mAh to KWH conversion

Status
Not open for further replies.

1koolkat

Member
Location
Littleton
I have a battery recharger that requires 26,800 mAh to charge. How many KWH is that? I would guess 2.6 but it could also be .26 or .026. Help
 

mgookin

Senior Member
Location
Fort Myers, FL
/1,000,000

because it's 1/1,000 of an amp per hour and we want kWh

edit --> Sorry Dennis: I see you stated in terms of Wh. OP asked for kWh.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
What is the battery voltage? you need 26,800 milli-amp hours, which is 26.8 amp hours.

If the battery voltage is 12 volts, multiply that by 12. Then divide by 1000 to get kilo-watt hours.

So if your battery voltage is 12 volts, you need approximately 0.32 kwh to recharge one battery. I say approx., because the battery voltage will be a little higher than 12 during charging.
 

Smart $

Esteemed Member
Location
Ohio
...So if your battery voltage is 12 volts, you need approximately 0.32 kwh to recharge one battery. I say approx., because the battery voltage will be a little higher than 12 during charging.
Actually the kWh should work out the same even at the higher charging voltage because kWH is a measure of energy transfer. Higher voltage required for charging, which also means slightly higher amperes, but less time... and it should amass to the calculated kWH.
 

1koolkat

Member
Location
Littleton
Travel charger

Travel charger

No one has had a respectful answer. Everyone wants to modify their replys. I gave the data from the device label. 26800 mAh/96.48wh, input at 5V and 2Amps: output at 5V and 4Amps. Model A1210 2nd Gen Astro E7 Anker model A210. What amount of electricity does it require to charge this device from a full discharge? I do not know how to calculate it. Is their someone out there who would be willing to give me a simple answer? I don't think I am far off, but someone out there has a better idea of the relationships between amps, volts, and mAh's. Hopefully. I just wonder if there is someone out there who can give me at least a glimmer of how to find this answer. Guess I am just an ass to even ask. Thought the internet was where you could find answers. Just leave me as ignorant, and unable to ask questions so that people will consider answering..... Cool....
 

Smart $

Esteemed Member
Location
Ohio
No one has had a respectful answer. Everyone wants to modify their replys. I gave the data from the device label. 26800 mAh/96.48wh, input at 5V and 2Amps: output at 5V and 4Amps. Model A1210 2nd Gen Astro E7 Anker model A210. What amount of electricity does it require to charge this device from a full discharge? I do not know how to calculate it. Is their someone out there who would be willing to give me a simple answer? I don't think I am far off, but someone out there has a better idea of the relationships between amps, volts, and mAh's. Hopefully. I just wonder if there is someone out there who can give me at least a glimmer of how to find this answer. Guess I am just an ass to even ask. Thought the internet was where you could find answers. Just leave me as ignorant, and unable to ask questions so that people will consider answering..... Cool....
Divide by 1000 (i.e. 1000wH = 1 kWH) and you have your answer in kWH.... 0.09648 kWH


And next time, check da 'tude at da door.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
No one has had a respectful answer. Everyone wants to modify their replys. I gave the data from the device label. 26800 mAh/96.48wh, input at 5V and 2Amps: output at 5V and 4Amps. Model A1210 2nd Gen Astro E7 Anker model A210. What amount of electricity does it require to charge this device from a full discharge? I do not know how to calculate it. Is their someone out there who would be willing to give me a simple answer? I don't think I am far off, but someone out there has a better idea of the relationships between amps, volts, and mAh's. Hopefully. I just wonder if there is someone out there who can give me at least a glimmer of how to find this answer. Guess I am just an ass to even ask. Thought the internet was where you could find answers. Just leave me as ignorant, and unable to ask questions so that people will consider answering..... Cool....

The 26800mAh figure is not one which properly relates to a charger. It describes the capacity of a battery.

The input and output voltage and amperage figures are appropriate for either a charger (which is not possible since the output power is greater than the input power) or to a battery with charging circuit included in one package.

I suspect that what you have is similar to one of the combination external battery and battery charger units designed to work with a cell phone through its USB port, only larger.

The combination of 5V and 26800mAh is all that you need to calculate the energy stored in the battery.
26.800 Ah times 5V (note the change in units in the capacity value) = 134Wh = .134kWh.
[Ah, or Amp-hours, times volts equals (Volt times Amp)-hours equals Watt-hours.]
The actual energy required to charge it will be perhaps 20% greater than that.

BTW, the input and output specifications are a bit misleading, since the input voltage must be at least slightly greater than the battery voltage to be able to charge it, while the output voltage must be appropriate to the needs of the electronics plugged into it or the battery it is going to charge.
The 5V is a "nominal" value and can only be used for approximate calculations. I do not know of any battery chemistry which gives a voltage of exactly 5V.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
No one has had a respectful answer. Everyone wants to modify their replys. I gave the data from the device label. 26800 mAh/96.48wh, input at 5V and 2Amps: output at 5V and 4Amps. Model A1210 2nd Gen Astro E7 Anker model A210. What amount of electricity does it require to charge this device from a full discharge? I do not know how to calculate it. Is their someone out there who would be willing to give me a simple answer? I don't think I am far off, but someone out there has a better idea of the relationships between amps, volts, and mAh's. Hopefully. I just wonder if there is someone out there who can give me at least a glimmer of how to find this answer. Guess I am just an ass to even ask. Thought the internet was where you could find answers. Just leave me as ignorant, and unable to ask questions so that people will consider answering..... Cool....

You're welcome.
 

steve66

Senior Member
Location
Illinois
Occupation
Engineer
Actually the kWh should work out the same even at the higher charging voltage because kWH is a measure of energy transfer. Higher voltage required for charging, which also means slightly higher amperes, but less time... and it should amass to the calculated kWH.

Some power will be lost in the internal resistance of the battery. I'm not sure the charging process is 100% efficient either.

Newer batteries types may be better, but some of the older types get pretty warm when you recharge them. That is obviously quite a bit of power lost in heat during charging.
 

Barbqranch

Senior Member
Location
Arcata, CA
Occupation
Plant maintenance electrician Semi-retired
I believe also the capacity of a battery is a function of how it is discharged. Discharge it quickly, and I believe you get less energy out of it than if the process discharges if more slowly.
 

Johnnybob

Senior Member
Location
Colville, WA
No one has had a respectful answer. Everyone wants to modify their replys. I gave the data from the device label. 26800 mAh/96.48wh, input at 5V and 2Amps: output at 5V and 4Amps. Model A1210 2nd Gen Astro E7 Anker model A210. What amount of electricity does it require to charge this device from a full discharge? I do not know how to calculate it. Is their someone out there who would be willing to give me a simple answer? I don't think I am far off, but someone out there has a better idea of the relationships between amps, volts, and mAh's. Hopefully. I just wonder if there is someone out there who can give me at least a glimmer of how to find this answer. Guess I am just an ass to even ask. Thought the internet was where you could find answers. Just leave me as ignorant, and unable to ask questions so that people will consider answering..... Cool....

I think part of the problem is the mis-leading info on the device itself. Power in equals power out. In other words, you cannot create power out of thin air. If you have 5v input @ 2A = 10 watts. Now you have 5v output @ 4A = 20 watts. Where on earth is that extra 10 watts coming from? If that were possible, the power company's would be out of business pretty quickly (or would buy up the technology)!
I'm not trying to be disrespectful, just can't get my head around where that extra 10 watts is coming from?
 

mike_kilroy

Senior Member
Location
United States
I have a battery recharger that requires 26,800 mAh to charge. How many KWH is that? I would guess 2.6 but it could also be .26 or .026. Help

Kitkat, realize WHY you received varying answers: You did NOT give all the info, nor even enough info, for a correct simple answer. There is NO answer possible to this question with only 26,800mAh given. Next time you ask a technical question, be sure to give ALL the info - as you finally did in your follow up post.


No one has had a respectful answer. Everyone wants to modify their replys. I gave the data from the device label. 26800 mAh/96.48wh, input at 5V and 2Amps: output at 5V and 4Amps. Model A1210 2nd Gen Astro E7 Anker model A210. What amount of electricity does it require to charge this device from a full discharge? I do not know how to calculate it. Is their someone out there who would be willing to give me a simple answer? I don't think I am far off, but someone out there has a better idea of the relationships between amps, volts, and mAh's. Hopefully. I just wonder if there is someone out there who can give me at least a glimmer of how to find this answer. Guess I am just an ass to even ask. Thought the internet was where you could find answers. Just leave me as ignorant, and unable to ask questions so that people will consider answering..... Cool....

Some will look at this info and say "not possible" since it does not explain well that the INPUT is 2amps MAX - for charging, and the OUTPUT is 4amps MAX for your loads. It assumes you charged it for some time at upto 2 amps, then you can pull upto 4 amps out of the (heehee funny specs) "26,800mah" battery inside - for upto 1/2 time it was on 2amp charge.

Perhaps you are not aware you did NOT give the data from the device in your post?

NOW you give enough info for real answers. BUT your question is PROBABLY wrong, so you will - and did - continue to get varying answers. You originally asked "how many KWH is that?" Now that you gave the missing info, you can have, and in fact received THAT answer: 0.09648 KWH.

I am curious: what are you going to use that answer for? Do you have some device that needs to know this number? Will this number tell you how long it will take to charge a fully discharged battery? What good is this answer to you? I suspect you asked a WRONG question to begin with, in addition to not giving enough info for an answer....

Now you ask a different question "What amount of electricity does it require to charge this device from a full discharge?" I again respectfully ask you kit, what use is this answer? How will you use it to benefit your use of this device? I cannot imagine ANY useful purpose to have this answer, even as ambiguous as IT is. What do you mean by "how much electricity?" In amps? In volts? In power? In time? In energy? As an engineer, I am VERY interested to learn how you will use the various answers you got/will-get from this ambiguous question?

I suspect your questions have all been geared to find out how to charge this? It is charged by plugging into a std computer or wallwart USB port. All USB ports are 5vdc. It clearly states it is rated at a max input of 2amps. So there is the USEFUL answer for you? 2 amps. Says so right in the device spec shown here:

http://www.amazon.com/Upgraded-Capacity-Anker-Ultra-High-Technology/dp/B00M3073L4

Maybe you wish to know how LONG it would take to charge a fully discharged unit? Now THAT info would be useful to know... So its simple answer is 96.48wh/(5v*2a)= 9.648 hours. Being a lithium battery, you will put in an extra 3-4wh, so in reality, it will be closer to 10.048 hours. Assuming the device is really capable of being discharged fully to 0wh or 0v, actually the recharge time will likely be infinite, as discharging 'fully' will kill the lithium battery and you can buy a new device.

BTW, this device is 'rated' 26,800mah @ 5vdc... Just realize this is not possible. Someone was smoking something funny, or has a special marketing degree, to have written that rating on it. A REAL 26,800mah @ 5v battery takes up a space at least double this units 18.7 cubic inches.
 
Last edited:

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
I think part of the problem is the mis-leading info on the device itself. Power in equals power out. In other words, you cannot create power out of thin air. If you have 5v input @ 2A = 10 watts. Now you have 5v output @ 4A = 20 watts. Where on earth is that extra 10 watts coming from? If that were possible, the power company's would be out of business pretty quickly (or would buy up the technology)!
I'm not trying to be disrespectful, just can't get my head around where that extra 10 watts is coming from?
As mentioned in post #10, the extra power comes from the battery!

It is not uncommon to have a battery/charger combination system which cannot recharge at as high a rate as it can discharge.
If you follow the OP's wording and think of it only as a charger, then your point would be valid.
 

mike_kilroy

Senior Member
Location
United States
Another 'proof' that this is not a real 26800mah battery unit is their statement:

Giant Capacity: Charges the iPhone 6 ten times, the iPhone 6 Plus or Galaxy S6 over six times

The S6 battery is 2550mah; so if you have 26800 available, you SHOULD be able to charge it 26800/2550 = 10.5 times. So they admit in its capability it is NOT a 26800mah device. So charge time will likely be LESS than 8-10hours sinced it is a smaller battery inside, so even the mathematical answers given are incorrect. This lends credability to the fact it is 2x smaller than physically possible to house that size battery.

We can only give correct answers when the data is factual and creative marketing is left out.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Another 'proof' that this is not a real 26800mah battery unit is their statement:

Giant Capacity: Charges the iPhone 6 ten times, the iPhone 6 Plus or Galaxy S6 over six times

The S6 battery is 2550mah; so if you have 26800 available, you SHOULD be able to charge it 26800/2550 = 10.5 times. So they admit in its capability it is NOT a 26800mah device. So charge time will likely be LESS than 8-10hours sinced it is a smaller battery inside, so even the mathematical answers given are incorrect. This lends credability to the fact it is 2x smaller than physically possible to house that size battery.

We can only give correct answers when the data is factual and creative marketing is left out.
On the other hand, if the internal battery is also Li chemistry, it will be able to power a phone all the way down to its cutoff point, but it will not be able to charge the phone battery fully once its open circuit terminal voltage has dropped below that of the phone battery.

(Unless there is more going on inside in terms of DC to DC buck/boost circuitry than there appears to be.)

But there certainly is a dearth of well specified technical information available to back up the marketing puff. :)
 

meternerd

Senior Member
Location
Athol, ID
Occupation
retired water & electric utility electrician, meter/relay tech
OK...I'll be the one to ask. WHY do you want to know? If you're using utility power at 13 cents per Kwh, you'll be spending chump change anyway. Was it just to get a response? If it's a question of just mathematics, maybe another forum would be more useful.

If I'm misinterpreting your motives, I apologize. Now charging your new Tesla car and trying to convince yourself it's a good investment, now there's a question!
 
Status
Not open for further replies.
Top