4160 or not

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patpappas

patpappas

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Looking for a good way to explain to the brass why operating a chiller 990HP motor is better at 4160 than 480, the bulding does have a primary rate from the POCO. I have been searching for liteature that can explin it better than me.
 
A motor that large would draw over 1000 amps at 480 volts. You would need at least three sets of 500 MCM conductors for the branch circuit. At 4160 volts, the current draw would be under 150 amps, and the wires would be 1/0 or smaller. There is a major savings here, both in installation costs and in the costs of power losses (and heating) within the branch circuit conductors. There is also the issue of the capacity of the existing 480 volt distribution equipment (i.e., can it support the addition of a 1000+ amp load?). I would start with this. I would also check with the motor manufacturer, to see if they even have a 480 volt version, and if so what size of wire it needs.
 
I would think the in-rush on a 990HP motor would be rather large and a breaker or 3 upstream might not like that. If there is any length to the conductors/feeder you may have voltage drop issues that would cause even larger sized wire then charlie already said.
 
I don't know what the service configuration is to your facility (type, size, service voltage, distribution, etc.). At this size you might consider possible voltage dip when starting...could affect your 480V distribution. Connecting upstream 4160V could help mitigate this.

Depending on your situation, 480V service level could be a more costly capital investment than a 4160V service level. You might have to provide cost estimates for 480V vs. 4160V scenario.

If you don't have an existing 4160V service, first check with your utility company for the max motor HP limit on 480V, 3PH.
 
patpappas said:
Looking for a good way to explain to the brass why operating a chiller 990HP motor is better at 4160 than 480, the bulding does have a primary rate from the POCO. I have been searching for liteature that can explin it better than me.
Click to expand...

What service do you have available? Sure, 4160v is a good idea but if you don't have 4160 available then it's a different issue.
Also, it is quite common to use 2300v with motors of that HP.
But with a motor of the HP 480v is a bit of a stretch. What size motor starter would you be considering if y ou stayed with 480v? And then the cable that would be required.
 
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Not having to pay for the power losses in the step down transformer.

And the POCO rates for a 4160 V service are probably much less than 480V rates.

Edit: If you talk to the POCO rep. I'll be they can convince eveyone 4160 is the way to go.
 
HP = HP = HP = HP, and 1HP = .746kW, nothing about that changes with voltage.

But in a motor circuit, consumed power = power to do the work (which also does not change) + losses in the system. Some losses are fixed, meaning they will not change regardless of voltage, but a big chunk of losses in a power system are what are called "Copper Losses", or "I2R" losses because they are associated with the resistance in the copper conductors and the current flowing through them. So the more current you have, the more I2R losses you have. This has nothing to do with voltage, only current.

At 460V, 990HP means roughly 1200A FLC, vs. 990HP at 4160V = means approximately 140A FLC. You will have almost 1/10th the I2R cable losses at 4160V. I2R losses extend to the motor windings and transformer windings as well.

Not to mention the huge up front COST of 4 x 500kCMIL cables per phase at 480V, compared to 1 x 1/0 MV cables per phase at 4160V.

Then factor in the risk of all that copper in the 480V system being a juicy target for copper thieves... It would be like waving around a bag of crack on a street corner in Oakland.

I did a 900HP wood chipper at 460V once, it was a nightmare.
 
Jraef said:
Then factor in the risk of all that copper in the 480V system being a juicy target for copper thieves... It would be like waving around a bag of crack on a street corner in Oakland.
Click to expand...

Hahaha, well put.

The increased cable size, associated equipment costs (unless they have a MCC with over 1200 amps of spare capacity) and I2R losses with 480V should be enough for them to go with 4160V.
 
Patpappas,

Ignoring the obvious disparity in installation cost, consider only cable losses, i.e., LV (480V) vs MV (4,160V). Since this is a 3-ph circuit, consisting of 3 conductors per circuit, then the loss per cable system is :

DP = 3 x Ic? x ( Rw/Np ) x Lc x 10??, kW, where,
o Ic = Cable current per phase, Amps
o Rw = Conductor resistance, W per k-feet of circuit length,
o Np = Number of conductors per phase (2 for LV and 1 for HV)
o Lc = Circuit length, ft.

Comparing LV & MV cases, the Loss ratio between LV and MV reduces to,

DPLV / DPMV = ( ILV / IMV )? x [ ( Rw(LV) / 2 ) / Rw(MV) ]

Using the current values you cited, 2x500-kcmil (and its Rw) for LV, and 1x250-kcmil (and its Rw) for MV, the LV losses are about 18 times greater than MV losses!

Regards, Phil Corso
 
Patppas...

I inadvertently left out two concerns if you pursue the LV case and you chose Wye-Delta Starting means:

1) Economics.
Three (3) single-phase circuts, involving doubling the number of conductors, are needed! Although somewhat smaller than 500-kcmil, an additional de-ratng factor must be used (0.8), which could increase the conductor-size (kcmil) as well as the conduit size (if used!) And, cable-loss could increase by as much as 40%, depending on final cable-size selected!

2) Voltage-Drop.
During "START" two sets of conductors are in series, doubling the circuit distance, Lc. And for "RUN" the Lc, is still double, thereby exacerbating V-Drop requirements!

Regards, Phil Corso
 
Patpappas...

I erred in my earlier post! Following is the correct formula and conclusion:

For any 3-ph circuit, loss per phase is:
DP = Np x (Ic / Np)? x Rw x Lc x 10??, kW, where:
o Np = Number of conductors per phase (2 for LV and 1 for HV)
o Ic = Cable-current per phase, Amps
o Rw = Conductor resistance per k-feet of circuit length, W
 o Lc = Circuit length, ft.

Comparing LV (1,000A, 2x500-kcmil/ph Rw, 0.0305W/k-ft); to MV (150A, 1x250-kcmil/ph, Rw, 0.0572W/k-ft), then LV losses are about 12 times greater than MV losses, as follows:

Ratio, LV:MV = [ (ILV/IMV)? x (RLV/RMV) ] / NLV = [ (1,000/150)? x 0.0305 / 0.0572) ] / 2 =

Regards, Phil Corso
 
Christoff84 said:
I would think the in-rush on a 990HP motor would be rather large and a breaker or 3 upstream might not like that. If there is any length to the conductors/feeder you may have voltage drop issues that would cause even larger sized wire then charlie already said.
Click to expand...

When it comes to upstream being the POCO, this doesn't matter, you still have the same kVA in rush either way, only place where current is different is on the secondary of the transformer, where the voltage is different.

........................

To add to what others have said about cost of copper, not only will 480 volts require more copper conductor, a transformer with 480 volt secondary will also have more copper than one with a 4160 secondary of the same kVA rating, probably the same with the motor.
 
kwired said:
... a transformer with 480 volt secondary will also have more copper than one with a 4160 secondary of the same kVA rating, probably the same with the motor.
Click to expand...

Neglecting differences in insulation, the transformer secondary will be the same amount of copper.

Look at it this way:
Suppose the xfm is 1000kva, 4160Y. That's three 2400V coils, 139A. Consider that each 2400V coil is made of of 5 480V coils. Parallel connect the 5 coils in each 2400V coil. Reconnect the 480V coils Delta. Transformer is good for 1000kva.

I suspect it is the same for the motors as well in that the volts per turn has to be the same

ice
 
Jraef said:
HP = HP = HP = HP, and 1HP = .746kW, nothing about that changes with voltage.
Click to expand...
Agreed. But go with SI and you don't need that conversion factor.:p

But in a motor circuit, consumed power = power to do the work (which also does not change) + losses in the system. Some losses are fixed, meaning they will not change regardless of voltage, but a big chunk of losses in a power system are what are called "Copper Losses", or "I2R" losses because they are associated with the resistance in the copper conductors and the current flowing through them. So the more current you have, the more I2R losses you have. This has nothing to do with voltage, only current. [/quote]
Well, yes, but the current changes with voltage for the same power all else being equal

Jraef said:
At 460V, 990HP means roughly 1200A FLC, vs. 990HP at 4160V = means approximately 140A FLC. You will have almost 1/10th the I2R cable losses at 4160V. I2R losses extend to the motor windings and transformer windings as well.
Click to expand...
Reducing the current by a factor of ten for the same conductor size would actually reduce the losses by a factor of 100 - remember it's i2R.
But that isn't really relevant given that conductor sizes will be different for the different voltages.
For the supply conductor losses using appropriately rated cable sized I came up with a figure of about five to one per unit length in favour of the 4160V.
At around 750kW for the motor and maybe 1MVA for a unit transformer, the efficiencies might not be greatly different at the different voltages so maybe those can be taken out of the equation.
As far as I can tell, the big differentiator is the cost of the cabling.
 
iceworm said:
Neglecting differences in insulation, the transformer secondary will be the same amount of copper.

Look at it this way:
Suppose the xfm is 1000kva, 4160Y. That's three 2400V coils, 139A. Consider that each 2400V coil is made of of 5 480V coils. Parallel connect the 5 coils in each 2400V coil. Reconnect the 480V coils Delta. Transformer is good for 1000kva.

I suspect it is the same for the motors as well in that the volts per turn has to be the same

ice
Click to expand...

I'm not a transformer builder or a motor builder. I do know that if you had 1000 amp 480 volt load, it would draw about 115 amps to do same work at 4160.

A conductor used to wind the motor or transformer would not need to nearly as large for the higher voltage. I understand there will be a different number of turns, but I kind of have the feeling there still will be less copper in the higher voltage winding than there will be in the lower voltage winding. Happens that way all the time with other parallel circuit conductors.

Your mentioning of parallel connecting shouldn't matter. Each 480 volt coil still operates with same voltage and current across it when it is operated at full load regardless if the output is 480 volts or 4160. These coils would be made from smaller conductors than they would be if the winding were a single continuous conductor designed to carry all 1000 amps.
 
kwired said:
..., but I kind of have the feeling there still will be less copper in the higher voltage winding than there will be in the lower voltage winding. Happens that way all the time with other parallel circuit conductors. ...
Click to expand...

View attachment coils.pdf
Transformer design is out of my areas of expertice. However, I have a feeling transformers are designed to minimize costs, including copper. Happens all the time with most other commodity equipment.

Then again, when I get those feelings - its usually just indegestion. You are likely luckier than I am in that respect.

ice
 
iceworm said:
View attachment 7803
Transformer design is out of my areas of expertice. However, I have a feeling transformers are designed to minimize costs, including copper. Happens all the time with most other commodity equipment.

Then again, when I get those feelings - its usually just indegestion. You are likely luckier than I am in that respect.

ice
Click to expand...

Ok, I see where you are coming from. Chances are to save costs, they may build such a transformer the way you have drawn the schematic, so it will use less copper, and likely would do the same with a motor.

I generally only deal with 480 volts max and usually no larger than 250 Hp for motors. Seldom run into multi voltage motors in anything over 60 Hp. Dual voltage is more obvious how/why they would use the same amount of copper, but a single voltage unit, you don't pay as much attention to that kind of detail unless you are a person who builds/rebuilds those kinds of items.
 
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