Voltage Drop Calculation Through Transformer

spraymax6

Senior Member
Location
USA
So I was reading over this forum thread: https://forums.mikeholt.com/threads/transformer-voltage-drop.101920/

It basically talks about an inspector requiring someone to do a voltage drop calculation and to include the voltage drop across the stepdown transformer.

My question is when doing a voltage drop calculation where you have, lets say, a 480V bus distribution system, and then a 480-208/120V step down transformer and you are doing the calculations by hand, do you include some sort of voltage drop across the transformer?

From the thread above it was nicely explained that the secondary voltage is going to be less than the nominal due to "voltage regulation" (if you have a leading PF it can actually even be higher) but as far as VD calculations, how does this get captured? By using SKM it shows the VD before the transformer and the VD on the bus after and it is very significant. For my particular transformer/case it goes from 1.4% or so to 3.8% VD by looking at the bus before the transformer and the bus immediately after.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
Retired Electrical Engineer - Power Systems
It is not uncommon to run a hand calculation using an assumed primary side voltage drop, like 3 or 5% in addition to the calculated VD.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
220914-2006 EDT

spraymax6:

One can use various different equivalent circuits for a transformer. But for a simple 60 Hz applications you probably only need to consider the transformer as a series inductor and resistor. The magnitude of this impedance is such that you usually see only a fraction of a percent to several percent voltage drop thru the transformer at transformer full load. Usually you will reflect the total transformer series impedance to the secondary side because this where you are most likely interested in doing circuit calculations.

If you were were considering the transformer load as seen at the primary side, then you would be concerned with determining the total impedance of the transformer and its load as seen looking at the primary of the transformer.

.
 

spraymax6

Senior Member
Location
USA
220914-2006 EDT

spraymax6:

One can use various different equivalent circuits for a transformer. But for a simple 60 Hz applications you probably only need to consider the transformer as a series inductor and resistor. The magnitude of this impedance is such that you usually see only a fraction of a percent to several percent voltage drop thru the transformer at transformer full load. Usually you will reflect the total transformer series impedance to the secondary side because this where you are most likely interested in doing circuit calculations.

If you were were considering the transformer load as seen at the primary side, then you would be concerned with determining the total impedance of the transformer and its load as seen looking at the primary of the transformer.

.
What I am trying to do is make sure that the voltage at load is sufficient. If I assume that the transformer produces a secondary voltage just equivalent to the turns ratio, I will under estimate the VD.

From my SKM example I see that the VD can be pretty significant.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
220915-2355 EDT

spraymax6:

I think you lack an understanding of electrical circuit theory, and the relation to practical circuits.

Do you have constant source voltage at the input to your transformer assuming you have a constant supply voltage? In other words is the source impedance very much lower for your source at the transformer input than the transformer internal impedance. In many cases you can make this assumption, or at least may need to assume this for analysis purposes.

Given this assumption, and are interested in the output voltage from the transformer, then you need to look at how your load current will vary with time.

If you need very good output voltage regulation from 0 to maximum secondary current, then you will need a transformer with a lower internal impedance, than if you only load the transformer with a single constant load.

In any case the transformer will be required to run continuously at whatever average load RMS current you require. You adjust turns ratio to give you whatever output voltage you need. If you need a lower internal impedance you may need to use a physically larger transformer than would be required from a power perspective. Or you may need to use more expensive materials, or other design concepts for the transformer.

.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
In a voltage drop calculation you pick a starting point which you _assume_ is an ideal voltage source, and you pick an end point which is your load current.

Then you calculate the voltage lost through each impedance going back to the load. Add all the voltage drops to get the total at the load. If you want to be more accurate you consider how the reduced voltage at the load changes the load current and redo the calculation.

For most hand voltage drop calculations we only consider the resistance of the circuit conductors from the source to the load.

For long, large feeders we need to take into account the reactive impedance of the wires, and do a voltage drop calculation that includes the load impedance. This calculation is more complex because you are doing vector math; the reactive voltage drop is not in phase with the resistive drop. But it is totally doable by hand.

When the load is a large fraction of the transformer capacity, we need to consider the impedance of the transformer itself.

When we say that a transformer has 5% impedance, that really means a 5% voltage drop between no load and full load, with a very important caveat: the transformer impedance is nearly pure inductance, so this voltage drop is reactive and out of phase with any resistive drop.

So when you want to include the transformer it is simply an additional impedance in the 'complex' (meaning vector, not meaning hard) voltage drop calculation.

Jon
 

spraymax6

Senior Member
Location
USA
220915-2355 EDT

spraymax6:

I think you lack an understanding of electrical circuit theory, and the relation to practical circuits.

Do you have constant source voltage at the input to your transformer assuming you have a constant supply voltage? In other words is the source impedance very much lower for your source at the transformer input than the transformer internal impedance. In many cases you can make this assumption, or at least may need to assume this for analysis purposes.

Given this assumption, and are interested in the output voltage from the transformer, then you need to look at how your load current will vary with time.

If you need very good output voltage regulation from 0 to maximum secondary current, then you will need a transformer with a lower internal impedance, than if you only load the transformer with a single constant load.

In any case the transformer will be required to run continuously at whatever average load RMS current you require. You adjust turns ratio to give you whatever output voltage you need. If you need a lower internal impedance you may need to use a physically larger transformer than would be required from a power perspective. Or you may need to use more expensive materials, or other design concepts for the transformer.

.
wow, not sure you can judge my level of understanding of electrical circuit theory. I look to ask a question in order to get clarification on a topic which has apparently been previously asked on this forum, and because of that I lack understanding of electrical theory?

Any sample voltage drop calculation I have seen has not included translation of the voltage drop through a transformer, I think asking how people typically deal with this is a solid question as I would venture to guess not many people take this into consideration. Take a look at the original thread to help support that notion.

Sample voltage drop calculations always assume a voltage source and give you the formulas depending on the system 3 phase or 1 phase.

In addition the question is geared toward the practical application of all this. Typically you choose a transformer based on your anticipated load. You don't know what that load will be all the time. Maybe the transformer is 25% loaded, maybe 50%, maybe 75%. Playing with taps to combat VD may lead to over voltage at times of light loading.

This is where the question stems, if people don't typically take into consideration transformer voltage drops and load profiles can change significantly so that using taps isn't practical, how aren't more VD issues occurring?

I have typically done VD calcs by hand and just translated the VD from the primary to the secondary, SKM is showing me this is not the best route.
 

spraymax6

Senior Member
Location
USA
In a voltage drop calculation you pick a starting point which you _assume_ is an ideal voltage source, and you pick an end point which is your load current.

Then you calculate the voltage lost through each impedance going back to the load. Add all the voltage drops to get the total at the load. If you want to be more accurate you consider how the reduced voltage at the load changes the load current and redo the calculation.

For most hand voltage drop calculations we only consider the resistance of the circuit conductors from the source to the load.

For long, large feeders we need to take into account the reactive impedance of the wires, and do a voltage drop calculation that includes the load impedance. This calculation is more complex because you are doing vector math; the reactive voltage drop is not in phase with the resistive drop. But it is totally doable by hand.

When the load is a large fraction of the transformer capacity, we need to consider the impedance of the transformer itself.

When we say that a transformer has 5% impedance, that really means a 5% voltage drop between no load and full load, with a very important caveat: the transformer impedance is nearly pure inductance, so this voltage drop is reactive and out of phase with any resistive drop.

So when you want to include the transformer it is simply an additional impedance in the 'complex' (meaning vector, not meaning hard) voltage drop calculation.

Jon

Yes, I assume the utility gives me the ideal voltage, say in this case 480V 3PH. I have a whole 480V distribution system and then a step down transformer for my lighting and receptacle loads. I need to make sure that my 480v distribution system which could be made up distribution to many locations brings me to an appropriate voltage drop to the primary of my 480-208/120V step down transformer and then confirm that the voltage drop across the transformer and the route all the way to the 208V (or 120V) utilization equipment is within reasonable bounds.

I avoid using taps on the transformer as the loading on the transformer can vary greatly.

I typically use the IEEE 141 "Exact" voltage drop formula because Mathcad does the heavy lifting of the numbers crunching so why not (I still consider that "hand calc".

Yes, I can see that the xfmr is an additional impedance in the circuit, that's why the light bulb went off when I saw SKM giving me larger VD numbers. It makes total sense that there should be a VD across that component. Just like we model all those components for short circuit calculations, it should be similar for VD calculations.

Your suggestion about using a larger xfmr to reduce the affect of the xfmr VD is a really good one. I will take that into consideration in the future as I take the VD across transformers a little more seriously in my hand calculations.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
So the 'Exact' voltage drop formula is what you need to consider both resistance and reactance of the supply circuit, and you are already doing that. So the only thing you need to add is the R and X values of the transformer in series with the R and X of the circuit wires so you can run your calculations.

This is one of those things that I do rarely enough that I only know the general concept but don't have the formulas on the tip of my mind. The key is that % impedance is reported at full load current, so a transformer with a larger KVA rating (at the same voltage) will have lower R and X given the same % impedance. The other thing you will need besides the % impedance is the transformer X/R ratio, something else that should be tabulated.

I agree with you on taps; they are there to adjust steady state voltages, and cannot compensate for the varying voltage drop of varying loads. If the supply from the utility is a _steady_ 5% low (or high) then it makes sense to use taps. But not to compensate for voltage drop in the circuits. (Well, I'd consider using taps for loads that are _always_ on, of if you have a tap changer, but we are talking building wiring here, not utility scale distribution...)

-Jon
 

spraymax6

Senior Member
Location
USA
So the 'Exact' voltage drop formula is what you need to consider both resistance and reactance of the supply circuit, and you are already doing that. So the only thing you need to add is the R and X values of the transformer in series with the R and X of the circuit wires so you can run your calculations.

This is one of those things that I do rarely enough that I only know the general concept but don't have the formulas on the tip of my mind. The key is that % impedance is reported at full load current, so a transformer with a larger KVA rating (at the same voltage) will have lower R and X given the same % impedance. The other thing you will need besides the % impedance is the transformer X/R ratio, something else that should be tabulated.

I agree with you on taps; they are there to adjust steady state voltages, and cannot compensate for the varying voltage drop of varying loads. If the supply from the utility is a _steady_ 5% low (or high) then it makes sense to use taps. But not to compensate for voltage drop in the circuits. (Well, I'd consider using taps for loads that are _always_ on, of if you have a tap changer, but we are talking building wiring here, not utility scale distribution...)

-Jon
Thank you for this. This clears things up a bit more.

thanks again
 

paulengr

Senior Member
In the “exact” formula you convert %Z to an actual impedance. Then you have to know X/R of the transformer. That’s generally not known. Most Brooke assume X/R=10. Then if you convert the low side to an equivalent impedance it will equal the square of the transformer ratio times the impedance as seen on the high side. Or you can divide the source impedance to convert to the low side. So in a 480:120 V transformer if you are working on the low side the supply is impedance as seen on the 120 V side is 1/16th as large. Then the transformer impedance appears as an impedance in series with the source This stands to reason since it is usually the biggest source impedance especially if the supply is a relatively stiff bus. That’s why often we just ignore it (infinite bus assumption). SKM etc. is using this exact same model. This is the exact model. With ANSI and most IEC equivalents we assume that we are looking at a distribution system. Resistance is negligible so we just ignore jt and only use X. So in that case we can assume that X is still 90% of %Z but since you really don’t know it is common practjce to just assume Z = X (R is zero) for the transformer and model VD on the low side using this and a high side X equal to the rest of the circuit divided again by the square of the transformer ratio. Good enough for the slide rule days. Even today this is what SKM, ETAP calculate for an ANSI short circuit. Arc flash uses what is usually called “IEC comprehensive” which is just using the full R+iZ values. Also with about #8 or smaller wire R is larger than X so on small loads the assumption that X/R is large so we can ignore R is violated. That is still covered in the point to point VD method though,
 

JEFF MILLAR

Senior Member
Hand calculation is not that complex. I can help if you can post your one line, complete with your load details.
" Everything is easy when you understand it " I learn plenty on this forum and pleased to help out when my knowledge permits.
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
The open thread says:

“My question is when doing a voltage drop calculation where you have, lets say, a 480V bus distribution system, and then a 480-208/120V step down transformer and you are doing the calculations by hand, do you include some sort of voltage drop across the transformer?”

The base standard for voltage drops it is ANSI C84.1 where the service supply voltage[Utilities] is limited up and down and also the utilization voltage limits [ at user ultimate terminal].

At first, there are to ranges: one for normal operation-as designed [Range A ] and one for outage cases Range B. The second has to be limited in time-usually no more than 2 hours.

The standard makes difference between 120 up to 600 V rated voltage and above 600 V.

The voltages recommended for 600 V take into consideration transformer voltage drop.

If I resume the graphic the voltage drops are as following:

For up to 600 V

The minimum service voltage at Range A [based on 120 rated]=114/120=95% [0.95]

The minimum service voltage at Range B=110/120=91.7% [0.917]

The minimum utilization voltage at Range A [for non-lighting equipment]=108/120=90%[0.9]

The minimum utilization voltage at Range A [for lighting equipment]=110/120=91.7%[0.917]

The minimum utilization voltage at Range A [for non-lighting equipment]=108/120=90%[0.9]

voltage drop recommended: 10% for non-lighting 3.33% for lighting

The minimum utilization voltage at Range A [for lighting equipment]=110/120=91.7%[0.917]

The minimum utilization voltage at Range B [for non-lighting equipment] =104/120= 86.7%[0.867]

The minimum utilization voltage at Range B [for lighting equipment]=106/120=88.3% [0.883]

For more than 600 V:

The minimum service voltage at Range A [based on 120 rated also]=117/120=97.5% [0.975]

The minimum service voltage at Range B 114.5/120=95.4% [0.954]

Utilization voltage remains the same.

Then voltage drop for non-lighting Range A [(117-108)/120=7.5%

Then voltage drop for lighting Range A [(117-110)/120=5.83%
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
See Voltage Disturbance as per ANSI C84.1
 

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