Synchronous Generator Design Assignment Engineering 1931F Spring 2019
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Synchronous Generator Design Assignment
Engineering 1931F Spring 2019
Select an arbitrary amount of power between 30 and 250 MW and design a two-pole,
smooth-rotor generator for 60 Hz power generation. As a fundamental constraint on your
design, use a nameplate power factor of 0.9 for the output under proper operating condi-
tions. (I chose that specification to match that of a large Hitachi generator, the design of
which is discussed in a paper 1 posted on the class website 2. This also makes sense as giv-
ing stable, high-efficiency, startup conditions.)
In class, we made a very simple argument that the output power of a generator is directly
proportional to the volume of the rotor. The easiest approach to the design process is to
exploit this relationship. Instead of starting with the power, start with the size and calcu-
late the power. Adjust size and turns counts to get the desired generator.
It also turns out that any given mechanical arrangement, that is, rotor and winding design,
will have an optimum output current and voltage that you can only tweak rather than
specify in advance. The optimization comes from the constraint of the power factor
through a phasor calculation of the output power at fixed angle and full excitation E.
The design style shall be what we have discussed in class, namely both stator and rotor
windings are crude discrete stepwise approximations to a sinusoidal distribution. The
rotor induces a stepwise approximate sinusoid in each turn of a stator winding and
spreading the stator turns sinusoidally smooths the output further. The number of turns in
each phase of the stator is usually quite small. The design process for this exercise does
not ask you to work out the real spacing for with set of windings. Instead it uses equa-
tions based on continuous sinusoidal distributions of current. For the general idea behind
this approach, see the Appendix to the “Generators Description and Analysis” from the
class website under Class Notes and PowerPoints. At the end, you draw a couple of ap-
proximate graphs of how the stepwise approximation changes the output.
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 prod-
uct, 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 all the details of coolant flow. The three places where heat is gener-
ated are the rotor and stator windings, and the armature iron where there are hysteresis
and eddy current losses. Each is a different cooling problem. You will calculate the
amount of gas or water needed to carry off each component of the heat but do not have to
figure out the pipe and port sizes or required pressures. (Should I teach this course again,
I will fix that lost design opportunity!)
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.htmlENGN1931F – Spring 2019 Generator Design
To make this exercise feasible, I have written an Excel spreadsheet that lays out the re-
quired inputs and outputs of the calculation and has some of the data that you need in it
already. I will post it on the class website shortly and will also post a pdf file of my own
design of a 100 MVA generator 3. Table I gives a list of the inputs of the calculation.
There is a sheet called “Parameters” in the Excel workbook from which Table I was tak-
en that lists output and working parameters as well.
The Excel file will be missing some strategic cell formulae so that there will be some-
thing for you to enter by way of formulae. Table II below lists the missing formulae
along with some indication of their functions and units.
I have put the datasheets of two product data bulletins on the properties and usage of AK
electrical steels and of Carpenter core irons on the class website. You will use the AK
data for the stator laminations and will choose a Carpenter alloy for the rotor. There is
section below on “Selecting Steel for the Stator” which is a short guide to where data
may be found in the AK Steel user guides. There is a place in the Excel spreadsheet for
you to enter the relevant material properties for the steel that that you select for the stator
laminations. You will decide on the best choice of stator material by first estimating the
stator weight and then looking at the total losses in the stator.
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 and want to know what output voltage and cur-
rent the generator will give for nameplate ratings. The other macro 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 draw what the generator waveform will be.
A Few Rules of Thumbs and Pedagogy:
1. Do not choose an MVA rating between 85 and 115 MVA because that is my terri-
tory.
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 1.6 m.
3. One cell (H16) requires choosing the cooling gas and the rough rule of thumb is
hydrogen above 100 MVA and air below. The only place in calculations this af-
fects is the cooling gas flow calculations. I have not automated calculating opera-
tion with hydrogen. The unfortunate truth is that the cooling calculations are very
crude and inaccurate.
4. Limit the peak field on the rotor in rated operating conditions to 1.0 T to limit the
required rotor magnetomotive force. Very likely, this will not be a real limit as I
have found lower than that was natural in several calculations.
3
I DO NOT guarantee the correctness of any results in my design.
2ENGN1931F – Spring 2019 Generator Design
5. Results for many of the calculations depends sensitively on the gap between the
rotor and stator. That gap is proportionately larger than it would be for a motor as
stator inductance has to be minimized. The gap in this spreadsheet is maximized
based on the limit of rotor heating.
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 outside the stator slots about equal to the radius of the rotor.
This assures that the peak flux in the rotor and in the return path are approximate-
ly equal. That width is a parameter of the machine and you are free to change it by
a reasonable amount because the working B_r is much lower than the unloaded
value.
7. In calculating the stator core loss, it is necessary to estimate the peak armature B
field since the field induced by the output current in the stator coils tends to re-
duce 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 ro-
tating field with an angular distribution 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. 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. Since the net rotor field is near 1.0 T, the mean permeability at that field
strength is a reasonable value for this correction.
9. There is no correction for fringing in either the gap or the stator and rotor end
plates. Hence the coupling coefficient K = 1.0.
10. The lamination thickness is a matter of your choice but I think the range is lim-
ited. Read the AK product brochure on selecting materials for a sophisticated and
informed discussion of the factors that go into the choice.
11. The loss in the stator from hysteresis and eddy currents is based on the aggregate
loss per pound for the material at peak magnetization of 15 kG (1.5 T) as given in
the AK product brochures. Revise this for your actual peak under operating con-
ditions as the square of the ratio of B to 1.5 T.
12. 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 tried to use that notation in all refer-
ences to cells in column H, e.g., $H$6 for the angular velocity of the rotor.
13. If you are confused by something, ask me! I pretty much guarantee some confu-
sion as this is only the third time I have ever tried an exercise like this and I have
completely rewritten the spreadsheet completely. Very likely there is an outright
mistake somewhere.
Structure of the Spreadsheet:
The Excel spreadsheet is called “SynchronousGeneratorDesign2019.xlsm” on the class
website from which you can download it. As Table I shows, I have tried to organize the
3ENGN1931F – Spring 2019 Generator Design
‘Design Sheet’ tab so that inputs to the problem, e.g., stator turns, rotor size, etc., are near
the top of column B in cells that are green. There are a three more inputs similarly la-
beled with green further down column B. Similarly materials properties and other con-
stants are in column H and the calculated outputs are in column B starting at line 14. I
have duplicated the most important output values in a compact table starting at column G
line 27.
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. All the missing pieces are covered in the class
notes on the class website on generator design.
The spreadsheet does two distinctly different types of design calculations. The first done
on the ‘Design Sheet’ tab is the calculation of rotor current, machine dimensions, output
voltage, output currents, open-circuit voltage, etc. These are the discrete numbers that
characterize the machine operation. Most of these calculations are done with simple ap-
proximate formulae embedded in the output cells of column B. The optimum nameplate
output current is the most complicated calculation and requires maximizing output power
by adjusting the output current. There is a macro called OptimizeLoadCurrent that in-
vokes the “Solver” in Excel to find the maximum power. It depends on the output per
phase voltage being calculated correctly and the formula for that calculation is missing.
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 calcula-
tion scaled to a nominal 1 volt peak 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 ‘SineApprox-
imation’. The result is graphed on the ‘Open-circuit Waveform’ tab.
Selecting Steel for the Stator:
You need to choose a grade of non-oriented electrical steel and the lamination thickness
from the materials supplied by AK Steel. I have copied two Product Data Bulletins from
AK Steel’s website: “Selection of Electrical Steels for Magnetic Cores” and “AK Nonor-
iented Electrical Steels.” I urge you to read these through but to help you along here are
some of the most important points in these guides.
In “Selection of Electrical Steels…”
• Page 7: Composition by grade, particularly silicon weight percent. (Cost is affect-
ed heavily by silicon and generally permeability decreases with increasing silicon
but loss decreases.)
• Page 8: Gauge (lamination) thickness
• Page 12: Aggregate core loss, hysteresis and eddy current, per pound of steel by
grade and lamination thickness. This simplifies and improves the accuracy of sta-
4ENGN1931F – Spring 2019 Generator Design
tor loss calculations. (Cost is affected by thickness for reasons discussed in this
section of the PDB.)
• Page 19: Industry practice in grade selection. Your most important consideration
is loss at fairly high peak magnetization.
In “AD Nonoriented Electrical Steels…”
• Page 4: Interesting information on insulating the laminations
• Page 12: Magnetic properties by grade, including saturation magnetization, hyste-
resis loss, and maximum permeability at 10 kilogauss.
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 try changing the number of your stator turns
leaving other inputs untouched. What effect does this have on the major output? Does it
affect loses? Nameplate ratings?
Strategic Derivations:
There is really no upper limit to the rotor excitation or mmf you would like to get but
there is a practical limit. Whatever power you put into the rotor to raise the number of
ampere turns ( N ROT I ROT ) has to be removed by some cooling system. A generator is a
confined space and this limits the amount of power per square meter that can be removed
within the limits set by the materials of the system. Similarly, there is an upper limit on
the current density, the number of amperes per square centimeters of the rotor winding
wire. The power dissipated under the side of the rotor is just the resistance of the wires
along that side times the square of the rotor current. The wire resistance can be calculated
from the resistivity of the wire in the usual way. Let ρ be the resistivity and AC be the
2 ρ lROT N ROT I ROT
2
cross sectional area of the rotor wire, then PROT _ SIDE = . Let pMAX be the
AC
maximum power per square meter that can be removed from the surfaces of the rotor by
gas cooling. Divide the power PROT _ SIDE from the side of the rotor by the area of the rotor
PROT _ SIDE
to find pMAX ≥ . Substitute for PROT _ SIDE and rearrange terms to get
2π RROT lROT
N ROT I ROT pMAX π
≤
RROT ρJ
I ROT
where J = is the current density in the rotor wire. There is a practical minimum for
AC
current density set by the need to fit enough turns onto the rotor. This implies that the
number of ampere turns around the rotor circumference is limited to a certain constant
5ENGN1931F – Spring 2019 Generator Design
pMAX π
. I have made my best guess as to this practical number and have put it into cell
ρ J MIN
H11. You will need this result because the mmf cell is missing the formula.
Once you have found the mmf, you can trade off the selection of the peak open-circuit
voltage (emf) against rotor-stator air gap. In my design flow, you choose E first and see
what the resulting stator inductance and output current/voltage/and power will be. To
some extent you can adjust the optimum output voltage with this choice.
In class I derived simple formulas for the inductances and mutual inductances of wind-
ings that are spread about the rotor or stator in approximately sinusoidal distributions.
That information is collected in the “Generator Analysis Notes2019.doc” WORD file on
the class website. These led to the equivalent circuit shown below. The factor of 3/2 the
stator inductance as the principal part of the generator output impedance is the result of
the mutual inductances of the stator coils at 120 degree spatial rotation relative to each
other.
By inspection generator of the equivalent circuit below, the open-circuit voltage relates to
the terminal voltage and current as: eO −C = vG + iOUT ( RS + jX LEQU ) . You know the phase
angle of the output current relative to vG from the nameplate power factor. The effect of
RS on the calculated generator voltage under load is negligible and you can neglect it.
The real output power is the product of the real component of iOUT with vG . If you use
the peak values in this product you need to divide by 2 to get RMS and multiply by 3 be-
cause all calculations are done on a per phase basis. You select E first and then must do a
phasor calculation to find the relation between output voltage and current. Optimize to
find the best nameplate rating by varying the current to maximize power with power fac-
tor and E constant.
LEQU = 1.5 LST RS I OUT cos(ωt − θ ) LGRID
vG = VG cos(ωt ) vGRID
Open-circuit voltage
=e E cos(ωt + δ ) Generator - Grid
There is a “Phasor Spreadsheet” on the website along with the other generator materials
under Class Notes. (I like spreadsheets for poking-at-things calculations.) It has phasor
calculations for joining a generator to the grid on one tab and the current-voltage relation
of the generator itself on another tab. The phasor diagram below shows the critical con-
struction at optimum. See spreadsheet for more details including phasors for over and un-
der optimum current load.
6ENGN1931F – Spring 2019 Generator Design
In this phasor diagram, the vector marked “|I| is Optimum” is the voltage vector across
the source inductance of the equivalent circuit. The dotted vertical line is the magnitude
of the component of the voltage across the “source” inductance that is exactly out of
phase with the output voltage. Dividing that magnitude by the inductive reactance gives
the magnitude of the in-phase current. In turn, multiplying that by the generator output
voltage gives the real output power. In the design flow, the rotor mmf is set by heat re-
moval. You set E somewhat arbitrarily and use this trigonometric construction to deter-
mine the current and voltage output ratings that optimize the output power. The algebraic
result of this construction is what goes in cell B27. You are free to adjust the stator turns
to trade off voltage for current (insulation for heat removal) somewhat. The ultimate per-
formance is not very sensitive to the initial choice of E.
In the analysis of generator operation that we did in class (see write-up under class notes),
we showed that the peak voltage induced in the stator by this rotor flux is always
πω RROT lROT BPEAK N STAT
VSTAT _ PEAK = . You substitute a hypothetical, open-circuit
2
BPK _ OPEN to calculate E and the rotor-stator gap. Then work backwards to find the peak
rotor field in full power operation from the generator voltage. The two values of BPEAK ,
open-circuit and in operation, come from the way the B field in the rotor from the output
current subtracts from the excitation mmf field caused by the rotor current. [Note: the
7ENGN1931F – Spring 2019 Generator Design
generator never runs with the hypothetical BPK _ OPEN value because one never runs it with
full rotor mmf when there is no load. One of the results of our phasor calculations of the
generator is a specification of how the generator excitation and load must be adjusted to-
gether during operation.]
You can calculate the maximum B field throughout the armature iron by using conserva-
tion of flux. The design uses an annular shape for the return path. Let T be the thickness
of that annulus from the bottom of a wire slot to the outer edge and equate the total flux
in the two sides of the return path to the flux on the armature surface to get:
2TlROT BARM = 2rlROT BPK (1.1)
The peak field in the right hand side of this expression is the net peak field under rated
operating conditions. 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. There is some possibility to trade off higher arma-
ture field for a smaller armature here. The armature field is what causes eddy and hyste-
resis losses so this calculation determines those losses when combined with the stator
magnetic properties.
AK Steel specifies the total loss per pound of their products at 15 kG peak magnetization
for different lamination thicknesses. You may estimate losses at lower fields as varying
with the square of the peak field. You calculate the armature B field from equation 1.1.
In doing so you, you have to compensate for the net effect of both the rotor excitation and
the opposing field induced by the output current on BPK . The easy way to do that is to
realize that the generator output voltage is in fact the product of the number of stator
turns with the actual flux over half the rotor at whatever orientation angle the net field
peaks
Finally, in approximating sinusoid fields with steps, the simplest thing to do is to select
the rotor bar positions such that the sinusoid passes through the center of each step transi-
tion. You set a peak field strength for the total stepwise approximation slightly smaller
than the peak value of the sine wave (cell F11 on Rotor Winding Design tab). You calcu-
late 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.
Table 1: Input Variables Units Cell Number
Rotor radius meters B5
Rotor length meters B6
Nameplate power factor -- B9
N - stator turns -- B8
8ENGN1931F – Spring 2019 Generator Design
Peak open circuit rotor B with full rated excitation - used
to calculate rotor mmf and the theoretical equivalent
open-circuit voltage, E. Tesla B7
Number of rotor slots: (also used on Rotor Winding De-
sign worksheet cell B14) -- B34
Number of Rotor turns -- B35
Width of stator from end of slot to outer radius; possible
tradeoff of size of machine versus stator core loss. This is
narrowest width of the flux return path meters B69
AK Steel material properties for your choice of material various H14 et seq.
Loss in the stator windings as a percentage of the output
power; stator bar size calculated from this % B54
Table II: Missing Formulae and Data
From Design Sheet Item: Cell Units
Open circuit V_gen as peak Y- equivalent
voltage. This is related to the power fac-
tor argument in this write-up. Another
reference point is the spreadsheet on
phasor relations posted on the class web
site under Class Notes. That has under
and over excited cases as well as the op-
timum excitation. It shows what is being
optimized by the solver. B27 V peak
Selection of AK steel and all its properties
from AK product selection bulletins Column H rows 14 to 21 Various
Number of rotor ampere turns as limited
by heat and current density. Very simple
if you follow the argument in this write-
up. B36 Tesla
Rotor current B37 ohm
Rotor self-inductance (The rotor self-
inductance is related to the mutual in-
ductance by a constant.) B38 H.
Ohms per kilometer of the stator winding
bar/wire. (Stator might be wound with
either type of copper material.) B59 Ohm/km
9ENGN1931F – Spring 2019 Generator Design
Stator core losses from mass of stator,
AK’s loss factor, and peak B in the return
flux path. Correct for peak B relative to
15 kG of the AK specification using the
assumption that those losses will scale as
the square of B over this range. B75 W
Cooling gas flow in cubic feet per minute,
the unit most American vendors of cool- B96
ing equipment use. CFM
From 'Rotor Winding Design:
Angle from base plane of the rotor to
turn number B11 to B20 (Same formula all slots) deg.
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