ct sat calculator (psrc)

CONTENTS Sheet 1: CALCULATOR (this sheet) A document of Sheet 2: INSTRUCTIONS Excel Spread Sheet IEEE Power Systems Relay Sheet 4: BACKGROUND See IEEE publication C37.110: "IEEE Guide for the Application Contact: gswift@nxt of CurrentTransformers Used for Protective Relaying Purposes" Refer also to "CT SAT Th ASSUMPTIONS: CT core losses and sec'y reactance zero (thru-hole primaFrequency: 60 Hz CT primary current is zero for t<0. CT is 5 amp nomina Time step = 1/12,000 second. INPUT PARAMETERS: ENTER: CALCULATED: Inverse of sat. curve slope = S = 22 --- Rt = Total bur voltage at 10A exc. current = Vs = 400 volts rms pf = Total bur Turns ratio = n2/1= N = 240 --- Zb = Total bur Winding resistance = Rw = 0.000 ohms Tau1 = System ti Burden resistance = Rb = 4.000 ohms Lamsat = Peak flux Burden reactance = Xb = 2.000 ohms Radian fr System X/R ratio = XoverR = 12.0 --- RP = Rms-to-pe it offset in primary current = Off = 1.00 -1<Off<1 A = Coefficie unit remanence (based on Vs) = 0.00 --- vers trical primary fault current = Ip = 12,000 amps rms dt = Time step = Lb = Burden in Thick lines: Thin lines: To determine the effect of saturation on a particular digital relay, one must have "models" for the blo Calculation of Mag. Of Fund'lCalculati is = IDEAL step time pri cur/N fl-linkages exc curr sec cur d/dt(is) number t i1(t)/N ie(t) i2(t) I1(t)*sin I1(t)*cos I1MAG I2(t)*sin seconds amps Wb-turns amps Wb-turns amps amps/s TIME IDEAL ACTUAL -200 -0.0167 0.000 0.00010 0.000 -0.00166 0.000 -3750 0 -199 -0.0166 0.000 0.00010 0.000 -0.00128 0.000 -2903 0 -198 -0.0165 0.000 0.00010 0.000 -0.00091 0.000 -2057 0 -197 -0.0164 0.000 0.00010 0.000 -0.00054 0.000 -1212 0 -196 -0.0163 0.000 0.00010 0.000 -0.00016 0.000 -370 0 -195 -0.0163 0.000 0.00010 0.000 0.00021 0.000 469 0 -194 -0.0162 0.000 0.00010 0.000 0.00058 0.000 1304 0 -193 -0.0161 0.000 0.00010 0.000 0.00094 0.000 2133 0 w = lrem Ideal (blue) and actual (black) secondary current in amps vs time in secon Ideal (blue) and actual (black) secondary current extracted fundamental rms l(t) Dl(t) -0.01 6 -0.012 -0.008 -0. 0 0 3 0.0 0 1 0.005 0.010 0.0 14 0.018 0.023 0.027 0.031 0.036 0.040 0.044 0.049 0.053 0.0 5 7 0.062 0.066 0.070 0.075 0.079 0.083 0.0 8 8 0.09 2 0.096 0.101 0.10 5 0.109 0.1 14 0.118 -100 -50 0 50 100 150 CT Saturation Calculator ACTUAL SECONDARY CURRENT (FR O M ABOVE) A NTI-A LIA SING FILTER D IS C R ETE FO UR IER TRANSFORM? PEA K DETECTIO N? DC -O FFSET REM O V A L? RM S DETERM INA TIO N? RELA Y FUNCTIO N A LG O RITHM ULTIM A TE EFFECT CORRECT TR IP? V s a m p s rm s slo pe = 1 /S lo g -lo g plot, equal decade spacing vo lts rms I e m fgr's da ta V e 10 Saturation C urve

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Saturación de TC's

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CT SAT Calculator

Sheet1CONTENTSSheet 1:CALCULATOR (this sheet)A document of theSheet 2:INSTRUCTIONSExcel Spread SheetIEEE Power Systems Relaying CommitteeSheet 4:BACKGROUNDSee IEEE publication C37.110: "IEEE Guide for the ApplicationContact: [email protected] CurrentTransformers Used for Protective Relaying Purposes"Refer also to "CT SAT Theory (PSRC)".ASSUMPTIONS:CT core losses and sec'y reactance zero (thru-hole primary).Frequency: 60 HzCT primary current is zero for t>>>>>>>>>>>>

3. Enter other parameters such as the CT winding resistance, the burden, the degree of dc-offset in the primary current waveform (up to 1 per unit), the primary system X/R ratio, the remanence**, and the primary symmetrical rms current.

Once any change to an INPUT parameter is made, a new plot appears automatically. The scale adjustment is automatic and the plot is believed to be self-explanatory._______*This is a definition used in the Guide that is not strictly correct but is used here to be consistent with the Guide. If the ideal (maximum rated) secondary current is 100 amps rms and the exciting (error) current is 10 amps rms, then the actual secondary current is typically 99+ amps rms, not 90 amps rms. This is because the exciting current is both out of phase and non-sinusoidal, so simple subtraction does not apply. So when we specify 10 amps error current at 100 amps secondary current, the error is actually less than 1% typically, but includes a phase error of, say 5 degrees, leading.

**Note that the per unit remanence is defined relative to Vs. If the knee-point voltage (45 degree slope point) is 80% of Vs, then the maximum remanence value is 0.8. Note that the polarities for this simulation are such that a positive remanence is the "worst-case" condition for premature saturation.

Sheet4BACKGROUNDThis program is intended to show LARGE CURRENT or HIGH BURDEN behavior only; it does not accurately represent the conditionsat the low end portion of the iron saturation curve. Consequently, hysteresis loss and eddy current loss are not modeled becausethey have negligible effect on the saturation region behavior.Another assumption is that the usual log-log plot of rms voltage versus rms current for the saturation curve is correct when plottedas a straight line for the saturated portion of the curve. This implies a pure power law relating instantaneous voltage toinstantaneous current, and the slope of the upper portion of the commonly-used log-log plot is in the neighborhood of 1/20 for amodern toroidal CT. The value for Vs, the "saturation voltage," is assumed to be at a properly measured rms value of 10 ampsexcitation current, the voltage being sinusoidal and the current being related to flux by an Sth(only) order function. If the 10-ampvalue is measureed using a 'not-true-rms' meter, the error is not great enough to cause significant error.The program has been checked against a variety of examples, including those of the recent IEEE paper "Mathematical Models forCurrent, Voltage, and Coupling Capacitor Voltage Transformers," Working Group C5 of PSRC, IEEE Trans on Power Delivery, Jan2000, pp. 62-72. (Note that there is an error in figures 4a and 5a of that paper: there was actually non-zero remanence for thiscase, as confirmed with the authors.)

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