continuous load for heat pump w/ heat strips

Status
Not open for further replies.
R

RustyShackleford

Senior Member
Location
NC
Occupation
electrical engineer
How does one compute the continuous load (which must be supported at 125%) for a heat pump with electrical resistance heat strips (that can operate simultaneously) ?

The heat strips are simple, assume they're fully on (if there are multiple segments used for staging, assume they are all energized).

But what about the heat pump ? Manufacturers quote a LOT of numbers. I'm thinking you don't need to consider numbers as high as the MCA or RLA as "continuous". There are various performance tables, which quote wattage for a plethora of indoor and outdoor temperatures; so use the highest such number in the tables ?
 
The only numbers you normally need to take into consideration are MCA for the conductor size and MOCP for the overcurrent protection size.
The 125% when needed, has been included in those numbers.
{The type OCO (fuse or breaker) and the withstand current can come into play}
 
augie47 said:
The only numbers you normally need to take into consideration are MCA for the conductor size and MOCP for the overcurrent protection size.
The 125% when needed, has been included in those numbers.
{The type OCO (fuse or breaker) and the withstand current can come into play}
Click to expand...

I usually install two separate circuits, one for the inside fan coil, and one for the inside heat strip. Then a third circuit for the outside condensing unit. All based on the manufacturer's requirements. Also, the MOP is the maximum recommended by the manufacturer, you can use a smaller size if the unit doesn't trip. I like using a fused pull out at the outside unit.
 
Last edited:
RustyShackleford said:
How does one compute the continuous load (which must be supported at 125%) for a heat pump with electrical resistance heat strips (that can operate simultaneously) ?

The heat strips are simple, assume they're fully on (if there are multiple segments used for staging, assume they are all energized).

But what about the heat pump ? Manufacturers quote a LOT of numbers. I'm thinking you don't need to consider numbers as high as the MCA or RLA as "continuous". There are various performance tables, which quote wattage for a plethora of indoor and outdoor temperatures; so use the highest such number in the tables ?
Click to expand...
If the compressor can run while the heat strips are on, then you have to use both the heat strip load and the motors.
 
Little Bill said:
If the compressor can run while the heat strips are on, then you have to use both the heat strip load and the motors.
Click to expand...

With my In-Service Training with Carrier, and the supplied documents supplied from the manufacturer, they recommend separate circuits. Other manufacturers are similar. And a separate disconnect at the inside fan coil, with in sight, for both the fan coil and the heat strip. The disconnects can be a close by sub-panel breaker.
 
augie47 said:
The only numbers you normally need to take into consideration are MCA for the conductor size and MOCP for the overcurrent protection size.
The 125% when needed, has been included in those numbers.
Click to expand...
So you're saying 80% of the heat pump's spec'd MCA plus the heat strip amperage = the continuous load ? Or, equivalently, the heat pump's MCA plus 125% of the heat strips' amperage = the required ampacity of the circuit ?
 
Little Bill said:
If the compressor can run while the heat strips are on, then you have to use both the heat strip load and the motors.
Click to expand...
I'm doing that. Just wasn't sure what number to use for the heat pump.

I'm sizing a feeder for a subpanel which will have two branch circuits: for the heat pump, and for the heat strips. There's already 6awg NM-B going to the system location.
 
mtnelect said:
With my In-Service Training with Carrier, and the supplied documents supplied from the manufacturer, they recommend separate circuits. Other manufacturers are similar. And a separate disconnect at the inside fan coil, with in sight, for both the fan coil and the heat strip. The disconnects can be a close by sub-panel breaker.
Click to expand...
Separate branch circuits are required nearly all the time. I think he wants to know how to do feeder/service calculations for the combined load of these items.
 
kwired said:
Separate branch circuits are required nearly all the time. I think he wants to know how to do feeder/service calculations for the combined load of these items.
Click to expand...
Thanks, that's exactly right; I should clarify OP, but it's un-editable now.

I am sizing the feeder to a subpanel which will have two branch circuits - one for the heat pump compressor and one for the heat strips (and they can operate simultaneously). The indoor unit is usually a small'ish 120vac load and will be powered separately.

So I need to know the continuous load for that feeder. The heat strips are simple. But I'm not sure what amperage to use for the compressor; hard to imagine locked-rotor number would be considered continuous, and the numbers in the performance tables are far lower.
 
Go to the "Product Data" sheet for each manufacturer of unit(s).
 

Attachments

  • Carrier FY5B-05PD #1 .pdf
    198.4 KB · Views: 8
Also, have you done a "Heat Load" engineering study for the structure ?
 

Attachments

  • Heat Load - Example.pdf
    65.5 KB · Views: 6
  • Mountain Electric DuctReport .pdf
    9.5 KB · Views: 4
mtnelect said:
Go to the "Product Data" sheet for each manufacturer of unit(s).
Click to expand...
Per your link, which is for the heat strips in the indoor unit, it's pretty straightforward. They draw a certain current when they're turned-on, period.

It's the compressor that's the question. Product literature tends to state many amperages: FLA, RLA, LRA, MCA, and various numbers in the performance tables. It looks to me like MCA is the right one (as mentioned above), and that it includes the 125% factor.
 
Last edited:
RustyShackleford said:
Per your link, which is for the heat strips in the indoor unit, it's pretty straightforward. They draw a certain current when they're turned-on, period.

It's the compressor that's the question. The literature quotes many amperages: FLA, RLA, LRA, and various numbers in the performance tables; which one is the correct one to use to compute the continuous load ?
Click to expand...

The first step is to do a "Heat Load" study, so you know what size of unit you will need. The next step is to look at the engineering data that the manufacturer displays. There is the sensible heat, latent heat and total heat. In southern California we have hardly any latent heat, so we disregard it. So, it's your job to interpret this information and select the proper size unit for your structurer.
 
In California we don't use the "Rule of Thumb" anymore. You can't take a 2,000 square foot home and divide 400 square feet to determine the size of unit. No more guessing, you have to do a "Heat Load" engineering study, like my attachment.

How can you determine the feeder size if you don't know what size unit you need ?
 
mtnelect said:
In California we don't use the "Rule of Thumb" anymore. You can't take a 2,000 square foot home and divide 400 square feet to determine the size of unit. No more guessing, you have to do a "Heat Load" engineering study, like my attachment.
Click to expand...
Next you're gonna tell us we can't figure load calcs by area, too. :sneaky:
 
LarryFine said:
Next you're gonna tell us we can't figure load calcs by area, too. :sneaky:
Click to expand...

This is an example I use ... You have two cabins on a lake. One is on the North side of the lake and the other on the South side of the lake. Both have the same floor plan, but they have different heat loads. Because the sun comes up and sets down on different sides of the cabins.
 
mtnelect said:
This is an example I use ... You have two cabins on a lake. One is on the North side of the lake and the other on the South side of the lake. Both have the same floor plan, but they have different heat loads. Because the sun comes up and sets down on different sides of the cabins.
Click to expand...
The one on the north would need more heat
The one on the south would need more AC
Don't believe me, ask a person in Minnesota and a person in Florida :cool:
 
Status
Not open for further replies.
Top