3 phase load calculation

Status
Not open for further replies.

Greg1707

Senior Member
Location
Alexandria, VA
Occupation
Business owner Electrical contractor
I need help with a three phase load calculation. The load is 5-240 volt dryers and 5-120 volt washers. The feeder will be three phase 120/208. I assume these appliance will be operating at the same time.

According to the specifications the dryers are 30 amps and the washers are 10 amps at 120 volts.

Where do I begin? I apologize in advance for my ignorance.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
I need help with a three phase load calculation. The load is 5-240 volt dryers and 5-120 volt washers. The feeder will be three phase 120/208. I assume these appliance will be operating at the same time.

According to the specifications the dryers are 30 amps and the washers are 10 amps at 120 volts.

Where do I begin? I apologize in advance for my ignorance.

You say three phase load calculation, and you describe a 208Y/120 service, to use the standard notation. That in itself poses a problem as the 240V dryers will be working on 208 volt phase to phase instead, with the motor getting 120. So they will draw less than their rated current and will dry slightly slower. (Most likely the on part of the thermostat cycle will just be greater to compensate and so the change in drying speed may not be noticeable.)

Once you get past that, your next problem will be that 5 does not divide by three very well, so you will be dealing with an unbalanced load. That by itself is not a problem. You can partially correct for that by deploying the washers on the appropriate phases.

Then the final problem will be that the reduced current (reduced voltage from 240 to 208, remember) will be drawn from the two phase lines and so the current in each phase line (should you be able to balance things) will be that current divided by the square root of three. It has no effect on the load power calculations, but it does affect the way the power gets converted to line current.

Does that give you a decent start?
 

Smart $

Esteemed Member
Location
Ohio
Assuming a non-dwelling location, (5 ? 30A ? 240V) + (5 ? 10A ? 120V) = 42,000VA

Are there any other loads on the panel?
 

Smart $

Esteemed Member
Location
Ohio
I realize I used 240V for the dryers even though they will be supplied with 208V. Code requires specific appliances to be calculated using nameplate ratings. Unless it specifically has 208V ratings, have to use what's there...
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
I realize I used 240V for the dryers even though they will be supplied with 208V. Code requires specific appliances to be calculated using nameplate ratings. Unless it specifically has 208V ratings, have to use what's there...
Thank you for the explanation. :)
Although the OP left the question open, if these serve dwelling units (e.g. shared laundry room) rather than commercial or industrial use, he can take advantage of the diversity factors listed in 220.54. But since they represent single phase loads, he may not actually end up with a lower number in that case.
 
Last edited:

Greg1707

Senior Member
Location
Alexandria, VA
Occupation
Business owner Electrical contractor
non dwelling unit

non dwelling unit

This is a non dwelling unit. I assume all the appliances may be in use at the same time. Since 120/208 is available I assumed installing a sub panel to the laundry area using three phase would be appropriate?
 

kwired

Electron manager
Location
NE Nebraska
This is a non dwelling unit. I assume all the appliances may be in use at the same time. Since 120/208 is available I assumed installing a sub panel to the laundry area using three phase would be appropriate?
It will help with load balancing on the service/upstream feeders at the very least as compared to only running two phases to this area.

If you figure 30 amps for dryers and 10 amps for the washers you should have plenty of capacity as I would be willing to bet the dryers don't draw much more than 23 amps if connected to 240 volts, and probably would reduce to about 20 amps @ 208 volts. The washers may draw up to 10 amps, but probably only during the spin cycle, the rest time it may be significantly less.

Next step is realizing that if you have 30 amps of load between each phase, the actual phase currents are 1.73 times each individual load current so if you only had three 30 amp loads equally balanced you would have 52 amps of current on each phase.

Now consider if the load really is 20 amps as suggested it may be that drops to only 35 amps. That is just for three dryers, you still need to add the other two - 20 (or 30) amps for two phases and 35 (or 52) to the phase that is common to both of the unbalanced portion of the load, plus your washers need added yet, which will be 10 amps per unit to each phase they are connected to.

I come up with best balance (assuming all are fully loaded at one time and loads of 10 and 30 amps as mentioned in OP) being three dryers balanced across all phases, put the remaining two from (for example) a to b and b to c , at this point you will have phase loads of 82, 104, 82. Then put two washers on each of the 82 amp phases to get the balance to 102, 104, 102. Put the remaining washer on any phase you wish.
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
I'd like to ask a question. Let's say I have three single-phase 480-V loads that I'm going to be serving from a 480Y feeder.

The single-phase loads are:

A-B = 155A
B-C = 320A
A-C = 210A

What would the currents be on the 3 phases of the 480Y feeder?

If you have the time, how does the math work?

Thanks
 

kwired

Electron manager
Location
NE Nebraska
I'd like to ask a question. Let's say I have three single-phase 480-V loads that I'm going to be serving from a 480Y feeder.

The single-phase loads are:

A-B = 155A
B-C = 320A
A-C = 210A

What would the currents be on the 3 phases of the 480Y feeder?

If you have the time, how does the math work?

Thanks

May be an easier way but let me give it a shot:

A phase has 155 x 1.73 coming from both B and C plus additional 55 coming from C giving you 323 on A

B phase has 155 x 1.73 coming from both A and C plus additional 165 coming from C giving you 433 on B

C phase has 210 x 1.73 coming from both A and B plus additional 110 coming from B giving you 473 on C

Even if this were 480 volt delta the numbers should be the same, but if you throw some line to neutral load in on the wye system it will get more complex.
 

Smart $

Esteemed Member
Location
Ohio
I'd like to ask a question. Let's say I have three single-phase 480-V loads that I'm going to be serving from a 480Y feeder.

The single-phase loads are:

A-B = 155A
B-C = 320A
A-C = 210A

What would the currents be on the 3 phases of the 480Y feeder?

If you have the time, how does the math work?

Thanks
Cannot be determined with any accuracy without knowing the the loads' power factor or phase angle of the load currents. All you can do with the information provided is approximate. The actual current may end up higher or lower. However, NEC compliance will be satisfied with this type of approximation.

Kwired provided one method of approximation.

Another is one used by EE's through a panel schedule determination. They put one-half the load in kVA in each phase column to which the load is connected. Sum the columns, then divide by the line-to-neutral voltage (or L-L voltage divided by the square root of three if delta).

For example, we'll assume you are sizing each feeder conductor the same size, so we know the largest current is going to be on C.

210A ? 480V ? 2 = 50.4kVA
320A ? 480V ? 2 = 76.8kVA

50.4kVA + 76.8kVA = 127.2kVA

127.2kVA ? 277V = 459A

Here's the shortcut method for the preceeding...

(320A + 210A) ? .866* = 459A

*sqrt(3) ? 2 = .866


FWIW, the method depicted by kwired is more conservative.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
kwired: I see one problem with your calculation:
Start by assuming resistive loads. Or at least all the same PF. The current you calculate using the 1.73 factor will be in phase with the voltage on that line. But the extra current coming through the side with the higher current will not be in phase and so will not add arithmetically.

One double check for any analysis will require (for resistive loads anyway) that the sum of 480 times the line to line current must equal 277 times the sum of the line currents.
The general solution (for different PF on each line to line load) will require calculating the phase angle of each of the known currents and doing vector addition. Difficult and tedious, but necessary.
Figuring out the total power over all three phase lines is easier: it is just the sum of the three line to line power numbers.

Sent from my XT1080 using Tapatalk
 
Status
Not open for further replies.
Top