Different slant on the refrigerator question

[Adelie]

Originally Posted by hellosailor

As stu said, you've murdered your batteries. Apologize to the inverter, and start battery shopping. Even if you have real deep-cycle batteries, that means they can provide the rated power about 50 times, and that's all. They can provide 50% of that power more like 800 times, and 1/3 of the rated power maybe 2000 times. So you really never want to pull more than 1/3 of the rated capacity out of them.

Your fridge and inverter together are pulling more than 100 amps at 12 volts, more like 110-120amps. So in 24 hours, they would pull something like 2700 amp hours IF the fridge really took that much power on a continuous basis. It may only pull full power1/4-1/2 of the time, as it cycles on and off.

I'm on decaf today, someone correct me if my math has gone wrong.

There are plenty of threads online, plenty of web sites, couple of very good books, all of which can tell you how to do all the math. But the bottom line is, you murdered those batteries and you'll need a very much bigger set if you want the next ones to last longer. In any case, you need to run those numbers and find out just how big a battery bank you will need, factoring in whatever power the solar array puts out

Yep your math is wrong. He said the fridge pulled .8a at 120v. At 12v is should be pulling 8a plus a little more for inverter inefficiency. Call it 9amps. Assuming 33% duty cycle that's 72a-hr per day.
Assuming 50% duty cycle that 108 a-hr/d

[Adelie]

Originally Posted by Rocinante33

Thanks for the replies,

My following questions/thoughts are designed to illustrate my thought process and not an argument. Since it is not working right, obviously there is a flaw. I would appreciate help highlighting the flaw in my analysis.

First off, the solar charger is 120 watts and not 120v. Thanks for the quick reference StuM

So my thought process went like this. The dorm fridge will consume a little over 2 amps/hr, which means about 48 Amp Hours per day. This is a little high as the thing cycles on and off, but this assumes constant operation.

The solar panel generates 5.5 amps/hr (@ 18 volts) and the sun shines on average in South Florida for 12 hours. Assuming all of the conversion to 12v is lost, that is 66 amps per day into the batts.

First, on a continuing power consumption basis (not power from storage)

Assume that I lose half of that for sun angle, cloudy, etc. I am still at 2.75 amps/h and not pulling nearly that amount, so why would it shut down on a bright sunny day? A 'murdered' battery should not make a difference at that point, right?

On the power storage issue;

I can see that putting in 66 amps/day and then assuming 50% loss is going to put consumption ahead of production. I can see that I will need to increase the capacity of the batts.

What rule of thumb ought I use? If my calculations start at 2.5 amps/hr and lets say for argument that I want the fridge on all night, then I will be using 60 amps per day.

If that is right, then how much solar power do I need to produce 60 amps in 12 hours of sunlight with shade, angles, clouds, etc. Is assuming 50% loss reasonable?

If so, then will 2 solar panels, which generate 66 amp hrs/day each (total 132) and then I lose half due to inefficiency, take me back to 66. That seems to be 10% over with some generous assumptions for loss.

Now on the question of storage, I clearly do not have enough. Thinking of a batt as a pail of water is probably not the right methodology, but it is the one I used.

How many amp/hrs should I plan to store in order to meet the storage needs for the above consumption and generation? If I use 60, and I never want that ot be more than a third, then 180 amp hrs of storage (round up to 200) be enough?

The only forum post regarding 'sailorchic' that I could find seems to indicate that she has 400 amp hrs of storage for what I have proposed above.

I do not want a genset, and I do have the alternator already.

I estimated 90 amp hrs of storage, but I have 125 on paper, was just assuming that as teh batts age, I lose capacity. You are, if I understand you correctly, clearly telling me that a reading sub 12 indicates dying (murdered) batts?

StuM, thanks again, but are you telling me that a reading of less than 12v on my screen multimeter tells me that the batts are dying per se? Guy I knew told me that until I read 10.5, things were fine. He was probably wrong, but one thing is clear, it reads 13v when alternator is on an 11.5-12 when off.

I appreciate the input here, thanks again

Cheers

The text book definition of a battery fully discharged is when it reaches 10.5v at some specific rate of discharge. This is a somewhat arbitrary number but consistent for most manufacturer's so it is useful as a benchmark.

If you discharge to 0% regularly you will only get 10s of charge-discharge cycles out of the battery. 50% is generally considered to be the most cost effective depth of discharge for regular use. For a Flooded Lead Acid (FLA) battery you might get 700-1000 cycles. For AGM batteries less, for Gel batteries more. There are trade offs for those batteries but the best bang for the buck is FLA.

At 50%State of Charge (SoC) the resting voltage of a battery in good condition should be about 12.20v. Resting means it's been sitting for 24hr. Probably 4hr is enough. Under 8amps load the voltage for your batteries will be somewhere about 12.0v or a bit less.

The rule of thumb for panel output is that output in amp-hours is equivalent to 25% of nameplate capacity of the panel. You have a 120w panel. Laid flat, with minimal shading you can expect to generate 30a-hr/day from that panel. Closer to the tropics you might average a bit better.

You can up the output by getting a better charge controller and by moving the panel throughout the day so it tracks the sun.

If you are at anchor leaving the panel flat is the best way to go, any wind or current shift then the boat swings and the panel is facing the wrong way. On passage when boat direction is better controlled if you have a tracking mount you can adjust it regularly when you check on the boat's progress.

What kind of charge controller do you have between the panel and the batteries? Follow the wiring from the panel all the way to the battery and then give us makes and models for anything you find. Pictures would be helpful. There may be other things in the circuit so follow the wires all the way to batteries and document everything. In addition to not liking to be discharged too far, batteries also like to be recharged at very specific voltages for specific periods of time. The charge controller does that. If you don't have a controller then that's something else to deal with.

What are the makes and models of your batteries? Photos again.

How did you come to a 2amp average draw for the fridge?

The other items on your boat do not draw as much as the fridge but together are a significant draw. Accounting needs to be made for their power demands.

Also, the sun shines on average 12hours per day everywhere. The limiting factor is what is the minimum it shines and that's about 10.5hr/day in the middle of the winter at Florida's latitude. Better to use the rule of thumb which gives you a conservative number that will account for things like cloudy or especially hot days.

[FlySideways]

Ok, I registered for the sole purpose of trying to help clean up the numbers in this thread. With any luck, and the correct information, Rocinante33 can get some help developing this system. He isn't an electrical engineer, but from what I've read, neither is anyone else in this thread.

First: None of these things are shipped with good specs, so all of the numbers are derived from one test.

The fridge is a 2.5 cu ft Black and Decker, and has been tested to consume 212,000 Watt hours per year at 120 Volts AC

That's 581 Watt Hours per day at 120 Volts AC

That's 24 Watt Hours Per Hour.

That's 0.2 amp hours per hour@ 120VAC
2 Amp hours per hour @ 12vDC

...And finally, 48 Amp Hours Per day at 12 Volts DC... Average. Peak (running) is unknown as yet, but given the observed compressor duty cycle, it can't be high.

The solar panel is rated at 100 Watts. @ 18 VDC

Originally Posted by StuM

But did you actually read and understand it? It seems from the rest of this post that you didn't.

Again no such thing as Amp/hr. I presume you mean that it will average 2 Amps and therefore consume 2 Amp hrs per hour.

Dropping "hours" is like dropping the axis labels off of a graph. It's not exactly punishable by death. The numbers still work and you know full-well what he was trying to say. "amp-hr/hr ~ amp/hr"

Originally Posted by StuM

The important thing is those Amps are at 110 Volts. So those 48 Amp hours total 5280 Watt hours. (48 x 110) which is 440 Amp hours at 12 V ( 5280/12)

440 AH/Day? It's not a walk-in industrial meat locker on a crew boat, it's a 2.5 cubic foot refrigerator.

Originally Posted by StuM

Again, Amps, not Amps/hr and Amp hours per day, not Amps per day!

5.5 Amps @ 18V = 99 Watts. OK, that's slightly low for a nominal 120W panel when the sun is directly overhead on a clear day. But lets work with 100 W as a round figure.

Since the sun will rarely if ever be in that situation, you can expect less than 99 Watts all the time. When the sun is lower in the sky in the morning and afternoon, a lot less. A good rule of thumb is that you may get about 5 hours "full sun equivalent" out of your panels, so plan on 500 Watt hours per day. At 12 Volts, that is about 40 Amp hours per day.

It's a 100W panel. The sun does not need to be directly overhead for a panel to produce rated output. The drop in output at the lowest visible angle is only (nominally) 40%

Originally Posted by StuM

You really need to understand the relationship between Watts, Amps and Volts. You can't compare Amps at 110V with Amps at 12 V.

That's what we are trying to teach here. And yes you can, you just have to do math to convert.

Originally Posted by StuM

It's easiest to think in Watts, because then the Voltage doesn't come into the equation.

Doesn't matter what terms you think in, you still have to do all the V*A=W equation at every point in the equation. You have to solve for one of the 3 variables for every device in the system.

Originally Posted by StuM

Your panel will at most generate 100 Watts, and will average about 50 Watts during day light hours. In the early morning and late afternoon, it will be a lot less than 50 Watts.

See above.

Originally Posted by StuM

If your fridge draws 2 Amps at 110 Volts, that 220 Watts....

We already established the fridge numbers were wrong so I'm just going to delete all of that.

Originally Posted by StuM

Let's go back to basics (and ignore all the other factors such a Peukert, inverter losses, other house loads etc which further increase your requirements). As a theoretical minimum:

You leave Peukert alone, the Germans have enough to answer for. ... But in serious terms, as above mentioned, we're talking about 2 amps. Applying the Peukert equation to a draw of 2 amps is staunchly pedantic. It's not a Tesla Roadster, it's a mini-fridge.

Originally Posted by StuM

You need enough solar to provide about 5000 Watt hours to keep up your 2+ Amps @ 110 V for 24 hours.

Half of those Watt hours are needed at night, so you need to store and then supply 2500 Watt hours.

From the above discussion, based on 5 hours "full sun equivalent", you need 1000W of solar panels to generate your 5000Wh of total energy.

Those 2500 Watt hours overnight is about 100 Amp hours or an average draw of about 8 Amps.

That suggests that about 300 Ah of battery would be a good start, assuming that all of your other house loads are fairly minimal.

In the final proofreading of anything mathematical, a "logic test" is usually applied, to catch things like dropped decimals and other errors that are otherwise easy to miss, but let's do a logic test of that statement. What appliance uses 5 Killowatt Hours per day? How big is it? More importantly, how much dead livestock will it hold?

Originally Posted by StuM

Guy you know knows very little. A lead acid at 10.5 Volts is absolutely dead. When the alternator is on, you are reading the alternator output, not the battery voltage. Once a battery gets down to 12V it is well below a 50% "state of charge". Keeping batteries at those levels will destroy them very quickly. If you can't get them back up to around 12.5/12.6 and have them hold that voltage, then they are indeed "stuffed".

I'm that guy. Interesting assumption. I did tell him that a 12 Volt Lead-acid battery is completely dead at 10.5 Volts, which is true. It's the textbook definition of dead.


This system is not woefully insufficient. It was self-sustaining in the summer. The batteries are ancient and cycled to near-death, but they still work (they need replacement). Another panel of the same size wouldn't hurt, a better charge controller is mandatory, and a small increase in battery capacity (to the tune of one more battery) will help extend the life of the new batteries, but is not strictly necessary for basic system viability.

Originally Posted by Ivansgarage

Here is one more option, I was a truck driver for years and used these
kind of refrigerators 12v, a inverter waists energy, I have had a truck frig run for days on three batteries 300ah and still start the truck.

That is a good idea.

Originally Posted by AndreBruijntjes

Hallo guys, don't know about your maths qualities on the other side of the ocean but here in Europe the consumption of the refridgerator is a factor 10 lower.
This means that production and consumption, taking only the refridgerator into account, are nearly balanced.
...

If there is enough room on board and Rocinante installs a second solar panel of the same size he will never have to drain his batteries and the solarpanels wil maintain his batteries in a perfect state. This on the condition that his solar charging is driven by a proper regulator.
If there is not enough room for a second solar panel Rocinante has to make a few hour trip every week charging his batteries by his Alternator, here again on the condition that the Aternator is driven by a smart(3 step) regulator.
Not clear from the posts, but I assume Roccinante has a set up where he uses a dedicated battery for starting his motor? If this is the case be sure to have a separator which copensates the voltage drop, if not consider installing a second battery and a separator.

Correct.

There is plenty of room for a second panel.

The regulator needs replacement.

All 3 batteries are wired in parallel. They shouldn't be, but needs must, as the system is a mess.

Originally Posted by PaulMarion

"the meter reads 13v when alternator is on an 11.5-12 when off." The alternator output, should read 13.9 - 14.2 Volts if working correctly! With the batteries in an obvious poor state, they could be preventing proper alternator output via resistance.

Likely the case. At steady RPM, the voltage starts at 12.9, and slowly rises to 13.6. The volts read ~12.3 after a long engine run, and quickly (hour or two) drop to 11.8~12.0. The drop to 11.5 takes several more hours. (At night. Voltage is stable during the day, but occasionally drops below the inverter alarm or cutoff threshold when the fridge kicks on... Depending on how low they got overnight.)

Originally Posted by StuM

Not if there is a high load such as the fridge and stuffed batteries - they will pull the voltage down. Or if he is running the engine at insufficient revs when charging

Above 800 RPM, the measured voltage does not vary with RPM. Once warm, the engine is seldom run below 2000 RPM.

[StuM]

Originally Posted by FlySideways

Ok, I registered for the sole purpose of trying to help clean up the numbers in this thread.

Welcome!

Dropping "hours" is like dropping the axis labels off of a graph. It's not exactly punishable by death. The numbers still work and you know full-well what he was trying to say. "amp-hr/hr ~ amp/hr"

If you've seen the amount of confusion that has occurred here on CF over the last few years through people miixing up Amps and Amp hrs, Watts and Watt hrs, you'd know why I keep stressing the use of correct units.

440 AH/Day? It's not a walk-in industrial meat locker on a crew boat, it's a 2.5 cubic foot refrigerator.

Using the OPs quoted figures there!

It's a 100W panel. The sun does not need to be directly overhead for a panel to produce rated output. The drop in output at the lowest visible angle is only (nominally) 40%

Sorry, I can't accept either of those assertions at all.

Under what real world conditions do you see a panel exceed or even reach it's STC rating, especially when not perpendicular to the sun. (We are assuming flat mounted panels on a boat here, not a domestic solar installation with latitude adjusted panels).

And there is no way that my panels generate anywhere near 60% of their STC rating in the early morning and late afternoon.

That's what we are trying to teach here. And yes you can, you just have to do math to convert.

OK, maybe a "compare" was an unfortunate choice of words. Maybe "equate" would have been better. But the OP clearly thought that 2 Amps at 110/120V was the same as at 12V.

Doesn't matter what terms you think in, you still have to do all the V*A=W equation at every point in the equation. You have to solve for one of the 3 variables for every device in the system.

No you don't, Start with V*A = W for any necessary inputs, do all the working in Watts and the do A = W/V on the final outputs, it avoids the errors made in equating Amps at different voltages.

You leave Peukert alone, the Germans have enough to answer for. ... But in serious terms, as above mentioned, we're talking about 2 amps. Applying the Peukert equation to a draw of 2 amps is staunchly pedantic. It's not a Tesla Roadster, it's a mini-fridge.

And a small battery. Peukert definitely comes into play when you want to use the available solar to get that battery up to 100% SOC.

> > Guy I knew told me that until I read 10.5, things were fine.
> Guy you know knows very little. A lead acid at 10.5 Volts is absolutely dead.

I'm that guy. Interesting assumption. I did tell him that a 12 Volt Lead-acid battery is completely dead at 10.5 Volts, which is true. It's the textbook definition of dead.

Ah-ha! So did he misquote you, misunderstand you or did you really tell him that things were fine until he reached 10.5?

This system is not woefully insufficient. It was self-sustaining in the summer. The batteries are ancient and cycled to near-death, but they still work (they need replacement). Another panel of the same size wouldn't hurt, a better charge controller is mandatory, and a small increase in battery capacity (to the tune of one more battery) will help extend the life of the new batteries, but is not strictly necessary for basic system viability.

That's a matter of interpretation/opinion. To my way of thinking, needing to double the solar, add 50% to storage and upgrade the charge controller makes the current system woeful

[FlySideways]

Originally Posted by StuM

Welcome!

Thank you.

Originally Posted by StuM

If you've seen the amount of confusion that has occurred here on CF over the last few years through people miixing up Amps and Amp hrs, Watts and Watt hrs, you'd know why I keep stressing the use of correct units.

Using the OPs quoted figures there!

Ok, but from the questions I'm getting back from OP, it doesn't seem like your point is getting across. It's not necessarily your fault, but that's feedback.

Originally Posted by StuM

Sorry, I can't accept either of those assertions at all.

Under what real world conditions do you see a panel exceed or even reach it's STC rating, especially when not perpendicular to the sun. (We are assuming flat mounted panels on a boat here, not a domestic solar installation with latitude adjusted panels).

And there is no way that my panels generate anywhere near 60% of their STC rating in the early morning and late afternoon.

It's Ok, you don't have to, I don't want to argue solar panel theory. The relevant point is, it ran for an extended period of time during the summer and there was no drop in the battery voltage over time. This means, on average, there was necessarily more charge going in than out. The refrigerator draw, maximum daylight hours, and solar panel peak output are known quantities. The minimum total output of the actual system in the real world can therefore be derived mathematically. ...And knowing that, and knowing the maximum variation in the length of daylight, you don't have to be a mathematician to know that doubling the solar capacity will solve the charging problem.

Originally Posted by StuM

OK, maybe a "compare" was an unfortunate choice of words. Maybe "equate" would have been better. But the OP clearly thought that 2 Amps at 110/120V was the same as at 12V.

No you don't, Start with V*A = W for any necessary inputs, do all the working in Watts and the do A = W/V on the final outputs, it avoids the errors made in equating Amps at different voltages.

The relevant fact is, you have to understand the concepts well enough to do translations with the basic function, and convert to like units before comparison. You rewrote the same equation in different terms. That demonstrates your understanding, but doesn't teach much. It doesn't matter which units you convert to, the understanding of the units is what's important.

Originally Posted by StuM

And a small battery. Peukert definitely comes into play when you want to use the available solar to get that battery up to 100% SOC.

We can have that argument separately, but you're wrong in this application. If you can prove otherwise, I'll come back and revise that statement... But the edit limit is 1/2 hour, so what I'm really saying is I'm not going to have to revise that statement.

Originally Posted by StuM

Ah-ha! So did he misquote you, misunderstand you

Both, in fact. Lucky me.

Originally Posted by StuM

That's a matter of interpretation/opinion. To my way of thinking, needing to double the solar, add 50% to storage and upgrade the charge controller makes the current system woeful

True, but it's the OP's understanding and opinion that matters, because it's his boat/system/money. "Woefully inadequate" is a bit melodramatic, as it indicates that he needs to throw the whole lot away and go for a total redesign, or abandon the quest altogether.

In reality, he has to make relatively minor revisions to the existing system. There is space to double the panel area in the existing location, the existing wiring can support an additional panel wired in parallel, and the existing battery bank area has sufficient space to hold 4 12-volt batteries of standard size, so it can easily accommodate over 300 Amp-Hours' worth of batteries.

The charge controller replacement is not strictly necessary, but there needs to be a component in the system that can be easily read and interpreted, that will warn of inadequate charge before hundreds of dollars worth of batteries have been "murdered."

We aren't designing a perfect system. The objective is to design an adequate system that will fit in the available space with an acceptable lifespan, at the minimum cost, knowing that there is no extra space available to vastly over-engineer it...