5.5 Análisis Estadístico ELDI
5.5.2 Baja Frecuencia
5.0 kVA / hp 5.6 By [6.90],
locked-rotor 1000 kVA / hp hp 1000 5.6 10
max 140.5 A
3 L 3 230
I V
locked-rotor 1000 5.0 10
min 125.5 A
3 230
I
The value of input current is read from Fig. 6.26 as approximately 125 A, which agrees with the lower end of expected full-voltage locked-rotor current.
6.24 Attempt to read the speed for which the output power is 1/4 hp will result in an inaccu-rate determination. If sp_perf.m is modified so that npts=5000, then the data point at location 4867 in each array is the point for 1/4 hp output. Then, the speed, efficiency, PF, and current arrays can be polled to find nm 1752 rpm, 67.6%, PF0.566, and
1 4.07 A
I for rated conditions.
6.25 For the 284T frame, Table 6.6 gives P D f 14.12 in. Allow a frame thickness 0.25 in
tf . Then, the lamination stack outside diameter is
5250 67.31 ft lb
1170
0.647 13.62 0.647 9.63 in 1.175 1.03/o 1.175 1.03/ 6
Use D9.6 in. The stator lamination stack length is then
In order to have a stator slot pitch in the range suggested by [6.102], select S136 from Table 6.7, giving
113
Estimate the maximum value of flux per pole by use of [6.109].
1/ 2
1/ 20.00300 60 0.00300 60
0.00145 0.00145 15 7.94 mWb
6 60
m p f Ps
From the formulas of Appendix A,
1 1 36 2 slots per pole per phase
s S 6 3
N pm
Since there are S p1/ 36/ 6 6 slots per pole , choose a pitch of 5 stator slots. This 5/6 pitch is a commonly used value for its favorable reduction of 5thand 7thharmonics.
1 0.9659 0.9659 0.933
w p d
k k k
A single circuit winding is chosen. Based on [6.108],
1 3
1 1
0.97 1 460/ 3
0.97 18.9 conductors per slot
2.22 2.22 0.933 6 2 60 7.94 10
Rounding to the nearest even integer, Cs 18 conductors per slot, or the coil has 9 turns.
The actual value of maximum flux can now be calculated using [6.108].
The full-load current is estimated by [6.113].
A design goal full-load efficiency has not been specified, but 89% is a typical value for this power range motor. For a current density of 4000 A /in2, the stator conductor cross-sectional area is
The closest satisfactory standard square wire from Appendix C.2 is No. 13 square. This wire size will be used but with film and glass insulation, giving a maximum conductor width over insulation of 0.0730 in. Each coil side will be made up of a 3 by 3 stack of conductors. The slot dimensions are set, based on the following work.
Slot Width
In order to allow for irregularities, use bs10.320 in. Slot Depth With irregularity allowance, use ds10.800 in.
Use [6.115] and [6.116] to make flux density checks.
1 1 1 0.838 0.320 0.518 in
9.6 10 97.2 kilolines/in 2m 13.62 9.60 2 0.800 4.25 0.96
Bstm is within the suggested limit of 120 kilolines /in2. Bscm is about 2.3 percent above the suggested upper limit of 95 kilolines /in2. However, this small variance is judged to not be significant enough to justify an iteration of the design.
The air gap is sized by [6.117].
0.0016D 0.001 a 0.0072 0.0016 9.6 0.001 4.25 0.0072 0.02681 in
Use 0.027 in.
115
The number of rotor slots is selected as S231, giving S S1 236 31 5 which avoids the undesirable combinations of [6.19]. The rotor slot lip is sized by [6.120] and [6.121].
2 0.01 0.045 0.01 9.60 0.045 0.141 in
br D
2 0.677 2 0.677 0.141 0.095 in
r r
d b
The rotor winding factors are calculated based on Appendix A formulas.
2 2 31 1.722 slots per pole per phase
s S 3 6
N mp
2 2
6 180
180 34.84 per rotor slot 31
2 0.9987 0.9698 0.9685
w pr dr
k k k
The rotor bar and end ring currents are calculated by [6.131] and [6.132].
1 1 1 2 2
1.8 9 0.933 36
1.8 18.56 336.37 A
1 0.9685 31 The cross-sectional areas are then found as
336.37 0.1103 in2
3050
b b b
s I
553.19 0.1649 in2 e 3355
r e
s I
Somewhat arbitrarily choose rdw2.5. Hence,
1 / 0.1103/ 2.5 0.210 in
Use a bar depth of dr1dr20.525 in. Then the rotor slot depth is
1 0.525 2 0.525 0.095 0.620 in
r r
d d .
The end ring design of Fig. 6.54 is to be used with the end ring depth equal to the bar depth, or der 0.525 in. Hence, the end ring width is
The flux density checks are made by [6.137] and [6.138].
This value is within the acceptable limit.
Use a 2.50-in diameter shaft through the rotor lamination stack and 1.00-in axial cooling holes.
61.8 kilolines /in2
Brcm
As is typically the case, rotor core flux density is not a problem area.
The above determined values are entered in the data file imdesign.m and imeval.m is executed. The screen display that results is as follows.
117
Inspection of the results shows three concerns. First, the required full-load speed is not satisfied. This can be addressed by reduction of the secondary resistance. Since the design has no constraints on cost or materials, it is decided to change the rotor bar and end rings to a copper casting without dimension change. The other two concerns are a low PF and a less-than-desirable efficiency. It is decided to add 0.50 in to the lamination stack length to reduce flux densities and, consequently, core losses to improve the effi-ciency. This action will also serve to improve the PF. As a final effort to improve the PF, the air gap was reduced to 0.020 in, thereby increasing the magnetizing inductance so that the reactive power requirement is lowered. After these three corrective actions, the resulting screen display from execution of imeval.m is shown below. The motor now meets the full-load speed requirement and thus satisfies the design requirement. Notice that the PF has improved from approximately 0.73 to approximately 0.80. The efficiency is also about 2.6 percent greater than for the original design.
Chapter 7 Problems Solutions
7.1 For any particular value of If , / 3
s L L
X V
I
where VL is read from the OCC and IL is read from the SCC.
The OCC and SCC data from Fig. 7.16 are used in the following MATLAB program Xs1.m to calculate and plot Xs vs. If .
%% Xs1.m - Uses OCC/SCC data from Fig 7.16 to solve
% Problem 7.1.
clear;
VL=[0 2.85 5.7 8.55 10.9 12.9 14.7 16.2 17.5 ...
18.6 19.3 19.8 20.2]*1000;
If=[0 100 200 300 400 500 600 700 800 900 1000 ...
1100 1200];
IL=21500/1200*If;
m=length(VL);
for i=2:m; Xs(i)=VL(i)/sqrt(3)/IL(i); end plot(0,0,If(2:m),Xs(2:m)); grid
xlabel('Field current, A'); ylabel('Xs, ohms');
119
7.2 Since insufficient information is available to determine Er, use Er Van. Enter the OCC curve of Fig. 7.16 with VL 13.8 kV to find Ifu 490 A and Ifs550 A. By [7.26],
0.15 0.85 0.15 0.85 0.891 0.898 0.815
s su s su
X X k X
This value for Xs is slightly larger than the value of Example 7.3 due to the fact that the saturation factor ks is only applied to X, the portion of Xs that is affected by satura-tion. This problem has emphasized that the method used to determine Xs in Example 7.3 is an approximate method. Although the error was about 2.4 percent in this case, the error becomes greater as the level of saturation increases.