heat pump calculation ?'s

Status
Not open for further replies.

dwellins

Member
Location
Punta Gorda, FL
I am bidding a job to install 5 geothermal (pool) heatpumps
RLA 23.2, Max OCD 40, Min circuit 29.3
Voltage is 120/208 3-phase.

Question is finding the subpanel that needs to be installed.
430.24 states 125% FLA plus other FLA
so total load is 23.2 x125% +23.2+23.2+23.2+23.2=121.8A conductors
breaker for subpanel is
430.62 125% Max OCD plus other FLA
so breaker is 40 x 125% +23.2+23.2+23.2+23.2=142.8A so 150A breaker

Would a 150A panel be sufficent with 150 amp main with #1 copper conductors, or do I add an addtional 125% on each item due to 215.3,etc states that you must size equipment/conductors 125% of continous load (80% of equipment capacity)? This would then make this a 200A breaker/Panel with 2/0 copper conductors.

Any input is greatly appreciated. Thank you
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
I find Art 440 to be a little lax on feeder calculations, but based on 440.33,for your conductors, I think you use the sum of the rated load or branch circuit selection current whichever is greater.
In your case, if I am correct, that would be 29.3 X 1.25 + 29.3 X 4 or 153.8 amps for the circuit conductors.

For the SCGF protection, 440.22 is similar.
29.3 x 1.75 + 29.3 x 4 or 168.8

With little valid reason to do so, I reference back to 430.62 and 52 and assume (I know) we can go to the next size up and select a 175 amp breaker. That point may be mute anyway as you will need a 2/0 to meet the MCA and it has a 175 amp rating.

This is subject to a LOT of controversy since 440 does not directly address feeders and to get where I am requires some liberal mixing of Art 440 and 430.
Therefore it is definitely an opinion.
 

dwellins

Member
Location
Punta Gorda, FL
thank you for the input which now brings up another question...as these are Geothermal Heat pumps for pool heating, Does this fall under article 440 (A/C and refrigeration) or 430 (motors)?
 

bob

Senior Member
Location
Alabama
total load is 23.2 x125% +23.2+23.2+23.2+23.2=121.8A conductors
breaker for subpanel is
430.62 125% Max OCD plus other FLA
so breaker is 40 x 125% +23.2+23.2+23.2+23.2=142.8A so 150A breaker

Augie, you are close. I would do the following: FLA = 23.2
23.2 x 1.75 + 4 x 23.2 = 132.8 for the breaker or 150 amps
conductor = 23.2 x 1.25 + 4 x 23.2 = 121.8 amps
Circuit for each unit would be 29.3 amps.
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
Bob, you very well may be right, I was basing my numbers on the wording throughout Art 440 that states:
protective device having a rating or setting not exceeding 175 percent of the motor-compressor rated-load current or branch-circuit selection current, whichever is greater,

As stated, I personally don't think there is a universally accepted answer so, as always, it will left up to his AHJ.
The fact that we have nameplates that give MCA, MOCP, etc as opposed to HP leads me to believe it's a Art 440 installation as opposed to Art 430, but 440.3 references you back to Art 430 so I took the liberty of using the Art 430 rules for feeders.
There may be as many opinions as there are posts :D
 

bob

Senior Member
Location
Alabama
It seems to me that you are applying the 1.25 factor twice. In your post you said "In your case, if I am correct, that would be 29.3 X 1.25 + 29.3 X 4 or 153.8 amps for the circuit conductors." The 29.3 amps is the min ckt size which has the 1.25 included.

Also you said "For the SCGF protection, 440.22 is similar.
29.3 x 1.75 + 29.3 x 4 or 168.8".
This has the 1.75 x 1.25 factor. I do not understand why that is necessary. Is it coming for the articles you quoted?

I am one of many who respect your knowledge of the book. I am just trying to understand.
 

bob

Senior Member
Location
Alabama
After looking at your post again are you referring to the 29.3 amps as the branch-circuit selection current
 
Last edited:

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
That's a very good point, and my math might definitely result in "overkill".
If I look at it from a Art 430 standpoint, which might be the correct route as Art 440 does not directly address feeders, your number looks good in that we are substituting RLA for FLA.
My logic may be flawed in that I am equating MCA as "branch circuit selection current", but 440.4(C) tells us that the branch circuit selection current must be listed on the nameplate and the MCA is the likely equivalent. I am concerned that in selecting conductors Art 430 overload is based on 125% whereas, according to Art 440.52, the overload current can be based on 140% so I am not comfortable with the 125% factor you are using.
That and the ever-present "whichever is larger" is where I based my logic, but your approach definitely causes me to re-think the position.
We need a Art 440 guru to step in.
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
After looking at your post again are you referring to the 29.3 amps as the branch-circuit selection current

Yes..which may be the flaw, but, as stated the nameplate is supposed to list the branch-circuit selection current and thats the only number given and it does come close to the 140% allowed by 440.52

added: This thread makes my head hurt
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
but what numbers ? RLA, MCA, something else ???

please give us you calculations

Gus I was just responding to the op question as to whether a geotherm unit would fall under 430 or 440. IMO, it is 440-- I am too tired to do the math as I thought you guys did it already.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
My logic may be flawed in that I am equating MCA as "branch circuit selection current", but 440.4(C) tells us that the branch circuit selection current must be listed on the nameplate and the MCA is the likely equivalent.

This is flawed logic. MCA and "branch circuit selection current" are not the same thing.

There seems to be some missing information that would be needed to provide a correct answer. The info given is RLA, MCA and MOCP. The RLA would seem to be for the compressor only, and 440.33 would tell us that MCA would be 125% of RLA plus the FLA of the other motors. 1.25*23.2=29. So it would appear that there is an additional 0.3A of load. So lets assume each heat pump has a compressor of 23.2 RLA and a fan of 0.3 FLA.

The minimum ampacity of a feeder for all 5 heat pumps would be 125% of the largest motor plus the full load of the other motors, per 430.22. So: 23.2*1.25 + 4*23.2 + 5*0.3=123.3.

The feeder OCPD would be determined by 430.62: not to exceed the value for the largest motor (compressor in this case) plus the sum of the remaining currents. The largest allowable value for a compressor is 1.75*23.2=40.6A or a 45A maximum OCPD.

So for the feeder the max OCPD size is 45 + 4*23.2 + 5*0.3 =139.3. Since this is a not to exceed number from 430.62, you have to go down to 125A.

So I'd say your feeder has to be #1awg (min) on a 125A c/b. (this assumes a 23.2A and a 0.3A load in the heat pump. This is not known from the given question.)

Of course nothing would prevent you from making the feeder #1/0awg with a 150A c/b
 
Last edited:

kwired

Electron manager
Location
NE Nebraska
440.3 Other Articles.
(A) Article 430. These provisions are in addition to, or amendatory of, the provisions of Article 430 and other articles in this Code, which apply except as modified in this article
.

Since 440 does not cover feeders you pretty much are entirely in 430 when doing feeder calculations. RLA of a hermatic compressor is synonomous to FLA of a motor. David took us through a pretty good example so I'm not going to go through it again.
 

kwired

Electron manager
Location
NE Nebraska
thank you for the input which now brings up another question...as these are Geothermal Heat pumps for pool heating, Does this fall under article 440 (A/C and refrigeration) or 430 (motors)?

440.1 Scope.
The provisions of this article apply to electric motor-driven air-conditioning and refrigerating equipment and to the branch circuits and controllers for such equipment. It provides for the special considerations necessary for circuits supplying hermetic refrigerant motor-compressors and for any air-conditioning or refrigerating equipment that is supplied from a branch circuit that supplies a hermetic refrigerant motor-compressor.

A heat pump contains a hermatic refrigerant motor-compressor, therefore 440 applies. But remember 440 requirements are in addition or amendatory to 430 requirements.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
According to this the MCA is the branch ckt sel amps which is the FLA x 1.25 as I understand it.

This is not correct. MCA is NOT branch circuit selection current.

For a single hermetic motor compressor, MCA is either 1.25*RLA, or 1.25*BCSC, whichever is larger. See 440.32.
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor

Yep and it comes down to what is meant in art. 440.32. David and I had a long discussion on this. The question is whether they mean 125% of the RLA or 125% of the branch circuit selection current. OR whether it means 125% of the RLA or the branch circuit selection current. IMO, now, I think David is correct. The placement of the word "either" does it for me-- I think. :lol:

440.32 Single Motor-Compressor. Branch-circuit conductors supplying a single motor-compressor shall have an ampacity not less than 125 percent of either the motorcompressor rated-load current or the branch-circuit selection current, whichever is greater.
 

kwired

Electron manager
Location
NE Nebraska
440.2 Definitions.
Branch-Circuit Selection Current. The value in amperes to be used instead of the rated-load current in determining the ratings of motor branch-circuit conductors, disconnecting means, controllers, and branch-circuit short-circuit and ground-fault protective devices wherever the running overload protective device permits a sustained current greater than the specified percentage of the rated-load current. The value of branch-circuit selection current will always be equal to or greater than the marked rated-load current.

Not all units will have a BCSC. In fact I can't say I have ever seen one that has one marked on it.
 
Status
Not open for further replies.
Top