Percent Voltage Drop, respect to which voltage? 2

[corvalan]

I was asked the following simple question that made me evaluate what I always thought a as a given.

When the NEC suggests in the informational notes to keep feeders and branches percent voltage drop below 3% and bus percent voltage drop below 5% is it with respect to the NOMINAL voltage of the system (I.e. 480 V) or to the voltage at the secondary terminals of the transformer feeding the system?

The difference of percentage voltage drop considering the NOMINAL voltage or the transformer secondary terminal voltage as the reference starting voltage could be substantial.

Let's assume a transformer has a nameplate voltage rating of 13,200 V to 480 V and is connected at the primary to a system that has an ACTUAL voltage identical to the NOMINAL voltage of 13,200 V.

Due to INTERNAL impedance of the transformer, the transformer secondary terminals will experience a voltage drop from the NOMINAL secondary voltage.

If we calculate PERCENT voltage drop downstream, the results are going to be less than if we calculate PERCENTAGE voltage drop based on the transformer secondary nominal voltage.

The question is:

Is there a standard that explicitly recommends calculating PERCENT voltage drop using the NOMINAL voltage as a base or the diminished transformer secondary voltage as a base?

 

[waross]

In Canada the code states;
"When calculating currents that will result from loads, expressed in watts or volt amperes, to be supplied by a
low-voltage ac system, the voltage divisors to be used shall be 120, 208, 240, 277, 347, 416, 480, or 600 as
applicable.".

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

NEC 2014 art.210.19(A)(b) Informational Note No. 4 recommends 3-5% voltage drop [this is an information note only-according to art.90.5- is not enforceable].
ANSI C84.1 Range A is the optimal voltage range. Range B is acceptable, but not optimal.
The percent is referring to rated voltage. The rated voltages are 120,240,480 V. See:

http://www.powerqualityworld.com/2011/04/ansi-c84-1-voltage-ratings-60-hertz.html

 

[7anoter4]

The rated voltage is rated high voltage and the no-load voltage resulted when the primary voltage supplied is rated.
Let’s say for this 1000 kVA 13,200/480 V the impedance voltage is 5.75%[5.35% with permissible error of 7.5% that means uk%=5.35*(1+7.5/100)=5.75%
The impedance [in ohm]Z=0.480^2/1*5.75%=0.013251 and X/R=11 [ANSI C37.010] X= 0.013196; R= 0.0012 ohm. Let’s say pf[cos(fi)=0.85 then sin(fi)=0.5268
Irated=1000/sqrt(3)/480*1000=1202.8
DV=SQRT(3)*Irated*[R*cos(f)+X*sin(f)]= 16.60707 V [3.46%]
According to NEC 5% drop will be 8.46% total
480*(1-8.46/100)=440 V
ANSI C84.1 Range A minimum utilization voltage is 110 V for 120 V then 440 for 480 V.

 

[waross]

Thank you for the interesting information, 7anoter4.
However, ANSI recommended system voltages are not pertinent to code requirements for voltage drops in feeders and branch circuits.

The code requirement is for maximum voltage drop, NOT minimum allowable voltage.
Example: On a nominal 120 volt circuit, the maximum voltage drop allowed by the code is 5% of 120 Volts or 6 Volts.
Whatever the supply voltage variation, and whether within ANSI limits or not, The maximum voltage drop is calculated based on 120 Volts and rated current of the load. The CEC has voltage drop tables that consider the impedance of the conductors (including the reactance) rather than just the resistance.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

Thanks, waross, I agree with you.
In IEC World the IEC 60038 Standard recommends IEC standard voltages.
Concerning supply voltage range, under normal service conditions, it is recommended that the voltage at the[Utility] supply terminals should not differ from the nominal[standard] voltage of the system by more than ±10 %.
[in USA +7/-10 for Range B-Range B is acceptable, but not optimal]
For the utilization voltage range, in addition to the voltage variations at the supply terminals, voltage drops may occur within the consumer's installations. For low-voltage installations, this voltage drop is limited to 4 %, therefore, the utilization voltage range is +10 %, –14 %.This utilization range should be taken into account by Product Committees.

 

[Mbrooke]

The rated voltage is rated high voltage and the no-load voltage resulted when the primary voltage supplied is rated.
Let’s say for this 1000 kVA 13,200/480 V the impedance voltage is 5.75%[5.35% with permissible error of 7.5% that means uk%=5.35*(1+7.5/100)=5.75%
The impedance [in ohm]Z=0.480^2/1*5.75%=0.013251 and X/R=11 [ANSI C37.010] X= 0.013196; R= 0.0012 ohm. Let’s say pf[cos(fi)=0.85 then sin(fi)=0.5268
Irated=1000/sqrt(3)/480*1000=1202.8
DV=SQRT(3)*Irated*[R*cos(f)+X*sin(f)]= 16.60707 V [3.46%]
According to NEC 5% drop will be 8.46% total
480*(1-8.46/100)=440 V
ANSI C84.1 Range A minimum utilization voltage is 110 V for 120 V then 440 for 480 V.

Great post, but can you break this down for me? Are you calculating the voltage drop across the transformer itself from no load to full load?

 

[7anoter4]

Thank you, Mbrooke. Actually, it is not so sophisticated. In IEEE 141/1993 ch.3.11.1 General mathematical formulas I used the approximate formula for the voltage drop which is:
(1) V=IRcosf+IXsinf
where:
V is the voltage drop in circuit, line to neutral
I is the current flowing in conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
You may take z% ,r% and x% from the ANSI C37.010 .See:

http://arcadvisor.com/faq/motor-transformer-xr-data

From ANSI C57.12.01 ch.9 Tolerances 9.2.(1) impedance of two-windings transformer = 7.5%
Then R=VL-L^2/S*r%/100; X=VL-L^2/S*x%/100
(2) You may use the "exact" formula from the same chapter.
(3) From ch.3.11.4 Transformer voltage drop-in the same standard it is
Figure 3-12 Approximate voltage drop curves for three-phase transformers, 225-10 000 kVA, 5-25 kV
However the curves are based on constant impedance of 5.5% and a resistance percent from 0.6% to 1.35%.
(4) IEEE Std C57.12.90-2006 ch.14.4.4.2 General expression for calculation of transformer regulation 14.4.4.1 Exact formulae for the calculation of regulation
(5) DIN/VDE presents another way-close to IEEE C57.12.90.
See: ABB Switchgear Manual vol.12 ch.12.1.3:
voltage variation [for load different from rated]:
uf=a*u'f+1/2*(a*u'f)^2/100+1/8*(a*u"f)^4/10^6 where:
u'f=uRr*cosf+uXr*sinf
u"f=uRr*sinf-uXr*cosf
where:
a=S/Sr=actual apparent power/rated apparent power
uRr=losses at rated load[kW]/rated power[kVA]
ukr=rated short-circuit voltage[V]/rated voltage[V]*100[percent]
uXr=sqrt(ukr^2-uRr^2)
There are differences of less than 1%, so I'd prefer the approximate (1).

 

[waross]

Clarification:
7anoter4's post is the answer to Mbrooke's request for an explanation.
This is all valid information in regards to the final utilization voltage.
However the answer to the OP's question is not as difficult.
The OP's question has to do with the maximum voltage drop in feeders and branch circuits as mandated by the codes.
Transformer impedance does not enter into the feeder voltage drop.
Common practice is to use the load current and the voltage drop from the voltage drop tables.
The voltage drop is then expressed as a percentage of the nominal system voltage (120V, 208V, 240V, 277V, etc.).
I have always found this method to be accepted by the AHJ.
I did a small project where long runs and low voltage made voltage drop an issue.
The AHJ requested that I provide copies of my worksheets showing the conductor size, current and percentage voltage drop for every conductor run.
I doubt that the calculations were checked but the AHJ wanted to be assured that the calculations had been done.
I was doing the calculations in any event so it was not burdensome to supply a copy of my spread sheets.
This method may be subject to errors due to varying supply voltages, load currents varying with applied voltage, phase angles and supply conductors that do not match the voltage drop tables.
However, the intent is to avoid large voltage drops in feeders and branch circuits.
Although in practice the actual voltage drop in a feeder may be more than that shown by the calculations, it will still be considered acceptable.
HOWEVER, there are trouble shooting issues where it is necessary to calculate the actual utilization voltage or possible range of voltages.
In such an instance the formula posted by 7anoter4 is the way to go.
lps

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jghrist]

Is there a standard that explicitly recommends calculating PERCENT voltage drop using the NOMINAL voltage as a base or the diminished transformer secondary voltage as a base?

Example voltage drop calculations in the NEC Handbook (following 2.15.2(A)(3), FPN No. 3, and following Table 9 in Chapter 9) use the nominal voltage as the denominator in calculating percent voltage drop.

 

[JensenDrive]

The NEC says voltage drop, not voltage regulation.
A side matter is voltage regulation is horribly complex. You cannot calculate voltage at the end load without a knowing voltage, current, and power factor at some location in the circuit. However, you do not typically know the load. If load given in kVA, this is insufficient in that the amps are unknown if voltage at the load is unknown, and typically power factor changes with voltage.