Application Questions

Our Watt Meter will work on any battery bank from 4 to 60 volts. If I understand your question correctly, I see no reason why our Watt Meter would not work on your battery bank system:

http://www.windynation.com/products/accessories/measurement/100-amp-dc-watt-meter

If you have any other questions about our Watt Meter, you can post them in this forum thread.

 

In my opinion, that meter should be only used at those current levels for period of a fraction of a minute. The wires are far too small to carry 50 amps continually. Having 2 controllers at 50A each driving the same battery bank is OK in many cases but there are some controllers that don't "stack" well. Ideally you would use only one controller. perhaps by doubling the battery bank voltage to keep the current at 50A max.

 

"I have a solar set up and currently have one 50 amp controller. I would like to have a meter that shows me my power being charged to the battery from my controller. I would also like to double my power and have a second 50 AMP controller that controls another set of more solar panels. My question is, first, is it possible to have two 50 amp controllers, where the outputs are joined together and then connect to the battery bank? Second, due to the 50 AMP power restriction of this meter, Would it be possible to get two of these, one for the output of each controller, and just join the two outputs together in parallel so that they each only get a max of 50 amps, but allows me to get 100 amps at once to the battery?"

14 AWG has a resistance of 2.53 milliOhms/foot.

Let's assume you leave 1.5 inches of wire hanging out the meter on both sides. That is a total of 6 inches of 14 AWG wire. Let's also take into account that these two wires travel through the meter for three inches. This gives us another 6 inches of 14 AWG wire. That is a total of 12 inches or 1 foot.

Using data from above:

(1 foot 14 AWG) x (2.53 milliOhms/foot) = 0.00253 ohms

Now we calculate the voltage drop through the wire using V=IR where I= current through wire (say 50 amps) and R = 0.00253 Ohms from above

V = 50 amps x 0.00253 = 0.1265 voltage drop through 1 foot of 14 AWG wire at 50 amps

Now, let's assume this is hooked up to a 12 volt battery bank which is at 13 volts and we can calculate voltage drop percentage:

(0.1265 voltage drop/13 volts) x 100 = 0.97% which is just about 1%

We can also calculate how many Watts are consumed in the meter wire when we have 50 amps traveling through the meter because we know Power = volts x amps:

Power loss = voltage drop in meter x amps running through meter = 0.1265 volts x 50 amps = 6.33 Watts

 

A 10 watt ceramic or wire-wound resistor is pretty large and can get very hot even though they are designed to dissipate heat. Not sure the digital meter would survive very long under the same conditions. Perhaps a life test is called for. There is also a shunt in the meter that is used for measuring the current ... it's not just the wires.

 

But the wire in a 10 ohm resistor has a resistance of 10 ohms. The resistance in the 14 AWG wire from above is 0.00253 ohms.

P= volts x amps = resistance x (amps)^2

so there is about 4000 times more power consumption in the 10 ohm resistor compared to the 1 foot of 14 AWG wire, if the current is held constant for both.

I was told the shunt has a resistance in the milliohm range.

 

Sorry ... I meant 10 watt ... I changed my post.

 

Good thing we can edit posts now