Power savings calc

[hilltop]

What is the calulation used when trying to figure out cost savings i.e. switch from 120V lighting to 277v lighting, or replacing a 230V motor with a 480V motor.

Thanks!

 

[LarryFine]

As far as 'power' goes, there's absolutely no difference. Power is the same at any voltage.

However, the higher voltage will have less voltage drop, because the current is lower.

However, if the higher current is accomodated by a larger conductor, we're back to no difference.

All other things being equal, I generally opt for the higher voltage. Insulation is cheaper than conductor.

For lighting, for example, the higher voltage allows for more fixtures for a given circuit ampacity.

 

[billyzee]

Lighting Energy Savings

Lighting Energy Savings

I agree with Larry. There is no cost savings associated with increasing lighting voltage.

These days the savings associated with lighting have to do with getting rid of T12 Fluorescents and Incandescents. Replacing them with the greener eco friendly T8/T5 and soon-to-come induction lamps is all the rage.

 

[mull982]

At the lower voltages the increased current will cause a greater (I^2)(R) power loss in the cables. This loss may be minimal depending on the size and arrangement of the system

 

[mull982]

My statement made me curious if due to increased losses if the system then had to supply additional power.

For instance lets say we had a 100W light at 100V. The current in the line would then be 1A. But lets say due to the impedance in the line we had 10W loss through I^2R losses. So does this mean then that the light only recieves 90W at its input, or does it increase its current to draw the rated 100W and thus cause the system to supply more than 100W to supply the losses and the light?

 

[K8MHZ]

My statement made me curious if due to increased losses if the system then had to supply additional power.

For instance lets say we had a 100W light at 100V. The current in the line would then be 1A. But lets say due to the impedance in the line we had 10W loss through I^2R losses. So does this mean then that the light only recieves 90W at its input, or does it increase its current to draw the rated 100W and thus cause the system to supply more than 100W to supply the losses and the light?

No,as the voltage drops, so does the power (watts) as the load (ohms) does not change.

Also, 10 watts loss in the conductor would equate to a pretty decent voltage drop.

 

[markstg]

My statement made me curious if due to increased losses if the system then had to supply additional power.

For instance lets say we had a 100W light at 100V. The current in the line would then be 1A. But lets say due to the impedance in the line we had 10W loss through I^2R losses. So does this mean then that the light only recieves 90W at its input, or does it increase its current to draw the rated 100W and thus cause the system to supply more than 100W to supply the losses and the light?

Yes the system will have to provide any losses in addition to the load requirements all things being equal

The light bulb will still draw 100W (for the same voltage at its terminals) and the wire losses of 10W will be drawn, so that the total power delivered is 110W.

 

[LarryFine]

So does this mean then that the light only recieves 90W at its input, or does it increase its current to draw the rated 100W and thus cause the system to supply more than 100W to supply the losses and the light?

The light, as a linear load, would see only the voltage across it, reducing the light output.

 

[LarryFine]

The light bulb will still draw 100W (for the same voltage at its terminals) and the wire losses of 10W will be drawn, so that the total power delivered is 110W.

That would only occur if the supply voltage was increased as compensation. That also means the terminal voltage would rise above nominal if the load was decreased.

That's what's wrong with using a voltage boost to compensate for voltage drop.

 

[K8MHZ]

Yes the system will have to provide any losses in addition to the load requirements all things being equal

The light bulb will still draw 100W (for the same voltage at its terminals) and the wire losses of 10W will be drawn, so that the total power delivered is 110W.

I disagree.

How can adding resistance increase the power?. Please, show me the math. If what you were saying was correct, putting two 100 watt bulbs in series would give 200 watts of power. (Remember, the conductors are in series with the load, not parallel - important difference.)

 

[markstg]

I disagree.

How can adding resistance increase the power?. Please, show me the math. If what you were saying was correct, putting two 100 watt bulbs in series would give 200 watts of power. (Remember, the conductors are in series with the load, not parallel - important difference.)

What I was saying, and what I thought the OP stated, that for 100V accross the lamp (which would put out 100W) and a 10W loss in the line would the source supply 110W or 100W. He didn't state the source voltage which would be 110V for this case.

If what he said, and I misinterpted, was that the source couldn't produce the voltage neccessary for the lamp to see 100V and would see something less than 100V, then yes the lamp will absorb something less than 100W.

I think he was asking more what ya'll answered than what I did.

 

[billyzee]

I2R in the wires

I2R in the wires

K8MHZ said it right.

Not only that,

Messing around with Power losses in conductors is a silly if the end game is associated with energy conservation. If the system is designed properly (i.e. NEC) these loses won't add up to anything significant. And if they do start adding up then you have crazy voltage drops and bigger problems than low-efficiency distribution.

 

[david luchini]

If the system is designed properly (i.e. NEC) these loses won't add up to anything significant.

I agree with this, and with what Larry said in post #2 about current and conductor size balancing out any potential cost savings.

However, I would say there is some potential cost savings from the reduction in transformer losses by shifting the lighting and motor loads to the 480/277V side.

 

[Jraef]

...However, I would say there is some potential cost savings from the reduction in transformer losses by shifting the lighting and motor loads to the 480/277V side.

But who owns the transformer? For the most part, the transformer losses are absorbed by the PoCo because you are metered off of the secondary side. That's partially why they assess PF penalties on large users. They have to size the transformer to supply all of the kVA, but they can only bill you for the kW you use from it and losses in the transformer are based on the kVA, so they eat them too.

 

[Jraef]

Oh wait, I see what you mean. If you have 480/277 and you have a transformer to give you 120V, YOU own those transformer losses. Point taken.

Still, is the ROI there to justify re-wiring everything and replacing all the ballasts? If it were a new installation, the point is more valid because the initial cost is probably equal. But the OP was about retrofitting.

...i.e. switch from 120V lighting to 277v lighting, or replacing a 230V motor with a 480V motor

I can't see even the transformer losses being worth it.

 

[david luchini]

Oh wait, I see what you mean. If you have 480/277 and you have a transformer to give you 120V, YOU own those transformer losses. Point taken.

Still, is the ROI there to justify re-wiring everything and replacing all the ballasts? If it were a new installation, the point is more valid because the initial cost is probably equal. But the OP was about retrofitting.

I can't see even the transformer losses being worth it.

I thought of the cost of rewiring, and I also thought of whether the question is 'is it better to have a 480V or 208V service.' I'm not sure what the OPs meaning was, but the discussion seemed to focus on energy cost savings. I took it to be a comparison of energy cost for an initial installation, not for a retrofit.:-?

 

[gar]

101028-1855 EDT

A slightly different perspective.

A single phase 120-0-120 service. A DeWalt radial arm saw. Panel to saw is 100 ft. The present wiring is #12 with 120 V to this outlet. With a reasonable size blade on the saw the performance is poor. Multi-seconds of start-up time. A few short on-off cycles and the 20 AMP QO breaker trips.

Simply reconnect the supply as 240 V and change the connections in the motor for 240 V. Performance greatly improved and a small power saving.

Have I done it? No. Not a sufficient problem. The saw only sees limited use.

.

 

[rattus]

Fwiw:

Fwiw:

The rated power of a lamp is dissipated only when rated voltage is applied. Current through the resistance of the distribution system causes the applied voltage to drop. The lamp is not a constant power device.