Circuit for Replacement HVAC System

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wwhitney

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Berkeley, CA
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topgone said:
I think there's something amiss there in that calculation.
Click to expand...
The only difference in your calculations is that you put the single phase fans on different phases, while david luchini assumed the worst case that they were wired on the same phase. So your answers differ by 2 * (fan current, 7 amps) = 14 amps.

Also you wrote 1.25*51.3 + 51.3, while david luchini combined that as 2.25*51.3

Cheers, Wayne
 
I

Ingenieur

Senior Member
Location
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topgone said:
I think there's something amiss there in that calculation. Here's how I see the distribution (USING RLA = FLA):
PHP: 
EQUIPMENT                 PHASE      FLA               PHASE "A"       PHASE "B"         PHASE "C"
COMPRESSOR 1               3          51.3              51.3             51.3             51.3
COMPRESSOR 2               3          51.3              51.3             51.3             51.3
CONDENSER FAN 1            1           7.0              7.0               -                -
CONDENSER FAN 2            1           7.0               -                7.0              -
CONDENSER FAN 3            1           7.0               -                -                7.0
SUPPLY FAN 3               3           16.7             16.7             16.7             16.7
                                          LINE AMPS =  126.3A          126.3A            126.3A
Given those figures above, the Minimum circuit amps:
MCA = 1.25 x 51.3 + (51.3 + 7.0 + 16.7) = 139.12A
Click to expand...

he put the 7 A all on the same phase
which is how it is likely wired

regardless, if balanced it is not 7 per phase, it is 3 x 7 x 208 / (1.732 x 208) = 12.1 per phase
 
I

Ingenieur

Senior Member
Location
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MyCleveland said:
Ingenieur
Your trane link reference
Shouldn't you be using middle column "rooftops with electric htg"
( sum of all loads ) x 1.25 = MCA
Click to expand...

it says to note that the condensor fans and compressors do not run
1.25 (main fan + heater) = 1.25 (16.7 + 45100/(1.732 208)) = 177

I was commenting on his clg calc

they say do both and pick larger
 
I

Ingenieur

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I would consult with the owner and inspector
if no electrical change is he even involved?

V drop is ~1% per 100', if less than 3
use the 2/0 and 175 A cb
run the htg
measure i
if ~ 177
call it a day
 
T

topgone

Senior Member
MyCleveland said:
topgone
Have you abandoned your post from earlier....now RLA = FLA ?

Can you respond to 11:50am post regarding your earlier calc
Click to expand...

Nope. I am trying to follow the poster's calculation process regarding the minimum circuit amperes!

Sorry for not responding to your earlier query,re 74%. It is actually an arbitrary factor which is the inverse of 1.35 (135%): 1/1.35 = 0.74. As discussed by others, the "maximum continuous amperes" is the current the compressor carries without tripping the motor thermal protection. Remember that refrigeration compressors use thermal protection integral to the motor and the max allowed current (for motors with currents above 20A) is 140% (430.32(A)(2). I picked 135% and reverse-computed the FLA from the "maximum continuous amps"; hence the multiplication of 0.74 (or you can do it by dividing the max continuous amps by 1.35!
 
david luchini

david luchini

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Staff member
Location
Connecticut
Occupation
Engineer
topgone said:
I think there's something amiss there in that calculation. Here's how I see the distribution (USING RLA = FLA):
PHP: 
EQUIPMENT                 PHASE      FLA               PHASE "A"       PHASE "B"         PHASE "C"
COMPRESSOR 1               3          51.3              51.3             51.3             51.3
COMPRESSOR 2               3          51.3              51.3             51.3             51.3
CONDENSER FAN 1            1           7.0              7.0               -                -
CONDENSER FAN 2            1           7.0               -                7.0              -
CONDENSER FAN 3            1           7.0               -                -                7.0
SUPPLY FAN 3               3           16.7             16.7             16.7             16.7
                                          LINE AMPS =  126.3A          126.3A            126.3A
Given those figures above, the Minimum circuit amps:
MCA = 1.25 x 51.3 + (51.3 + 7.0 + 16.7) = 139.12A
Click to expand...
I don't believe there is anything amiss in my calculation. Your calculation, on the other hand, includes three 7A, 120V single phase motors, where the OP information states that they are 7A, 208V single phase motors.
 
I

Ingenieur

Senior Member
Location
Earth
Is the unit rated 208 only or 208-240 like most?
is the mca calc'ed on 208 or 240?
is the 45.1 kw based on 208 or 240?

the htg calcs seem to indicate 208
 
T

topgone

Senior Member
david luchini said:
I don't believe there is anything amiss in my calculation. Your calculation, on the other hand, includes three 7A, 120V single phase motors, where the OP information states that they are 7A, 208V single phase motors.
Click to expand...

Well, I have yet to see a 1HP, single-phase motor that will draw a load of 7A at 120 volts (FLA tables will tell you FLA=13 to 16A). I was looking at those OP's figures before I posted here. If there is one 1-HP, 120V with that current draw, the power factor must be dang low at 0.51!
Still, the nearest figure I see from FLA tables is a 3HP, 3-phase which draws 8A at 220V!
 
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