Fuse Sizing

[heirloom hamlet]

Does anyone know the general rule of thumb for determining you fuse size for your battery bank. Will talks about 125 percent of an appliance's rating... But what if it is for your entire system. For example, I will be using a 24v All-in-one MPPT system with an 80a SCC. It will be charging a 280aH 24v bank, and feeding a home via an 8 to 10 awg wire feed. The entire home will probably never pull more than 20 amps. So should a 30amp fuse be used at the main positive on the bank? Any insight would be appreciated.

 

[JoeHam]

Look here:

Fuse/Circuit breaker Sizing and placement.

I am starting this thread to discuss Fuse/Circuit Breaker Sizing and placement. Edit: Updated the rule of thumb based on feedback in a later post from @JoeHam My rule of thumb is the protection Device should be the lower of 1) The capacity of the power source or 2) 1.25 x the expected current...

diysolarforum.com

 

[Dzl]

Does anyone know the general rule of thumb for determining you fuse size for your battery bank. Will talks about 125 percent of an appliance's rating... But what if it is for your entire system. For example, I will be using a 24v All-in-one MPPT system with an 80a SCC. It will be charging a 280aH 24v bank, and feeding a home via an 8 to 10 awg wire feed. The entire home will probably never pull more than 20 amps. So should a 30amp fuse be used at the main positive on the bank? Any insight would be appreciated.

I am not an expert but I have some thoughts.

So I think the general design 'flow' should be:

1. Determine the greater of charge current or discharge current, in most cases it will be discharge current (DC loads + Inverter/AC) that might run at one time (or determine the maximum you want to design for).
2. Size your wire based on this (accounting for both Ampacity and Voltage Drop)
3. Size your fuse greater than the maximum designed for load, and less than ampacity rating of the wire.

Wire Ampacity > Fuse > Maximum Total Current flowing in or out of the battery

(note: in OR out not in AND out, choose whichever is larger don't add them together)

In your case it sounds like charge current will be the greater of the two (but are you sure of this? 20-30 amps seems pretty low to me (are you certain you won't ever be using corded power tools, a blender, toaster, microwave, vacuum, electric stove/oven, dishwasher, air compressor, hair dryer, or other high current loads?) Regardless of this, since your charge current is greater than your load current, your fuse should be sized at least large enough to handle your maximum charge current of 80A (so 100A-150A).

And just to be clear, is your home wired in 24VDC or 120AC, when you say 20-30 Amps are you talking about 24V or 120V?

 

[heirloom hamlet]

I am not an expert but I have some thoughts.

So I think the general design 'flow' should be:

1. Determine the greater of charge current or discharge current, in most cases it will be discharge current (DC loads + Inverter/AC) that might run at one time (or determine the maximum you want to design for).
2. Size your wire based on this (accounting for both Ampacity and Voltage Drop)
3. Size your fuse greater than the maximum designed for load, and less than ampacity rating of the wire.

Wire Ampacity > Fuse > Maximum Total Current flowing in or out of the battery

(note: in OR out not in AND out, choose whichever is larger don't add them together)

In your case it sounds like charge current will be the greater of the two (but are you sure of this? 20-30 amps seems pretty low to me (are you certain you won't ever be using corded power tools, a blender, toaster, microwave, vacuum, electric stove/oven, dishwasher, air compressor, hair dryer, or other high current loads?) Regardless of this, since your charge current is greater than your load current, your fuse should be sized at least large enough to handle your maximum charge current of 80A (so 100A-150A).

And just to be clear, is your home wired in 24VDC or 120AC, when you say 20-30 Amps are you talking about 24V or 120V?

Thanks for that.
The 20 amps I'm referring to would be the 120v ac load of my home in its entirety.
I will be setting up the panel arrays, running them a few feet to a powerhouse where they will be converted to AC and then shipped out varying distances (30 to 300 ft) to each cabin (max 20 amp draw).

 

[offgriddle]

Thanks for that.
The 20 amps I'm referring to would be the 120v ac load of my home in its entirety.
I will be setting up the panel arrays, running them a few feet to a powerhouse where they will be converted to AC and then shipped out varying distances (30 to 300 ft) to each cabin (max 20 amp draw).

Perk .. cabins? I dream of building a few air B&B cabins for eco tourism .. in the planning stages now ..

 

[heirloom hamlet]

Perk .. cabins? I dream of building a few air B&B cabins for eco tourism .. in the planning stages now ..

Great minds think alike brother. A few years ago my wife and I felt like the internet and YouTube was loaded with folks building tiny, sustainably and homesteading...which is incredibly helpful. But there is almost nowhere you can go and experience such living. We started brainstorming and planning that very thing, sold everything last spring and bought land near Savannah. Got land prep done and started building in November. Very fulfilling... plus, in hindsight, we timed the market and corona virus and won! So it's a win, win.
IG: @theinnconvenients @ucreativething

 

[heirloom hamlet]

I am not an expert but I have some thoughts.

So I think the general design 'flow' should be:

1. Determine the greater of charge current or discharge current, in most cases it will be discharge current (DC loads + Inverter/AC) that might run at one time (or determine the maximum you want to design for).
2. Size your wire based on this (accounting for both Ampacity and Voltage Drop)
3. Size your fuse greater than the maximum designed for load, and less than ampacity rating of the wire.

Wire Ampacity > Fuse > Maximum Total Current flowing in or out of the battery

(note: in OR out not in AND out, choose whichever is larger don't add them together)

In your case it sounds like charge current will be the greater of the two (but are you sure of this? 20-30 amps seems pretty low to me (are you certain you won't ever be using corded power tools, a blender, toaster, microwave, vacuum, electric stove/oven, dishwasher, air compressor, hair dryer, or other high current loads?) Regardless of this, since your charge current is greater than your load current, your fuse should be sized at least large enough to handle your maximum charge current of 80A (so 100A-150A).

And just to be clear, is your home wired in 24VDC or 120AC, when you say 20-30 Amps are you talking about 24V or 120V?

I'm sorry, but I'm s a bit confused. For my setup, my full circle distance between my array and SCC/Inverter will be 50 to 60 ft.
The SSC/Inverter and battery bank will be 6 ft or less round trip.
Now... the distance between the Inverter and the cabin, the AC load, will vary but the longest round trip could be as much as 600 ft.
Which length am I using to select wire and fuse size. Which would you use based on this info?

 

[Dzl]

Which length am I using to select wire and fuse size. Which would you use based on this info?

Yeah it definitely gets a bit confusing doesn't it.

First, for fuse size, don't worry about lengths and distances and losses. Just choose a fuse that is > your maximum total simultaneous loads < your wires rated current carrying capacity.

The 20 amps I'm referring to would be the 120v ac load of my home in its entirety.

So when thinking about the battery fuse/wiring you need to 'translate' everything into your batteries banks voltage. 20-30A @ 120V is 100-150A @ 24V. So your fuses and 24V wiring needs to be sized for this load (1.25 x 150A)

I'm sorry, but I'm s a bit confused. For my setup, my full circle distance between my array and SCC/Inverter will be 50 to 60 ft.
The SSC/Inverter and battery bank will be 6 ft or less round trip.
Now... the distance between the Inverter and the cabin, the AC load, will vary but the longest round trip could be as much as 600 ft.

Wow those are some substantial distances. So basically you can (and should) calculate all the legs separately, then add them together. If "round trip" is a confusing way to think of it, just think of it as 'total wire length' (red + black) for each leg.

1. SCC --> All-in-one (@ the voltage of your solar array)
2. All-in-one --> Batteries (@ 24V)
3. Batteries --> All-in-one (@ 24V)
4. All-in-one --> House (@ 120V)

There are calculators like this one so you don't have to think about the math. If you do want to know the math, you can calculate voltage drop using the formula V = I x R (Voltage = Current x Resistance of the wire), and calculate power loss using the formula P=I^2 x R (Power = Current x Current x Resistance of the Wire).

What are the details of your PV array (total wattage, voltage, series/parallel configuration) and what is the maximum PV input voltage of your All-in-one?

 

[Dzl]

The advice I got in this thread helped me understand voltage drop and power loss a little better.

 

[FilterGuy]

Which length am I using to select wire and fuse size. Which would you use based on this info?

You need to think of this as several different circuits:
1) Circuit connecting the solar panels and the charger.
2) Circuit connecting the Batteries, Charger and Inverter
3) Circuit(s) connecting the inverter to the cabins.

Each of these will be operating at different voltages and currents, so the wire and fuse sizing must be calculated separately for each circuit.

The circuit that jumped out at me from your description is the 300' @ 20A, 120V.

....and then shipped out varying distances (30 to 300 ft) to each cabin (max 20 amp draw).

I assume that is 20amp max per cabin...???

That is a pretty long run for 120V and you could get some hefty voltage drops. In normal house wiring the distances are short enough and current low enough that we just don't worry about voltage drop, but in this case it is worth running some numbers.

EDIT for clarity: You have to consider the round trip distance on the wires so for the 300 ft distance to the cabin I used 600 ft round trip.

In normal house wiring, we would use 14 gauge wire for 20 amps. 14 gauge wire has ~2.5 ohms/1000ft so your round trip of 600 ft would be ~1.5ohms.

Voltage drop = 1.5 ohms * 20 amps = 30 volts drop. That means you would only see 90 volts at the cabin if you start with 120 volts at the inverter.

Clearly, that is not acceptable. One solution is to use bigger wire. Lets try 8 gauge.
.6282 ohms/1000ft
.377 ohms/600ft
V=.377*20=7.538 Volts drop. (112.5V at the cabin) From a voltage point of view, this would probably work. However, even with this large wire size you are looking at 150W of line loss! Heavier gauge wires would reduce this loss but would also get pretty expensive.

With the kind of distances you are talking about, you may want to look into a 240V distribution of the AC and put a step-down transformer at each cabin. With 240 volt, you would be running 10 amp and your line loss would be cut to 38W if you stayed with 8 gauge, but you will have some amount of loss in the transformer.

Note: I have not tried to do something like a 240V distribution with step-down transformers. Consequently it is getting out of my ability to give much practical advice on implementation. I am a big proponent of DIY, but this is getting into the world of a micro-grid and you may want to get some professional engineering involved in the system design.

Alternatively, can you have separate systems for each cabin and eliminate the long runs from the inverter?

Note: I used this table for looking up the wire resistance: https://fiskalloy.com/products/conductor-facts/american-wire-gauge/

 

[heirloom hamlet]

You need to think of this as several different circuits:
1) Circuit connecting the solar panels and the charger.
2) Circuit connecting the Batteries, Charger and Inverter
3) Circuit(s) connecting the inverter to the cabins.

Each of these will be operating at different voltages and currents, so the wire and fuse sizing must be calculated separately for each circuit.

The circuit that jumped out at me from your description is the 300' @ 20A, 120V.


I assume that is 20amp max per cabin...???

That is a pretty long run for 120V and you could get some hefty voltage drops. In normal house wiring the distances are short enough and current low enough that we just don't worry about voltage drop, but in this case it is worth running some numbers.

EDIT for clarity: You have to consider the round trip distance on the wires so for the 300 ft distance to the cabin I used 600 ft round trip.

In normal house wiring, we would use 14 gauge wire for 20 amps. 14 gauge wire has ~2.5 ohms/1000ft so your round trip of 600 ft would be ~1.5ohms.

Voltage drop = 1.5 ohms * 20 amps = 30 volts drop. That means you would only see 90 volts at the cabin if you start with 120 volts at the inverter.

Clearly, that is not acceptable. One solution is to use bigger wire. Lets try 8 gauge.
.6282 ohms/1000ft
.377 ohms/600ft
V=.377*20=7.538 Volts drop. (112.5V at the cabin) From a voltage point of view, this would probably work. However, even with this large wire size you are looking at 150W of line loss! Heavier gauge wires would reduce this loss but would also get pretty expensive.

With the kind of distances you are talking about, you may want to look into a 240V distribution of the AC and put a step-down transformer at each cabin. With 240 volt, you would be running 10 amp and your line loss would be cut to 38W if you stayed with 8 gauge, but you will have some amount of loss in the transformer.

Note: I have not tried to do something like a 240V distribution with step-down transformers. Consequently it is getting out of my ability to give much practical advice on implementation. I am a big proponent of DIY, but this is getting into the world of a micro-grid and you may want to get some professional engineering involved in the system design.

Alternatively, can you have separate systems for each cabin and eliminate the long runs from the inverter?

Note: I used this table for looking up the wire resistance: https://fiskalloy.com/products/conductor-facts/american-wire-gauge/

This is very helpful, maybe helping me avoid disaster.
I could use 6awg for the main long run (500 ft) and maybe be okay?
Yes, I considered using 2 separate power houses, 1 close to each set of cabins (more like 75ft one way instead of 250). But the sun is full and unshaded for at least 6 hours in a particular location for a couple of the cabins, but probably less than half that near the other 2. I can only connect a 2.25kw array to my SSC. So I thought maybe it would be better to stage in great sun, convert to AC close by, and run 120v AC a long way to the two.

 

[FilterGuy]

This is very helpful, maybe helping me avoid disaster.
I could use 6awg for the main long run (500 ft) and maybe be okay?
Yes, I considered using 2 separate power houses, 1 close to each set of cabins (more like 75ft one way instead of 250). But the sun is full and unshaded for at least 6 hours in a particular location for a couple of the cabins, but probably less than half that near the other 2. I can only connect a 2.25kw array to my SSC. So I thought maybe it would be better to stage in great sun, convert to AC close by, and run 120v AC a long way to the two.

I was using 300 ft for the calculations on the long run. At 500 ft you would need at least 6AWG and would still have substantial line loss.

It sounds like you have 2 sets of two cabins. Are you planning on running one set of wires to the two far cabins and then split it at the cabins? If so, will that line have 40A max? At 40 amps 500Ft, even 6 AWG would not be enough.

 

[heirloom hamlet]

I was using 300 ft for the calculations on the long run. At 500 ft you would need at least 6AWG and would still have substantial line loss.

It sounds like you have 2 sets of two cabins. Are you planning on running one set of wires to the two far cabins and then split it at the cabins? If so, will that line have 40A max? At 40 amps 500Ft, even 6 AWG would not be enough.

Yes two sets of cabins. Two close to the powerhouse, two far away.
I was planning on running each cabin it's own wiring...each has its own array and SCC/inverter. Looking at decent pricing, and assuming I would use 10 awg regardless of the situation, it looks like it would only cost less than $400 more to do it with 6 vs. 10. To me, much easier and cheaper than changing the entire system, building a second powerhouse, purchasing additional panels, etc. That is, if it can be done.
I sad 600 feet total distance before. But the AC line would be about 250 feet one way to each cabin. The other figure included in the footage before added in the solar panel array wire (10awg) to the powerhouse and the SCC.
In a cabin, if everything electrical was running at the same time, it would probably be less than 20a continuous. That would rarely be the case... usually 10a or less.
I'm sorry if this is complicated or confusing, I appreciate the advanced help. Don't want any regrets.

 

[FilterGuy]

But the AC line would be about 250 feet one way to each cabin.

OK.... Now I understand the various numbers. I used 300 feet one way (600ft round trip), so 8 gauge would be minimum size to make it work, 6 gauge would help reduce the line loss.

Lets run the numbers for 6AWG
.3951 ohms/1000ft
.237 ohms/600ft
V=.2371*20=4.741 Volts drop. (115V at the cabin)
Line loss = Ohms * Amps * Amps = .2371*20*20 = 95W.

95W is a lot of loss, but it will only happen when the amperage is maxed out. Typically it will be less. However, you should still account for this when you are doing sizing of your batteries and solar array.

 

[heirloom hamlet]

OK.... Now I understand the various numbers. I used 300 feet one way (600ft round trip), so 8 gauge would be minimum size to make it work, 6 gauge would help reduce the line loss.

Lets run the numbers for 6AWG
.3951 ohms/1000ft
.237 ohms/600ft
V=.2371*20=4.741 Volts drop. (115V at the cabin)
Line loss = Ohms * Amps * Amps = .2371*20*20 = 95W.

95W is a lot of loss, but it will only happen when the amperage is maxed out. Typically it will be less. However, you should still account for this when you are doing sizing of your batteries and solar array.

Okay, I see what your doing here. So 6awg gets my voltage back to acceptable for use, but I'm still losing wattage. Which is okay if I have more than I need, which I can control with design. 4awg would be even better, losing like 60w.
What is this 4/0, 2/0 wiring on these charts?

 

[John Frum]

Okay, I see what your doing here. So 6awg gets my voltage back to acceptable for use, but I'm still losing wattage. Which is okay if I have more than I need, which I can control with design. 4awg would be even better, losing like 60w.
What is this 4/0, 2/0 wiring on these charts?

its pronounced 4 ought and 2 ought.
They are thick, heavy and expensive.

Southwire (By-the-Foot) 2/0- 2/0- 2/0 Gray Stranded CU SER Cable 27083599 - The Home Depot

www.homedepot.com

That over $12.00USD per foot its not even direct burial rated.

 

[heirloom hamlet]

Yeah it definitely gets a bit confusing doesn't it.

First, for fuse size, don't worry about lengths and distances and losses. Just choose a fuse that is > your maximum total simultaneous loads < your wires rated current carrying capacity.



So when thinking about the battery fuse/wiring you need to 'translate' everything into your batteries banks voltage. 20-30A @ 120V is 100-150A @ 24V. So your fuses and 24V wiring needs to be sized for this load (1.25 x 150A)



Wow those are some substantial distances. So basically you can (and should) calculate all the legs separately, then add them together. If "round trip" is a confusing way to think of it, just think of it as 'total wire length' (red + black) for each leg.

1. SCC --> All-in-one (@ the voltage of your solar array)
2. All-in-one --> Batteries (@ 24V)
3. Batteries --> All-in-one (@ 24V)
4. All-in-one --> House (@ 120V)

There are calculators like this one so you don't have to think about the math. If you do want to know the math, you can calculate voltage drop using the formula V = I x R (Voltage = Current x Resistance of the wire), and calculate power loss using the formula P=I^2 x R (Power = Current x Current x Resistance of the Wire).

What are the details of your PV array (total wattage, voltage, series/parallel configuration) and what is the maximum PV input voltage of your All-in-one?

My array is x9, 250w, 37.6v, 8.26a each panel...wired three in series, and those 3 series in parallel, totalling 2.25kw, 113v, 24.8a. It is an 2000w 80a dc 60a ac SSC, 3000w inverter.

 

[heirloom hamlet]

its pronounced 4 ought and 2 ought.
They are thick, heavy and expensive.

Southwire (By-the-Foot) 2/0- 2/0- 2/0 Gray Stranded CU SER Cable 27083599 - The Home Depot

www.homedepot.com

Holy smokes! Wow.