Eng-Tips is the largest engineering community on the Internet
Intelligent Work Forums for Engineering Professionals
-
Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!
VOLTAGE DROP AT XFMR 4
- Thread starter anbm
- Start date
-
2
- #2
You're going to need to know the loading (either kVA or amps), the power factor and the transformer impedance.
Volt Drop (in volts) is approximately equal to I (R pf + X qf)
I is current in amps
R & X are transformer impedance components in ohms
pf = power factor
qf = sqrt(1-pf*pf)
Since you want secondary voltage, I, R & X are secondary quantities.
Volt Drop (in volts) is approximately equal to I (R pf + X qf)
I is current in amps
R & X are transformer impedance components in ohms
pf = power factor
qf = sqrt(1-pf*pf)
Since you want secondary voltage, I, R & X are secondary quantities.
For homework, use 479.47 V x (208 V/ 480 V)
For an exact solution, what is the transformer regulation and X/R ratio. What is the impedance and X/R ratio of the feeders, and what is the magnitude and power factor of the load.
In the real world, use the transformer regulation, feeder impedance and loading. Close enough for most work. Be aware that when the feeders are appreciably inductive and/or the load has a poor power factor, the errors will increase.
Bill
--------------------
"Why not the best?"
Jimmy Carter
For an exact solution, what is the transformer regulation and X/R ratio. What is the impedance and X/R ratio of the feeders, and what is the magnitude and power factor of the load.
In the real world, use the transformer regulation, feeder impedance and loading. Close enough for most work. Be aware that when the feeders are appreciably inductive and/or the load has a poor power factor, the errors will increase.
Bill
--------------------
"Why not the best?"
Jimmy Carter
-
1
- #4
I second what magoo2 says but would add that the secondary voltage drop would be from the primary voltage referred to the secondary. 479.47·(208/480) = 207.8
Although your statement of actual primary volts to two decimals is very suspect. Your basic meter will have a ±1% accuracy (4.8 volts). Also, every time I've tried to measure a voltage, it varies second by second by at least a volt. I'd just use 479 volts.
Although your statement of actual primary volts to two decimals is very suspect. Your basic meter will have a ±1% accuracy (4.8 volts). Also, every time I've tried to measure a voltage, it varies second by second by at least a volt. I'd just use 479 volts.
magoo2 and jgrist; The impedance of a transformer determines the current under short circuit conditions.
Under loading, the regulation of the transformer, (determined primarily but not completely by the resistance of the transformer), is the predominant factor.
Calculating the voltage drop based on the impedance of the transformer will give show greater voltage drop than calculations based on the transformer regulation.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Under loading, the regulation of the transformer, (determined primarily but not completely by the resistance of the transformer), is the predominant factor.
Calculating the voltage drop based on the impedance of the transformer will give show greater voltage drop than calculations based on the transformer regulation.
Bill
--------------------
"Why not the best?"
Jimmy Carter
waross is right. Regulation is what defines the % change in secondary voltage from no-load to full load and it will vary with the load pf.
Percentage impedance applies to short circuit state.
The regulation detail is not included in the trafo name-plate unlike the percentage impedance.
Percentage impedance applies to short circuit state.
The regulation detail is not included in the trafo name-plate unlike the percentage impedance.
Sorry, guys. Although the % impedance is obtained by a short circuit test, it is this leakage impedance what determines the secondary voltage under load conditions. It is what determines the regulation.
With 480 V on the primary, you'd get 208 V on the secondary for an unloaded transformer. It's a simple ratio as waross points out, so if it's slightly less than 480 V on the primary you get slightly less than 208 V on the secondary. With load on the secondary, the voltage drop will vary in accordance with the approximate but generally used equation that I gave.
With 480 V on the primary, you'd get 208 V on the secondary for an unloaded transformer. It's a simple ratio as waross points out, so if it's slightly less than 480 V on the primary you get slightly less than 208 V on the secondary. With load on the secondary, the voltage drop will vary in accordance with the approximate but generally used equation that I gave.
Once quickly.
The current is determined by the impedance of the circuit.
Under short circuit conditions, the impedance of the circuit is the impedance of the transformer and determines the current.
Under normal conditions, the impedance of the circuit is the vector sum of the reactances of both the load and the transformer and the resistances of both the load and the transformer. At normal transformer loading resistance generally predominates and the PU regulation is lower than the PU impedance.
I learned this on Eng-Tips. Thank you to the friends who taught me.
Short circuit, transformer impedance.
Normal load, transformer impedance and load impedance. Resistance often predominates and PU regulation does NOT equal PU impedance.
Bill
--------------------
"Why not the best?"
Jimmy Carter
The current is determined by the impedance of the circuit.
Under short circuit conditions, the impedance of the circuit is the impedance of the transformer and determines the current.
Under normal conditions, the impedance of the circuit is the vector sum of the reactances of both the load and the transformer and the resistances of both the load and the transformer. At normal transformer loading resistance generally predominates and the PU regulation is lower than the PU impedance.
I learned this on Eng-Tips. Thank you to the friends who taught me.
Short circuit, transformer impedance.
Normal load, transformer impedance and load impedance. Resistance often predominates and PU regulation does NOT equal PU impedance.
Bill
--------------------
"Why not the best?"
Jimmy Carter
waross,
Go back to the equation
I (R pf + X qf)
Now replace R & X by %R and %X. Then let load equal 1.0 or 1 per unit.
Under load conditions, power factor dominates, so %R dominates. Since %R is usually less than %Z, the drop or regulation will be less than %Z.
Does this help?
Go back to the equation
I (R pf + X qf)
Now replace R & X by %R and %X. Then let load equal 1.0 or 1 per unit.
Under load conditions, power factor dominates, so %R dominates. Since %R is usually less than %Z, the drop or regulation will be less than %Z.
Does this help?
Back to the OP, you also need to know the transformer %Z and the X/R ratio to calculate the calculate what you want.
If you have the transformer equivalent resistance and the inductance (R and X) that would make things easy as magoo2 replied earlier.
Back in school (that was a long time ago), we test transformers to get these parameters: Open circuit and short circuit (or a back-to-back test if you have another similar transformer- Sumpner test). We plot the results in a circle diagram to predict transformer performance at any load condition.
Also, see thread238-147658.
If you have the transformer equivalent resistance and the inductance (R and X) that would make things easy as magoo2 replied earlier.
Back in school (that was a long time ago), we test transformers to get these parameters: Open circuit and short circuit (or a back-to-back test if you have another similar transformer- Sumpner test). We plot the results in a circle diagram to predict transformer performance at any load condition.
Also, see thread238-147658.
I think that we are saying the same thing in different words. I think that we both agree that at full load the % regulation will be less than % impedance. I misread or misunderstood one of your posts. Sorry.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Bill
--------------------
"Why not the best?"
Jimmy Carter
Vdrop = IZ
Where all are vector quantities
Z = Transformer impedance, commonly stated as Z as a scalar quantity, commonly in %. The mfg. can probably tell you X/R. From those you can calculate X and R for the transformer.
I = load current
Where all are vector quantities
Z = Transformer impedance, commonly stated as Z as a scalar quantity, commonly in %. The mfg. can probably tell you X/R. From those you can calculate X and R for the transformer.
I = load current
Thanks for the vector sketches, jghrist. That is what I was trying to express. I guess I wasn't very good at it. Sory for any confusion I may have caused.
The point is that the OP does not have enough information to determine the voltage drop, and calculating the voltage drop of a transformer under load is more complex than many folk realize.
Bill
--------------------
"Why not the best?"
Jimmy Carter
The point is that the OP does not have enough information to determine the voltage drop, and calculating the voltage drop of a transformer under load is more complex than many folk realize.
Bill
--------------------
"Why not the best?"
Jimmy Carter
- Thread starter
- #16
alehman,
Unfortunately that's wrong.
The vector IZ is used to get the receiving end voltage and it is the magnitude difference between the sending end and receiving end voltage that we're trying to compute.
jghrist tried to point this out earlier.
Unfortunately that's wrong.
The vector IZ is used to get the receiving end voltage and it is the magnitude difference between the sending end and receiving end voltage that we're trying to compute.
jghrist tried to point this out earlier.
- Status
Similar threads
- Locked
- Question
- Locked
- Question
- Locked
- Question