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« on: September 10, 2009, 06:48:52 PM »
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Understanding Electric Power Systems Part 1

It seems that a lot of people have trouble understanding electric power systems. If you are confused about how electric motors work then follow along this series and I am sure by the end of this article you would have a better understanding of the electric power system. I am presenting this article assuming that you have no prior experience or knowledge in this field and are complete newbie. I will start with a simplified system and then go onto more complex ones including real time factors like resistances, back emf of motors etc. What I am going to present was learned by me over the years through internet & other sources etc. I had a soft copy but unfortunately lost it…now all is left a hard copy with me.

Lets imagine a simple electric motor which we call the ideal motor. We call it ideal because it has extremely simple characteristic and is 100% efficient in operation. Our motor has a single quality: for every volt of energy applied to it, it turns exactly 1000 rpm. On 5 volt the motor will turn 5000 rpm, on 25 volts the motor will turn 25000 rpm. The rpm will always be equal to the amount of volt applied to the motor times 1000

Now we need a power source for the motor. Let’s imagine that we have an ideal cell to match our ideal motor. For now, our ideal cell can be defined by a single characteristic: the amount of voltage it produces. Lets assume for simplicity a cell produces exactly 1 volt of electricity.

So no we can make up a pack of ideal batteries and attach to our motor. Lets see what the results would look like:

Cell count to RPM Relationships for our Ideal Motor
No of Ideal CellsRPM
11000
22000
33000
44000

Our ideal motor and ideal cell make life very easy for us. We can almost use the words cell and volts interchangeably because of our contrived values. The real world isn’t that simple, but its not entirely unlike our ideal world either. Keep that in mind as we go along.
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« Reply #1 on: September 10, 2009, 06:50:05 PM »
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Our ideal motor and ideal cell make life very easy for us. We can almost use the words cell and volts interchangeably because of our contrived values. The real world isn’t that simple, but its not entirely unlike our ideal world either. Keep that in mind as we go along.

The relationship between volts (cell) and rpm for our power system can work backward as well as it does forward. So if I measure that my ideal motor is spinning at 4000 rpm I can bet that the input voltage is 4 volts. That in turn means that 4 cells are being used.

But you may notice that our motor is just happily spinning away and doing no work. We need to add something to output shaft so that it can twirl around and move lots of air. So lets put a propeller on our motor and watch what happens to the RPM. In fact, lets compare two different propellers with the same motor and cell count.
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« Reply #2 on: September 10, 2009, 06:56:43 PM »
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If you are really new to RC then you might need to know how props are specified. Each prop has a diameter and a pitch. I assume you know what diameter means, but the word pitch can be a little confusing. The pitch is defined as the distance the prop would travel forward in one revolution in a perfect medium. The higher the pitch the more angled the blades of the prop are and the farther it would travel in single revolution. A high pitch is usually used on a fast plane while a low pitch prop is usually used on a slow plane. Out of the two props we will use one will have a 5 inch dia and 5 inch pitch, designated as 5x5. The other will have a 12 inch dia and 8 inch pitch, designated as 12x8.

Cell count to RPM relationship for our ideal motor and two different props
No of ideal cellsPropRPM
15x51000
25x52000
35x53000
45x54000
And on and on...
112x81000
212x82000
312x83000
412x84000
And on and on...
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« Reply #3 on: September 10, 2009, 06:57:26 PM »
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Nothing in the above table should surprise you because our ideal motor always turns 1000 rpm for every volt regardless of what kind of load is placed on the output shaft. This is a crucial point and its real world analogy is one of the hang-ups that keep many beginners from understanding electric power.

Of course we must also realize that it takes far more energy to spin a 12x8 prop at 4000 rpm than it does to turn a 5x5 prop at the same rate in the real world. There must be something missing from our simplified motor model or all those Speed 400 pylon racers out there would be flying with 18x18 props.

Indeed, the thing we are missing is called current. Current is the other half of the energy equation. Sadly, we can no longer go on further without introducing some kind of formula into the discussion.

Power (Watts) = Volts x Amps

The formula relates power (energy/time) as a product of Volts and Amps. You see a volt is just a part of the energy equation. Without amps, you don’t have any energy at all.
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« Reply #4 on: September 10, 2009, 06:58:12 PM »
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The formula relates power (energy/time) as a product of Volts and Amps. You see a volt is just a part of the energy equation. Without amps, you don’t have any energy at all.

The word watt is just a unit of power. Lets put watts to work right now. Remember the last chart that showed our ideal motor mated with a variable number of ideal cells to turn two different props?? Remember how the rpm was always dependent solely on how many volts were applied to the motor, even though it takes a lot more effort to turn a big prop than it does to a small one? Lets use watts to show just how much effort is involved:
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« Reply #5 on: September 10, 2009, 07:03:51 PM »
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Amount of energy required to spin two different props
Rpm5x512x8
10001 watt10 watts
20004 watts40 watts
300010 watts100 watts
400025 watts250 watts
note these are mythical values

Please note that these values are not actual values. I am just using simple values (that mirror realty in a few key ways) to demonstrate a point. And that point is that it takes a lot more power (energy) to turn a 12x8 prop at 4000 rpm than it does to spin a 5x5 prop. Also it takes more than twice as much power to spin a prop at twice as much rpm.
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« Reply #6 on: September 10, 2009, 07:13:22 PM »
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Since we now know how to express energy (watt= volts x amps), we can pull an example out of the table above and see what is going on with our ideal motor. Lets concentrate on the table entry which shows that it takes 100 watts to spin a 12x8 prop at 3000 rpm. Since we know that watts is the product of volts and amps this means that we would need a combination like the following examples:

Cell count to RPM relationships for our ideal motor and two different props
Cells/voltsCurrentPropRPMPower
115x510001watt
225x520004watts
335x5300010watts
465x5400025watts
Cells/voltsCurrentPropRPMPower
11012x8100010watts
22012x8200040watts
33312x83000100watts
46312x84000250watts
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« Reply #7 on: September 10, 2009, 07:17:20 PM »
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So here’s what we have learned so far:

•Our motor turns exactly 1000 rpm for every volt regardless of load.
•Our motor draws the current necessary to make the watts of electrical energy equal to the watts of power it takes to turn the prop at the rate demanded by the voltage.
•Watts is the product of volts and amps.
•A big prop requires a lot more power to spin at a specific RPM than a small prop does.
•Given a fixed prop and motor, increasing voltage will increase current at exponential rate

Based on what we have learned in this part, you can try and answer the following questions:

•At what RPM will an ideal motor run if I apply 17 volts?
•How much current will an ideal motor draw if a 12x8 prop is attached to it and run on an ideal cell? 2 cells? 3 cells?
•Suppose I had an ideal airplane that was lacking I power but I didn’t want to add cells, what would you suggest I do?


That’s enough for part one. I will continue this topic in next discussion. Next part will have the answers to the above question and we will add a little more real world effects until finally we have a complete understanding of the power system.
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« Reply #8 on: September 12, 2009, 06:05:59 PM »
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Understanding Electric Power Systems Part 2

Here are the answers to the last parts questions.

At what RPM will an ideal motor run if I apply 17 volts?
17000 RPM. This is true regardless of what prop I put on the motor because the motor always turns exactly 1000 rpm for every 1 volt applied to it.

How much current will an ideal motor draw if a 12x8 prop is attached to it and run on an ideal cell? 2 cells? 3 cells?
According to tables in the article, the current values are 10, 20 and 33 amps respectively. Note that the current goes higher and higher as the voltage increases. These values are mythical and are simply chosen for demonstration purposes.

Suppose I had an ideal airplane that was lacking I power but I didn’t want to add cells, what would you suggest I do?
Switch to a bigger prop. Our ideal motor will turn the new prop at the same rpm as the smaller one, which will provide the airplane with more power. The new prop will draw more current than the old prop.
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« Reply #9 on: September 12, 2009, 06:48:05 PM »
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Current Consumption and Cell Capacity

Today we are going to expand our ideal power system by improving our definition of our ideal cell. Thus far, we have described an ideal cell based on a single characteristic: the fact that it only produces only 1 volt of electricity.

We have also explored how much power is demanded by the motor based on the kind of prop we stick on the shaft and the number of cells we use. But it is at this point that reality rears its ugly head and forces our way into our discussion. Alas, we cannot have all the energy we want forever. Even our ideal cell has a limit to how much energy it can store and, therefore, provide to the motor. This limit is called cell’s capacity.
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« Reply #10 on: September 12, 2009, 06:56:04 PM »
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Let’s fix our ideal cell capacity at a nice round number of 1 amp-hour. This means that our cell can store enough energy that it can produce 1 amp for 1 hour. If we take out more than 1 amp then the cell will run out of energy earlier than 1 hour. For example a 2 amp draw would deplete the cell of energy in half an hour and a 4 amp draw will deplete the cell of energy in 15 minutes.

Length of time a 1 amp-hour cell can Supply Energy at Various Current Draws
CurrentVoltage
1 amp60 minutes
2 amp30 minutes
3 amp20 minutes
4 amp15 minutes
........
20 amp3 minutes


The Duration Equation

Lets express the above as an equation. Our ideal cell makes this easy for us:

Duration= 60/Current

The duration is the length of time (in minutes) that a cell can provide energy, given the measured current draw. Note that the above formula is only valid for ideal cells. The real world formula is slightly complicated.
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« Reply #11 on: September 12, 2009, 07:01:55 PM »
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Adding cells together

In last part, I discussed using more than one cell to power a motor. These cells were assumed to be wired together in a chain, with the positive end of each cell connected to the negative end of the next cell. Series winding adds up the voltage but keeps the capacity same as that of a single cell.

Voltage and Capacities of cells Connected in Series
No of CellsVoltageCapacity
11 volt1 amp-hour
22 volt1 amp-hour
33 volt1 amp-hour
44 volt1 amp-hour

As you can see in the above table, wiring cells in series does not change the capacity of the battery.
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« Reply #12 on: September 12, 2009, 07:56:16 PM »
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Computing Duration of Various Power Systems Combination

Now that we know how to compute duration, we can take another look at the power systems presented in 1st part and compare them in more detail.

Duration of Various Power Systems
No of Ideal CellsCurrentPropRPMPowerDuration
115x510001 watt60 mins
225x520004 watts30 mins
335x530009 watts20 mins
465x5400024 watts10 mins
No of Ideal CellsCurrentPropRPMPowerDuration
11012x8100010 watts6 mins
22012x8200040 watts3 mins
33312x8300090 watts2 mins
46312x84000240 watts1 mins


In each row of the table, note that the value in the duration column is equal to 60 divided by the value in the current column. The duration starts off at 60 minutes but drops dramatically as current goes up or the prop size is increased.
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« Reply #13 on: September 12, 2009, 08:45:42 PM »
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One thing that escapes many newcomers to electric fliers is that adding cells will decrease the full throttle duration of a power system. Take a look at the above table and see for yourself how the duration drops as cells are added. This is because the current draw goes up as the voltage increases. The result is that adding a cell will give you more power for less time. Of course, if you have an Electronic speed controller to vary the throttle setting in-flight then this is not an important concern.

Watt Hours

Since our ideal cell can deliver 1 volt and I amp-hour, and since we know that volts and amps combine to make a watt, we can also describe our ideal cell using the term watt-hours. A watt-hour is a unit that describes how much total power is available from a cell over a given amount of time.

Watt-Hours  = Volts x Amp-Hours

Naturally, I’ve thought ahead on this and chosen really easy values that make the Watt-Hour rating easy to compute. Our cell is exactly rated at 1 Watt-Hour because it provides 1 volt and has a 1 Amp-hour capacity.
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« Reply #14 on: September 12, 2009, 08:51:49 PM »
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If we combine cells into battery through a series connection then we are adding up watt-hours. Let’s jump back to an earlier table and add a Watt-Hour column

Voltage and Capacities of cells Connected in Series
No of CellsVoltageCapacityWatt-hours
11 volt1 amp-hr1 Watt-hr
22 volt1 amp-hr2 Watt-hrs
33 volt1 amp-hr3 Watt-hrs
44 volt1 amp-hr4 Watt-hrs

Imagine for a moment that our ideal cell supplied two volts. This would double the amount of energy in the cell and thereby double how much power we could get out of it in a hour. This kind of cell would have a watt-hour rating of 2. Like-wise, doubling the capacity of the ideal cell would double the watt-hour rating.

Based on idea of watt-hour, we can start to check our work. Since our perfect motor system has no losses, we can double check the watt-hour rating coming out of the system based on the watt-hour going into the power system. From now on I’ll be abbreviating watt-hours and amp-hours as Wh and Ah respectively.
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« Reply #15 on: September 12, 2009, 10:24:40 PM »
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So, referring to table attached (sorry..its really difficult to put tables in the posts), we can see that a 3 cell battery pack contains 3 Wh of power. This is because it is composed of 3 cells, each of which contains 1 Wh.

This 3 cell pack produces 9 watt for 20 minutes when powering our ideal motor fixed to a 5x5 prop. This is the same as 180 Watt-minutes, or 3 Wh. As you can see, the number of Wh going into the system is the same as the number of Wh coming out of the system.

With the 12x8 prop, our 3 cell pack produced 90 watts for 2 minutes. Again, this is 180 Watt-minutes, or 3 Wh.

table.jpg
Re: Understanding Electric Power Systems
* table.jpg (57.5 KB, 800x297 - viewed 3259 times.)
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« Reply #16 on: September 12, 2009, 10:28:18 PM »
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Summary

Let’s summarize our model of an ideal power system:

•The ideal motor spins at 1000 rpm for every volt of energy supplied to it, regardless of load.
•The ideal cell produces exactly 1 volt of energy and has a capacity of 1 Ah.

And let’s summarize what we’ve learned in this part:

•The duration of an ideal cell is equal to 60 divided by the current being drawn.
•Capacity is specified in Amp-hrs (Ah). A cell rated at 1 Ah can provide one amp for one hour.
•Ideal cells connected in series to form a battery have their voltages added, but their capacities do not add together.
•The amount of power a cell can produce over time is indicated by its Watt-hour rating. This is the product of the cell’s voltage and Amp-hour rating. Our ideal cell is rated at 1 Watt-hr meaning it can provide 1 watt for 1 hour.
•The no watts going into our ideal system is equal to the number of watts coming out. Likewise the number of watt-hrs going into our perfect motor system is equal to the number of watt-hours coming out.
•A battery can deliver a little power for a long time or a lot of power for a little time. Either way the total amount of energy coming out is fixed by its Wh rating.


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« Reply #17 on: September 12, 2009, 10:29:50 PM »
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Questions for this part

•What happens to the full throttle duration of a model if I add a cell and change nothing else? Why?
•How long can an ideal cell produce 15 watts of power?
•Suppose I had a cell that provided 2 volts of electricity and had a capacity of 2 Ah. How many watt-hours would that be? How long the cell could produce 20 watts? How long can the cell provide 8 amps of current?
•Suppose you were designing a plane to compete in a distance task. What could you do to the battery to make the plane travel faster?
•What does increasing prop size do to power? Duration ? How can I increase both power and duration of an electric power system?
•I need a power system that provides 300 watts from a battery composed of ideal cells. List two different volt/current combinations to accomplish this and duration they will provide at full throttle?
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« Reply #18 on: September 12, 2009, 10:56:54 PM »
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common guys you can post your responses too.. Grin
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« Reply #19 on: September 13, 2009, 08:20:31 AM »
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i haven't read the complete posts but nice effort n good info  Clap
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« Reply #20 on: September 15, 2009, 08:45:29 PM »
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Ankur,

Well, one thing I am still trying to understand is the cell count and amp rating..

So if I have a 3S rated at 3000mAH (or 3AH),  then does this imply that the battery is capable of supplying 11.1 x 3 amps for 1 hour that is 33.3 amps for one hour?

Which would mean that if I have a motor that draws 30 amp current, then can I run this for 1 hour?

Pankaj
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« Reply #21 on: September 15, 2009, 11:28:28 PM »
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pankaj

voltage X current = watts (power) and in your case

a 3 cell lipo with a capacity of 3000mamps will give = 11.1X3000X10power-3
                                                                     = 11.1X3
                                                                     = 33.3 watts
so the battery can deliver 33.3 watts of power for an hour where in the voltage is 11.1volts and current is 3 amps
across the load which can be motor,bulb etc
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« Reply #22 on: September 16, 2009, 03:54:17 PM »
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Agreed, VA = W.

Now in choosing a battery for the model, what should be preferred increasing the number of cells or the amp hour rating?

Or is the battery chosen primarily by weight considerations for a certain cell rating?

Pankaj
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« Reply #23 on: September 16, 2009, 08:04:59 PM »
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Ankur,

Well, one thing I am still trying to understand is the cell count and amp rating..

So if I have a 3S rated at 3000mAH (or 3AH),  then does this imply that the battery is capable of supplying 11.1 x 3 amps for 1 hour that is 33.3 amps for one hour?

Which would mean that if I have a motor that draws 30 amp current, then can I run this for 1 hour?

Pankaj



well pankaj you got a little confused there...the "3s" means that there are 3 lipos connected in series and if you have read the earlier posts...connecting lipos in series does not change the capacity but the voltage.(so it means essentially that all the three lipos in your cell have 3amp-hr rating) now each lipo is generally rated at 3.7 volts so you have 3x3.7=11.1 v. 3Ah itself means that the battery can supply 3 amps for 1 hour so there is no question of 33 amps being supplied by your lipo

to find the maximum discharge current that the lipo can give...look at the "C" rating. the product of "c" rating and the amp-hours give you the max safe current draw for your lipo...above this draw your lipo will be overloaded and probably damaged.


i hope your query is resolved..keep reading the next parts for better understanding Grin
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« Reply #24 on: September 16, 2009, 08:16:10 PM »
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ideally increasing the number of cells results in increased voltage to motor which in turn leads to the motor spinning at higher rpm..which means more power draw..which means more current draw...which means less full throttle duration.( and if your motor is not capable of handling that much amps..it means burnt winding Cheesy)

but since we use an ESC increasing the no of cells will be decided by weight factor. by increasing the no of cells you actually limit the max throttle on your stick that you can apply..and if you keep on pushing the throttle up..the ideal condition will set..and probably you will burn your motor...so it is favorable to have a higher amp-hour rating if you want long duration of flights rather than adding cells.
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