How do you guys counting that MH/W efficiency?

[BCMan]

 For example I'm spending ~100 kwh per month on my 5770 and 6870. Both are delivering 541 mhash/s.
 
 So, 100 kwh per month, or 3,333 kwh per day/24h, or ~0,139 kw per hour or 139 watt*h. And I'm stuck here. I should convert 139 watt*h to J (watt per sec) now? But then I'm getting crazy numbers, 500400 Joules. 541/500400 = 0,00108 Mh/J or Mh/W.  

Or 541/139 = 3,89 MH/W? But then it's still incorrect Watt*H is not J. I'm something messed up here? MH/W is (Mhash per sec)/(W*h)? Whats then Mh/J on hardware wiki?

Or 1947600 mhashes per hour / 500400 J = 3,89 MH/J . Correct? Then MH/J (or MH/W) means (Mhashes per hour)/(J or W)?

[1714169814]

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[1714169814]

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[1714169814]

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[DeathAndTaxes]

kWh & watt hours  = units of energy (power over time)

vs

KW and watts  =  units of power

100 kWh (how do you know?) * 1000 = 100,000 watt hours

30*24 = 720 hours

100,000 watt hours / 720 hours = 139 watts hours / hours = 139 watts

(not watt hours, you divided kwh by h leaving KW and then converted to watts)

Note you can express this more simply as 100 kWh * 1000 / 30 / 24  = 139 watts but the above helps to show the conversion of units.

(541 MH/S) / (139W) = 3.89 MH/WS = 3.89 MH/J

On edit: corrected for accuracy.

Often you will see this expressed as
541 MH / 139W = 3.89 MH/W.  It is technically wrong but I have to admit I do it to.  Most people are more familiar with the concept of watts and kWh than Joules.

IF someone is saying 3.89 MH/W they likely mean 3.89 MH/J.

Or 1947600 mhashes per hour / 500400 J = 3,89 MH/J . Correct? Then MH/J (or MH/W) means (Mhashes per hour)/(J or W)?

Yeah you lost me here.

Alternatively you simply use a kill-a-watt mete and measure the current.  Take hashrate / current.

BTW I doubt you have a rig which only uses 139W.  I am not sure where you "know" you are using 100 kWh per month but that likely is the source of your error.

[PulsedMedia]

From used KWh to hourly usage:

Month kWh/30.33 (Average month is 30.33 days)/24
That's it, you got your hourly usage
So you were correct.

You are using ~137W only, that's amazing for those cards --- and can't be correct most likely. Many computers use @ idle that much.
But you would be getting ~3.95Mhash/W at those figures tho

[tucenaber]

541 MH / 139W = 3.89 MH/W

I think you mean 541 MH/s / 139 W = 3.89 MH/Ws = 3.89 MH/J

[DeathAndTaxes]

541 MH / 139W = 3.89 MH/W

I think you mean 541 MH/s / 139 W = 3.89 MH/Ws = 3.89 MH/J

True although almost nobody has a clue what a Joule is and most people understand a watt.  Saying MH/W isn't accurate but I have to admit it is a bastadization that I use.  Still your right it should me MH/J and since this is a thread about calculating the efficiency it should be done right.  Post updated.

[BCMan]

 Thanks for explanation, guys. Yeah, 100 kWh isnt correct. Need to wait a month to get correct numbers (counting like that: kwh usage in next month minus year's average kwh/m usage before mining. should be more less correct). Plus I'm shutting down the rig for 2 nights per week, so its not a pure 24/7 work.

[tucenaber]

Just in case someone else needs a lesson about units and how not to confuse Wh and W/s, I'll elaborate a little bit more

Watts per hour is meaningless (unless you mean accelerating power usage)! Since power usage is enegy per time unit (Joules per second) or watts, one Joule equals 1 watt*second. (*not* 1 watt/second). kWh/day also is not kW/hour, because h/day=24, a constant so 24kWh/day is just 1kW.

Learn to use units correctly, they are your friend Imagine you have a multiplication sign between the units, so that 1kWh is 1*k*W*h = 1*1000*W*h and 1J/s = 1*W*s/s = 1*W after simplification. That way the original questions becomes

hasrate / power usage = (541*M*H / s) / (139*W) = 541/139 * M * H / s / W = 3.89 M*H/(W*s) = 3.89 MH/J,
since power usage = 100kWh/30days=100*k*W*h/(30*24*h) =100*k*W/(30*24)=0.139*k*W = 0.139*1000*W = 139W

Edit: corrected my mistake

[DeathAndTaxes]

kWh/day also is not kW/hour, because h/day=24, a constant so 1kWh/day is just 24kW.

Something is not right here.

24kW > 1 kWh per day.  Hell 24 kW > 1 kWh / hour

I think you mean 1 kW = 24 kWh per day.

[tucenaber]

kWh/day also is not kW/hour, because h/day=24, a constant so 1kWh/day is just 24kW.

Something is not right here.

24kW > 1 kWh per day.  Hell 24 kW > 1 kWh / hour

You're right, of course. I should learn to handle my units!

What I should have said is 1kWh/day equals 1kWh/24h or 1/24 kW. There!

[MrTeal]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

[BCMan]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v and european Schuko.

[MrTeal]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v with european Schuko.

Do you have a multimeter? A modern PSU has a power factor so close to 1 that you can safely just measure current and multiply it by your line voltage to get a power draw that will be quite accurate.

[BCMan]

For example I'm spending ~100 kwh per month on my 5770 and 6870. Both are delivering 541 mhash/s.
 
 So, 100 kwh per month, or 3,333 kwh per day/24h, or ~0,139 kw per hour or 139 watt*h. And I'm stuck here. I should convert 139 watt*h to J (watt per sec) now? But then I'm getting crazy numbers, 500400 Joules. 541/500400 = 0,00108 Mh/J or Mh/W.  

Or 541/139 = 3,89 MH/W? But then it's still incorrect Watt*H is not J. I'm something messed up here? MH/W is (Mhash per sec)/(W*h)? Whats then Mh/J on hardware wiki?

Or 1947600 mhashes per hour / 500400 J = 3,89 MH/J . Correct? Then MH/J (or MH/W) means (Mhashes per hour)/(J or W)?

Don't use MH/W.  Use $/BTC mined.  Easier to see if you are underwater.

In your case, you probably generated 8+ BTCs and paid $10+ (per month).
Measure power consumption at the wall for the whole system.

6870 is only 2 weeks old, so no idea how much Btc will be. Estimated around 10 with current difficulty. This month overall cost is 14,38$ (248 kwh) and higher on 80 kwh than month without mining.

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v with european Schuko.

Do you have a multimeter? A modern PSU has a power factor so close to 1 that you can safely just measure current and multiply it by your line voltage to get a power draw that will be quite accurate.

Found this one.
 http://gadgets.masterkit.ru/product/42.htm

[DBordello]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v with european Schuko.

Do you have a multimeter? A modern PSU has a power factor so close to 1 that you can safely just measure current and multiply it by your line voltage to get a power draw that will be quite accurate.

I don't think you want to push 10A through a multimeter....

[MrTeal]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v with european Schuko.

Do you have a multimeter? A modern PSU has a power factor so close to 1 that you can safely just measure current and multiply it by your line voltage to get a power draw that will be quite accurate.

I don't think you want to push 10A through a multimeter....

Why not? Obviously it will depend on your model, but even a cheap Motomaster one I picked up 5 of for $10 each have a 10A range. If he's drawing more than 10A @ 240V with a 6870 and a 5770, he's doing something wrong.

[n4l3hp]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v and european Schuko.

There are cheap "made in china" digital watt meter in eBay. You can choose if you want the 110v unit or the 220v one.

[BCMan]

Your method of estimating power draw is going to have huge error in it. It's really going to be better to buy something cheap like a P3 Kill-A-Watt for $20 and use that to actually measure it. The Kill-A-Watt is actually useful elsewhere, so it's not like it's lost money to buy one.

Well, problem is it doesnt support 220v and european Schuko.

There are cheap "made in china" digital watt meter in eBay. You can choose if you want the 110v unit or the 220v one.

Well, Ive found few already made in Russia. Will be cheaper to buy it here, than order from ebay.
http://gadgets.masterkit.ru/product/42.htm
http://www.feron.ru/catalog/catalog.php?SECTION_ID=537&ELEMENT_ID=7480