The set up

The set up
5.46mm jet delivering 0.68 l/s to the pelton which is rotating at 900 rpm and generating 135 watts into the grid.

Friday, 1 January 2016

The relationship between power, speed and voltage.

Note added after this post was published: the comments section at the end corrects some of what I write below.

2016 and the start of a new year, - but the science theory I write about in this entry dates from 1831. How true it is that we stand on the shoulders of the giants of former times.

In April 2015 I installed a tachometer, (which you can read about here) and ever since I've been measuring the Powerspout's rpm and dc voltage at different levels of power output. 

Gathering this data has produced some interesting results which have helped me understand better how shaft speed and operating voltage* vary according to the power generated.  In this post I want to write about what I've found, and to explain my understanding of the physics behind why the results are as they are.  

I am not someone who is a little qualified to write about the matter, - I am not qualified to write about it at all ! - and so some may find my explanation incomplete and insufficiently technical. But others like me, trying to work it through for themselves might be helped by an attempt at a simple explanation. For the subject is complicated,  and it gets more complicated the deeper you venture into it.  Reasoning it all through carries a serious headache warning. You have been warned !

When operating voltage and rpm are plotted on the same graph against power output to grid, this is what the two plots look like:

Take note that the scales on the two vertical axes, rpm and dc volts, are numerically the same. Because the scales are the same, we can make an observation, a valid one, about the slopes of the two plots: the slopes are not the same.  To put it another way, operating voltage does not increase proportionately with rpm. 

This seems a strange finding.  The first axiom of a pma (permanent magnet alternator) is that its voltage rises as the speed of rotation rises.  What can be happening to make voltage rise more slowly?

When 'volts per rpm' (v/rpm**) is plotted at different levels of power output, this is what we find:

Note that the scale along the horizontal axis is the same as in the first graph. This allows us to say that over the same range of power as is depicted in the first graph, the v/rpm decreases. 

Putting the findings of these two graphs together, as power increases we can conclude that: rpm increases, v/rpm decreases and the overall effect of the two together is an increase, albeit a modest one, in operating voltage. 

The second axiom of a pma says that voltage is inversely proportional to load, ie voltage goes down as load goes up; having established that voltage (expressed as v/rpm) goes down as power increases, it should follow that load increases as power increases, which is to say: more power is more circuit load. Is this the case or have we violated the second axiom?

The way through to answering this is best reached by thinking about the current flowing in the pma stator coils as power increases.  

When more water is made to strike the pelton (from having changed the nozzles for larger ones), the torque on the pelton is increased.  This increase in torque translates through to the electrical side mostly as increased current flowing through the stator coils. We have to say 'mostly' because it is evident from the first graph that rpm increases too and that must mean some increase in voltage, but the main effect nevertheless is to increase current: more hydro power is more current.

Going back to the second axiom in its re-arranged form, it said "more power is more circuit load". Is this the same as saying "more power is more current ? - and the answer is "yes, it is", because load and current equate to each other.  (load is resistance; more load is less resistance; less resistance (at constant voltage) is more current (Ohms Law): therefore more load equates to more current).  The second axiom is thus seen to remain intact.

Why should v/rpm decrease as power output increases? It's because of an effect of the higher current flowing when hydro power is more, and that effect is of a magnetic field being created by the current as it passes through the coils of the stator.  The field so created opposes the magnetic field which caused the current in the first place, - so it opposes the field created by the spinning magnets of the rotor. The opposing field induces a correspondingly opposing voltage in the coil, called a 'back voltage'***.  And because the back voltage opposes the voltage induced by the rotor field, the 'net' voltage actually leaving the pma gets to be reduced. 

Well, net voltage (= operating voltage) would be reduced if rpm remained constant. But as we saw above, rpm did not remain constant: it rose with increased power generation to make operating voltage go up, albeit modestly. So the unit we have to use to see that the 'net' voltage does indeed go down is the unit of v/rpm. 

The laws of physics that describe these phenomena are Lenz's and Faraday's Laws.  For those interested, they can be looked up under the links given, - although some descriptions I found quite difficult to follow because they usually talk about these effects in motors rather than generators. For me, working it out from observed data and relating it all specifically to a Powerspout, has been the way to greater understanding. But as I said at the start, there is much more to it than this basic run through.

They do say it's therapeutic to keep the older brain thinking !

*in an earlier version of this post I used the terms 'operating voltage' and 'MPPv' (maximum power point voltage) to be the same thing. But as Hugh has pointed out in a comment, if I am using a WindyBoy in turbine mode, the term MPPV is not correct: the mode does not seek a maximum power point.  So I have dropped the term MPPV as of today: 27Jan 2016.

** v/rpm is simply obtained by dividing the value for MPPv at a given level of power output by the turbine rpm at that power level. The value for v/rpm in open circuit (Voc) is used to designate stators with different coil configurations.  When measured in open circuit no current flows, so the v/rpm at Voc will be the highest v/rpm ever possible for that stator.

*** back voltage is more properly termed back electromotive force, or back emf.


Hugh said...


I am guessing that you are using wind mode and so the power vs. voltage is determined by the table. As such it’s not necessarily MPPv but it’s the voltage the inverter is using based on the table it has in it. It seems to give good power output but as I noted before that is partly good luck and a good choice of stator. I would rather call it the operating voltage point or just “voltage".

"The first axiom of a pma (permanent magnet alternator) is that its voltage rises as the speed of rotation rises.” Well this is only true of open circuit voltage. Actual voltage equals open circuit voltage less internal voltage drop. Internal voltage drop equals current x impedance. Internal impedance is what slows the delivery of current. It partly comprises resistance (ohm’s law says volts = amps x ohms) but partly a reactive element that relates to distortion of the magnetic field in the stator windings.

It doesn’t really make sense to look at V/rpm except in the context of open circuit voltage. Voltage on load can fall as low as zero depending on the load current. At low revs the impedance is mostly resistive and it works in a similar way to the resistance of the cable and causes a loss of both voltage and power in the copper wires. At high revs the reactive part of the impedance is strong enough to limit current without so much loss of power in a short circuit situation.

"The second axiom of a pma says that voltage is inversely proportional to load, ie voltage goes down as load goes up” It is true that the open circuit voltage is the maximum and it will be reduced by a load current at mentioned above. The drop in voltage is more or less proportional to current although it will be steeper at high revs than at low revs. Strictly speaking it is not an inverse proportion relationship since doubling current will not halve the voltage but the voltage drop is roughly proportional to current. Actual operating voltage equals open circuit voltage minus voltage drop.

"(load is resistance; more load is less resistance; less resistance (at constant voltage) is more current (Ohms Law): therefore more load equates to more current).” Yes more load equates to more current. The resistance part is a bit garbled unless we are speaking of a resistive load. (such as a heater). The inverter is far more complex and although it may have some internal resistance this needs to be very small as it is using the current (converted) to feed power into the grid. I would use ohms law to work out the voltage drop in the cables and (to an extent) in the stator but I would not apply ohms law to the load side of the circuit.

I am not sure that you have quite got the hang of back emf. When a motor spins it generates back emf (being effectively a generator) that counters the supply voltage. When up to speed with no torque, the back emf more or less matches the supply voltage so that very little voltage is left to drive current through the coils. Only a very small current is needed to overcome windage in the free spinning motor. As a torque is applied the motor slows, the back emf drops, a voltage difference appears that drives a current and this creates torque which meets the mechanical load and the motor is working. Back emf is the voltage induced by the movement of the coils and is similar to the voltage produced by a generator except that being a motor is is “back” against the supply voltage and current. That’s my take. I have made a bit of a study of this stuff since childhood.

The current does interfere with the magnets by distorting their field and this is variously called armature reaction and reactive impedance and stuff like that but it basically pushes the magnetism off sideways until it effectively chokes the current off at a certain speed and limits current to a flat maximum limit.

A bit garbled but maybe you get my drift.

Bill said...


When I started asking myself questions about why my results looked as they did, I knew the answers were a bit beyond me. But I also knew the answers would be located in the minds of those who understand these things better. You being one such person, I’m really glad for your input here to set me straight. Thank you so much.

To try to summarise the form my understanding now takes, some several weeks after you posted your comment, I offer the following in case it helps others. What I’ve written is born out of your explanations and incorporates further explanations your “guru” has helped me with since you put me in touch with him:

1. MPPv is, indeed, not the correct term when the inverter is in table mode; I have changed it throughout the post to ‘operating voltage’.
2. The term ‘back voltage / emf’ is not a good way of describing the cause of the reduced operating voltage seen at progressively higher levels of power output. Such a term is best restricted to motors not generators. Rather, the reduced operating voltage does not have a single cause but several causes, one of which is weakening of the flux field of the rotor magnets caused by the opposing field created by the current flowing in the stator coils. This effect, as your guru pointed out, is significant when the magnet field is relatively weak (as it is when created by the ferrite magnets of a SmartDrive). When the magnet field is strong, as it is with neodymium magnets, such an effect is less significant.
3. This weakening of the magnet field by an opposing field I now understand as being called ‘armature reaction’. Incidentally, at first I found this term a bit confusing; to me ‘armature’ had meant only the revolving part of a generator or motor and in this sense it could not apply to the fixed stator of a SmartDrive; then, from the dictionary, I discovered that armature can refer to any part of an electric machine within which an emf is induced: -so the term does make sense after all.
4. ‘Armature reaction’ is a major reason, but only one reason, for operating voltage to fall as the level of power output rises: another reason is ‘flux leakage’. Flux leakage describes, as the term implies, a failure of the stator coils to take full advantage of the revolving magnetic field. How great its effect depends on the level of power output, being less at modest power levels and greater at higher levels. To the load connected to the alternator, in our case an inverter, the effect is like an inductor being connected in series with the stator coils, adding an inductive reactance (impedance) to the purely resistive component inevitably contributed by the resistance of the coils themselves. The material effect is for voltage to fall and to fall progressively as power output rises.
5. In the interests of completing the nomenclature used for what happens in an alternator under load, the term often used to describe these two effects is “synchronous reactance”. This is used because it is not easy in practice to separate the effect on voltage drop contributed by flux leakage from that contributed by armature reaction, and so the two tend to get lumped together under the term “synchronous reactance”, which is the same as your term: “internal voltage drop”.
...continued in next comment

Bill said...

6. Operating voltage then, as you say above, will be open circuit voltage less the voltage drop resulting from synchronous reactance.
7. The open circuit voltage will be the open circuit voltage at the speed which the turbine / alternator settles to for each level of power output.
8. This speed is determined ultimately by the point at which the torque output of the pelton is exactly met by the torque requirement of the alternator.
9. In reaching this equilibrium, a factor which needs to be brought in to consideration is that the torque output of the pelton itself will be changed by the load imposed by the alternator.
10. Such a change in torque output of the pelton comes about when it comes under load, not because of any change in the primary determinants of pelton torque, namely net head and flow, but by the speed of the pelton moving away from, or toward, that speed at which it is maximally efficient at converting the available head and flow into useable torque.
11. This point is the ‘sweet spot’ for the system.
12. The ‘sweet spot’ happens when the pelton’s speed, measured at its pcd, is about half jet velocity (more precisely 0.46 x Vjet).

I’m intending to write a new blog post under the title “More thoughts about inverters” bringing in some of the understanding touched on above. It might not be any more sound, scientifically speaking, than my earlier efforts so feel free to comment further!
As ever thanks for taking the time to put your mind into gear about all this.