|
Post by tonyqj on Oct 31, 2016 21:41:39 GMT
Β A spritsail rig is a Thames sailing barge rig where the "sprit" is a long spar running diagonally from the foot of the mast to the peak of the mainsail You can clearly see the sprit in this photo 079 by mudlarker2, on Flickr Today I learnt... what a Spritsail is π
|
|
|
Post by tonyqj on Oct 31, 2016 20:58:42 GMT
Not boat but house. We need to add an understair light with it's own dedicated switch. Nearby (other side of stud wall)Β is a simple switch which operates a simple ceiling bulb on the other side of stairs (this has an L2, L1, Common and Earth connection but wires are only connectred to Common, Earth and L1). Need to know wiring to extend from the existing switch to new switch so the new switch operates the new light independantly. And the old switch just operates the old light it has always done. In other words we don't want one switch to operate both lights - but each switch to switch on only it's dedicated light. Even ignoring the 'Part P Building Regs' which says you can't do it, if you're unsure how to do this then I'd say you can't do it. Get an electrician in, he'd probably do it for Β£50.
|
|
|
Post by tonyqj on Oct 31, 2016 20:53:46 GMT
Yes, I got that. But so what? It has virtually no bearing on boating. The point is that if you are going to explain why you need to put more charge back into a battery than you take out, you might as well explain it the correct way rather than a fallacious way if both are equally simple concepts to understand. Simplest is to say "due to inefficiencies". That's more than enough for the average boater.
|
|
|
Post by tonyqj on Oct 31, 2016 20:44:10 GMT
What is it that won't end up 'girted' if it uses a 'snotter' ?Β That's a new one on me Tony ...... I know a snotter as the heel fitting for a sprit on a spritsail rig Β but I'm damned if I can put it together with girting I understood 'rig'...
|
|
|
Post by tonyqj on Oct 31, 2016 20:40:00 GMT
It's back to the point that charge (AH if you like) doesn't have intrinsic energy. It only has energy in the presence of potential gradient. In a circuit, energy is lost due to the reduction in the voltage part of Power = VI, not the I part. So by reciprocity if power is dissipated whilst charging, it cannot be due to a loss of AH (or current if you like), only by loss of voltage (from internal resistance). Therefore CEF is not explained by energy loss due to heating. Yes, I got that. But so what? It has virtually no bearing on boating.
|
|
|
Post by tonyqj on Oct 31, 2016 19:56:26 GMT
Nicks got a good point though... You don't neeed voltage to move charge around, merely to overcome the voltage drop caused by passing current through a resistive material. Yes, but so what?
|
|
|
Post by tonyqj on Oct 31, 2016 19:54:17 GMT
Tony - I think this works just fine. It offers good advice to real world boaters. However, could you edit your first para into shorter sentences to make it easier to read? Can I ask one thing that the tail current is measured for 30 minutes and checking there is no further reduction. And that the charger is checked that it has not gone to float and is actually in absorption mode. I completely missed both points. Yes, well caught. I'd even say '30 to 60 minutes' (in fact I believe I already did in my 'Charger Primer' post.)
|
|
|
Post by tonyqj on Oct 31, 2016 19:46:08 GMT
Like...
Such meters usually have 5 or 6 display modes: 1. Amps 2. Volts 3. Amp hours into the battery (but note this - NOT Amp hours actually converted into chemical energy and stored within the battery) 4. Amp hours out of the battery 5. percentage of charge 6. and finally maybe time left before charging (or time left until flat)
In the case of the first two (amps & Volts) they are accurate so would agree with an ammeter and voltmeter or a multi-meter. Just because these two are accurate however does not mean the others are. Although pure Amp hours in and Amp hours out will be accurate they are somewhat problematical, and as for the percentage of charge and time left they are often a work of fiction.
|
|
|
Post by tonyqj on Oct 31, 2016 19:41:12 GMT
I did this for someone on the other side and thought it might do for the Advice section. Please comment so I can edit it and may be get it moved once it is approved by the members. Such meters usually have 5 display modes. Amps, Volts, Amp hours into the battery (but note this - NOT Amp hours actually converted into chemical energy and stored within the battery), Amp hours out of the battery, percentage of charge, and finally maybe time left before charging. In the case of the first two (amps & Volts) they are accurate so would agree with an ammeter and voltmeter or a multi-meter but just because they are accurate does not mean the other are although pure Amp hours in and Amp hours out will be but they are somewhat problematical, as for the percentage of charge and time left they are often a work of fiction. The percentage of the Amp hours into a battery that is converted to stored chemical energy alters hour to hour and day to day as the temperature, charging voltage, and battery condition but it will NEVER be 100%. It will also vary from battery type/design to battery type. Just to substantiate this feel some batteries that have been on charge for a while, they will be warm or hot. Where did this heat come from? The answer is the charging current so from that you can deduce that some charge went to make heat, not convert chemicals. Batteries that are on charge tend to vent hydrogen and oxygen so once again the question is where did the energy come form to break down the electrolyte? The answer is still the charging current. Just to complicate things further the actual capacity of the battery reduces over time means that the meter can not know how much electricity is actually stored in the battery - even if it is 100% charged. Within days or at the most a few weeks it WILL be less that the number on the label. The upshot of the above paragraph is that any display that results from Amp hour counting will over record the amount of charge put into the battery and once that is wrong the percentage of charge and time left must also be wrong. The makers try to compensate for this by putting "fiddle factors" into the algorithms the gauges use to work out the percentage charge and time left. If you have understood some of the topics that are around concerning battery charging you will know that when the charging current drops to a very low level - often quoted as 1% of bank capacity - the battery can be CONSIDERED as fully charged. It probably won't be but it will be near enough for practical use. The meterβs electronics knows this and at that point resets itself to 100% charged. All that sounds fine and dandy but the meters are not usually set at 1% by the factory. Probably more like 2% to 3% so they resynchronise early. This causes them to overstate the degree of charge and that in turn persuades many boaters to destroy their batteries by constant and ongoing undercharging. The problems get worse if users do not continue charging until the current drops to such a low percentage. This is why solar charging has such advantages, it provides many hours at a low charge rate to get the final few percentage of charge back in. If you really understand the limitations in such meters, reset them to synchronise at a very low percentage (known as tail current) AND very regularly charge for long enough to cause them to resynchronise then they will be fairly accurate on all scales. If you do not all you can rely upon is the amps and volts. Even so as the battery capacity has dropped over time the percentage of charge and the time left might still be incorrect. The basic rule for most boaters with this type of meter is to only use the amps scale while charging and keep going at least once a week until the charge drops to 1% of bank capacity or less, then manually resynchronise the meter. Then you can note the Amp hours out and percentage of charge and calculate the current capacity of the bank. e.g. 50Ah out that takes it to 75% of charge tells you the battery bank capacity is 200Ah because 50Ah is 1/4 of the bank capacity. Such meters usually have 5 display modes. Amps, Volts, Amphours into the battery (but note this - NOT Amphours actually converted into chemical energy), Amphours out of the battery, percentage of charge, and finally maybe time left before charging. In the case of the first two (amps & Volts) they are accurate so would agree with your multi-meter but just because they are accurate does not mean the other are although pure Amphours in and Amphours out will be but they are somewhat problematical, as for the percentage of charge and time left they are often a work of fiction. Tony - I think this works just fine. It offers good advice to real world boaters. However, could you edit your first para into shorter sentences to make it easier to read?
|
|
|
Post by tonyqj on Oct 31, 2016 18:13:48 GMT
Gibbo v Nick, so glad I joined this site, sits back to enjoy the show..... Where's Β Chris W when you need him ? π Oh God no, please. Not broken maths as well!
|
|
|
Post by tonyqj on Oct 31, 2016 16:54:17 GMT
There is a guy in the other place who knows a lot about propellers, admittedly aircraft, but it is still a fan in a fluid. So if you don't get a good answer here, you could try there, or even entice him over here with your question.................... Deslandia? I guess he's be good to invite π Edit - snap!
|
|
|
Post by tonyqj on Oct 31, 2016 15:22:23 GMT
Calculations for the above (not my work): The reaction is 2H2O -> 2H2 + O2 The bond energies (correct to 3 sf) are H-O : 459 kJMol-1 H-H : 432 O=O : 494 So the total energy is 459*4 - 494 - 432*2 = 478 kJ But that is for two moles of H2O, so the energy per mole is 239kJ You are correct - I forgot the energy released by O+O becoming O2 Β and H+H becoming H2 Β - very embarrassed. Hey, it's not me that's correct, all I did was copy/paste. π
|
|
|
Post by tonyqj on Oct 31, 2016 15:18:28 GMT
N has left the hustings. Time for a break Thank Gawd for that ...... I'm afraid my days of thinking quickly through things like this are long gone ..... it takes me a little while to A) get my thoughts straight and B) get my typing straight edit to add .... only the funnies come quickly !!! Always π Anyway, all the atomic energies calculations are a bit beyond me - but it's fun to watch Gibbo & Nick get into a scrap. Reminds me of the old days with Chris W except Nick tends not to get his maths wrong.
|
|
|
Post by tonyqj on Oct 31, 2016 13:48:41 GMT
Welcome π
|
|
|
Post by tonyqj on Oct 31, 2016 13:35:54 GMT
Calculations for the above (not my work): The reaction is
2H2O -> 2H2 + O2
The bond energies (correct to 3 sf) are
H-O : 459 kJMol-1 H-H : 432 O=O : 494
So the total energy is 459*4 - 494 - 432*2 = 478 kJ
But that is for two moles of H2O,
so the energy per mole is 239kJ
|
|