc − 4 P Ra 2
= 226 V
part (b): Increase the dc-motor field excitation until Ea = Va,dc = 230 V, in which case the dc motor input current will equal zero and it will produce no shaft power. The ac machine will operate at a power angle of zero and hence its terminal current will be
Ia,ac =
Eaf − V a, ac Xs
= 12.2 A
part (c): If one further increases the dc-machine field excitation the dc machine will act as a generator. In this case, defining the dc generator current as positive out of the machine,
P = EaIa,dc = 30 kW
Ea = Va,dc + Ia,dcRa
Thus,
Ea^2 =
Va,dc +
V (^) a^2 ,dc + 4P Ra 2
= 226 V
and
Ia,dc =
Ea − Va,dc Ra
= 128 A
Using MATLAB and its ‘spline()’ function to represent the fan character- istics, an iterative routine can be written to solve for the operating point (the intersection of the motor and fan characteristics). The result is that the motor will operate at a speed of 999 r/min and an output power of 8.39 kW.
Problem 7. part (a): Assuming negligible voltage drop across the armature resistance at no load, the field current can be found from the 1300 r/min magnetization curve by setting Ea = 230 V. This can be most easily done using the MATLAB ‘spline()’ function. The result is If = 1.67 A. This corresponds to Nf If = 2500 a·turns of mmf. part (b): At rated load, Ea = Va − IaRa = 230 − 46. 5 × 0 .17 = 222.1 V. From the no-load, 1300 r/min magnetization curve, the corresponding field current is 1.50 A (again obtained using the MATLAB ‘spline()’ function). Thus, the effective armature reaction is
Armature reaction = (1. 67 − 1 .5) A × 1500 turns/pole = 251 A · turns/pole
part (c): With the series field winding, Rtot = Ra + Rs = 0.208 Ω. Thus, under this condition, Ea = Va −IaRa = 220.3. This corresponds to a 1300 r/min generated voltage of 236.7 V and a corresponding field current (determined from the magnetization curve using the MATLAB ‘spline()’ function) of 1.84 A, corresponding to a total of 2755 A·turns. Thus, the required series field turns will be
Ns =
or, rounding to the nearest integer, Ns = 11 turns/pole. part (d): Now the effective field current will be
Ieff =
2500 − 251 +20 × 46. 5
= 2.12 A
From the 1300 r/min magnetization curve, Ea = 246.1 V while the actual Ea = Va − RtotIa = 220.3 V. Hence the new speed is
n = 1300
= 1164 r/min
Problem 7. part (a): At full load, 1185 r/min, with a field current of 0.554 A
Ea = Va − IaRtot = 221. 4 V
where Rtot = 0.21 +0 .035 = 0.245 Ω.
An 1825 r/min magnetization curve can be obtained by multiplying 230 V by the ratio of 1185 r/min divided by the given speed for each of the points in the data table. A MATLAB ‘spline()’ fit can then be used to determine that this generated voltage corresponds to a field-current of 0.527 A. Thus, the armature reaction is (0. 554 − 0 .527)2000 = 53.4 A·turns/pole. part (b): The full-load torque is
T =
EaIa ωm
221. 4 × 34. 2
1185(π/30)
= 62.8 N · m
part (c): The maximum field current is 230/310 = 0.742 Ω. The effective field current under this condition will therefore be
Ieff =
2000 × 0. 742 − 160
= 0.662 A
From the 1185 r/min magnetization curve found in part (b), this corresponds to a generated voltage of 245 V. Thus, the corresponding torque will be
T =
EaIa ωm
245 × 65
1185(π/30)
= 128 N · m
part (d): With the addition of 0.05 Ω, the total resistance in the armature circuit will now be Rtot = 0.295 Ω. The required generated voltage will thus be
Ea = Va − IaRtot = 219. 6 V
This corresponds to 219.6(1185/1050) = 247.8 V on the 1185 r/min magnetiza- tion curve and a corresponding effective field current of 0.701 A. As can be seen from the data table, a no-load speed of 1200 r/min corre- sponds to a field current of 0.554 A. Thus the series-field A·turns must make up for the difference between that required and the actual field current as reduced by armature reaction.
Ns =
Nf (If,eff − If ) +Armature reaction Ia
=
= 9.8 turns
Problem 7. part (a): From the demagnetization curve, we see that the shunt field current is 0.55 A since the no-load generated voltage must equal 230 V. The full-load generated voltage is
Ea = Va − IaRa = 219. 4 V
and the corresponding field current (from the demagnetization curve obtained using the MATLAB ‘spline()’ function) is 0.487 A. Thus the demagnetization is equal to 2000(0. 55 − 0 .487) = 127 A·turns.
Operation in the fourth quadrant means that the motor torque is positive while the speed is negative. In this case the motor is acting as a generator and absorbing energy from the lowering load, which would otherwise accelerate due to the effects of gravity. part (c): -473 r/min
Problem 7. part (a): At rated load, Ea = 230 − 122 times 0 .064 = 222 V. Thus, rated- load speed is
n = 1150
= 1133 r/min
part (b): The maximum value of the starting resistance will be required at starting.
230 Ra + Rmax
= 2 × 122 = 244
and thus Rmax = 0.878 Ω. part (c): For each value of Rtot = Ra + Rext, the armature current will reach its rated value when the motor reaches a speed such that
Ea = 230 − 122 Rtot,old
At this point Rtot will be reduced such that the armature current again reaches 122 A. Based upon this alogrithm, the external resistance can be controlled as shown in the following table:
Step number Rext[Ω] Ea,min [V] nmin [r/min] [V] Ea,max [V] nmax [r/min] 1 0.878 0 0 115 587 2 0.407 115 587 173 882 3 0.170 173 882 202 1030 4 0.051 202 1030 216 1101 5 0 216 1101 - -
Problem 7. part (a): At no load, Ea,nl = Kmωm,nl = Va. Thus
ωm,nl = Va Km
= 405 rad/sec
Hence, the full-load speed is ωm,nl(30/π) = 3865 r/min. part (b): At zero speed, the current will be Ia = Va/Ra = 44.7 A and the corresponding torque will be T = KmIa = 9.4 N·m. part (c):
T = KmIa =
Km(Va − Ea) Ra
Km(Va − Kmωm) Ra
Here is the desired plot, obtained using MATLAB:
Problem 7. part (a): At no load, ωm,nl = 11, 210(π/30) = 1174 rad/sec and Ea,nl = Va − Ia,nlRa = 4.94 V. Thus
Km =
Ea,nl ωm,nl
= 4. 21 × 10 −^3 V/(rad/sec)
part (b): The no load rotational losses are
Prot,nl = Ea,nlIa,nl = 62 mW
part (c): At zero speed, the current will be Ia = Va/Ra = 1.09 A and the corresponding torque will be T = KmIa = 4.6 mN·m. part (d): The output power versus speed characteristic is parabolic as shown below.
An iterative MATLAB scripts can easily find the two desired operating points: 2761 r/min for which the efficiency is 24.3% and 8473 r/min for which the efficiency is 72.8%.
