Gauge Wire Calcuation for 1.5KW device

[goodoboy]

Hello, happy to be here. I am a bit new to electrical study and been awhile since study.

I need to calculate the gauge wire for a a 120V 1 phase, 1.5kW device.

Details:
12.5 amps load
Ambient temp here is 104
insulation: RHW
copper wire
I am using table 310.15(B)(16)


I decide 14AWG rated for 20amps will work. After ambient correction 310.15(B)(2) my rated current becomes 17.6amps (.88 x 20).

Now I use 240.4(D) to determine if my circuit breaker limitation, correct?

I see that for 14AWG, I can use a 15amp circuit breaker to protect 14AWG rated at 17.6 amps.

Questions:

1. Did I perform the above corrections correct?
2. Did I use 240.4(D) correct?
3. What if I only have a 20amp circuit breaker, do I have use 12AWG? But if I use 12AWG, this will not protect my 17.6amps wire because 12AWG requires circuit breaker of 20amps.

I am bit confused and want to make sure I am on the right track.

 

[goodoboy]

3. What if I only have a 20amp circuit breaker, do I have use 12AWG? But if I use 12AWG, this will not protect my 17.6amps wire because 12AWG requires circuit breaker of 20amps.

Correction: If I use 20amp circuit breaker and 12AWG, I think this will work cause my rated wire current is 22amps (.88 x 25).

I only have 20 amp circuit breaker, so I think 12AWG will get it done and also within guidelines of 240.4(D)

 

[david luchini]

Questions:

1. Did I perform the above corrections correct?
2. Did I use 240.4(D) correct?
3. What if I only have a 20amp circuit breaker, do I have use 12AWG? But if I use 12AWG, this will not protect my 17.6amps wire because 12AWG requires circuit breaker of 20amps.

1) You have the basic calculations correct. You didn't indicate if the 12.5A load was continuous or non-continuous. This may change the overall picture.

2) 240.4(D) tells you that you can't protect #14 with a larger than 15A OCPD unless it falls under 240.4(E) or 240.4(G).

3) If you are using a 20A circuit breaker, 240.4(D) would require that you use a #12 minimum conductor (again, unless it fell under 240.4(E) or 240.4(G)). IF you used a #12 conductor with a corrected/adjusted ampacity of 17.6A, the 20A circuit breaker could properly protect the wire in accordance with 240.4(B).

 

[goodoboy]

1) You have the basic calculations correct. You didn't indicate if the 12.5A load was continuous or non-continuous. This may change the overall picture.

2) 240.4(D) tells you that you can't protect #14 with a larger than 15A OCPD unless it falls under 240.4(E) or 240.4(G).

3) If you are using a 20A circuit breaker, 240.4(D) would require that you use a #12 minimum conductor (again, unless it fell under 240.4(E) or 240.4(G)). IF you used a #12 conductor with a corrected/adjusted ampacity of 17.6A, the 20A circuit breaker could properly protect the wire in accordance with 240.4(B).

Thank you.

The 12.5A load is continuous.

I only have a 20A circuit breaker, and 240.4(D) requires me use #12 conductor. So I will proceed with #12.

How do I if my #12 conductor will work for certain voltage drop? Lets just say for example, my disctance from circuit breaker to load is about 1200 ft.

Thanks for the help.

 

[david luchini]

Thank you.

The 12.5A load is continuous.

I only have a 20A circuit breaker, and 240.4(D) requires me use #12 conductor. So I will proceed with #12.

How do I if my #12 conductor will work for certain voltage drop? Lets just say for example, my disctance from circuit breaker to load is about 1200 ft.

Thanks for the help.

With a 12.5A Continuous load, 210.20(A) would require a minimum 20A OCPD. You could not use a 15A c/b even if you had one available.

You could find several voltage drop calculators on the web that you can use. Mike Holt has a free one on his main page. http://www.mikeholt.com/freestuff-menu.php

 

[scott minter]

Hi Voltage drop is VD=2KID/CMIL

so voltage drop is equal to 2*12.9(for copper)*amps*distance/circular mils.

 

[goodoboy]

Hello, I am back.

I calculate my distance as 260ft.

And I use the calculator at http://www.mikeholt.com/freestuff-menu.php and http://www.calculator.net/voltage-drop-calculator.html and both give me

#10 as the less voltage drop. Can I please get a confirmation from someone as this my first time performing voltage drop calculation.


What do I put for Number of conductors or number of wires in parallel? Its single phase, so I can thinking 3 conductors (hot, netrual and ground)?


Thank you

 

[JDBrown]

What do I put for Number of conductors or number of wires in parallel? Its single phase, so I can thinking 3 conductors (hot, netrual and ground)?

Number of conductors in parallel would be 1, not 3. Parallel conductors would be if you ran, say, 2 hots and 2 neutrals, with both hots connected at both ends (in parallel) and both neutrals connected at both ends (in parallel). You will not have parallel conductors, as NEC 310.4(A) limits their use to conductors size #1/0 AWG and larger except in very special circumstances (which do not seem to apply to your situation).

 

[goodoboy]

Number of conductors in parallel would be 1, not 3. Parallel conductors would be if you ran, say, 2 hots and 2 neutrals, with both hots connected at both ends (in parallel) and both neutrals connected at both ends (in parallel). You will not have parallel conductors, as NEC 310.4(A) limits their use to conductors size #1/0 AWG and larger except in very special circumstances (which do not seem to apply to your situation).

Thank you kindly.

From the spreadsheet, I calculate a voltage drop of 13.33V for #12. This will give me a 11% voltage drop which is is not in line with 210.19, FPN No.4, recommending 3%. I am running cable from the circuit breaker panel to the load.

To get within 3%, I would need atleast #6 wire. WOW, I am glad did this voltage drop test before buying and running #12 cable.

Does my conclusion seems correct for a #6 wire for my application? Can I still use a 20amp circtuit breaker to protect the #6 wire?

Thank you kindly

 

[suemarkp]

Just watch the distance numbers. Is the 260ft from source to destination, or total wire length (i.e. 130' there and 130' back)? You need to bump up a wire size about every 75' at 120V.

 

[goodoboy]

Just watch the distance numbers. Is the 260ft from source to destination, or total wire length (i.e. 130' there and 130' back)? You need to bump up a wire size about every 75' at 120V.

260ft is from Circuit breaker to the load. Do I need to calculate distance to and from source to destination?

Are there any conflicts from using a #6 with a 20 amp circuit breaker? The breaker should protect the wire anyway.

 

[goodoboy]

Do i need to measure voltage drop both supply and return line? The distance from panel to load is 260ft.

I am using 1 phase 120V

 

[goodoboy]

Do i need to measure voltage drop both supply and return line? The distance from panel to load is 260ft.

I am using 1 phase 120V

Nevermind this question. The answer is yes, "duh goodoboy". THink my brain fried.

 

[suemarkp]

Most voltage drop calculators want the distance between the source and destination. Then internally, they double it so both the source and return wires are calculated. But you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

Most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. You also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). This will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. You need a #6 grounding conductor if the circuit is 20A. A #8 cable is no better, as it will have a #10 grounding conductor too. But most 20A breakers will take a #8 directly.

In your related post, you didn't say what type of load this is. If it is a light or heater, voltage drop won't be a big issue. If this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. If this is a motor, knowing if the motor is under much load during startup will help answer the voltage drop effect (a saw is easier to start than a compressor).

 

[goodoboy]

In your related post, you didn't say what type of load this is. If it is a light or heater, voltage drop won't be a big issue. If this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. If this is a motor, knowing if the motor is under much load during startup will help answer the voltage drop effect (a saw is easier to start than a compressor).

The load is a 1.5kW water heater requiring 120V single phase.

Can you please explain why voltage drop is not a big issue for the water heater?

Thanks for you help.

 

[GoldDigger]

The load is a 1.5kW water heater requiring 120V single phase.

Can you please explain why voltage drop is not a big issue for the water heater?

Thanks for you help.

The heater is a resistive load. It will not be harmed by a reduced.voltage,, it will just produce less heat in a given time and therefore will end up running longer.
The heat produced will be V²/R, so a 5% voltage drop will cause a roughly 10% drop in heat output.

 

[JDBrown]

Can you please explain why voltage drop is not a big issue for the water heater?

It might not be a big issue for a water heater -- you'll need to contact the manufacturer to be sure. Just give them a call, ask for the technical department, and explain your issue; they're usually really helpful.

As for why voltage drop isn't too big a deal for a resistive heater, let's consider your 1.5kW, 120V water heater. We can calculate the resistance of the heating element using the Power equation P=V²/R, or R=V²/P. In your case, R=(120V)²/1500W=9.6ohms.

Now let's assume that you've decided to connect your heater using #12 copper wire. Your voltage drop would be about 12.5V (give or take, depending on what method or calculator you used), so the voltage at your water heater is only 107.5V.

Now we can return to our Power equation using our new calculated voltage: P=(107.5V)²/9.6ohms=1204W. So the power delivered to your heating element is now only 1.2kW instead of 1.5kW, which is a 20% reduction. At this point, you have to determine if 1.2kW will get the water hot enough fast enough for your customer's needs (you may need to go to the plumbing designer to determine this).

All of this assumes that a simple resistive heater is your only load. If there are any controls involved, especially electronic controls, then all bets are off -- that's why it's important to contact the manufacturer and find out exactly what you're dealing with.

 

[suemarkp]

A resistive load draws less current as the voltage is reduced. This generally just makes it dissipate less heat, make less light, etc.

A motor loas tried to do a given amount of work. If you reduce the voltage, it will draw more current to do the same work. This is especially true with motors whose speed is determined by frequency and not voltage. A universal motor (speed is determined by voltage) will just run slower and may do less work. If it is a power tool, the operator may become frustrated at the slower speed (less work) and push the work piece harder which to increase the work which increases the current. All this increased current draw will make the motor winding hotter and comporise the insulation sooner rather than later.

Controls are the tough one. If they have a relay coil, at some point below the rated coil voltage the relay will no longer operate. Each relay is different as to what minimum voltage it wil pull in. You'll see this on dual rated 208/240V HVAC equipment when it is run at 208V and the voltage drop is excessive. On a 240V circuit, you could have a lot of drop and the controls would be fine.

 

[goodoboy]

At this point, you have to determine if 1.2kW will get the water hot enough fast enough for your customer's needs (you may need to go to the plumbing designer to determine this).

All of this assumes that a simple resistive heater is your only load. If there are any controls involved, especially electronic controls, then all bets are off -- that's why it's important to contact the manufacturer and find out exactly what you're dealing with.

Thank you JDBrown for responding. This makes so much since now. I was thinking that a big voltage drop will make the heater not work at all, but like you said this is not electronics where a certain voltage is required. For the heater case its needs power to heat up the water.

Thanks

 

[JDBrown]

Glad to help.