$50.00 to the first person who answers this correctly.I need to solve this problem?
Answer:
You might take a close look at power curve chart and be sure the information is actually relevant, for example one web site I found (http://www.survivalunlimited.com/windpow... says:
"Since we do not know the torque of the driver, this data is given in absolute separate peaks.
Amps are read as a short in the circuit closed condition. Then we read Volts in the circuit open condition. These charts do not reflect working conditions only the absolute separate readings."
if your chart is constructed similarly, you may have to run somewhat faster to actually get the output you need.
Also the comment made earlier about a rectifier is valid, if you are feeding the batteries with AC, a net _negative_ current might be expected as the discharge on the negative half of the cycle will probably be faster than the charge on the positive half
i don't know...that answer is correct.if you don't give me my $50, I'll see you in court.
the permanant valgulator is unregulated configuring it to make it sustaianbly impossible to with held anything at hand in desire to make it combatibly un cooperative
Your alternator is the problem. Change it. They don't last forever. Perhaps a diode has blown.
IT IS THE DIODE that allows AC current from an alternator to flow only in one direction for DC charging.
With a new alternator, your 48 volt system will recharge the batteries as designed.
Give the 50 bucks to children with bad skin burns. Shriner's Burn Institute.
I don't understand even half of what you said. This is wild guess, but is it possible you have the polarity reversed? If the battery if fighting the alternator, that could be your problem.
You can't contact me though YA, but if you send money c/o 21 E State St, Columbus, OH 43215, it will find its way to me. That is NOT my address, but someone there will know who I am.
Have you tried more than one instrument for each of your measurements? Could a faulty instrument be leading you astray? Something must not be what it seems to be.
solve it by kirchoff's law;
all are in parallel and there's a circulating current in two loops, you measure dc current at load and current in the loop of genset and battery. observe polarities.
I'm assuming that when you say you are connecting the alternator direct to the battery you mean that you are connecting the rectified (DC) output - not trying to charge a DC system from AC. A fully charged 12v lead-acid battery is 13.2 volts. 4 of them is thus 52.8 volts. You need to exceed this voltage at the battery terminals to charge the batteries. There is a voltage drop across your rectifier diodes and your ammeter as well as the cables.You probably are not getting a high enough voltage to supply your charging current, especially if more than one alternator gives you the same results. Connect the alternator to the load AND the batteries, crank up the RPM until you see 53 volts at the battery terminals with full electrical load applied. This will be the minimum RPM you can run at. I could be more specific if supplied with your alternator specifications or model.
You say "...permanent magnet alternator unregulated direct to the battery...", but most PMAs I came across deliver AC and not DC.
Do you have a rectifier between PMA and Battery?
Because if not, then you only make use of one half wave which would explain the 17Amps somehow.
And the other half wave DIS-charges your battery, which would also explain "...letting my battery run down..."
Your PMA will slow down, and output voltage will drop with high current demand, but not by half...
First Have you checked the diodes of the alternator to ensure you are running a DC out on all three legs of the alternators' internal windings?
Second, You should regulate the voltages, at the higher voltages, you are running 20 to 25% above nominal, which if you follow Ohm's law, will actually reduce the current by the same percentage given the same load.
Third, the higher internal resistance of a depleted battery, against a higher than normal charging voltage, will increase the internal temperature of the battery, and ultimately increase the internal resistance, ie, less current flow, less charge.
Last , recalculate your load, At the higher voltages you are running a strictly resistance load will drop the current by about 8 to 9 volts, but If you have any inductive loads included in your total load, reactive power will reduce the total load current inverse to the increase in reactance.
Oh one other item that may put all else to moot, Are you sure the battery is good?
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: