Don, You are confusing charge with energy. That is what several have been trying to get across. Remember that power is the product of voltage and current. The energy, in this case, is the product of power and time. This is where the Watt-hour or kilowatt-hour comes from.
Consider two ideal cells each with a capacity of 1Ah. One cell is a 1V cell and the other is a 2V cell. Hook these in series and you have a 3V battery. Suppose that the two cells are each at 0%SOC. Hook the battery to a powersupply and supply 1A of current at 3V for 1 hour. (I know that there has to be a voltage difference but bare with me for a moment.) During charging, 3V * 1A = 3W of power is being put into the battery pack. The first cell is receiving 1V * 1A = 1W of power and the second cell is receiving 2V * 1A = 2W of power. At the end of 1 hour each cell has received 1A * 1h = 1Ah of _charge_ but the first cell has received 1W * 1h = 1Wh of _energy_ and the second cell has received 2W * 1h = 2Wh of _energy_. The total energy into the battery pack is 3Wh. Energy is conserved. Note that one cell received twice the energy but exactly the same number of Ah. In keeping a battery pack balanced we are not concerned with keeping the energy in each cell the at a matched state, we are concerned with keeping the _charge_ in each cell at a matched state. We are concerned with State of Charge, not State of Energy. This is why, as Cor pointed out, two cells with vastly different IR but the same charge capacity, when starting at the same State of Charge, can run out of charge, and thus energy, at the same time even if the energy each delivers is vastly different. HTH, On Wed, Jun 19, 2013 at 5:11 PM, Cor van de Water <[email protected]> wrote: > Don, > > Please read my explanation and pay attention to the fact that > - the charger needs to provide a higher voltage (so, more energy) > to feed power into the cell PLUS the internal resistance. > - the discharge delivers LOWER voltage, due to the energy lost > across the internal resistance, reducing the output voltage > and thus the delivered power, due to the power lost in the > internal resistance. > > The info that you are looking for is already in the answer I provided > earlier. > > Regards, > > Cor van de Water > Chief Scientist > Proxim Wireless Corporation http://www.proxim.com > Email: [email protected] Private: http://www.cvandewater.info > Skype: cor_van_de_water Tel: +1 408 383 7626 > > > -----Original Message----- > From: [email protected] [mailto:[email protected]] On > Behalf Of [email protected] > Sent: Wednesday, June 19, 2013 5:08 PM > To: [email protected] > Subject: [EVDL] Resistance > > It still does not answer or address where the energy comes from that was > > loss in raising the cell or cells with the higher resistance over > ambient > temperature. > > Yes the energy delivered to each cell in series is equal, but the loss > of > energy as heat is not. This would be on both charging and discharging. > > As an example NiMH cells in the past have a very high self discharge > rate. You can charge up a pack and they lose energy while reaching a > full > charge. Cells that have higher resistance or are in a higher state of > charge > lose more energy as heat then the others. If you let the pack sit in a > fully > charged state they self discharge and the energy loss produces heat. > Enough > that it actually helps keep the pack warmer in the winter. > > Once a NiMH pack has reached a full charge the only way it will retain > the > same amount of energy is if you replace the energy lost as heat. > > Your saying "Still, if the heat does no damage and does not affect > efficiency of > accepting charge" > > It does affect the efficiency of accepting a charge and that added heat > > also causes a higher rate of degradation to the warmer cells. > > Since energy is not created but only changes form, some energy is lost > as > heat. > > Yes if all the cells had the same resistance and no other factors > involved > the energy retained in each cell would be the exact same amount. > > Having heat without a loss of energy in the cells would be the > equivalent > of a perpetual motion machine. > > Don Blazer > > > > In a message dated 6/19/2013 7:01:51 A.M. Pacific Daylight Time, > [email protected] writes: > > Message: 1 > Date: Tue, 18 Jun 2013 13:04:28 -0700 > From: "Cor van de Water" <[email protected]> > To: "Electric Vehicle Discussion List" <[email protected]> > Subject: Re: [EVDL] Resistance > Message-ID: > <a73bc4b8b3218642a56a2c9eb01b44e001ce1...@exchange.corp.proxim.com> > Content-Type: text/plain; charset="us-ascii" > > Indeed. Current in a series string is equal (by definition). > Only thing that is suffering from the internal resistance is the > *Voltage*. > When charging, the bad (high resistance) cell will cause a voltage drop > across its resistance, causing extra heating and higher voltage (this > is > the loss that you were expecting, turning energy into heat) but the > cell > is still charged with the same current, so it just runs hotter > (depending on charge current, cooling and other factors). > When discharging (driving the EV) the internal resistance causes a > voltage drop that *reduces* the apparent cell voltage (Bill sketched > the > model: a resistor in series with an ideal cell, we call the value of > that resistor the internal resistance). This voltage drop again causes > heating of the cell, which can be excessive if the voltage drop is > large > - if the internal resistance is large enough, the output voltage can > even become *negative* which means that the voltage drop across the > resistance is larger than the output of the cell. In those cases it is > better to remove the cell from the string, not only due to the bad > efficiency but more due to the risk to set fire to the battery pack. > One > example to illustrate: > Say we have Lithium cells (any chemistry, but say the cell is at 3.5V > rest voltage). > Due to construction or abuse, the internal resistance of the cell has > increased to 10 mOhm and you try to pull 500A from the string of cells. > The resistor drops 0.01 (Ohm) * 500A = 5V while the cell tries to > deliver 3.5V so if you measure the terminals of the cell under this > load, you will see the cell at 3.5 -5 = -1.5V. > The ideal cell is delivering a power of 3.5V * 500A = 1750 Watts. > The internal resistance is sucking up and producing heat to the tune of > 5V * 500A = 2500 Watts. > Total power delivered by the damaged cell is -750 Watts (it is > consuming > 750 Watts of power from the adjacent cell by dropping part of the > adjacent cell delivered voltage across its internal resistance) > > Still, if the heat does no damage and does not affect efficiency of > accepting charge, then the high-resistance cell will stay in balance, > it > will just be inefficient and possibly disastrous in its operation if > the > internal resitance causes dangerous heating to occur. It is comparable > in electrical effect to a bad (corroded) terminal on a lead-acid > battery, which can (and has, on my truck) heat up to the point of > burning itself off the battery. However, in case of Lithium, an > overheating cell can be quite dangerous while it is rare that a > lead-acid battery burns. > > Hope this clarifies, > > Cor van de Water > Chief Scientist > Proxim Wireless Corporation http://www.proxim.com > Email: [email protected] Private: http://www.cvandewater.info > Skype: cor_van_de_water Tel: +1 408 383 7626 > > > -----Original Message----- > From: [email protected] [mailto:[email protected]] On > Behalf Of Bill Dube > Sent: Monday, June 17, 2013 8:54 PM > To: Electric Vehicle Discussion List > Subject: Re: [EVDL] Resistance > > Intuition would make you think so, but your intuition turns out to be > wrong in this case. > > Reread Lee Hart's post on this subject. He has it correct. > > All cells get/produce the same current because they are in series. The > cells all are charged and discharged at the identical rate. Thus, have > the identical state of charge. Any imbalance is caused by unequal > self-discharge, which is a strongly influenced by temperature. > > The variations in temperature are indeed caused by variations in > internal resistance. You can visualize that resistance as a separate > resistor in series with the (ideal) cell. It does not influence the > state of charge because the current is the same in all cells. > > It is the fact that the current is identical that is the key. All > electrons that enter one end of the string emerge on the other end. > None > > are lost. Each electron flips an ion in each cell. Whatever voltage is > needed is what there _will_ be, or electron flow will stop. > > True fact. > > Bill Dube' > > > -------------- next part -------------- > An HTML attachment was scrubbed... > URL: > <http://lists.evdl.org/private.cgi/ev-evdl.org/attachments/20130619/ede5 > 4a51/attachment.htm> > _______________________________________________ > UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub > http://lists.evdl.org/listinfo.cgi/ev-evdl.org > For EV drag racing discussion, please use NEDRA > (http://groups.yahoo.com/group/NEDRA) > > _______________________________________________ > UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub > http://lists.evdl.org/listinfo.cgi/ev-evdl.org > For EV drag racing discussion, please use NEDRA > (http://groups.yahoo.com/group/NEDRA) > -- David D. Nelson http://evalbum.com/1328 http://www.levforum.com Nokia Lumia 920 Windows Phone 8 _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA)
