8/8/2019 How to Perform Short
How to perform short-circuit calculations, part 2.Dec 1, 1999 12:00 PM, Mercede, Frank J.
0 Comments| Related ContentShare1Performing short-circuit calculations requires an understanding of various system components and theirinteraction.
In Part 1 of this article, which was featured in the June 1995 issue, we discussed the types of networks tocalculate short-circuit current (i.e., symmetrical rms current). In Parts 2 and 3 (April 1996 issue), we'lldescribe the per-unit method of performing short-circuit calculations in accordance with ANSI/IEEE 141-1993, IEEE Recommended Practice for Electric Power Distribution for Industrial Plants (the Red Book). We'lluse a simple example of an industrial power system to show the data preparation steps necessary indetermining the appropriate per-unit reactances and resistances of power system equipment for first-cycle(momentary) and contact-parting (interrupting) networks.
Overview of per-unit analysis
Per-unit analysis is based on "normalized" representations of the electrical quantities (i.e., voltage, current,impedance, etc.). The per-unit equivalent of any electrical quantity is dimensionless and is defined as theratio of the actual quantity in units (i.e., volts, amperes, ohms, etc.) to an appropriate base value of theelectrical quantity. This is expressed by the following equation.
per-unit value = actual value [divided by] base value (eq. 1)
The actual value can be a phasor or complex number (i.e., magnitude and phase) in units, whereas the basevalue is simply a real number in units.
The key factor of a per-unit normalization procedure is the selection of the base values of the electricalquantities. In practice, the base values of 3-phase apparent power in MVA (i.e. base MVA) and line-to-linevoltage in kV (i.e. base kV) are assigned on each side of every 3-phase power transformer in accordancewith the following simple rules.
First, a convenient base MVA is chosen (e.g., base MVA = 1, 10, or 100) and is common throughout theentire power system. Second, base kV on any side of a 3-phase power transformer is designated to beequal to the nominal, line-to-line nameplate kV rating of the transformer.
The base values of per-phase base impedance in ohms (i.e., base Z) and line current in kiloamperes (i.e.,
base kA) are then derived from base MVA and base kV, according to the following equations. base Z =[(base kV).sup.2] [divided by] base MVA (eq. 2) base kA = base MVA [divided by] ([square root of 3] x basekV) (eq. 3)
You gain an important advantage by assigning the base values per the above selection rules and usingthem to normalize (i.e., "per-unitize") the electrical quantities. This advantage can be seen in the per-phaseequivalent circuit model of any 3-phase transformer connection: It is simply a per-unit series impedance thataccounts for the conductor losses and leakage fluxes.
The following equation is often used in [TABULAR DATA FOR TABLE 1 OMITTED] the data preparationstage to adjust the per-unit impedance of power system apparatus whenever the 3-phase nameplate ratingsare different than the power system's 3-phase base quantities.
adjusted [Z.sub.pu] = unadjusted [Z.sub.pu] x (base MVA [divided by] MVA rating) x [(kV rating [divided by]base kV).sup.2](eq. 4)
Data preparation for a simplified example In the absence of nameplate information or data from equipmentmanufacturers, typical data can be referenced from tables and figures included in the IEEE Red Book. Thedata preparation steps to determine the appropriate per-unit reactances and resistances for first-cycle(momentary) and con-tacting-parting (interrupting) networks for the simplified industrial power system shownin Fig. 1 are as follows.
Base values of voltage. The base values of line-to-line voltage are simply the nominal nameplate, line-to-linevoltage ratings of the 3-phase transformers. Base MVA is chosen to be 10 MVA and is constant throughoutthe system.http://ecmweb.com/mag/electric_perform_shortcircuit_calculations_2/#disqus_thread%23disqus_threadhttp://ecmweb.com/mag/electric_perform_shortcircuit_calculations_2/#relatedarticle%23relatedarticlehttp://ecmweb.com/mag/electric_perform_shortcircuit_calculations_2/#relatedarticle%23relatedarticlehttp://ecmweb.com/mag/electric_perform_shortcircuit_calculations_2/#disqus_thread%23disqus_threadhttp://ecmweb.com/mag/electric_perform_shortcircuit_calculations_2/#relatedarticle%23relatedarticle
8/8/2019 How to Perform Short
Utility. The per-phase equivalent circuit model of the utility tie at the plant is a voltage source in series withan impedance. This source is the nominal per-unit line-to-neutral voltage at the service entrance point. Theper-unit impedance is the same for both first-cycle and interrupting networks; thus, no superscript "f" or "I" isnecessary for its symbol. (In the equations that follow, superscripts "f" and "I" refer to the per-unit reactanceand resistance, respectively, for the first-cycle and interrupting networks.)
The magnitude of the impedance ([Z.sub.u] = 0.01 per unit) is calculated by using the following equation.
[Z.sub.u] = base MVA [divided by] (SCA MVA)(eq. 5)
Here, SCA MVA is equal to 1000 and represents the available short-circuit apparent power delivered by theutility from all sources outside the plant.
To resolve [Z.sub.u] into reactive ([X.sub.u]) and resistive ([R.sub.u]) components, the following equationsare used.
X = Z sin ([tan.sup.-1] X/R)(eq. 6a)
R = Z cos ([tan.sup.-1] X/R)(eq. 6b)
In this event, a conservative assumption is to let [X.sub.u] = [Z.sub.u] = 0.01 per unit and [R.sub.u] = 0.
In our example, as shown in Fig. 1, the short-circuit X/R ratio at the utility tie is unavailable. Thus, weassume [X.sub.u] = 0.01 per unit and [R.sub.u] = 0.
Transformers. The per-phase equivalent circuit model of any 3-phase transformer [TABULAR DATA FORTABLE 2 OMITTED] connection is simply a per-unit series impedance. This impedance is the same for bothfirst-cycle and interrupting networks; thus, no superscript "f" or "I" is necessary for its symbol.
The unadjusted impedance of the transformer is provided on its nameplate and is expressed as apercentage of rated impedance. In other words, you simply divide it by 100% to arrive at its unadjusted per-unit value. This per-unit impedance is adjusted with respect to the system base quantities per Equation 4,and the adjusted per-unit impedance ([Z.sub.t]) is resolved into reactive ([X.sub.t]) and resistive ([R.sub.t])components per Equations 6a and 6b. (The typical short-circuit X/R ratios of the transformers in Fig. 1 aretaken from Fig. 4A-1 of the 1993 IEEE Red Book.) The results for the transformers in Fig. 1 are shown inTable 1.
Cable. The per-phase equivalent circuit model of the cable in Fig. 1 is simply a per-unit series impedance.This impedance of the cable is the same for both first-cycle and interrupting networks; thus, no superscript
"f" or "I" is necessary for its symbol.
To find the cable's per-unit reactance ([X.sub.c]) and resistance ([R.sub.c]), the following equations areused.
[X.sub.c] = [(X ohms per 1000ft) x (length of run in ft)] [divided by] [number of parallel conductors per phasex base Z in ohms] (eq. 7a)
[R.sub.c] = [(R ohms per 1000ft) x (length of run in ft)] [divided by] [no. of parallel conductors per phase xbase Z in ohms] (eq. 7b)
Using the above equations, [X.sub.c] is 0.0030 per unit and [R.sub.c] = 0.0043 per unit.
(The approximate reactance and resistance data (in ohms per 1000 ft) noted alongside the cable in Fig. 1are taken from Table 4A-7 in the 1993 IEEE Red Book.)
Turbine-generator. In general, the per-phase equivalent circuit model of a rotating machine is a voltagesource in series with an impedance that varies with time during the fault. Based on Table 4-1, Chapter 4 ofthe 1993 IEEE Red Book, the unadjusted per-unit reactance of the turbine-generator for both the first-cycleand interrupting networks is 1.0 [X.sub.d]", where [X.sub.d]" is the saturated direct-axis subtransientreactance of the generator in per-unit. You adjust this reactance with respect to the system base quantitiesby using Equation 9, with the corresponding adjusted resistance calculated by using Equation 8.
adjusted R = adjusted X [divided by] short-circuit X/R (eq. 8) [X.sup.f]=(1.0 [X.sub.d]") x (base MVA [dividedby] MVA rating) x [(kV rating [divided by] base kV).sup.2] (eq. 9)
8/8/2019 How to Perform Short
The calculation results are shown in Table 2. (The typical machine reactance and short-circuit X/R data arefrom Table 4A-1 and Figs. 4A-2 and 4A-3 of the 1993 IEEE Red Book. Incidentally, there is a typographicalerror in Table 4A-1 of the first printing of this book: the left-most column of Table 4A-1 should be labeled[X.sub.d]" and the right-most column [X.sub.d]'.)
Large motors. Based on Table 4-1, Chapter 4 of the 1993 IEEE Red Book, the unadjusted per-unitreactances for the first-cycle and interrupting networks are 1.0 [X.sub.d]" and 1.5 [X.sub.d]" respectively.
Equation 9 is used to adjust the first-cycle reactance, where the 3-phase kVA rating is approximately equalto the horsepower (hp) rating for induction motors and 0.8 power factor (PF) synchronous motors. Thecorresponding adjusted first-cycle resistance is calculated by using Equation 8. The following equations arethen used to calculate the adjusted interrupting reactance and resistance from the corresponding first-cyclevalues.
[X.sub.[M.sup.I]] = 1.5 x [X.sub.[M.sup.f]] (eq. 10a)
[R.sub.[M.sup.I]] = 1.5 x [R.sub.[M.sup.f]] (eq. 10b)
The results of these calculations also are shown in Table 2.
Small horsepower induction motors. The unadjusted per-unit reactances of the small horsepower (less than250 hp) induction motors shown in Fig. 1 for the first-cycle [TABULAR DATA FOR TABLE 3 OMITTED] andinterrupting networks are taken from the footnotes of Table 4-2 in the 1993 IEEE Red Book. Specifically, you
should refer to the footnotes of the row entitled "All others, 50 HP and above" for the 150 hp induction motor.Also refer to the footnotes of the row entitled "All smaller than 50 HP" for the group of small-horsepowerinduction motors whose ratings are less than 50 hp. The following equation is then used to adjust both theper-unit first-cycle and interrupting reactances, where the 3-phase kVA rating is approximately equal to thehp rating for an individual induction motor or the sum total of hp ratings for a group of motors.
[X.sub.M] = (unadjusted [X.sub.M]) x (base MVA [divided by] MVA rating) x [(kV rating [divided by] basekV).sup.2] (eq. 11)
The corresponding adjusted first-cycle and interrupting resistances are calculated by using Equation 8. Theresults of the calculations are listed in Table 3.