Electrical Math Help

AnOldUR

Looking for a little electrical math help with testing my inverters capabilities.

At 115V the Air8 is drawing between 6.5 and 7 amps.

So that's 6.7 amps x 115 volts = 770 watts

So, that converts to 770 watts / 12 volts = 64 amps

When I measure the current at the battery with the Air8 running on the inverter I get about 72 amps.

Here's where I'm confused.
If 770 watts / 72 amps = 10.7 volts

Does that represent 1.3V in inefficiencies or voltage drop over the 8' run of 2 AWG wire between the battery and the inverter?
(or more if the battery is at a higher state of charge than 12V)

Am I wrong in thinking that the Air8's wattage is constant at 770W when going from 120V to 12V?

(What might be skewing the numbers a little is that the inverter is putting out close to 120V, but my house is at 115V.)

Yoshi_TAB

Hi,  Here is a voltage drop calculator  Voltage Drop Calculator - Inch Calculator I've used to determine voltage drop across a length of wires.  I plugged in your numbers (is the 8' one way or round trip) , using 150 amps for your inverter (I don't know the size) and only get about 0.4 V, about 3%, which I believe is an acceptable number.  Your measured 10.7 V seems too low.  How did you measure current (shunt, multimeter)?

Your calculated numbers are right..but don't know why the difference

Your calculated  power consumption looks close to the spec. number.

MuttonChops

AnOldUR said:
So, that converts to 770 watts / 12 volts = 64 amps {calculated}
When I measure the current at the battery with the Air8 running on the inverter I get about 72 amps.

The Renogy 2000W Pure-Sine Wave Inverter has a Nominal Efficiency Rating of 90%

     Air8  calculated demand     770W
     Battery measured demand  864W

     Inverter eff (770/864)          89%

Your measured value seems correct for this inverter.

AnOldUR

@MuttonChops thanks for that. How does wattage play into this? I was mistaken in thinking that the 770 watts while running on 120 volts could be used to calculate voltage drop. The voltage must still be in the 12 volt range, but it translates to a higher power requirement at the battery because of the amperage?

MuttonChops

AnOldUR said:
. . . How does wattage play into this?

Time for you to visit the library, search the web, or take a class as there are lots of resources far better than I to explain the physics of electricity.

That said:

Electricity is measured in units of power called Watt. A Watt is the unit of electrical power equal to one ampere under the pressure of one volt.  Watts are power and what we care about,  Voltage and Current are just helpful indicators.

The ampere is a measure of the amount of electric charge in motion per unit time ― that is, electric current.

Voltage is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop.

rfuss928

AnOldUR
The higher power requirement at the battery is because of the inefficiency of the inverter.  Nothing is 100% efficient.  The inverter is ~90% efficient so to get 770 watts out you need 856 watts in.  (770/.9=856)
As MuttonChops said, power management is what matters and voltage with amperage are factors used for that accounting.  It appears you system is working as expected.

AnOldUR

Thanks for the help. I took a semester of this about fifty years ago, but this foggy old brain needed a refresher. Research is helpful, but sometime it takes one on one discussion to drive the point home. Thanks again!

MarkAl

Your understanding is correct, only change would be I'd use the actual voltages present at the time you measure the current. The main difference in Input power to the AC and the input power to the Inverter is what is lost in converting the battery voltage to the AC needed input voltage (that's why it gets hot!). Also, depending on the rating of the inverter and how much power you are pulling from it, the efficiency changes. The closer you run to the rated output is usually the best efficiency (because many of the losses inside the inverter are there if running high or low). The wire loss contributes to the losses you may be measuring but they are easy to calculate and separate by knowing the wire gauge's Ohms/foot (both ground and power length) times the currents OR not a factor if measuring voltage and current at the inverter and AC box's inputs for inverter efficiency.

But overall, you have it correct, just be very detailed in where and how much you actually are using at the same time.
(edited to clear up some implied statements)

Yoshi_TAB

Hi, 

 I find this interesting and trying to get my head around it.  I follow everything, but so I'm clear, the 10.7 V reduced voltage is due to a combination  inverter inefficiency and wire line losses? 

Thank you

MarkAl

The "calculated" 10.7V is due to some assumptions about the way you did the power equates between 115V and 12V not a flaw in the calculations. If you want to know all the power losses you do need to account (carefully) for the wire losses though they probably only equal under 10-15 watts depending on length of wire run, wire gauge and current through each.

AnOldUR

@Yoshi_TAB the 10.7V was calculated, not measured. I was questioning that calculation because the numbers did not make sense. Later posts confirmed that my using the 770 watts in that formula was the problem. The wattage is higher due to the inefficiencies of the converter. As @rfuss928 pointed out, "The inverter is ~90% efficient so to get 770 watts out you need 856 watts in.  (770/.9=856)". So, 856 is the wattage that should have been used in that calculation.

edited to clarify

rfuss928

Yoshi_TAB said:

Hi, 

 I find this interesting and trying to get my head around it.  I follow everything, but so I'm clear, the 10.7 V reduced voltage is due to a combination  inverter inefficiency and wire line losses? 

Thank you

The 10.7 volts is a miscalculation.  
battery Watts = battery Volts x battery Amps
battery watts is actually 856 (calculated) so:
Vb = Wb / Ab  =>  Vb = 856/72  => battery Voltage = 11.9 
that seems like a reasonable value given the 72 amp load and represents what would be expected at the 12v input of the inverter (so includes any wiring losses from the battery to the inverter).
A couple of additional actual voltage or current measurements would help isolate each components contribution to the overall efficiency and eliminate some assumptions made for these estimates.

***For those following the details, power calculations for AC circuits require a power factor to correct for the alternating current waveform  or use of a true RMS meter***

Yoshi_TAB

Ah, makes sense..thank you

Bayliss

I don't know about others, but I have a headache.  I never did like math in school, but I admire that we have some folks on the forum that have figured out how to do the calculating.  I often keep notes of stuff like this, just in case I find a need for it down the road, so I appreciate it.  Impressive!

fstop32

I ditto your sentiments @Bayliss!  Always good to know you smart guys are out there and willing to share!  

Horigan

For me, a simple way to think about it is to divide the Air 8 power used (770W) by the inverter efficiency (0.90) to get the required power input to the inverter (856W).  Divide that power by 12V to get the approximate current.  If the wire gauge is appropriate for that current, then you're done and don't really need to calculate the losses in the wire.

Yoshi_TAB

Hi @AnOldUR,

How warm, if at all, did your inverter get when you ran the Air 8 test?  I'm just curious,  I believe you know from my other posts, I plan on placing an inverter next to the Air 8  (with lithium batteries).  I ask not because I'm concerrned about the Air 8, but maybe I should put the inverter where you did for more breathing space if it generates heat.

Thank you

rfuss928

Yoshi_TAB said:

Hi @AnOldUR,

How warm, if at all, did your inverter get when you ran the Air 8 test?  I'm just curious,  I believe you know from my other posts, I plan on placing an inverter next to the Air 8  (with lithium batteries).  I ask not because I'm concerrned about the Air 8, but maybe I should put the inverter where you did for more breathing space if it generates heat.

Thank you

The inverter uses 86 watts doing the conversion (856 - 770) as described above .  So you can think of the heat it generates quite similar to a 100 watt light bulb in that same space.

Yoshi_TAB

"heat it generates quite similar to a 100 watt light bulb in that same space."

Hmm, they can get pretty hot/warm to the touch.  But  the cooling fans disappate the heat?   Correct?  Do you think  that space is too confined and  the "electrical cubby" near the WFCO would be better?  The only negative is longer 2/0 cables to the batteries?

Thank you

AnOldUR

@Yoshi_TAB as @rfuss928 said the heat is related to the wattage so that should give you the answer you're looking for, but you made me curious. I had never bothered to even give it a touch test, but this motivated me to put a Bluetooth thermometer on top of the inverter, close the compartment back up and turn on the AC. After about 20 minutes of run time the temperature only went from 65 to 68 degrees. The inverter housing was cool to the touch. At higher ambient temperature the results may be different, but it looks like the internal fan does a pretty good job.

Yoshi_TAB

Is your TAB at room temperature (65-70)  or is it cold with heat off?  I see you are in NJ..it's 33 in DC right now.  I'm thinking twice about my intial location.  I think your spot in the electrical cubby may give it more ventilation.

AnOldUR

Yoshi_TAB said:

Is your TAB at room temperature (65-70)  or is it cold with heat off?

Our T@B is stuffed into a one car heated garage at about 65 degrees. Either inverter location that you've suggested looks good. Under the bed will keep your inverter closer to the batteries, but I'm not sure how much heat builds up around the Air8 or Alde when they're running.

MarkAl

@AnOldUR Caution, the inverter dissipates heat based on its quiescent power and the load it's delivering. Were you measuring at an equivalent load (high) and if the fan is ON, the heat is being dumped into the space around it (room? assuming a cool input air source), keeping the inverter cool. 

I have a 300 watt DC-CD charger that at 6-7 amps is barely warm but pump the load up to 250 and the heat sink really gets hot! Design it for worst case cooling and load.  Keeping it cool dramatically improves the life of the product.