More thoughts about inverters, power, speed and operating voltage.

Sometimes a writer writes just to get something off his chest.  Nothing could be more true of this blog post.  
I've still been mulling over the matter of understanding what is going on when my Powerspout generates electricity: 

• Why does it settle at a particular speed; 
• Why does the operating voltage settle at a particular point; 
• Why do speed and operating voltage change with different levels of power output; 
• How does the inverter control the output of the alternator; 
• How does the inverter have an effect on the speed of the pelton; 
• How significant is it if the system is not operating at the "sweet spot" speed.

If these are not questions that have ever kept you awake at night, this post is not for you.  Skip it and I'll try to write something more engaging next time.

For those as seriously obsessed as I find myself to be, see if you can follow my line of thought below.  I must give full credit to Hugh Piggott at Scoraig Wind for his corrections of my earlier efforts to articulate understanding in these matters, and by his corrections leading me on to a more clear-headed grasp; and also to "Flux" who writes on the Fieldlines discussion board and who has been giving me personal tuition to improve my comprehension.  These credits mentioned, any errors or plain wrong reasoning below are entirely mine.

1. A Powerspout in operation is a system in steady state; the speed is steady; the operating voltage is steady; the power output is steady.
2. The steadiness of this steady state happens because the energy available for conversion to electricity is un-altering; it is un-altering because for a water turbine having fixed nozzles, the flow does not change and neither does the net head.  
3. Fundamentally the steady state depends on an equilibrium being established between the torque output of the pelton and the torque requirement of the alternator.  
4. Not all the torque output of the turbine is available for generation of electricity.  Some is taken up in overcoming rolling resistance in bearings and some in overcoming ‘windage’: - the loss incurred as the pelton turns through the fog of spray inside the pelton housing, but since these losses are constant, the balance available to the alternator remains constant.
5. For an alternator whose output is fed to a grid tied inverter, this torque equilibrium demands that the alternator’s torque requirement be manipulated until it matches the torque available from the pelton, or more precisely that it matches the torque left over for the generation of electricity after bearing and windage losses are subtracted.
6. Such manipulation is carried out by an algorithm within the inverter. The algorithm can be either one which seeks the maximum power point (mppt mode) or one which ‘commands’ how much current is drawn from the alternator by reference to the voltage the alternator is producing (table mode). 
7. Whichever mode is employed, manipulation of the alternator torque is not without effects on the torque output and rotational speed of the pelton: the manipulation  actually changes that which it is trying to match. This comes about, not because of any change in the basic parameters, flow and head, which are respectively the primary determinants of torque and speed, but because the efficiency of the pelton in converting flow and head into useful torque is optimal at a certain rotational speed.  When the alternator imposes load on the pelton it changes the pelton's rotational speed, and this change, by moving the speed away from (or towards) the optimum speed, in other words improving or degrading the pelton’s efficiency, will have the effect of changing the pelton’s torque output.
8. So the steadiness of the steady state comes about by being the end point of a process of accommodation, between on the one side the alternator functioning under the influence of the inverter, and on the other side the pelton, which itself is ever responding to changes in the alternator. In effect there is a continuously operating feed back loop which comes eventually to settle at a point giving constant speed, constant voltage and constant power output.
9. The time it takes to reach this point varies depending which algorithm the inverter is using: mppt mode can take tens of minutes, much depending on the make of inverter, whilst turbine mode takes less than a minute.
10. Apart from the time  factor, another difference between the modes is that the point at which mppt mode settles should be the point where the rotational speed of the pelton is optimal.  It should truly be the ‘sweet spot’ for the system. Why - because it should be, by definition, the maximum power point, the point at which pelton efficiency is greatest and most power is extracted from the system’s head and flow. 
11. But if the algorithm is table mode, there is no such expectation.  The steady state rpm may, and almost certainly will, settle at a figure which is off from the ideal speed and the effect of this will be to make the system as a whole less efficient: it will generate less into the grid than the prevailing flow and head are capable of.  It will not be operating at the ‘sweet spot’ speed.
12. Two things need to be said to qualify these last mentioned differences between the two modes, leading to a deduction from the second:                                                                                                 (i) MPPT mode was devised to extract maximum power from solar panels.  There is a huge gulf between manipulating voltage and current from a photo-diode and manipulating the same parameters from a rotating generator. The complexities which apply to a pma alternator when it comes under load make it very different from a photo-diode, such that it may not be the case that the true maximum power point gets to be identified. Bench tests done by EcoInnovation do, however, lend support to mppt mode seeming to function with a pma in a satisfactory way.         (ii) The apparent draw back of table mode settling on a rotational speed which is not optimal may not be as dire as it might seem.  The fall-off in efficiency for a pelton turbine operating either above or below its optimum speed is not great: just a 3% fall in efficiency for operating 15% away from its optimum speed. So if the equilibrium point settles where operating speed is not further away than +/-15%, the loss in generating capacity will be slight. There will still only be a loss of 5% of attainable power when operating speed is 25% below optimal speed.
13. The deduction from this last point is that the one power curve programmed into a table mode inverter can probably be used to control turbine speed across a range of flows, without as has been suggested, the need to program in a new curve for each new flow regime.  As long as the output voltage of the pma falls broadly within the range of input voltage that the inverter is programmed to expect, the one power curve will cope with a range of flows albeit by fixing rotational speed differently for different flows.  So long as the different speeds are within + 15% and - 25% of optimal speed, the result will be acceptable.
14. Ensuring that the output voltage of the pma falls broadly within the range expected by the inverter is not an issue: it is the purpose of the EcoInnovation calculator tool to predict what the optimal rotational speed will be for a pelton at a site with a given head, and then to suggest which SmartDrive stators will give an output voltage of a chosen value at that rotational speed.

Well, that's got that off my chest: where I am at the moment in my understanding. But it could all change in the future ! 

For any one reading this far and wanting more, there is a description in the comments section following this earlier post, of what happens in a permanent magnet alternator to make the voltage rise as power output increases be less than would be expected from the increase in operating rpm; the piece includes definitions of the various terms used to describe the effects seen.

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, - Faraday and Lenz.

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 operating voltage 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.