Range calculation

[Jpflex]

Therefore, your range is well over that rating so we must apply Note number 1of that table.
Referring again to that note, it says to increase Column C by 5 percent for every additional kilowatt of rating or major fraction thereof for the individual ranges exceeding 17.3 kW. (drop the 3 fraction).

I’m with you on most of your calculations, but not all of it. This is because, if you drop the fraction 0.3 kVA at the beginning of the equation before calculating you receive a different answer than if you drop a fractional ampere 0.3 or at the end of the equation. Also 0.3 kva is actually 300 Va or 300VA / 208 E (single phase?) = 1.44 amperes and over 1 ampere which you do not drop

Therefore, 17.3 KVA = 17,300 - 12,000 = 5,300 VA / 1,000 VA = 5.3 x 0.05 (5%) = 0.265% increase or 1.265 multiplier for column C value of 8,000 VA (1 range) = 10,120 VA / 208 E (volts) = 48.65 i amperes. Since end result 0.65 i amperes is larger than 0.3 or even half an ampere, i think its best to round up to 1 ampere for a total of 49 amperes

People on this post have told me to not consider ranges as a continuous load because the burners are cycled. With this in mind you would not need to multiply this by 125% to size the branch circuit for the appliance but what about conductor correction for ambient temperature and those if current carrying conductors in a conduit exceed 3 or more?


what’s confusing is that the book from what I remember said both the branch circuit shall be the name plate rating 17,300 VA (larger wire) = 83.17 i but also says the branch circuit shall be permitted to be based on the table 8,000 VA for 1 range plus 5% increase to column C over 12k, which yields a smaller 49 I ampere conductor I calculated

 

[LarryFine]

People on this post have told me to not consider ranges as a continuous load because the burners are cycled.

Also, how often does one simultaneously use all surface burners on high and preheat the oven?

 

[Dennis Alwon]

I’m with you on most of your calculations, but not all of it. This is because, if you drop the fraction 0.3 kVA at the beginning of the equation before calculating you receive a different answer than if you drop a fractional ampere 0.3 or at the end of the equation. Also 0.3 kva is actually 300 Va or 300VA / 208 E (single phase?) = 1.44 amperes and over 1 ampere which you do not drop

Therefore, 17.3 KVA = 17,300 - 12,000 = 5,300 VA / 1,000 VA = 5.3 x 0.05 (5%) = 0.265% increase or 1.265 multiplier for column C value of 8,000 VA (1 range) = 10,120 VA / 208 E (volts) = 48.65 i amperes. Since end result 0.65 i amperes is larger than 0.3 or even half an ampere, i think its best to round up to 1 ampere for a total of 49 amperes

People on this post have told me to not consider ranges as a continuous load because the burners are cycled. With this in mind you would not need to multiply this by 125% to size the branch circuit for the appliance but what about conductor correction for ambient temperature and those if current carrying conductors in a conduit exceed 3 or more?


what’s confusing is that the book from what I remember said both the branch circuit shall be the name plate rating 17,300 VA (larger wire) = 83.17 i but also says the branch circuit shall be permitted to be based on the table 8,000 VA for 1 range plus 5% increase to column C over 12k, which yields a smaller 49 I ampere conductor I calculated

Why would you consider 1.25 (125%) for a range. Would there ever be a chance that the load would be on for 3 hours continuously? I don't believe it is possible to get it into that scenario.

 

[Jpflex]

Also, how often does one simultaneously use all surface burners on high and preheat the oven?

To answer your hypothetical question, not often but probably only on or before thanksgiving or large gatherings.

What would you say to my previous post on not rounding down fractions under 0.5 until the end of the equation pertaining to amperes

 

[infinity]

Why would you consider 1.25 (125%) for a range. Would there ever be a chance that the load would be on for 3 hours continuously? I don't believe it is possible to get it into that scenario.

Definitely not. Many fail to understand the actual wording in Article 100 when applying the actual definition of a continuous load.

 

[Jpflex]

Why would you consider 1.25 (125%) for a range. Would there ever be a chance that the load would be on for 3 hours continuously? I don't believe it is possible to get it into that scenario.

Yes we went over this and established this concept a while back on previous post. My original thought / post was based on occasions such as thanksgiving where burners and the stove would all be on. However, I was explained that burner cycling mitigated this

 

[infinity]

Yes we went over this and established this concept a while back on previous post. My original thought / post was based on occasions such as thanksgiving where burners and the stove would all be on. However, I was explained that burner cycling mitigated this

Yes one component of the definition of a continuous load is that it operates at maximum output for 180 minutes or more. A stove or range has thermostat(s) and will never operate continuously at full output for 3 or more hours.

 

[LarryFine]

To answer your hypothetical question, not often but probably only on or before thanksgiving or large gatherings.

Even then, the individual T-stats will cycle well before three hours.

What would you say to my previous post on not rounding down fractions under 0.5 until the end of the equation pertaining to amperes

"Round them down! Round them all down!"

 

[Dennis Alwon]

Years ago I have seen some stoves where the burners cycled on and off. I don't believe that is the case anymore or do they?

 

[Dennis Alwon]

Yes we went over this and established this concept a while back on previous post. My original thought / post was based on occasions such as thanksgiving where burners and the stove would all be on. However, I was explained that burner cycling mitigated this

I misread that last post. I thought you were saying that we needed to use 125%.. My bad

 

[LarryFine]

Years ago I have seen some stoves where the burners cycled on and off. I don't believe that is the case anymore or do they?

The only possibilities are T-stat on/off control, which means full voltage with a variable duty cycle, or a continuously-variable voltage, which means electronic (triac) control (which technically speaking is still variable duty cycle).

 

[roger]

I don't believe that is the case anymore or do they?

Mine does.

 

[TX+ MASTER#4544]

You need to read the other posts in a thread. The OP's installation is in a commercial kitchen so 220.55 does not apply.

Yes, I read that," but mostly commercial/industrial, sizing OCPD, he said".
And I used Article 220 Part III Feeder and Service Load Calculations based on the general or standard method verses the Optional Method found in Part IV 220.80. So that other interested readers know that there are two methods to calculate similar type loads.

Now, having said that I will still go to Part III Section 220.56 Kitchen Equipment-Other Than Dwelling Units.
Calculating the load for commercial cooking equipment and other kitchen equipment loads are found here.

retnoc, the op #39 said it was a 208 volt, 91 amp and 17.3 kW name plate rating.
Only one piece of kitchen equipment was mentioned.
Section 220.56 and T.220.56 Demand Factors for Kitchen Equipment Other Than Dwelling Units.
You have only one unit so there's no demand factor.

T.220.56 and number of units of equipment you'll find 1 In right side column under Demand Factor(%) is the demand factor(s).
However, you only have 1 unit, therefore there's no demand factor for only 1 unit.
You must have 3 or more to calculate the demand.
So, 17,000 watts or VA and 208 volts single phase 17,000 VA / 208 = 81.73 or 82 amps.
Section 240.6 and accompanying table, there we find an 80 amp or a 90 amp OCPD.

If this were an exam question and it stated 'what is the maximum size permitted by NEC'?
I would select the 90 amp OCPD based on section 240.4(B) Overcurrent Devices Rated 800 Amps or Less.
This is one the rules for under 800 amps vs over 800 amps.

There are 3 conditions allowing for using... "next higher standard rating (above the ampacity of the conductors being protected)......."
We met all three of them with this type of question.

What about the conductors? I'm sure that this size electric oven would be wired with a metallic raceway and not Type NM.

Assume terminals are rated 75 degrees and we'll use type THHN copper wire which is rated for a 90 degree terminal

but we will make an ampacity adjustment because it will land on a 75 degree terminal at 85 amps.

So our 17,000 W / 208 = 81.73 amps or 82 amps would call for a 4 AWG copper conductor at 85 amps.

Also, there are no more than 3 current carrying conductors and voltage drop is not a factor. And you cannot reduce the neutral had it required one because this is not a dwelling unit.

We use T.220.55 for household use. Electric ranges/ovens whether for dwellings or commercial are not considered to be a type of continuous load.
Thanks for reading
TX+MASTER #4544
Comments accepted.

 

[david luchini]

retnoc, the op #39 said it was a 208 volt, 91 amp and 17.3 kW name plate rating.
Only one piece of kitchen equipment was mentioned.
Section 220.56 and T.220.56 Demand Factors for Kitchen Equipment Other Than Dwelling Units.
You have only one unit so there's no demand factor.


If this were an exam question and it stated 'what is the maximum size permitted by NEC'?
I would select the 90 amp OCPD based on section 240.4(B) Overcurrent Devices Rated 800 Amps or Less.
This is one the rules for under 800 amps vs over 800 amps.

A 90A ocpd for a 91A load? I would think 150A would be the maximum allowable ocpd.

 

[augie47]

Wow! 50+ posts on what would seemingly be a cut-n-dry install.
91 amp load on a commercial range.
minimum conductor size would have to be 91 amp rating.
minimum breaker I would think would be 100 amp but, as david points out, it could be as high as 150 amp (based on 422.11) so the conductor size would have to be increased IF you went beyond the 100 amp breaker.

 

[TX+ MASTER#4544]

A 90A ocpd for a 91A load? I would think 150A would be the maximum allowable ocpd.

Yes, a 90 amp OCPD based on my calculations. This is not a continuous load.
Section 220.56 Kitchen Equipment Other Than Dwelling Units. And about the 3rd sentence in that paragraph....."These demand factors shall be applied to all equipment that has either thermostatic control or intermittent use as kitchen equipment." HVAC equipment not included.

Here's why: When a temperature setting is selected on an electric oven that's used as kitchen equipment, the circuit is activated to it until it reaches the desired temperature and then the circuit opens (turns off) until the next cycle and so on.

Therefore, the electrical conductors have time to cool off until the thermoset temperature calls for the circuit to open once again.

This cycle is constantly switching on and off. Thus the equipment is controlled by a thermostat setting.
This is true for household cooking equipment, too. Another reason why it's not considered to be a continuous load.

Thanks for reading,
Comments accepted
TX+MASTER #4544

 

[david luchini]

Yes, a 90 amp OCPD based on my calculations. This is not a continuous load.
Section 220.56 Kitchen Equipment Other Than Dwelling Units. And about the 3rd sentence in that paragraph....."These demand factors shall be applied to all equipment that has either thermostatic control or intermittent use as kitchen equipment." HVAC equipment not included.

I'm confused. Per 220.56, the demand factor is 100%.

91A times 100% is 91A.

A 90A OCPD is smaller than the load on the circuit. That seems unusual. I think that would be a violation of 220.18.

 

[infinity]

I'm confused. Per 220.56, the demand factor is 100%.

91A times 100% is 91A.

A 90A OCPD is smaller than the load on the circuit. That seems unusual. I think that would be a violation of 220.18.

Seems logical to me.

 

[augie47]

I'm confused. Per 220.56, the demand factor is 100%.

91A times 100% is 91A.

A 90A OCPD is smaller than the load on the circuit. That seems unusual. I think that would be a violation of 220.18.

and 210.19(A)(1)
(1) General. Branch-circuit conductors shall have an ampacity not less than the maximum load to be served.

 

[wwhitney]

I'm confused.

What's confusing is that the OP says "Nameplate is 208v, single phase, 91amps, 17.3kw." And 17,300/208 = 83A. Which is not 91A.

Given the discrepancy, obviously best to go with the higher number. Possible reasons for the discrepancy that occur to me are non-unit power factor (but on an electric range?), or perhaps it's European and "17.3 kW" is referring to the heat output, not the electrical input.

Cheers, Wayne