Hi, I'm planning to replace my tank-type water heater with a tankless model.

Electrical specs: 60a dp c/b; #6 AWG wire; 11.8kw; 54a max @220v

Available voltages: 208v/220v or 240v/277v

QUESTION #1: Why the range of voltages for each option rather than just 220v (voltage
Drop due to wire length)?


My installation requires a 90' wire run, which will cause a noticeable voltage drop.

QUESTION #2: How do I determine whether #6 wire is acceptable, or will I need #4?

THANKS for any info!

Tom

My understanding is larger wire will reduce voltage drop during in-rush, and the ampacity is also dependent on the type of insulation on the wire... heat under load...

Talk to the dealer of the water heater. The different voltage ranges may be how the heating elements are wired... wye or delta, a schematic would illustrate that.

Be safe...

Tom,

The cable formula I used set the cable at a #5 sized wire.

My concern is that I have not proven the accuracy of this formula yet, so please wait for verification from other responders.

Q!: The manufacturer has to specify the voltages that the appliance will operate at. 208, 277 and 240 are common voltages. If you had a motor, for instance, if it did not say 240V, you could not operate it at 240 V. Residential voltages are now 240/120, not 110/220.

Tom,

I left some information. You would be using 240VAC. 220 is no longer furnished to a household. The available voltages for household are 120 and 240.

Guessing that you would be using 6/3 (I wasn't sure if you would just need 240 VAC or a combination of 120/240)

I calculated using 6/3, which adds an extra conductor to the mix. I also use the value of 12 to identify the conductor type of copper. I figured you would prefer copper over aluminum.

Any way, #6/3 works out to 7.4%. The max you are allowed is 7.2? For 240.
that why I went to #5/3 which brought the voltage drop down to 5.4%.

I'll run it through at 6/2 just on the possibility that you only need 6/2 cable.

If you are using just 240, I get a number of approximately 6% using #6/2 cable.

Q2: Using 240V, 59A, 90', Copper, #6 gives a 2.2% drop.

Don,

Thanks for your response!

Yes on the copper, 240v only, and 6/2 (two hots + ground) wire.

< The max you are allowed is 7.2? For 240. >
I see your question mark after 7.2 - does the max voltage drop vary per local code?

< just 240, I get a number of approximately 6% using #6/2 cable. >
Apparently, #6 would be OK if 7.2 is the max...

If you don't mind, could you run the same calculations for a larger heater?
It requires two separate lines. Each line requires 240V; 40A c/b; #8/2 wire. The total is 18 kw; max load 77amps at 240v.

Would #8 copper be adequate for the same 90 foot run?

Again, thanks VERY MUCH for this info!

Tom



Thanks,

Tom

Tom,

The NEC restricts conductors to specific amperages. #8 is for 40 amp max. So no, it cannot be used for a 60 amp line.

However, please review KISS' posting that yielded a 2.X voltage drop for 60 amp 2 wire copper size 6 AWG.

Don,

< The NEC restricts conductors to specific amperages. #8 is for 40 amp max. So no, it cannot be used for a 60 amp line. >
This calculation would be for a different model heater. *EACH* of the TWO separate #8 lines would carry only a 40 amp load - not 60 amps.

See quote below.
<It requires two separate lines. Each line requires 240V; 40A c/b; #8/2 wire. The total is 18 kw; max load 77amps at 240v. >
This is the way the specs are listed by the manufacturer. I presume the total kw and max amp load numbers refer to the load SHARED by the two lines.

Tom

Thanks, KISS!

Quote:

---

Originally Posted by KeepItSimpleStupid

Q2: Using 240V, 59A, 90', Copper, #6 gives a 2.2% drop.

---

So, you're saying that with the 5.28v drop the #6 wire is adequate?

Tom

Yep.

Don's calculation was in volts. Not %. We are shooting for <3%.

For your other Q: #8, 90', 240 V, copper, 39 A (~77 A/2) = 2.3% drop, OK

Note I took 54 A/220 V = x A/240 V since the relationship is linear and got x = ~59A for your first case.

You don't have to derate the cable by multiplying the current x 1.25 since the load is not continuous (more than 3 hrs at a time in a day).

Tom,

Where in the devil did the second heater come from? Did I miss something?

Donf: You must be aging way to fast. In post 7:

Quote:

---

If you don't mind, could you run the same calculations for a larger heater?
It requires two separate lines. Each line requires 240V; 40A c/b; #8/2 wire. The total is 18 kw; max load 77amps at 240v.

---

So, I divided the 77/2, I was to lazy, so I used 78 and got an even number. Assumed load was shared equally between both services and checked Vd for #8 copper.

Still trying to convince donf that his formula is wrong.

Quote:

---

Originally Posted by KeepItSimpleStupid

Note I took 54 A/220 V = x A/240 V since the relationship is linear and got x = ~59A for your first case.

Thanks for the formula! I was planning to ask. You are (at least) one step ahead of me.

You don't have to derate the cable by multiplying the current x 1.25 since the load is not continuous (more than 3 hrs at a time in a day).

---

That's good to know. I wasn't thinking about factor.

THANKS to you KISS and to DonF, also!

Tom

Why would you use an electric tankless water heater? They are not efficient, they rarely work well and they are huge enegy hogs... as you know by the wiring requirements.

He has 12 kids and they all take 30 minute showers, I guess.

Doesn't the water heater require TWO 220 volt circuits, 4 breakers? I recently installed one and it needed 2 two pole breakers.

MB,

You can not get 220 VAC for households any more. You can only get 120/240.

You use two breakers because each 120 VAC will be 240 VAC apart. This is the normal way to convert 120 VAC to 240 VAC. This is what you do for water heaters, furnaces and the like.