Load Calculation questions

Status
Not open for further replies.
J

jotw

Member
Location
Texas
I should provide a little history for this one. I'm involved in a solar installation for a business with an existing 200A single phase service. Does this load calculation look right? If I'm doing this right, my math comes to 230A. As stated above, the main is a 200A. Obviously, the business isn't pulling 230A. Is there another load cal. I can use to show what the business is actually requiring for power? Or, do I go through and add up computers, light fixtures, ACs, furnaces, etc. in VA and divide by 240 to get total connected load? My goal is to be able to show an inspector that by replacing that 200A main with a 175A and supplying additional power from the solar system @ 60A, I'm not undersizing the service requirements and also not overloading the bus capacity. I hope that's not too confusing. :confused:

3416 sq. ft.

Lighting - 3416 X 3VA @ 125% (continuous use) = 12,810
Receptacles - 3416 X 1VA (unknown # of rec.) = 3,416
A/C (larger than heat) 4 units - 7200VA
7200VA
12000VA
9600VA = 36,000VA
Laundry (2 washer/ gas dryer combination units) - 3000VA

= 55,226VA/240V = 230Amps

Also, does the opt. calculation for feeders and service equipment apply to non dwelling units?
 
jotw said:
I should provide a little history for this one. I'm involved in a solar installation for a business with an existing 200A single phase service. Does this load calculation look right? If I'm doing this right, my math comes to 230A. As stated above, the main is a 200A. Obviously, the business isn't pulling 230A. Is there another load cal. I can use to show what the business is actually requiring for power? Or, do I go through and add up computers, light fixtures, ACs, furnaces, etc. in VA and divide by 240 to get total connected load? My goal is to be able to show an inspector that by replacing that 200A main with a 175A and supplying additional power from the solar system @ 60A, I'm not undersizing the service requirements and also not overloading the bus capacity. I hope that's not too confusing. :confused:

3416 sq. ft.

Lighting - 3416 X 3VA @ 125% (continuous use) = 12,810
Receptacles - 3416 X 1VA (unknown # of rec.) = 3,416
A/C (larger than heat) 4 units - 7200VA
7200VA
12000VA
9600VA = 36,000VA
Laundry (2 washer/ gas dryer combination units) - 3000VA

= 55,226VA/240V = 230Amps

Also, does the opt. calculation for feeders and service equipment apply to non dwelling units?
Click to expand...

Calculat duplex rec at 180 va, double duplexes as 360. The lighting calc. is correct , I'm not sure what you mean by "larger than heat" but your rec. calc is what is throwing your numbers way off.
 
What are the numbers listed under the AC? What type of business is this? Is this a office? Is the 4 unit AC total 7200 or is each unit 7200? Optional calcs are only for schools, dwellings and whatever else is listed in that section. You maybe could use historical data per 220.87 to support your idea of reducing the service size. I'm curious what does reducing the service to 175 amps gain you? Is there a cost reduction for that?
 
jotw said:
I should provide a little history for this one. I'm involved in a solar installation for a business with an existing 200A single phase service. Does this load calculation look right? If I'm doing this right, my math comes to 230A. As stated above, the main is a 200A. Obviously, the business isn't pulling 230A. Is there another load cal. I can use to show what the business is actually requiring for power? Or, do I go through and add up computers, light fixtures, ACs, furnaces, etc. in VA and divide by 240 to get total connected load? My goal is to be able to show an inspector that by replacing that 200A main with a 175A and supplying additional power from the solar system @ 60A, I'm not undersizing the service requirements and also not overloading the bus capacity. I hope that's not too confusing. :confused:

3416 sq. ft.

Lighting - 3416 X 3VA @ 125% (continuous use) = 12,810
Receptacles - 3416 X 1VA (unknown # of rec.) = 3,416
A/C (larger than heat) 4 units - 7200VA
7200VA
12000VA
9600VA = 36,000VA
Laundry (2 washer/ gas dryer combination units) - 3000VA

= 55,226VA/240V = 230Amps

Also, does the opt. calculation for feeders and service equipment apply to non dwelling units?
Click to expand...

I'm not familiar with solar systems, but I would say your feeders would still have to be sized for the demand to be severed. What happens when it is cloudy out and you don't produce and electricity?

Is this an office building?

The way I read 220.87 if you have a 230A demand you would have to up size your service to 250A.

Like I said I am not familiar with solar systems.
 
I think the problem you're running into is the classic issue of the NEC calculation resulting in a larger feeder/service size than the installation actually demands. Even with demand factors outlined in the NEC, you still end up with a very conservative calculated service size. You could likely obtain peak demand load during the past 12 month period from the utility, and use that number to justify a reduction in service size. But I agree with a previous post that the utility service should be sized for the load when your PV system isn't available (unless the NEC allows you to reduce it, possibly based on batteries in your PV system?).
 
Thanks for the replies. This is for a dog kennel. Probably only 1100 sq. ft. is for office use. The other 2300 sq. ft. is dog kennel (air conditioned). On the AC load in my calculation I used the maximum size breaker X 240V for VA. Is that right? There are 4 AC units so I used all four and they are different sizes. I don't know how many receptacles there are total so I used 1VA X sq. ft. and didn't use the demand factors. I've read that somewhere in 2005.

I understand that load calculations are conservative, but, really this place isn't pulling much off of this service. Everything is gas and they have CFLs in all of the light fixtures. The biggest load will be the AC in the summer. I just feel like I have more room to work with than what load cal. gives me.

The purpose for trying to install the smallest main breaker possible is to avoid overloading the bus in the main panel. It can only be loaded to 120% of the rated capacity. I'm putting a 60A breaker in for the solar generation. That means I can only have a main breaker of 175A.

I should have been a rock star instead of an electrician. I'm much better at that.
 
Is that 60 amp breaker and 175 main a manufuctures spec?

The way I see it; lets say your service is humming along at 200 amps and you produce 60 amps of solar power that is put on the grid through your panel. Now you are only using 140 amps from the Poco. This is base on pure, non-engineer, non-electrical theory, ideas so I would not build a system based on that. But I'd want to hear from the solar company and maybe some tech types off this site what really happens when you add amperage to that main of yours.

You added a lot of load on those AC's that you don't have to. Use the FLC based off the HP and voltage from the tables at the back of 430.
 
That's what I was looking for wawireguy. Thank you! And, that main panel will never see 60A from the solar. The inspector wanted OCPD sized to the maximum output current of the inverter (gotta love your local AHJ). The short circuit current on the entire system is 50A. If I could size the OCPD to the solar panels and what they'll actually produce, then this would have been much easier. Thanks again.
 
jotw said:
Thanks for the replies. This is for a dog kennel. Probably only 1100 sq. ft. is for office use. The other 2300 sq. ft. is dog kennel (air conditioned). On the AC load in my calculation I used the maximum size breaker X 240V for VA. Is that right? There are 4 AC units so I used all four and they are different sizes. I don't know how many receptacles there are total so I used 1VA X sq. ft. and didn't use the demand factors. I've read that somewhere in 2005.

I understand that load calculations are conservative, but, really this place isn't pulling much off of this service. Everything is gas and they have CFLs in all of the light fixtures. The biggest load will be the AC in the summer. I just feel like I have more room to work with than what load cal. gives me.

The purpose for trying to install the smallest main breaker possible is to avoid overloading the bus in the main panel. It can only be loaded to 120% of the rated capacity. I'm putting a 60A breaker in for the solar generation. That means I can only have a main breaker of 175A.

I should have been a rock star instead of an electrician. I'm much better at that.
Click to expand...

YOu should have gave more info to start with, I would calculate the dog kennel at 1/2 or 1 VA per sq ft, not the full 3 or 3 1/2 I allow for offices.
 
jotw said:
On the AC load in my calculation I used the maximum size breaker X 240V for VA. Is that right?
Click to expand...

No, it isn't. Use the MCA (minimum circuit ampacity) value on the nameplate. This will be quite a bit less than the max breaker size.
 
If you are humming along at 200 amps and the solar panels are putting out 60 amps, the

panel busses will see 260 amps, that is why the 20% code needs to be followed.

On the other hand, I don't see a need for load calcs, exept for formalites, to lessen the

service Main C.B. .
 
benaround said:
If you are humming along at 200 amps and the solar panels are putting out 60 amps, the

panel busses will see 260 amps, that is why the 20% code needs to be followed.

On the other hand, I don't see a need for load calcs, exept for formalites, to lessen the

service Main C.B. .
Click to expand...


Not sure I follow your logic.

If you have 200A actual current draw load on the panel you will not have any more than 200A anywhere on the panel bus.

Definately less if the PV input breaker is on the opposite end of the bus than the main.
 
MarkWhite said:
Not sure I follow your logic.

If you have 200A actual current draw load on the panel you will not have any more than 200A anywhere on the panel bus.

Definately less if the PV input breaker is on the opposite end of the bus than the main.
Click to expand...

He gave a pooor example, but the intent was right. If the Main ocpd is 200, and the solar is backfeeding 60, there is a POSSIBILITY of pulling 260 amps on a bus rated for 200. So, the formality is, if you can do a load calc and it comes in at 175, you can change the main, and your backfeed will not exceed the ratings of the bus.
 
I have finally got my head "wrapped" around why a main breaker would need to be reduced from 200A to 175A in size because of the potential in the increase load to the bus, even though common sense should prevail in that the chances of overloading the main bus are extremely remote.

But is it possible that at night when the solar system is not producing power that you would have a potential of the "reduced in size" main breaker being overloaded and tripping?
 
jotw said:
Thanks for the replies. This is for a dog kennel. Probably only 1100 sq. ft. is for office use. The other 2300 sq. ft. is dog kennel (air conditioned). On the AC load in my calculation I used the maximum size breaker X 240V for VA. Is that right? There are 4 AC units so I used all four and they are different sizes. I don't know how many receptacles there are total so I used 1VA X sq. ft. and didn't use the demand factors. I've read that somewhere in 2005.

I understand that load calculations are conservative, but, really this place isn't pulling much off of this service. Everything is gas and they have CFLs in all of the light fixtures. The biggest load will be the AC in the summer. I just feel like I have more room to work with than what load cal. gives me.

The purpose for trying to install the smallest main breaker possible is to avoid overloading the bus in the main panel. It can only be loaded to 120% of the rated capacity. I'm putting a 60A breaker in for the solar generation. That means I can only have a main breaker of 175A.

I should have been a rock star instead of an electrician. I'm much better at that.
Click to expand...

jotw - If I'm understanding you correctly, in regards to the 120% factor.
2005- 690.64(B)(2) says that the ratings of overcurrent devcices shall not exceed the rating of the bussbar. So the total of your backfeed breakers shall not exceed the buss bar rating of 2ooamps. By reducing the 200amp main to 175amps this leaves you only 25amps to back feed w/ PV not the 60amps you are planning.
 
mhanson said:
jotw - If I'm understanding you correctly, in regards to the 120% factor.
2005- 690.64(B)(2) says that the ratings of overcurrent devcices shall not exceed the rating of the bussbar. So the total of your backfeed breakers shall not exceed the buss bar rating of 2ooamps. By reducing the 200amp main to 175amps this leaves you only 25amps to back feed w/ PV not the 60amps you are planning.
Click to expand...

Your summary of 2005 NEC 690.64(B)(2) misses a key part of that section. This section specifically states a) exception for residential: sum of the above shall not exceed 120% the rating of the buss bar or conductors.

200 amp main panel - Max ampere rating of overcurrent devices supplying power: 240 amps

Breakers supplying power: 175A+60A = 235amps. Your kosher. The kicker is if this is a retrofit (Interconnecting a solar system to an existing electrical system) depending on where you are interconnecting you will need to check the conductors from the interconnection point to the meter using the same rule to make sure they are not overloaded. AL wires typically create problems.

Under 2008 NEC, I believe they are allowing for the 120% on commercial structures as well but they have added other interconnection requirements.
 
electrical designer

electrical designer

suemarkp said:
No, it isn't. Use the MCA (minimum circuit ampacity) value on the nameplate. This will be quite a bit less than the max breaker size.
Click to expand...

I have been an electrical designer for quite a while. One of the first lessons I learned was never use the MCA for sizing your circuit breaker. It can lead to under sizing. I use the FLA (full load amps) on the nameplate. On a split system and/or package unit (roof top unit). In the heat mode, the heater and indoor fan run simutaneously, when cooling, the indoor fan, outdoor fan and the compressor run simutaneously. Add the first two terms and then add the other three terms. The larger of the two is your largest A/C load. That is what the poster meant when they wrote A/C larger than heat. The other load is non-concurrent. Multiply the largest total FLA by the voltage and you have your wattage. As a practice I multiply by 125% for inrush to size the circuit breaker. For larger units (15 hp and above) I multiply 150%, as long as it is not larger than the max breaker size. That is how you calculate an accurate A/C load
 
Calculating the load is not the same as calculating the breaker size. MCA is solely for conductor sizing and load calculation. Typically, you have a branch circuit for the airhandler and another one for the outdoor compressor. Each unit should have MCA on its nameplate. You don't have to care how many fans, compressors, motors or anything else are inside it if it has MCA on the nameplate. There are control electronics too which draw power, and this is included in the MCA value.

You would just add the MCA of the outdoor unit(s) and the MCA of the indoor units (if they have do not heat strips). If they have heat strips, then its a bit more complicated at the air handler and you look at the components (fan(s), controls), know which is the largest motor and multiply it 125%, etc.

You can try to calculate each item manually, but it isn't always easy. And you must multiply the largest motor or compressor on the circuit by 125%. Usually, the MCA with no heat strips is what is installed by the factory. You then field install the heat strips and they give you new MCA stickers depending on which strip sets you installed. If you can find the heat stripless MCA value, then you don't have anything to calculate.

Sizing the breakers again should be as simple as reading the nameplate (Max Breaker Size). You don't have to size per the largest breaker value, but something larger than MCA and smaller than max typically works.
 
Status
Not open for further replies.
Top