Joules converted to amp hours

[djkwolf]

I just met a fellow who works in directional drilling. He has a problem when his dry cell lithium ion batteries die. It costs him $300,000 when it happens. What he needs to know is when to pull the batteries BEFORE this happens. These Double D cell batteries when connected, hold their 36 volts until the 3.2MJ are used up, then he gets an estimated .2 millivolt instantly.
I told him I knew a site that could tell us that for him.

SO...here is the question. He starts out with 3.2 MJ...the motor requires 1300 J/H to idle...pulses from his equipment cost 1282J every 6 seconds. How many amp hours does he have before it costs him? When should he pull the batteries for replacement or charge?

Is there a formula for this we can give him?

My personal interpretation of this equation is to mutiply the J/6 sec into hours and add to the hourly cost. Then divide into the 3.2 MJ for a time allowance. That was not his question though. Any takers?

[Danger]

Welcome to PF, Djkwolf.
I'm afraid that I can't help with your solution, but there are others here who can. I don't know what time zone you're in. It's coming up on 2:30 am here, and most respondents are in North America, so it might be a while before someone else gets back to you.
Now, I'm going to bed.

[djkwolf]

Thanks Danger!
I AM in Alberta, and it WAS after 2am that I posted.

[crank]

I just met a fellow who works in directional drilling. He has a problem when his dry cell lithium ion batteries die. It costs him $300,000 when it happens. What he needs to know is when to pull the batteries BEFORE this happens. These Double D cell batteries when connected, hold their 36 volts until the 3.2MJ are used up, then he gets an estimated .2 millivolt instantly.
I told him I knew a site that could tell us that for him.

SO...here is the question. He starts out with 3.2 MJ...the motor requires 1300 J/H to idle...pulses from his equipment cost 1282J every 6 seconds. How many amp hours does he have before it costs him? When should he pull the batteries for replacement or charge?

Is there a formula for this we can give him?

My personal interpretation of this equation is to mutiply the J/6 sec into hours and add to the hourly cost. Then divide into the 3.2 MJ for a time allowance. That was not his question though. Any takers?


I'll take a stab at it, but I'm no electrical engineer. It appears to be all unit cancellation.

So first, covert all of your units so they're the same. So:
3.2 MJ = 3,200,000J
1282J/6s = 769,200J/h

Then you solution should be J/(J/h) = h
So your estimated run time should be:
3,200,000J / (1300J/h + 769,200J/h) = 4.15h = 4h 9m 11s

I hope this helps.

[Jobrag]

Perhaps he should invest in a totalising power meter.

[nirax]

since you are saying 'pulses', it would be prudent to have a safety margin ... what exactly do you mean by pulses per six seconds. is it a discharge or flash of some kind ?

[djkwolf]

Pulse was his word. His equipment reads every six seconds using battery power, and sends that data costing battery power.

[vk6kro]

He must know what his average usage time is.

If he replaces the batteries every 2 weeks, why not change them every 10 days regardless of condition? Set up a schedule and have someone assigned to do it.

Considering the consequences of getting it wrong, wasting a few dollars on batteries that still have a bit of life in them is trivial.

[brewnog]

He must know what his average usage time is.

If he replaces the batteries every 2 weeks, why not change them every 10 days regardless of condition? Set up a schedule and have someone assigned to do it.

Considering the consequences of getting it wrong, wasting a few dollars on batteries that still have a bit of life in them is trivial.

Indeed, I'd approach this problem empirically and use some statistics to put a TPM programme in place.