Can you reduce KVAR without affecting KW?

[S_Dot]

Hello MH Forums,

I just have a simple question, imagine this scenario.


I have a purely resistive load, with hugh inductive losses along the way, the load is transformer coupled. The inductive losses were lowered by reducing the seperation between conductors, and bringing the transformer closer to the load.

This in fact reduces the KVARs involved, but I want to know if this does anything with the KW? not directly, but indirectly, KW is related to current. Isnt the inductance also related to current?

Let me paint two scenarios, 1) I dont make any changes to the inductive losses, and I compare my utility bill for the year (Indutrial bill, but only charged in KW)
2) I reduce inductive losses, and look at my utility bill in KW (Do I expect to see the same bill?)


Thank you. I need to understand this.

 

[Besoeker]

Hello MH Forums,

I just have a simple question, imagine this scenario.


I have a purely resistive load, with hugh inductive losses along the way, the load is transformer coupled. The inductive losses were lowered by reducing the seperation between conductors, and bringing the transformer closer to the load.

This in fact reduces the KVARs involved, but I want to know if this does anything with the KW? not directly, but indirectly, KW is related to current. Isnt the inductance also related to current?

Let me paint two scenarios, 1) I dont make any changes to the inductive losses, and I compare my utility bill for the year (Indutrial bill, but only charged in KW)
2) I reduce inductive losses, and look at my utility bill in KW (Do I expect to see the same bill?)


Thank you. I need to understand this.

High frequency application?

Can you define what you mean by "huge inductive losses"?
The fact that you reduced these losses by reducing separation between conductors suggest that reduced inductance only. The same conductors would have the same resistance thus the same i²R loss. The reduced inductance won't change that.

 

[charlie b]

I have a purely resistive load, with hugh inductive losses along the way, the load is transformer coupled. The inductive losses were lowered by reducing the seperation between conductors, and bringing the transformer closer to the load.

The inductive reactance of the wires that serve the load are not going to give you "huge inductive losses." So the setup of the problem statement is flawed. But if there are inductive losses along the circuit conductors, and if you reduce those losses, then there will be a slightly higher voltage level at the load. This will cause the purely resistive load to draw slightly more current. The product of the slightly higher voltage and the slightly higher current will result in a slightly higher value of KW. It would be a "second order effect," meaning that it is likely to be too small to notice.

 

[Besoeker]

The inductive reactance of the wires that serve the load are not going to give you "huge inductive losses."

Unless by the the OP means voltage drop and that could matter if it is a high frequency application.
That physically changing the proximity of the conductors made a measurable difference might be a clue.

To the OP, you need to provide more specific information if you want constructive responses.

 

[S_Dot]

High frequency application?

Can you define what you mean by "huge inductive losses"?
The fact that you reduced these losses by reducing separation between conductors suggest that reduced inductance only. The same conductors would have the same resistance thus the same i²R loss. The reduced inductance won't change that.

The inductive reactance of the wires that serve the load are not going to give you "huge inductive losses." So the setup of the problem statement is flawed. But if there are inductive losses along the circuit conductors, and if you reduce those losses, then there will be a slightly higher voltage level at the load. This will cause the purely resistive load to draw slightly more current. The product of the slightly higher voltage and the slightly higher current will result in a slightly higher value of KW. It would be a "second order effect," meaning that it is likely to be too small to notice.

This is an industrial application, and yes the inductive losses caused by seperation of conductors, in this case buss bar and high temperature cables. When I say losses, they translate to voltage drops because V = IZ. The overall impedance goes through the roof because of the inductance (magnetic field is produced by supply and return buss bars).

The result of reducing these inductive losses yield about 15-20% change in KVA required for the load. As I mentioned, the KW resistive LOAD does not change with the configuration. I just want to know if making changes to the KVAR losses will impact the utility readings --- since they only read KW.

Unless by the the OP means voltage drop and that could matter if it is a high frequency application.
That physically changing the proximity of the conductors made a measurable difference might be a clue.

To the OP, you need to provide more specific information if you want constructive responses.

Frequency is 50HZ/60HZ, but these are high power applications, so a change in impedance will result in a large KVA difference.

EDIT

From, what I understand the utility only charges on KW, but what I don't understand is that if they still need to deliver the Apparent Power needed, will reducing the Reactive Power (reducing inductance of conductors), affect my Real Power utility bill.

Again, this is industrial application, whatever numbers you are thinking multiply them by 10^6

 

[GoldDigger]

[/B][/I]
From, what I understand the utility only charges on KW, but what I don't understand is that if they still need to deliver the Apparent Power needed, will reducing the Reactive Power (reducing inductance of conductors), affect my Real Power utility bill.

Again, this is industrial application, whatever numbers you are thinking multiply them by 10^6

For perhaps the umpteenth time since the founding of the board (but who is counting) the utility must deliver the apparent power needed, and it must do that by delivering more current (at a less than ideal phase angle) than it would have to deliver to a purely resistive load of the same true wattage.
This apparent power costs the utility something to deliver, mostly in the form of higher I²R losses in their distribution lines and generating equipment. They try to minimize their infrastructure cost by installing power factor correction capacitor banks of their own.
But they (or the regulators) choose not to bill you based on apparent power, instead charging only by the kWh for real energy delivered and perhaps imposing a flat rate penalty for poor PF to commercial/industrial customers only.
There is a calculable increase in the amount of real power you consume based on the higher I²R losses in your own customer wiring because of the higher current. But that is a second order effect and will generally not be noticeable against the PF penalty if there is one.

I do understand your factor of 10^6, but all that means is that instead of looking at a difference of $.01 out of $100 you are looking at $10,000 out of $100,000,000. The absolute number gets bigger but the percentage (and the economies of trying to recover it) do not change.

 

[S_Dot]

For perhaps the umpteenth time since the founding of the board (but who is counting) the utility must deliver the apparent power needed, and it must do that by delivering more current (at a less than ideal phase angle) than it would have to deliver to a purely resistive load of the same true wattage.
This apparent power costs the utility something to deliver, mostly in the form of higher I²R losses in their distribution lines and generating equipment. They try to minimize their infrastructure cost by installing power factor correction capacitor banks of their own.
But they (or the regulators) choose not to bill you based on apparent power, instead charging only by the kWh for real energy delivered and perhaps imposing a flat rate penalty for poor PF to commercial/industrial customers only.
There is a calculable increase in the amount of real power you consume based on the higher I²R losses in your own customer wiring because of the higher current. But that is a second order effect and will generally not be noticeable against the PF penalty if there is one.

I do understand your factor of 10^6, but all that means is that instead of looking at a difference of $.01 out of $100 you are looking at $10,000 out of $100,000,000. The absolute number gets bigger but the percentage (and the economies of trying to recover it) do not change.

Hello Golddigger,

thanks for the response. This is what I would like you expound upon. So,if I have a system comprised of 30% losses (KVAR) and 70% (KW) true power used, compared to a system with 5% losses (KVAR) and 95% true power used (KW). Assuming utility does not charge any power factor penalty, what you are telling me is my utility bill at the end of the year will be the same for these two systems?

Even though my lack of efficiency in the first system is affecting how much Apparent Power is delivered, and is also forcing upsizing of utility equipment, eg wire sizing and so on.

Is this correct?

 

[gar]

140827-1100 EDT

S_Dot:

You list yourself as a power engineer. I would assume you could answer your own question, but I will try to clarify the question.

Assumptions:

The power company meter is an ideal meter. This means it only measures instantaneous real power (the resistive component of the load), and integrates the instantaneous power with respect to time to provide a kWH reading over some time period.

A fixed resistive load is place on the output of the power/energy meter with a connecting wire length of 0.

The voltage at the power/energy meter and the load remains at the same value.

An ideal inductor is available to put in parallel with the resistive load at the meter. Again wire length is 0.


The experiment:

Measure power/energy under exactly the same conditions with and without the inductor connected in parallel with the resistor. In both experiments the power or energy reading will be exactly the same.

.

 

[GoldDigger]

Hello Golddigger,

thanks for the response. This is what I would like you expound upon. So,if I have a system comprised of 30% losses (KVAR) and 70% (KW) true power used, compared to a system with 5% losses (KVAR) and 95% true power used (KW). Assuming utility does not charge any power factor penalty, what you are telling me is my utility bill at the end of the year will be the same for these two systems?

Even though my lack of efficiency in the first system is affecting how much Apparent Power is delivered, and is also forcing upsizing of utility equipment, eg wire sizing and so on.

Is this correct?

First, I would never use the word "losses" for reactive power, since it is energy moving back and forth between POCO's generators and your loads without any energy loss in the ideal situation.
And as for the meter reading, yes it would be identical in that same ideal situation.

 

[Jraef]

First, I would never use the word "losses" for reactive power, since it is energy moving back and forth between POCO's generators and your loads without any energy loss in the ideal situation.
And as for the meter reading, yes it would be identical in that same ideal situation.

Yes, your concept of "losses" is what's flawed here. Inductance does not EQUAL losses, but it can affect losses, by virtue of changing the copper (I²t) losses in the entire circuit.

But back to the voltage drop issue mentioned earlier; if you lower the VD, more voltage gets to the heaters, they draw more kW. But BECAUSE it is a heating system, that just means they raise the temperature faster, which usually results in them having to be on for less time, and higher kW for less time = the same kWH, which is really what you pay for. Bottom line, usually no net change in the energy of the circuit worth writing home about. The only tangible difference will be in the slight difference in circuit energy that was lost in the air around the conductors, and then if it was an air conditioned space, that reduces the heat gain in the building and thereby the HVAC loading anyway so again, no realistic changes.

 

[Besoeker]

Yes, your concept of "losses" is what's flawed here. Inductance does not EQUAL losses, but it can affect losses, by virtue of changing the copper (I²t) losses in the entire circuit.

But back to the voltage drop issue mentioned earlier; if you lower the VD, more voltage gets to the heaters, they draw more kW. But BECAUSE it is a heating system, that just means they raise the temperature faster, which usually results in them having to be on for less time, and higher kW for less time = the same kWH, which is really what you pay for. Bottom line, usually no net change in the energy of the circuit worth writing home about. The only tangible difference will be in the slight difference in circuit energy that was lost in the air around the conductors, and then if it was an air conditioned space, that reduces the heat gain in the building and thereby the HVAC loading anyway so again, no realistic changes.

No disagreement with your points.
But, as yet, we don't actually know what the load is other than that it is resistive and, evidently, very large.
If the OP could provide the actual application and rating, it might be easier to come up with constructive answers.

 

[Jraef]

No disagreement with your points.
But, as yet, we don't actually know what the load is other than that it is resistive and, evidently, very large.
If the OP could provide the actual application and rating, it might be easier to come up with constructive answers.

True. I'd hazard a guess that it is a process heating system that is using a 3 phase SCR phase-angle controller, so they are using a transformer down stream in order to get high current at the heating elements but allow the use of a smaller SCR controller ahead of it and keep the cost down. Very common way it is done in the industrial heating world, i.e. annealing ovens, sintering systems, electric autoclaves, sometimes even industrial baking systems. I once helped build large conveyorized ovens that made McDonalds' buns, we did that on those..

 

[Besoeker]

True. I'd hazard a guess that it is a process heating system that is using a 3 phase SCR phase-angle controller, so they are using a transformer down stream in order to get high current at the heating elements but allow the use of a smaller SCR controller ahead of it and keep the cost down. Very common way it is done in the industrial heating world, i.e. annealing ovens, sintering systems, electric autoclaves, sometimes even industrial baking systems. I once helped build large conveyorized ovens that made McDonalds' buns, we did that on those..

I wonder about a couple of points from the OP.

"magnetic field is produced by supply and return buss bars

Again, this is industrial application, whatever numbers you are thinking multiply them by 10^6"

"Supply and return" doesn't immediately suggest three phase.
Multiply what by 10^6?

One kW? That's probably at the low end for three phase. Multiply that by 10^6 and you get a thousand MW.
What purely resistive application uses that?

I think the OP needs to be a bit more specific about actual ratings and applications if he/she is requesting technical assistance?
There is a wealth of experience and willing helpers. But if you can't quantify the problem, you can't offer solutions.

On solving engineering problems, I got this piece of advice from one of my university lecturers.

1. Understand the problem.
2. Formulate it mathematically.
3. Solve the mathematical problem.
4. Interpret the solution.

I don't think we've yet got to stage one here.
Perhaps the OP can be more forthcoming.