Amps v Amp hours v Amps/Hr

[StuM]

Originally Posted by colemj

The real question is how many RPM's are in an inch/lumen?

Mark

Meaningless question. RPM is a one dimensional unit. inch/lumen is a two dimensional unit and all of the dimensions are different. So you can't compare them. It's like asking "How many furlongs in a firkin fortnight"

[wunderluster]

I have read all of this post but, I still can not calculate the amps for two batteries I want to use as a house bank. I believe they were intended to be used in a UPS.

They are Eaton 12V 500W batteries. That convert to be only 41.67 amps.

The Features are;

They are designed for "high power density applications".
Can be used for more than 260 cycles at 100% discharge
Capacity 500W @15-minute rate to 1.67V per cell@77F (25C)
VRLA Battery PWHR12500W4FR

Will these be good for a house bank?
How do I calculate the typical amp hour capacity?

David

[StuM]

Originally Posted by sailorchic34

Both one ampere and one coulomb are equal to 6.241×10^28 electrons per second.

No!. You are still apparently confused by the concept of dimensions. This is where you are going wrong every time.


An Ampere is equivalent to 6.241×10^28 electrons per second
A Coulomb is equivalent to 6.241×10^28 electrons.

Note that I used the word "equivalent" rather than equal.
You also again fail to differentiate between the particle called an electron and its charge. If I were to use the word equal, I would have to say: A coulomb is equal to 6.241×10^28 elementary charges.

See https://en.wikipedia.org/wiki/Elementary_charge

[hellosailor]

"knowing how many Amps your freezer, water pump etc uses when running...is exactly what you need when drawing up an energy budget. "


What, no one mentions an accumulating watt-meter? Or totalizing watt-meter?


You can buy inexpensive digital meters, for AC or DC, that keep a running total of how many volts, amps, watts, have been pulled through them (used) in a 24-hour period, or any other period.


With anything like that, you don't worry about "intermittent use" "Startup draw" or "cycling". You plug your fridge or other device into the totalizing meter, note the time (since not all of them have time averaging or tracking functions), and come back 24 hours later. Then you see how much power the fridge used during a whole 24-hour period, and that's generally all, and exactly, what you need to know to figure out the energy budget.


Power used in 24 hours, versus power made and supplied in 24 hours. No fancy spreadsheets and math needed that way, just a little simple stuff.

[StuM]

Originally Posted by wunderluster

I have read all of this post but, I still can not calculate the amps for two batteries I want to use as a house bank. I believe they were intended to be used in a UPS.

They are Eaton 12V 500W batteries. That convert to be only 41.67 amps.

The Features are;

They are designed for "high power density applications".
Can be used for more than 260 cycles at 100% discharge
Capacity 500W @15-minute rate to 1.67V per cell@77F (25C)
VRLA Battery PWHR12500W4FR

Will these be good for a house bank?
How do I calculate the typical amp hour capacity?

David

Although they call it a 12V 500W battery, that name actually tells us nothing about the capacity of the battery.

The datasheet: http://upserve.com.au/assets/downloads/eaton/pwhr12500w4fr-12v-500w.pdf
tells us that it has 6 cells so

500W @15-minute rate to 1.67V tells us that if we draw 42 Amps @ 12 V (500W) for 5 minutes, the battery will be down to 10.02V. (essentially dead) So at that rate of discharge, it can potentially deliver a total of 42/12 = 10.5 Amp hours

However the data sheet provides quite a bit more information. The table there tells us about different discharge rates. For example
182W @ 60 minute rate to 1.67V and 128W @ 90 minute rate to 1.67V.

Looking at the table we can get:
42 (500/12) Amps for 15 minutes giving us 10.2 Amp hours
26 (312/12) Amps for 30 minutes giving us 13 Amp hours.
15 (182/12) Amps for one hour giving us 15 Amp hours
10.6 (128/12) Amps for 90 minutes giving us 15.9 Amp hours.

So at fairly low current draws, we can expect 15 or more Amp hours from one battery, but put a heavy load on it and you won't get nearly that much.

Two of them will give you about 30 Amp hours of power. If that is enough for you, then you can use them.

But: they are as you said intended for use in a UPS where they stay fully charged for days/weeks/months at a time and occasionally get run down when the power goes out. They will last a long time if used for their intended purpose, but if you running them down and recharging them frequently, they won't last long.

Only you can tell whether these figures meet your style of boat usage and power requirements.

[StuM]

Originally Posted by hellosailor

You plug your fridge or other device into the totalizing meter
...
Power used in 24 hours, versus power made and supplied in 24 hours. No fancy spreadsheets and math needed that way, just a little simple stuff.

As long as you are prepared to rewire your whole boat to put those meters on your fridge, freezer, light circuits, autohelm, bilge pumps, fresh water pump, sea water pump, watermaker, entertainment system, radio, chartplotter, radar, etc etc etc

[sailorchic34]

Originally Posted by StuM

No!. You are still apparently confused by the concept of dimensions. This is where you are going wrong every time.


An Ampere is equivalent to 6.241×10^28 electrons per second
A Coulomb is equivalent to 6.241×10^28 electrons.

Note that I used the word "equivalent" rather than equal.
You also again fail to differentiate between the particle called an electron and its charge. If I were to use the word equal, I would have to say: A coulomb is equal to 6.241×10^28 elementary charges.

See https://en.wikipedia.org/wiki/Elementary_charge

Yet an elementary charge carrier is equal to one electron. A charge carrier is also equal to one electron. A charge carrier could also be protons, but then it would be a positive carrier. We don't use protons in wiring.

A Coulomb is equivalent to 6.241×10^28 electrons, which is the exactly the same as one Ampere second. Two units that measure the same quantity of electrons. They use the same basic unit, that is electrons as a measurement. Does not matter if you call it an elementary charge or a charge carrier. For DC and AC that everyone uses its all electrons.

I'm really not confused at all.

[sailorchic34]

Originally Posted by StuM

As long as you are prepared to rewire your whole boat to put those meters on your fridge, freezer, light circuits, autohelm, bilge pumps, fresh water pump, sea water pump, watermaker, entertainment system, radio, chartplotter, radar, etc etc etc

You could do that, I guess. Most of us would simply have one totalizing meter on the DC or AC mains. It does the same thing that way and without error rates creeping in from using too many different totalizing meters. And after 24 hours we would have ampere days. So cool.

[StuM]

Originally Posted by sailorchic34

I'm really not confused at all.

Dunning-Kruger.

[sailorchic34]

Originally Posted by StuM

Dunning-Kruger.

Nice insult and ad hominem there Stu.

That would apply if I was not skilled. Yet I've sat and passed a professional engineering "PE" exam. That at least allows that I have some small skill or perhaps I am really good at guessing, even with essay questions. That must be it.

[LakeSuperior]

Originally Posted by StuM

Nope. Stone is a dimensional unit of weight (mass). Dozen is dimensionless.

Sorry, weight is not mass. Weight is a unit of force, e.g. pounds. A unit of mass in the English system is a slug. Weight = mass x acceleration due to gravity. In the MKS system mass is kilograms and force is in Newtons.
Dunning-Kruger all over again.

[fryewe]

There is no reasonable practical meaning of Amps/hr or Amps per hour (which is actually a measure of how fast a current is increasing or decreasing.)

Nice insult and ad hominem there Stu.

v=d(phi)/dt; v=L x di/dt, and are practical relationships for evaluating counter emf in motors and inductance in electrical circuits for the engineer...but not for the layman. The hour is not a practical unit of time in most cases, but the time rate of change of current is certainly of practical interest.

So, SC, perhaps the insult was not directed at you, but was a simple self assessment.

[StuM]

Originally Posted by LakeSuperior

Sorry, weight is not mass. Weight is a unit of force, e.g. pounds. A unit of mass in the English system is a slug. Weight = mass x acceleration due to gravity. In the MKS system mass is kilograms and force is in Newtons.
Dunning-Kruger all over again.

I am well aware of the difference between weight and mass.

Are you aware of the fact that a pound can be either a pound-mass (officially defined in kilograms) or a pound-force (officially defined in kilograms multiplied by a standard gravity).

https://en.wikipedia.org/wiki/Pound_%28mass%29

https://en.wikipedia.org/wiki/Pound_%28force%29

In practical terms when you ask for a pound of butter, what you want is a pound-mass (0.453592 kg) not a pound-force (4.44822 Newtons). When they send a pound of butter to the moon, it is still a pound of butter even though it weighs less.

[fryewe]

In practical terms when you ask for a pound of butter, what you want is a pound-mass (0.453592 kg) not a pound-force (4.44822 Newtons).

If you ask for a pound of butter (if not already weighed and packaged), the merchant will weigh the mass of butter he will sell you on a scale that determines the weight (force of gravity) on the butter. What you want is a mass of butter that will deflect the scale to indicate a lb-f as weighed on a calibrated scale, which means you will get (for practical purposes) a lb-m of butter for your money.

When they send a pound of butter to the moon, it is still a pound of butter even though it weighs less.

If a lb-m of butter were sent to the moon, it would still be a lb-m of butter. The force of gravity on it when it is loaded aboard would be a lb-f. When a manned craft with the butter arrived on the moon it may not be weighable. If I were on that moon mission, the butter would be too widely distributed among my organs to have its weight evaluated.

[StuM]

Originally Posted by fryewe

v=d(phi)/dt; v=L x di/dt, and are practical relationships for evaluating counter emf in motors and inductance in electrical circuits for the engineer...but not for the layman.

And in the context of this thread on Cruisers forum - Electrical: Batteries, Generators & Solar it is hardly relevant.