electrical budget??

[Eyrie]

Posted on the 08SHTP web site is an excel worksheet to assist in calculating a boats energy consumption and recharging capabilities during the course of the race.
http://www.sfbaysss.org/TransPac/tra..._worksheet.xls
The worksheet originated from the Pacific Cup web site, has been in use for several years, and assumed to be correct, but after closer examination by Bob/Ragtime! and Rob/Tigerbeetle it seems the results of the worksheet can be misleading.

Below are the observations made by Rob/Tigerbeetle regarding the worksheet. Unfortunately this forum does not allow excel attachments, so to follow along with Rob's explanation you will need to go to the list of posted energy budgets that are on the web site and bring them up.
http://www.sfbaysss.org/TransPac/tra...l_budgets.html

The calculation for charging time is flawed as it does not take into account the ability (or lack thereof) of the battery bank to accept a charge.

For example, if I look at Ragtime!'s budget I see that a 30 amp alternator will charge the battery bank in 90 minutes. If I change the alternator to be a 1000 amp output I am informed the charge is accomplished in 3 minutes. However, the batteries will not accept a charge at that rate and will in fact overheat and crack, explode, or catch fire (or do all three more or simultaneously).

The flaw is that the spreadsheet does not consider the maximum charge rate that a battery can accept; as a rule of thumb a lead/acid wet battery can accept a *maximum* of 25% of capacity when battery charge is below 85% of capacity.

To follow up further, the spreadsheet calculations are misleading, and decidedly incorrect.

For example, Alchera's worksheet calculates that he can cycle the batteries through a 50% charge/discharge cycle (the last 15% will be very slow and inefficient), and can push 90 amps into a 270 amp battery wet cell lead/acid wetcell battery bank - which isn't possible.

At best a 270 AH bank can accept 270/4 = 67.5 amp charge current between 50% and 85% charge, and then the curve changes dramatically.

The correct calculation is to a) ask the user which battery type they are using, and if the answer is lead/acid wetcell, the spreadsheet should fix the charge/discharge at 50%-85% and size the battery bank accordingly. For Alchera, this turns into: 270AH/day * 0.5 day between charging = 135AH to charge every 12 hours. To do this efficiently requires that the 135AH be no less than 35% of battery bank, therefore bank should be (minimum) 135/0.35=386 AH capacity (not 270).

That 386AH bank should be charged with an alternator no larger than 25% of capacity: 386/4 = 97AH alternator capacity (not 90).

That 97AH alternator will take 99 minutes to push 135 amps back into the bank at a charge acceptance efficiency of 85% (not 106 minutes).

None of the above numbers appear on Alchera's spreadsheet.

The upshot is the value of the spreadsheet is to get people to at least think about the power consumption they may have on board; the spreadsheet in fact produces incorrect data on how to size the bank and charger to determine charge times, and therefore may
lead people to believe they have a suitable system with sufficient fuel reserves to complete the race, when in fact they do not.

As a race entrant it is your responsibility to determine that you can power your boat during the course of the race, the worksheet is just an aid. If you have any doubt about your energy budget or recharging system you might consider seeking some advice from a knowledgeable source. Or perhaps someone in the forum audience will redesign a better worksheet.

Synthia/Eyrie

[AlanH]

So, making this up close and personal for Yours Truly....I have two, 90 amp-hour lead-acid batteries. My solar charge controller allocates 50% of the current from the panels to both batteries if they're equally discharged, or favors the more deeply discharged battery if they're not equal. Let's take the theoretically possible case of battery #1 being topped up and battery #2 being pretty drained. The 25% charging capacity that Rob mentions would mean that battery #2 can accept a maximum charge rate of 22.5 amps. That will taper off a LOT as the battery nears fully-charged.

I have 140 watts of solar panel capacity on the boat. It's reasonable to assume that about 110 of those watts will be generating at any given time since at least one panel will be at least partially shaded. So my array will generate about 6-7 amps in full sun. .....and that's well under any charge rate that will toast batteries, though the batteries probably won't accept 6-7 amps when they get over 90% charged..

6 amps multiplied by 8 hours a day is 48 amp-hours. 7 amps multiplied by 8 hours of day of full sunshine is 54 amp hours. I'll be somewhere in between. About 85% of that will actually get into the battery every day, meaning something on the order of 40 amps/day will get shoved back into my batteries from the solar panels, as long as there's sun. That, right there is enough to replace the autopilot drain and the lights.

If I can manage to get all 140 watts of solar cranking for 8 hours, that adds up to about 56 amps a day going into the batteries, which will cover everything.

In addition to this I'm carrying an old 1990's' vintage Coleman "lightning" generator which according to the label shoves out 50 amps and has a charge regulator. I have my doubts about that amperage rating, but whatever the case, an hour with that thing shoving out electrons will give any battery a big boost, around 40 amps back into a well-drained battery.

[tiger beetle]

I have 140 watts of solar panel capacity on the boat. It's reasonable to assume that about 110 of those watts will be generating at any given time since at least one panel will be at least partially shaded.

If the solar panels are a typical parallel/series cell wiring, shading a cell will result in the panel output dropping by half. If you have 2x70 watt panels, shading one panel reduces output of the panel pair (under perfect test conditions) to 105 watts.

As regards solar panel efficiency, it's been suggested by several panel testers that a realistic output seen on a boat for a given panel is: (rated watts - 10%) / 2 = (140-14)/2 = 63 watts. Given that watts = volts x amps, you need to know the voltage *at the battery terminal* (not the panel) in addition to the panel watt output to calculate amperage available to recharge the batteries.

To know how well your boat's charge system works will require that you be a bit more precise in your expectations. A 'well-drained battery' should be defined in terms of voltage, and then the time to recharge to a higher voltage at a specific current can be measured.

The Coleman (originally Active Technologies' Lightning Charger) gas-engine battery chargers were good units, I had one which worked well until the ceramic in the alternator winding cracked - unfortunately Coleman Powermate stopped making the unit. The problem with the little Honda generators is they are battery replacements and do not have high enough voltage to act as battery chargers. The Coleman unit really does have output voltage that will charge a battery.

Trying to do someting as simple as charge a battery is really rather interesting.

- rob

[AlanH]

If the solar panels are a typical parallel/series cell wiring, shading a cell will result in the panel output dropping by half. If you have 2x70 watt panels, shading one panel reduces output of the panel pair (under perfect test conditions) to 105 watts.

As regards solar panel efficiency, it's been suggested by several panel testers that a realistic output seen on a boat for a given panel is: (rated watts - 10%) / 2 = (140-14)/2 = 63 watts. Given that watts = volts x amps, you need to know the voltage *at the battery terminal* (not the panel) in addition to the panel watt output to calculate amperage available to recharge the batteries.

To know how well your boat's charge system works will require that you be a bit more precise in your expectations. A 'well-drained battery' should be defined in terms of voltage, and then the time to recharge to a higher voltage at a specific current can be measured.

The Coleman (originally Active Technologies' Lightning Charger) gas-engine battery chargers were good units, I had one which worked well until the ceramic in the alternator winding cracked - unfortunately Coleman Powermate stopped making the unit. The problem with the little Honda generators is they are battery replacements and do not have high enough voltage to act as battery chargers. The Coleman unit really does have output voltage that will charge a battery.

Trying to do someting as simple as charge a battery is really rather interesting.

- rob

Rob, my panels all feed into one common bus bar. All the positive leads go together, all the negative leads all lead together. Voltage in full sun at about 4:30 with the panels all pointing straight up, on Sunday at Coyote Point was 19.08 volts at the bus bar. One single lead goes from the bar to the Morningstar two-bank charge regulator.

See, one 30 watt panel is across the top of the companionway and whenever the boom is near the centerline, it will be shaded. The two 40 watt panels are way out on the stern pulpit, so usually in full sun. The other 30 watt panel is mounted on a gizmo that stuffs into the winch handle holes on a winch, so it moves around and I can keep it out of the shade.. OK, so if the 30 that's on the cabintop is shaded, it's output is halved. That would leave me with 40 watts + 40 watts + 30 watts + 15 watts = 125 watts. 125 watts div by 17 volts (Kyocera's short-circuit voltage, I'll assume it's the same for the other panels) = 7.35 amps.

7.35 amps generated over 8 hours of sunlight = 59 amps in a day. Wooo-hooo!

Let's do some more....

You wrote that I need to know the panel voltage at the terminal. By that I assume you mean the battery terminals. The Morningstar charge regulator I'm using has dip-switch settings for Lead Acid and sealed batteries. If I remember rightly, the Lead Acid charge setting is 14.4 volts and the sealed battery charge setting is 14.1 volts, though I might have those backwards. Let's say it's 14.4 volts. Now, the rating for the panel array is to take the wattage of the whole shebang, subtract ten percent and divide by two. OK for my 140 watt array, that gives 63 watts, like you said. 63 watts div by 14.4 volts gives 4.3 amps. Eh? Ouch! That's' only 35 amps a day, for an 8-hour day of sunlight. If that's how it works out I'd better take 4-5 gallons of pre-mix for that Lightning generator and use the Navik windvane a lot!

Did I get that right? The shaded panel (unless it's *really* shaded) drops IT'S contribution in half, it doesn't drop the WHOLE ARRAYS output in half?

The Coleman Lightning generator is a champ. I'm almost tempted to ship it back from Hawaii, but it being a gas engine and all.....probably not practical.

[tiger beetle]

Rob, my panels all feed into one common bus bar. All the positive leads go together, all the negative leads all lead together. Voltage in full sun at about 4:30 with the panels all pointing straight up, on Sunday at Coyote Point was 19.08 volts at the bus bar. One single lead goes from the bar to the Morningstar two-bank charge regulator.

That's good voltage and seems excess to requirements. A top-end charge voltage should be 14.6 volts, almost never as high as 15 volts. Many panels will show a no-load voltage of 17 volts, and 19 is way up there. Now measure the voltage at the battery terminal - that's the voltage the charger can deliver to the battery. This takes into account loss through the wiring, the charge controller, and a blocking diode (if you installed one).

So, knowing that much, would shading of one panel seriously drop the output of the array as a whole? It sounds like you're suggesting that it will.

Yes, if the panel is a typically-wired panel. Each solar cell delivers a small portion of the total voltage, and the cells are wired in series/parallel to deliver the design voltage x amps to the panel output terminal. Shading a cell kills the section of cells that the shaded cell is part of. This is easy to test - place a small piece of wood on the panel such that it covers part of one cell, and see what the effect is. Continue with shading more and more cells to determine when the panel output drops below a charging voltage of perhaps 13.6 volts *at the battery terminal*.

EDIT: the shaded cell only affects output of the shaded panel; a shaded cell of panel A does not affect output of Panel B. The output of the array as a whole will drop, but not necessarily by 50%.

See, one panel is across the top of the companionway and whenever the boom is near the centerline, it will be shaded. If shading that one panel (the most likely one to get shaded) will kill the output of the whole danged thing, I'm inclined to put a little switch on the + lead of that panel.

That's a good idea, but most likely unecessary. In practice the shadows will be swinging all over the place, periodically shading different parts of different panels. The loss caused by a shaded panel as it draws from the unshaded panel is minimal in an environment in which the shading is partial and inconsistent.

The Coleman Lightning generator is a champ. I'm almost tempted to ship it back from Hawaii, but it being a gas engine and all.....probably not practical.

If you want to sell the generator after the race, let me know.

- rob

[AlanH]

What may be more helpful is to ask them what they will do if the on-board electrical system suffers catastrophic failure halfway there - then what??? They lose charging ability, storage capacity, or suffer an electrical fire.... A lightning strike, anyone???

Do they have a plan for that???? How exactly do they get the boat across a thousand miles of ocean and find a safe harbour - at that point, probably not even Hanalei Bay??

In the case of total electrical charging system failure; solar panels disabled, Coleman charger won't start:

1. all my GPS's are AA battery-powered and I'll have a LOT of spare batteries. With those, parallel rulers, a pencil, paper charts and four pairs of reading glasses (I can't read diddly without glasses any more) I ought to be able to find Hawaii. My oldest GPS lives in a Faraday cage, a little aluminum box I made out of roof flashing so that it has at least a snowballs chance in hell of surviving an electrical strike. And finally, I have three watches on the boat and a sextant and tables. God Forbid..

2. the running lights are a tiny little draw now that I've put on LED bulbs, so the batteries will keep them running for quite a while. i also will have several big 6 volt lantern batteries on the boat and the boats strobe lights work with those quite happily. They just don't flash quite as often. I also have three armband strobes that work on D-cell batteries for about 3 nights....and I have a lot of extra D-cell batteries. So I'm lit up at night.

3. the Navik Windvane will do a LOT of steering, which it's entirely capable of doing.

4. I'm renting a satellite phone and an extra battery. I won't be chatty, but people will know where I am, once a day.

How 'zat?

[AlanH]

Shading a cell kills the section of cells that the shaded cell is part of. This is easy to test - place a small piece of wood on the panel such that it covers part of one cell, and see what the effect is. Continue with shading more and more cells to determine when the panel output drops below a charging voltage of perhaps 13.6 volts *at the battery terminal*.

EDIT: the shaded cell only affects output of the shaded panel; a shaded cell of panel A does not affect output of Panel B. The output of the array as a whole will drop, but not necessarily by 50%.

If you want to sell the generator after the race, let me know.

- rob

That's a really good idea, and simple to do. OK, quick project for Sunday!

I would very much like to sell the generator to someone who will use it, I'll get back to you.

[Libations Too]

....As regards solar panel efficiency, it's been suggested by several panel testers that a realistic output seen on a boat for a given panel is: (rated watts - 10%) / 2....
- rob

Rob,

I always appreciate your perspective...but why does the formula above use a factor of 50% (divide by 2) to get to actual watts? I can understand the 10% factor as line and other losses, but a 50% reduction to rated output seems extreme. What am I missing here?

[tiger beetle]

Rob,

I always appreciate your perspective...but why does the formula above use a factor of 50% (divide by 2) to get to actual watts? I can understand the 10% factor as line and other losses, but a 50% reduction to rated output seems extreme. What am I missing here?

All I know is I read 3 different published reports derived that equation empirically. Based on observations of my Kyocera 120 watt polycrystalline panel, I'd say that's about right for my installation.

Manufacturer's rating tests are designed to make the panels perform their absolute best - the panel is temperature controlled, perfectly clean, the wiring to the panel junction box is brand new, and the light is unobstructed and at the best possible angle and wavelength.

What I find is that my panel usually has dust on it, the sun is other than directly overhead most of the time, the panel heats up in the sun (increasing resistance and reducing output), the panel is partially shaded at times, and I'm measuring output at the battery terminal where the panel can do useful work as opposed to the manufacturer whom measures output at the panel's junction box.

If you have a panel it's relatively easy to measure how it behaves in real life, which is always better than an empirical equation or rule of thumb. Until you buy one you're limited to calculations to forecast the work the panel will do. My goal was to have the panel supply the power the onboard refrigeration consumes (which it does). Ultimately I simply purchased the panel with the highest watt/square inch energy density that fit within the footprint I had available. Turns out the panel kept up with not only the fridge but also with the limited (3 hours/day) laptop power use - which was an extra bonus!

- rob

[BobJ]

I have two Kyocera KC-40T (43 watt) panels (20"x25"), each with a rated output of 2.48A, so a total of 4.96A. They are connected to the batteries via a Flex Charge PV7-D dual-bank controller.

My real-world observation, via my battery monitor, is that the batteries are receiving 4.8A from them if I make any attempt to aim them towards the Sun.

Sometimes a shadow of the upper lifeline cuts across one or both but because it's a thin shadow, it doesn't seem to affect the output much. I can pivot them fore 'n aft, from near vertical (if the Sun is in front of the boat) to flat against the transom (if the Sun is behind the boat).