P.U. impedance for a circuit

[mull982]

Lets say I have a circuit with a motor and from the motors given voltage and full load current we calculate a base impedance of 100ohms. Lets also say that at the motors rated kVA the manufacturer says that the motor has a p.u. impedance of 6%.

If the motor only contains 6% of the base impedance what in the circuit accounts for the other 94% impedance. I am somehow confusing myself.

 

[charlie b]

Short answer: The 6% does not mean 6% of the total impedance of the system, but rather 6% of the value of “base impedance.” Longer answer follows.

Yea, I am afraid you are missing the fundamental concepts here. There are formulas that relate a single components voltage, current, apparent power, and resistance to each other. Given any two of these values, you can calculate the other two. Similarly, within the per unit system, there are the concepts of base voltage, base current, base power, and base resistance. Here again, given any two base values, you can calculate the other two base values.

If the “base voltage” is assigned to be 480 volts, and if a given component has a voltage across it of 10%, or equivalently 0.1 per unit voltage, then the voltage across that item is 48 volts. So if the manufacturer says that an item has an impedance of 6%, then in order to convert that to ohms you need to know the base impedance.

It is common practice to assign base values to voltage and VA. Starting, for example, at a utility transformer that feeds a building, with a rated value of 500 KVA and a rated secondary voltage of 480, it would be common practice to assign the base voltage as 480 volts and the base apparent power as 500 KVA. Thus the transformer would have values of 1.0 per unit voltage and 1.0 per unit apparent power.

Back to your motor. You know the motor’s rated voltage and KVA, so we would presume the manufacturer would have used those values as the base values. You need to calculate the base impedance. My text book is not immediately available, so I can’t tell you the formula right now. Perhaps someone else can give you that. Once you calculate the base impedance, multiply that value to 6% (i.e., by 0.06), and you will have the impedance in ohms.

 

[markstg]

Base Impedance = (Base KV)^2/Base MVA, MVA 3 phase, KV Line to Line.

 

[jim dungar]

So, given your OP of Zmotor = Zbase*6%.

'PU' and '%' are similar, the difference is where the decimal point goes. 1PU=100%, .0575PU=5.75%

 

[mull982]

So, given your OP of Zmotor = Zbase*6%.

'PU' and '%' are similar, the difference is where the decimal point goes. 1PU=100%, .0575PU=5.75%

Yes I understand this.

But if the motor is only 6% of the base impedance what makes up the remaining impedance in the system or the other 94%.

In the case of a motor isn't it the total system impedance that makes up the full load current. If the motor is only 6% of this then what accounts for the rest?

For instance if I had (10) 1 ohm resistors in series I'd have a base impedance of 10 ohms. Each resistor would respectively have a p.u. impdeance of 10% and the other 90% of the system impedance would be make up by the other (9) resistors.

 

[markstg]

Yes I understand this.

But if the motor is only 6% of the base impedance what makes up the remaining impedance in the system or the other 94%.

In the case of a motor isn't it the total system impedance that makes up the full load current. If the motor is only 6% of this then what accounts for the rest?

For instance if I had (10) 1 ohm resistors in series I'd have a base impedance of 10 ohms. Each resistor would respectively have a p.u. impdeance of 10% and the other 90% of the system impedance would be make up by the other (9) resistors.

The Base Impedance Value of the System is NOT a total impedance value of a System. A piece of equipment that has 6% impedance does not imply there is 94% impedance someplace else.

Your 10 resistor example is not correct.

 

[jim dungar]

It is not PERCENTAGE of a total. It is a PERCENT expression of the ratio of two values (actual value / base value).

Per Unit is simply the ratio of an actual value to a base value.
Vpu = V/Vbase
Ipu = I/Ibase
Zpu = Z/Zbase

Therefore
%V is simply Vpu*100
%I is simply Ipu*100
%Z is simply Zpu*100.

 

[skeshesh]

I think the problem here is that OP is assuming that Z of motor is simply the same R = V / (I * 1.732) = V^2 / S which is not the case. This is leading him to believe that if S(base) = rated kVA and V(base) = rated V then the result of Z(base) would be the total impedance of the motor which is not correct. Mull982, I've seen you post a bunch of times and you have a good grip on electrical concepts, I'm sure if you google around and take a look at short hand calculations for motor impedance you'll realize where the error was made.

 

[mivey]

Mull,

Think of it like weight. Let your weight be the base with a pu value of 1. Your child might weight 0.5 pu. Your wife might weigh 0.8 pu. All combined, the weight is 2.3 pu. That does not mean you have exceeded 100%. To get the percentage weights, you must convert everyone's pu weight into lbs (kg for Bez).

If we make the combined weight the new base, and derive new pu values for everybody, THEN the pu values also equal the weight percentage of the group. A series circuit that has the total impedance as a base would work the same way because each pu value would also represent the percentage of the total impedance.

The base is just a reference value.

 

[mull982]

I think the problem here is that OP is assuming that Z of motor is simply the same R = V / (I * 1.732) = V^2 / S which is not the case. This is leading him to believe that if S(base) = rated kVA and V(base) = rated V then the result of Z(base) would be the total impedance of the motor which is not correct. Mull982, I've seen you post a bunch of times and you have a good grip on electrical concepts, I'm sure if you google around and take a look at short hand calculations for motor impedance you'll realize where the error was made.

This is starting to make sense now. So for the base impedance of a device what is it that makes up the base impedance? Is it the internal impedance of the device?

For instance with a motor is the base impedance made up for combining all of the resistances and reactances in the motor equivelent circuit? In other words if I took all the series and parallel combinations that make up the motor equivelent circuit model and combine these into one impedance is this then considered the base impedance? Or is this the P.U. impedance?

If I had a circuit where I wanted to install a reactor that was 50% of the circuits impedance then would I simply find the base impedance of the circuit and multiply it by .50 to determine what value of impedance the reactor should have to give half of the circuits impedance?

 

[jim dungar]

The "base" is simply a reference value for the entire electrical system. We usually derive the base Z, using Ohm's law, from the system's base V and the base VA. We almost never use a single device's values as the base for other devices.

Take your Zdevice and divide it by Zbase - the result is: Zpu, if you express it as a decimal number, and %Z if you multiply it by 100 and express it as a whole number.

 

[charlie b]

You are still missing the concept. Forget everything you have every heard about the word “base” and the notion of “percentage,” and let’s start over.

Suppose in my restaurant a “normal helping” of carrots is eight pieces. Why eight? BECAUSE I SAID SO! I am the restaurant owner and I get to decide what a normal helping is. Now suppose a person wants a half helping of carrots. How many do they get? Suppose someone wants a quarter helping. How many do they get? Suppose someone wants a double helping. How many do they get?

Now let’s change the language.

Suppose in my restaurant a “base serving” of carrots is eight pieces. Why eight? BECAUSE I SAID SO! Now suppose a person wants 50% of the base serving of carrots. How many do they get? Suppose someone wants 25% of the base serving. How many do they get? Suppose someone wants 200% of the base serving. How many do they get?

Do you see how the notion of "base value" is coming onto play here?

Now finally let us suppose that in my electrical shop the base impedance is 8 ohms. Why eight? BECAUSE I SAID SO! Now suppose a person wants a resistor with a 50% per unit impedance. What is the value of that resistor, in units of ohms? Now suppose a person wants a resistor with a 25% per unit impedance. What is the value of that resistor, in units of ohms? Now suppose a person wants a resistor with a 200% per unit impedance. What is the value of that resistor, in units of ohms?

We go through a process of selecting base values, then compare the actual value to the base value. There are four key parameters, as I mentioned earlier: voltage, current, apparent power, and impedance. Since there are formulas that relate the four parameters, we cannot select any base values we want for all four. We can use the “because I said so” as the reason for picking two base values, but then the other two base values have to be calculated on the basis of those choices. Tell me what you pick as the base voltage and the base KVA, and I will tell you the base current and the base impedance. Then tell me the per unit value of an item (e.g., the p.u. impedance of a motor), and I will use the base value of impedance to tell you the value of that impedance in units of ohms.

 

[jim dungar]

Part of the issue is determining what to use as the base values.

If you are dealing with a single device's data, like a motor information sheet, it is typical to calculate the motor's impedance using its rated voltage and kva just as you did in your OP. In that case you got a reference value of Zbase=100 ohms. You went on to say that the motor had %Z=6 (notice this is not the same as Z=6%). So to determine the motor's impedance in actual ohms, the formula is (%Z/100)*Zbase = (6/100)*100 = 6ohms.

In your next example you wanted to determine the size of a 50% reactance. You noted that you needed to include the impedance of the entire circuit. Using the above methodology you can now add the motor's internal impedance to that of the conductors to determine the circuit impedance.

The key is to decide what your are going to use as your base, or reference. It is not uncommon to use the actual service entrance values of V and KVA to derive bases values, although in the 'power world' of utilities and such we have typical bases that we use (i.e. 100MVA).

 

[mull982]

Part of the issue is determining what to use as the base values.

If you are dealing with a single device's data, like a motor information sheet, it is typical to calculate the motor's impedance using its rated voltage and kva just as you did in your OP. In that case you got a reference value of Zbase=100 ohms. You went on to say that the motor had %Z=6 (notice this is not the same as Z=6%). So to determine the motor's impedance in actual ohms, the formula is (%Z/100)*Zbase = (6/100)*100 = 6ohms.

O.K. so if I was measuring this circuit at rated conditions and was measuring both the resistance and reactance of the circuit would I measure the 6ohms or the 100ohms. I would think that you would measure the 6ohms since this is the ACTUAL impedance. If that is the case then the BASE impedance really has no relevance to the impedance of the circuit other then a reference as others have mentioned. This reference for the sake of consistancy is usually rated V/I or V^2/VA. Is this correct?

In your next example you wanted to determine the size of a 50% reactance. You noted that you needed to include the impedance of the entire circuit. Using the above methodology you can now add the motor's internal impedance to that of the conductors to determine the circuit impedance.

O.k. so with this example lets say I had a 100kVA single phase transformer with 480V primary and we are looking at the primary circuit. Then to find the base impedance you can take the 480V/208A (rated primary current) and arrive at a base impedance of 2.3ohms. Correct?

Now lets say that we want a reactor included in the circuit that is 50% of the circuit impedance (similar to 5% reactor used for choke on input to VFD). So this would mean the the impedance of the reactor would have a value of 2.3ohms *.50 = 1.15ohms. Is this correct.

Now as far as voltage drop in this circuit would it be true that half the voltage would drop across the reactor since it is 50% of the circuit overall impedance? Would the rest drop across the transformer even though it only had a rated p.u. impedance of 6%? Can someone clarify this part of it.

Thanks for all the help.

 

[jim dungar]

... the BASE impedance really has no relevance to the impedance of the circuit other then a reference ...

This is correct.

This reference for the sake of consistancy is usually rated V/I or V^2/VA. Is this correct?

V^2/VA is the preferred method.

...so with this example lets say I had a 100kVA single phase transformer with 480V primary and we are looking at the primary circuit. Then to find the base impedance you can take the 480V/208A (rated primary current) and arrive at a base impedance of 2.3ohms. Correct?

You are heading for trouble by using the amps (just take it on good faith), it is better to stick with the known values of 480V and 100kVA, Zbase =2.30 ohms. We do not know what the actual impedance is yet.

Now lets say that we want a reactor included in the circuit that is 50% of the circuit impedance (similar to 5% reactor used for choke on input to VFD). So this would mean the the impedance of the reactor would have a value of 2.3 ohms *.50 = 1.5ohms. Is this correct.

No.
2.3 is the value of the supply. You still need to add the impedance of the conductors and the motor, then you can size the reactor.

Now as far as voltage drop in this circuit would it be true that half the voltage would drop across the reactor since it is 50% of the circuit overall impedance? Would the rest drop across the transformer even though it only had a rated p.u. impedance of 6%? Can someone clarify this part of it.

You are getting caught up in percentages again.
You should say its impedance is 6% or .06pu. Note: pu is a dimensionless number, it is the base value that is ohms, volts, kva, etc.

Say you have a transformer with a nameplate of 480V, 100KVA and %Z=6.
We have already calculated Zbase = 2.3, so Zpu = %Z/100 = 6/100 = .06 and therefore Zactual = Zbase * Zpu = 2.3 * .06 = .138 ohms.

The transformer is a source, therefore it is a voltage rise, not a voltage drop. If 50% of the voltage drops across the reactor, the other 50% is dropped across the conductors+any other loads.

 

[dkarst]

Jim provided a good answer but I just wanted to point out something that may be contributing to the confusion. Once the system VA and voltage is determined at for example the primary of the transformer above, the base impedance = 2.3 ohms. (I think Jim has slight typo of Zbase =2.0 ohms).

It is very important to understand that base impedance is 2.3 ohms regardless of the transformer's %Z. If that 6% transformer fails and I replace it with a new transformer (same VA and voltage) BUT with a %Z = 4 or 0.04 pu, THE BASE IMPEDANCE AT THAT POINT IS STILL 2.3 OHMS. Yes the regulation and fault current capability will change but not the base values.

 

[mull982]

2.3 is the value of the supply. You still need to add the impedance of the conductors and the motor, then you can size the reactor.

You are getting caught up in percentages again.
You should say its impedance is 6% or .06pu. Note: pu is a dimensionless number, it is the base value that is ohms, volts, kva, etc.

Say you have a transformer with a nameplate of 480V, 100KVA and %Z=6.
We have already calculated Zbase = 2.3, so Zpu = %Z/100 = 6/100 = .06 and therefore Zactual = Zbase * Zpu = 2.3 * .06 = .138 ohms.

The transformer is a source, therefore it is a voltage rise, not a voltage drop. If 50% of the voltage drops across the reactor, the other 50% is dropped across the conductors+any other loads.

In my example I was refering to a circuit that is feeding the primary of a 100kVA transformer as opposed to looking at the transformer secondary which it appears you did.

Can you show me the similar results you would get when looking at the primary circuit of this transformer?

Thanks for the help.

 

[jim dungar]

In my example I was refering to a circuit that is feeding the primary of a 100kVA transformer as opposed to looking at the transformer secondary which it appears you did.

Can you show me the similar results you would get when looking at the primary circuit of this transformer?

Thanks for the help.

In the world of Per Units there is no difference in how a transfromer is handled on its primary versus its secondary. Once defined, Base values are Base values.

So the anaylsis of my transformer remains the same, its impedance is .138 ohms on the 480V side. This is the value you would use to size your reactor.

 

[mull982]

So the anaylsis of my transformer remains the same, its impedance is .138 ohms on the 480V side. This is the value you would use to size your reactor.

O.k. so to size my reactor on the primary of the transformer I would take .138ohms * .50 = .069ohms. Is this correct?

So would 50% of the voltage then drop across the reactor with the other 50% dropping across the primary of the transformer?

Would the .138ohms be what the impedance I would measure if I measured the impedance of the transformer primary under rated conditions?

 

[jim dungar]

O.k. so to size my reactor on the primary of the transformer I would take .138ohms * .50 = .069ohms. Is this correct?

So would 50% of the voltage then drop across the reactor with the other 50% dropping across the primary of the transformer?

Would the .138ohms be what the impedance I would measure if I measured the impedance of the transformer primary under rated conditions?

Think of two resistors in series.
To have 50% voltage drop across each resistor you need the resistors to be equal.
Sizing the reactor to 50% of the transformer's impedance will give you 1/3 of the voltage across the reactor and 2/3 across the transformer.