Second Homework Assignment - Engineering 1931F Spring 2015
Select an arbitrary amount of power between 20 and 500 MVA and design a two-pole,
smooth-rotor generator for 60 Hz power generation. As the basis for your design, you
will also need to choose a nominal power factor for the output and a delta-connection
nameplate terminal voltage. Based on the specification for a large Hitachi generator, the
design of which is discussed in a paper1 posted on the class website2, use a power factor
of   0.9 at full nominal output. The output voltage of generators varies from 14 to 28
KV for generators above 10 MVA. I believe that a 500 MVA generator might have a
terminal voltage of 22 KV and a 50 MVA might be 16 KV with some interpolation in
between. You are free again to choose your own value but remember this is the deltaconnected output of a machine that is wye-wound.
The design style shall be what we have discussed in class, namely three stator windings
that are each spaced with their turn slots close to one another though spread enough to
smooth the steps that are caused by the discrete turns of the rotor winding. The rotor is
wound with turns spaced as a discrete step approximation to a sinusoidal distribution.
The rotor induces a stepwise approximate sinusoid in each turn of a stator winding and
spreading the turns of those windings will smooth the output further. (Real generators
have a wider stator turns distribution and the rotor distribution is adjusted to optimize the
waveform but that is more work than I wanted you to try.)
By “designing” I mean choosing quantitative values for the size of rotor and armature
(stator), optimizing the number of stator turns, determining the rotor ampere-turns product, choosing a number of rotor turns, calculating required “wire” sizes for both stator
and rotor, etc. While you will calculate the losses in each part of the generator, you do
not have to work out any of the cooling problems. (The three places where heat is generated are the rotor and stator windings, and the armature iron where there are hysteresis
and Eddy current losses. Each is a different cooling problem.)
To make this exercise feasible, I have written an Excel spreadsheet that lays out the required inputs and outputs of the calculation and has some of the data that you need in it
already. I have put the datasheets of Carpenter core irons on the class website. There is a
place in the Excel spreadsheet for you to enter the relevant material properties for three
grades of core iron that might be used in the stator laminations. The Carpenter datasheets
give the electrical conductivity, mean permeability, hysteresis loss per cycle and coercivity at the peak rotor magnetization. You will decide on the best choice of stator material
by looking at the losses (Eddy and hysteresis) in the stator. Table I gives a list of the inputs and outputs of the calculation. There is a sheet in the Excel workbook that lists the
cell numbers where the input parameters go.
1
Seijiro Muramatsu, Kado Miyakawa, Mitsuru Onoda, Kazuhiko Takahashi, and Kengo Iwashige,
Completion of a 1,120-MVA Turbine Generator for Huadian International Zouxian Power Plant in
China, Hitachi Review Vol. 56 (2007), No. 4
2
http://www.brown.edu/Departments/Engineering/Courses/ENGN1931F/index.html
ENGN1931F – Spring 2015
Homework 2
I will post the Excel file on the class website shortly and will also include a pdf file of my
own design of a 100 MVA generator3. The Excel file will be missing some strategic cell
formulae so that there will be something for you to enter by way of formulae. I will also
delete some of the properties of the three Carpenter Technology core irons to make you
read the datasheets. There are two macros in the spreadsheet, one of which optimizes the
number of stator turns and the other calculates and graphs the output voltage. The first is
invoked when you have the system dimensions in place. The other is used after most of
the generator is designed to show what the output is like. In the latter calculation, you
select a number of rotor winding slots and optimize the spread of the stator windings to
smooth the output.
A Few Rules of Thumbs and Pedagogy:
1. Do not choose an MVA rating between 80 and 120 MVA because that is my territory.
2. The power of a generator is largely proportional to the volume of the rotor. The
textbooks say the rotor is usually between 1.0 and 1.5 m in diameter but I think
the range for our MVA ratings is a little wider, possibly as small as 0.8 m and as
large as 1.7 m.
3. One cell (H16) requires choosing the cooling gas and the rough rule of thumb is
hydrogen above 100 MVA and air below.
4. Limit the peak field on the rotor in open-circuit operation to 1.3 T to limit the required rotor magnetomotive force. The B field in the stator and hence the eddy
current losses will be lowered by the load current in the stator windings during
operation.
5. Results for many of the calculations depends sensitively on the gap between the
rotor and stator. That gap is larger than it would be for a motor as stator inductance has to be minimized.
6. In calculating the mean magnetic path length, the peak return path magnetization,
and the mass and kinetic energy of the system, assume that the armature iron is an
annulus with width about equal to the radius of the rotor. That assures that the
peak B field at the surface of the stator coils and in the return path are approximately equal.
7. In calculating the eddy current losses, it is necessary to estimate the peak armature
B field since the field induced by the output current in the stator coils tends to reduce the armature field below what would be induced by the rotor alone, that is,
in open-circuit operation. The B field on the surface of the armature is a single rotating field with a distribution with angle that is nearly sinusoidal. This net field
on the armature surface is what supplies the flux in the armature coils. Notice that
the output voltage of the generator is simply whatever is induced by this flux in
the stator coil. Estimate the peak armature surface field from the output voltage
magnitude.
8. The data from Carpenter Technology on hysteresis loss was measured for core
iron cycling to +/- 10 KG. The volume loss changes with the strength of the peak
field but I am not exactly sure what that variation is. For the purpose of this exer3
I DO NOT guarantee the correctness of any results in my design.
2
ENGN1931F – Spring 2015
Homework 2
cise we will assume that the variation is directly proportional to the peak field as
calculated for eddy current losses.
9. A correction to the gap length from the finite permeability of the iron is optional
but I encourage it. Doing so gives a better feel for the effect of high permeability
material.
10. There is no correction for fringing in either the gap or the stator and rotor end
plates.
11. The lamination thickness is a matter of your choice but I think the range is limited. At 1 mm the Eddy current losses are significant and below .5 mm the laminations would be too difficult to handle. Try 0.5 and/or 0.65 mm.
12. The output open-circuit waveform is calculated incorrectly in Excel if the angular
spread of the stator windings is the smallest spacing of rotor slots. This is an artifact of my way of graphing the smoothing.
13. In Excel the dollar sign is used in cell references to prevent automatic changes to
a formula as it is copied or moved. I have used that in all references to cells in
column H, e.g., $H$6 for the angular velocity of the rotor.
14. If you are confused by something, ask me! I pretty much guarantee some confusion as this is the first time I have ever tried an exercise like this and it turns out to
be more complicated than I thought it would be.
Structure of the Spreadsheet:
I have placed a spreadsheet called “SynchronousGeneratorDesign2015.xlsm” on the class
website from which you can download it. As Table I shows, I have tried to organize the
‘Design Sheet’ tab so that inputs to the problem, e.g., MVA, rotor size, etc., are near the
top of column B. The exception to that rule is the two percentages of power loss to allocate for resistance heating the stator and rotor windings. Those entries are in cells B59
and B47 respectively. Similarly materials properties and other constants are at the top of
column H and the calculated outputs are in column B starting at line 22. I have duplicated
the most important output values in a compact table starting at column G line 18.
To give you something to do, I have removed some of the formulae. Table II lists the
cells and functions that are missing. These are simple formulas and there are a couple of
derivations and listings in the Strategic Derivations section of this write-up that should
help you figure out the missing pieces. I also left out some of the materials properties of
the Carpenter core irons. Their data sheet uses very strange units for electrical conductivity so I left my conversions in place. They do not give a hysteresis loss for the C iron, so I
left my estimate. The datasheets will be on the class website shortly.
The spreadsheet does two distinctly different types of design calculations. The first done
on the ‘Design Sheet’ tab is the calculation of current, machine sizes, output currents,
open-circuit voltage, etc., that is, the discrete numbers that characterize the machine operation. Most of these calculations are done with simple approximate formulae embedded in the output cells of column B. The number of turns in the stator requires the most
sophisticated calculation. We showed in class that for a given output, the required rotor
3
ENGN1931F – Spring 2015
Homework 2
B field is inversely proportional to NSTATOR when that value is too small and is directly
proportional to NSTATOR when that value is too large. This means there must be an optimal value for NSTATOR. There is a macro in the spreadsheet called ‘OptimizeTurns’ that
minimizes the B field for operation at full output. You fill in all your inputs and missing
formulae and invoke that macro to finish the calculations.
The second type of calculation is the approximation of the output waveform to a sinusoid
by finding the best rotor slot placement and the optimal spread for the stator windings.
This calculation done on the ‘Rotor Winding Design’ tab and is a relative voltage calculation scaled to a nominal 1 volt output. You fill in the slot angle formula, the number of
slots and the spread of the stator winding and run a second macro called ‘SineApproximation’. The result is graphed on the ‘Open-circuit Waveform’ tab.
What to Hand In:
I would like both an electronic copy of the spreadsheet as you finish it up for your choice
of power, size, etc. and a printed copies of the output table from the Design Sheet tab and
the graph of the sine approximation. Finally I would like a short written description of
what you tried to optimize besides the stator turns. Might that be the mass of the system
($$), rotor mmf (smaller exciter), core losses,…?
Strategic Derivations:
In class I twice derived the simple formula for the inductance of closely spaced stator
0 (r  g )lROTOR N 2
windings as LSTATOR 
. (NOTE: This formula has an error and the g
l
g  MAG
 REL
in the denominator has to be multiplied by 2. I can’t change that in time to hand this out
in class but will correct on the website soon.) I also explained without derivation4 that the
effect of the mutual inductance between the three stator coils is to increase the effective
inductance in the single-phase equivalent circuit by a factor of 1.5. That equivalent circuit is given below. By inspection the open-circuit voltage relates to the terminal voltage
and current as: eO C  vG  iOUT ( RS  jX LEQU ) . You know the output current from the
power generated and you know the phase angle of the output current relative to vG from
the nameplate power factor.
There is one simple result that is the basis for calculating both the rotor excitation or mmf
and the maximum value of B in the stator iron. In both cases if you know the surface integral of the B field over half the surface of the rotor as a function of the peak value of B,
then you can calculate that peak value from the voltage it induces in the stator. We have
designed our system such that open-circuit or under load, the output is always sinusoidal
which implies that B is distributed sinusoidally around the rotor. In general we do not
4
See Appendix A of Grainger and Stevenson, Power System Analysis, for that derivation. The factor 1.5 is
closely related to the same factor in the strength of magnetic fields generated by three-coil structures that I
did derive in class.
4
ENGN1931F – Spring 2015
Homework 2
know the exact angle at which the field changes sign so we simply call that   0 and
assume that B  BPK sin( ) . Integrating over the rotor or stator surface yields

ur uur
B

dS

rl
Ò

 BPK sin( )d  2rlBPK
0
LEQU = 1.5 LST
RS
I OUT cos(t   )
vG  VG cos(t )
Open-circuit voltage
e  E cos(t     )
LGRID
vGRID
Generator - Grid
You solve for the two important peak values of B by using this result to calculate the
open-circuit voltage and the loaded output voltage and working the equation backwards
to B from the known magnitudes of voltage.
You can calculate the maximum B field throughout the armature iron by using conservation of flux. The design uses an annular shape for the return path. Let T be the thickness
of that annulus and equate the total flux in the two sides of the return path to the flux on
the armature surface to get: 2TlBARM  2rlBPK . When the thickness of the armature return
path is the same as the radius of the rotor, i.e., T  r , the peak field in the armature iron
is the same as the peak field on the surface of the armature. This is the basis for calculating eddy and hysteresis losses.
The formula for eddy current loss per cubic meter is still in the spreadsheet so a correct
value for BARM will lead to a value that only has to be multiplied by volume to get the
loss. Hysteresis losses involve a similar calculation with the loss parameter from the iron
datasheets multiplied by frequency and peak B (in Tesla) to get the volume loss. As discussed above, the B field multiplier is a scaling factor for the change in loss when the
field cycles less than the 1 T peak field of the datasheet number.
Finally, in approximating a sinusoid with steps, the simplest thing to do is select the rotor
bar positions such that the sinusoid passes through the center of each step transition. You
set a peak voltage for the total stepwise approximation slightly smaller than the peak value of the sine wave (cell F11 on Rotor Winding Design tab). You calculate the values of
the voltages halfway up or down each step and use the acos() function in Excel to find the
angles for the rotor slots. The spreadsheet is set up so you only have to calculate for half
the number of turns and it calculates the remaining positions based on the symmetry of
the rotor. The spreadsheet use the total number of slots as input (cell F14) and looks for
one fourth of that number for the starting slot positions.
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ENGN1931F – Spring 2015
Homework 2
Table I: Input and Output Parameters Input Variables Nameplate power Nameplate power factor Units MVA RMS ‐‐ Cell Number B11 B12 Nameplate voltage ‐ Delta‐wound terminal to terminal RMS voltage nominal Rotor radius Rotor length Air gap on radius V RMS meters meters meters B10 B5 B6 B7 Number of rotor slots: (also used on Rotor Winding Design cell B14) Lamination thickness in the stator structure ‐‐ mm B15 B8 H in Oe for B value required at the rotor dia‐
metrical plane Remnant magnetization Oe Tesla B16 B14 Acceptable loss in the rotor winding as a per‐
centage of the output power % B47 Acceptable loss in the stator windings as a percentage of the output power Materials Data: Hysteresis loss factor of iron Resistivity of the iron Relative permeability of the iron Density of magnetic iron % J/cc/cycle ohm*cm ‐‐ Kg/meter^3 B59 H11 H13 H14 H12 Resistivity of copper for wire cross section calculation ‐ @ 75 C Calculated Outputs Effective stator inductance N ‐ stator turns Stator slot depth Rotor slot depth Peak stator B Peak open circuit rotor B Output RMS current ohm*cm H. ‐‐ cm cm Tesla Tesla Amp. H14 B36 B35 B63 B54 B39 B42 B31 Rotor mmf Rotor current Core losses total Total losses Efficiency Ampere*turns
Amp. KW KW % B44 B45 (B71+B67)/1000 B73 B74 6
ENGN1931F – Spring 2015
Homework 2
Table II: Missing Formulae From 'Design Sheet: Open circuit Y‐ equivalent voltage Stator peak surface B value Cell B38 B39 Units V peak Tesla Rotor diameter B value from rotor wind‐
ing same as rotor peak B open‐circuit B42 Tesla Stator hysteresis loss based on volume of pi*l*((2r+g)^2‐
(r+g)^2)*k12*frequency*(Bstator/Btest) Stator eddy current losses B67 B71 W W Rotor stored energy = 1/2 Moment of Inertia * omega^2 B79 J From 'Rotor Winding Design: Angle from base plane of the rotor to turn number <n> B11 to B20 (Same formula all slots) deg. 7