MentDB WeakMQL Recipes
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Electrical engineer

Impedances, three-phase power and cable sizing with complex numbers and units.

10 recipes on this page · open in the interactive view

Recipes
  1. Series RLC circuit at 50 Hz
  2. Three-phase motor and power factor correction
  3. Voltage drop and losses in a cable
  4. Higher-order modes of a coaxial cable
  5. Symmetrical components and voltage unbalance
  6. ABCD parameters and Ferranti effect of a 400 km line
  7. Short-circuit and peak current at a LV switchboard (IEC 60909)
  8. Skin depth of a copper busbar at the mains harmonics
  9. Distance protection: apparent impedance in a mho zone
  10. All-day efficiency of a distribution transformer
executed on MentDB

Series RLC circuit at 50 Hz

A series RLC load (10 ohm, 50 mH, 100 uF) is fed by 230 V / 50 Hz. The script builds the complex impedance, gives its modulus and phase, the current and the resonance frequency.

Pipeline
01 · reactanceNet reactance X = Lw - 1/Cwmath calc→02 · impedanceZ = R + jXcomplex add→03 · polarModulus and phase of Zcomplex arg→04 · resonancef0 = 1 / 2 pi sqrt(LC)unit calc
Script
admin
#1. Reactances of the series RLC at 50 Hz: R = 10 ohm, L = 50 mH, C = 100 uF;
-> "[w]" (math calc "2*pi*50");
-> "[X]" (math calc "[w]*0.050 - 1/([w]*100e-6)");
#2. Complex impedance Z = R + jX;
-> "[Z]" (complex add "10" (complex mul "i" [X]));
#3. Modulus, phase and current under 230 V;
-> "[mod]" (complex abs [Z]);
-> "[phi]" (math rad_to_deg (complex arg [Z]));
-> "[I]" (math calc "230/[mod]");
#4. Resonance frequency f0 = 1 / 2 pi sqrt LC;
-> "[f0]" (unit calc "1/(2*pi*sqrt(50 mH * 100 uF))" "Hz");
concat "Z = " [Z] " ohm, |Z| = " (math round [mod] 2) " ohm, phase = " (math round [phi] 1) " deg, I = " (math round [I] 2) " A, f0 = " (math round [f0] 1) " Hz"
Result
mentdb
Z = 10-16.123025350430098i ohm, |Z| = 18.97 ohm, phase = -58.2 deg, I = 12.12 A, f0 = 71.2 Hz
Go further
  • Sweep the frequency with num table on the modulus to draw the resonance curve
  • Compute the quality factor Q = (1/R) sqrt(L/C) with math calc
  • Add a parallel branch with complex div for the product over sum formula
executed on MentDB

Three-phase motor and power factor correction

A 400 V three-phase motor draws 32 A at cos phi 0.85. The script computes P, Q and S, then sizes the delta capacitor bank that raises the power factor to 0.95.

Pipeline
01 · activeP = sqrt(3) U I cos phiunit calc→02 · reactiveQ and Qc = P (tan phi1 - tan phi2)math calc→03 · bankC = Qc / (3 w U^2) in uFunit calc
Script
admin
#1. Three-phase motor 400 V, 32 A, cos phi = 0.85;
-> "[P]" (unit calc "sqrt(3)*400 V*32 A*0.85" "kW");
-> "[S]" (unit calc "sqrt(3)*400 V*32 A" "kW");
-> "[Q]" (math calc "sqrt([S]^2-[P]^2)");
#2. Reactive power to compensate to reach cos phi = 0.95;
-> "[Qc]" (math calc "[P]*(tan(acos(0.85))-tan(acos(0.95)))");
#3. Capacitance per phase of a delta bank: C = Qc / 3 w U^2;
-> "[C]" (unit calc (concat [Qc] " kW / (3*2*pi*50 Hz*(400 V)^2)") "uF");
concat "P = " (math round [P] 2) " kW, Q = " (math round [Q] 2) " kvar, S = " (math round [S] 2) " kVA; bank: Qc = " (math round [Qc] 2) " kvar, C = " (math round [C] 1) " uF per phase in delta"
Result
mentdb
P = 18.84 kW, Q = 11.68 kvar, S = 22.17 kVA; bank: Qc = 5.48 kvar, C = 36.4 uF per phase in delta
Go further
  • Use complex from_polar to write the apparent power S = P + jQ
  • Compare a star connection: C = Qc / (w U^2) with the phase voltage
  • Estimate the yearly saving on reactive energy with math percent
executed on MentDB

Voltage drop and losses in a cable

A 40 m single-phase copper line of 2.5 mm2 carries 16 A. The script computes the loop resistance, the voltage drop against the 3 % lighting limit, the Joule losses and the energy lost per year.

Pipeline
01 · resistanceR = rho 2L / Sunit calc→02 · dropdU and percent of 230 Vmath calc→03 · lossesR I^2 and kWh over 2000 hunit calc
Script
admin
#1. Single-phase line: copper rho = 0.0225 ohm.mm2/m, 40 m, 2.5 mm2, 16 A, cos phi = 1;
-> "[R]" (unit calc "0.0225 ohm*mm^2/m * 2*40 m / 2.5 mm^2" "ohm");
#2. Voltage drop in V and in percent of 230 V;
-> "[dU]" (unit calc (concat [R] " ohm * 16 A") "V");
-> "[pct]" (math calc "100*[dU]/230");
#3. Joule losses and energy lost over 2000 h per year;
-> "[Pj]" (unit calc (concat [R] " ohm * (16 A)^2") "W");
-> "[E]" (unit calc (concat [Pj] " W * 2000 h") "kWh");
-> "[ok]" (if (<= [pct] 3) {"OK"} {"too high, use a larger section"});
concat "R = " [R] " ohm, dU = " (math round [dU] 2) " V = " (math round [pct] 2) " % -> " [ok] "; losses " (math round [Pj] 1) " W, " (math round [E] 0) " kWh/year"
Result
mentdb
R = 0.72 ohm, dU = 11.52 V = 5.01 % -> too high, use a larger section; losses 184.3 W, 369 kWh/year
Go further
  • Find the smallest standard section that passes with num solve on the drop formula
  • Correct rho for the conductor temperature with alpha = 0.00393 per K
  • Add the reactance of the cable for long three-phase lines
executed on MentDB

Higher-order modes of a coaxial cable

A PTFE coaxial cable carries only the TEM mode below the cutoff of the first higher modes. The script computes the characteristic impedance, the TM01 cutoff from the cross product of Bessel functions J0 and Y0, and the TE11 cutoff, the upper limit of single-mode operation.

Pipeline
01 · Z0TEM impedancemath calc→02 · TM01Root of J0 Y0 - J0 Y0num solve→03 · cutoffCutoff frequenciesbessel_y(nu, x)
Script
admin
#1. PTFE coaxial cable: inner radius a = 1.5 mm, outer b = 5 mm, er = 2.1;
-> "[a]" 1.5e-3; -> "[b]" 5e-3; -> "[er]" 2.1; -> "[c]" (math calc "[b]/[a]");
#2. Characteristic impedance of the TEM mode;
-> "[Z0]" (math calc "60/sqrt([er])*ln([c])");
#3. TM01 cutoff: first root of J0 of x * Y0 of c x - J0 of c x * Y0 of x, x = kc a;
-> "[x]" (num solve (concat "bessel_j(0, x)*bessel_y(0, " [c] "*x)-bessel_j(0, " [c] "*x)*bessel_y(0, x)") "x" 1.0 1.7);
-> "[fTM]" (math calc "[x]/[a]*_c/(2*pi*sqrt([er]))");
#4. TE11, the first higher mode, with the usual approximation kc = 2/a+b;
-> "[fTE]" (math calc "2/([a]+[b])*_c/(2*pi*sqrt([er]))");
concat "Z0 = " (math round [Z0] 1) " ohm, TE11 cutoff = " (math round (math calc "[fTE]/1e9") 2) " GHz, TM01 cutoff = " (math round (math calc "[fTM]/1e9") 2) " GHz: single mode below " (math round (math calc "[fTE]/1e9") 1) " GHz"
Result
mentdb
Z0 = 49.8 ohm, TE11 cutoff = 10.13 GHz, TM01 cutoff = 29.06 GHz: single mode below 10.1 GHz
Go further
  • Find the exact TE11 cutoff with the derivatives of J1 and Y1 and num solve.
  • Check the attenuation with unit calc and the skin depth of copper.
executed on MentDB

Symmetrical components and voltage unbalance

The three phase voltages of a feeder are measured in per unit with their angles. Compute the zero, positive and negative sequence components with the Fortescue operator a, then the voltage unbalance factor V2 / V1 checked against the 2 pct limit of EN 50160.

Pipeline
01 · phasorsVa, Vb, Vc and a = 1 at 120 degcomplex from_polar→02 · operatora squaredcomplex pow→03 · sequencesV0, V1, V2complex mul→04 · unbalanceModuli and VUFcomplex abs
Script
admin
#1. Measured phase voltages in per unit: unbalanced feeder;
-> "[Va]" (complex from_polar 1.00 0);
-> "[Vb]" (complex from_polar 0.85 (math calc "-125*pi/180"));
-> "[Vc]" (complex from_polar 0.92 (math calc "112*pi/180"));
#2. Fortescue operator a = 1 at 120 degrees, and a squared;
-> "[a]" (complex from_polar 1 (math calc "2*pi/3"));
-> "[a2]" (complex pow [a] 2);
#3. Zero, positive and negative sequence components;
-> "[V0]" (complex div (complex add (complex add [Va] [Vb]) [Vc]) 3);
-> "[V1]" (complex div (complex add (complex add [Va] (complex mul [a] [Vb])) (complex mul [a2] [Vc])) 3);
-> "[V2]" (complex div (complex add (complex add [Va] (complex mul [a2] [Vb])) (complex mul [a] [Vc])) 3);
#4. Voltage unbalance factor VUF = V2 / V1, limit 2 pct in EN 50160;
-> "[vuf]" (* 100 (/ (complex abs [V2]) (complex abs [V1])));
concat "V0 = " (math round (complex abs [V0]) 4) " pu, V1 = " (math round (complex abs [V1]) 4) " pu at " (math round (math calc (concat (complex arg [V1]) "*180/pi")) 2) " deg, V2 = " (math round (complex abs [V2]) 4) " pu | VUF = " (math round [vuf] 2) " pct" (if (> [vuf] 2) {" > 2 pct: out of EN 50160"} {" within EN 50160"})
Result
mentdb
V0 = 0.0765 pu, V1 = 0.9217 pu at -4.19 deg, V2 = 0.029 pu | VUF = 3.15 pct > 2 pct: out of EN 50160
Go further
  • Apply the same transform to the currents to find a single-phase fault (I0 = I1 = I2).
  • Write the Fortescue matrix and use matrix mul on real and imaginary parts.
executed on MentDB

ABCD parameters and Ferranti effect of a 400 km line

A 400 kV, 400 km overhead line has z = 0.03 + j0.33 ohm/km and y = j3.8 uS/km. The script computes the propagation constant and surge impedance, the exact long-line A and B parameters with the complex cosh and sinh, the no-load voltage rise at the receiving end and the surge impedance loading.

Pipeline
01 · propagationgamma l = l sqrt(zy), Zc = sqrt(z/y)complex sqrt→02 · AA = cosh(gamma l)complex cosh→03 · BB = Zc sinh(gamma l)complex sinh→04 · ferrantiVr = Vs / |A|, SILcomplex abs
Script
admin
#1. Line constants per km and length;
-> "[z]" "0.03+0.33i"; -> "[y]" "0.0000038i"; -> "[l]" 400;
#2. Propagation constant times length and surge impedance;
-> "[gl]" (complex mul (complex sqrt (complex mul [z] [y])) [l]);
-> "[Zc]" (complex sqrt (complex div [z] [y]));
#3. Exact long-line parameters;
-> "[A]" (complex cosh [gl]);
-> "[B]" (complex mul [Zc] (complex sinh [gl]));
#4. Receiving end voltage at no load and surge impedance loading;
-> "[Vr]" (/ 400 (complex abs [A]));
concat "gamma l = " (math round (complex re [gl]) 5) " + j" (math round (complex im [gl]) 5) " | Zc = " (math round (complex abs [Zc]) 1) " ohm | A = " (math round (complex abs [A]) 5) " at " (math round (math calc (concat (complex arg [A]) "*180/pi")) 3) " deg, |B| = " (math round (complex abs [B]) 2) " ohm | no-load Vr = " (math round [Vr] 1) " kV (+" (math round (* 100 (- (/ [Vr] 400) 1)) 2) " pct) | SIL = " (math round (/ 160000 (complex abs [Zc])) 0) " MW"
Result
mentdb
gamma l = 0.02034 + j0.44839 | Zc = 295.3 ohm | A = 0.90138 at 0.561 deg, |B| = 128.16 ohm | no-load Vr = 443.8 kV (+10.94 pct) | SIL = 542 MW
Go further
  • Size a shunt reactor at the receiving end that brings Vr back to 400 kV
  • Compare with the nominal pi model, A = 1 + ZY/2
executed on MentDB

Short-circuit and peak current at a LV switchboard (IEC 60909)

A 1000 kVA, 20/0.4 kV transformer (uk = 6 %, copper losses 10.5 kW) is fed by a 500 MVA network. The script builds the loop impedance at 400 V, computes the initial symmetrical short-circuit current Ik", the peak factor kappa and the peak current ip used to check the breaker making capacity.

Pipeline
01 · transformerZt, Rt, Xt from uk and lossesmath sqrt→02 · loop|Z| = hypot(R, X)math hypot→03 · angleImpedance anglemath atan→04 · peakkappa = 1.02 + 0.98 e^(-3R/X)math exp
Script
admin
#1. Transformer 1000 kVA, 400 V, uk 6 pct, copper losses 10.5 kW, voltage factor c = 1.05;
-> "[U]" 400; -> "[S]" 1000000; -> "[c]" 1.05;
-> "[Zt]" (/ (* 0.06 (* [U] [U])) [S]);
-> "[Rt]" (/ (* 10500 (* [U] [U])) (* [S] [S]));
-> "[Xt]" (math sqrt (- (* [Zt] [Zt]) (* [Rt] [Rt])));
#2. Network 500 MVA referred to 400 V, R/X = 0.1;
-> "[Zq]" (/ (* [c] (* [U] [U])) 500000000);
-> "[Xq]" (* 0.995 [Zq]); -> "[Rq]" (* 0.1 [Xq]);
#3. Loop impedance and initial symmetrical short-circuit current;
-> "[R]" (+ [Rt] [Rq]); -> "[X]" (+ [Xt] [Xq]);
-> "[Z]" (math hypot [R] [X]);
-> "[ik]" (/ (* [c] [U]) (* (math sqrt 3) [Z]));
#4. Peak factor and peak current;
-> "[kappa]" (+ 1.02 (* 0.98 (math exp (/ (* -3 [R]) [X]))));
-> "[ip]" (* [kappa] (* (math sqrt 2) [ik]));
concat "Z = " (math round (* 1000 [Z]) 3) " mohm at " (math round (math calc (concat (math atan (/ [X] [R])) "*180/pi")) 2) " deg, R/X = " (math round (/ [R] [X]) 4) " | Ik = " (math round (/ [ik] 1000) 2) " kA | kappa = " (math round [kappa] 4) " | ip = " (math round (/ [ip] 1000) 2) " kA -> " (if (< [ip] 63000) {"breaker 25 kA / 63 kA peak OK"} {"breaker too small"})
Result
mentdb
Z = 9.935 mohm at 80.07 deg, R/X = 0.1751 | Ik = 24.41 kA | kappa = 1.5996 | ip = 55.21 kA -> breaker 25 kA / 63 kA peak OK
Go further
  • Add the busbar and cable impedances to get Ik at the end of a feeder
  • Compute the minimum current with c = 0.95 and hot conductors to check the protection trips
executed on MentDB

Skin depth of a copper busbar at the mains harmonics

Harmonic currents increase the AC resistance of conductors. With mu0 and 2pi from the constant library, the script computes the skin depth of copper at 50 Hz and at the 5th, 11th and 20th harmonics, and the AC/DC resistance ratio of a 30 mm round conductor.

Pipeline
01 · constantsmu0 and 2 piconstant physics_perm_vac_magnetic→02 · omegaomega = 2 pi fconstant math_two_pi→03 · depthdelta = sqrt(2 rho / omega mu0)math sqrt→04 · ratioRac/Rdc, low or high frequency formulamath power
Script
admin
#1. Copper at 20 degC, mu0 and 2 pi from the constant library, conductor radius 15 mm;
-> "[rho]" 0.0000000172; -> "[r]" 0.015;
-> "[mu0]" (string regexp_extract (constant physics_perm_vac_magnetic) "[0-9.]+(E-[0-9]+)?" 0);
-> "[twopi]" (string regexp_extract (constant math_two_pi) "[0-9.]+" 0);
-> "[out]" "";
#2. Skin depth and Rac/Rdc for each harmonic order;
json load "rx_p4f_h" "[1, 5, 11, 20]";
json parse_array "rx_p4f_h" "/" "[h]" {
  -> "[delta]" (math sqrt (/ (* 2 [rho]) (* (* [twopi] (* 50 [h])) [mu0])));
  -> "[x]" (/ [r] [delta]);
  -> "[k]" (if (< [x] 2) {+ 1 (/ (math power [x] 4) 48)} {+ (/ [x] 2) 0.25});
  -> "[out]" (concat [out] "h" [h] " " (* 50 [h]) " Hz: delta " (math round (* 1000 [delta]) 2) " mm, Rac/Rdc " (math round [k] 2) " | ");
};
json unload "rx_p4f_h";
string left [out] (- (string length [out]) 3)
Result
mentdb
h1 50 Hz: delta 9.33 mm, Rac/Rdc 1.14 | h5 250 Hz: delta 4.17 mm, Rac/Rdc 2.05 | h11 550 Hz: delta 2.81 mm, Rac/Rdc 2.91 | h20 1000 Hz: delta 2.09 mm, Rac/Rdc 3.84
Go further
  • Use flat busbars or hollow conductors when the skin depth is smaller than the thickness
  • Weight Rac/Rdc by the harmonic spectrum of the load to get the transformer K factor
executed on MentDB

Distance protection: apparent impedance in a mho zone

A 110 kV line of 60 km (0.12 + j0.40 ohm/km) is protected by a distance relay whose zone 1 is a mho circle reaching 80 % of the line. During a fault the relay measures 38 kV and 2 kA lagging by 68 deg. The script computes the apparent impedance, tests it against the mho circle and estimates the fault location with the reactance method.

Pipeline
01 · lineLine impedance and anglemath atan→02 · measureZ = V / I from phasorscomplex from_polar→03 · divideApparent impedancecomplex div→04 · mho|Z - Zr/2| < |Zr|/2complex sub→05 · locateReactance methodcomplex im
Script
admin
#1. Line impedance, its angle and zone 1 reach 80 pct;
-> "[ZL]" (complex mul "0.12+0.40i" 60);
-> "[ang]" (math atan (/ (complex im [ZL]) (complex re [ZL])));
-> "[Zr]" (complex mul [ZL] 0.8);
#2. Phasors measured during the fault: 38 kV and 2 kA at -68 deg;
-> "[V]" (complex from_polar 38000 0);
-> "[I]" (complex from_polar 2000 (math calc "-68*pi/180"));
-> "[Z]" (complex div [V] [I]);
#3. Mho test: distance to the center of the circle smaller than the radius;
-> "[d]" (complex abs (complex sub [Z] (complex div [Zr] 2)));
-> "[rad]" (/ (complex abs [Zr]) 2);
#4. Fault location with the reactance method;
-> "[km]" (/ (complex im [Z]) 0.40);
concat "line " (math round (complex abs [ZL]) 2) " ohm at " (math round (math calc (concat [ang] "*180/pi")) 2) " deg | Z seen = " (math round (complex re [Z]) 3) " + j" (math round (complex im [Z]) 3) " ohm | margin " (math round (- [rad] [d]) 3) " ohm -> " (if (< [d] [rad]) {"zone 1 trip"} {"no zone 1 trip"}) " | fault at " (math round [km] 1) " km = " (math round (/ (* 100 [km]) 60) 1) " pct of the line"
Result
mentdb
line 25.06 ohm at 73.3 deg | Z seen = 7.118 + j17.616 ohm | margin 0.955 ohm -> zone 1 trip | fault at 44 km = 73.4 pct of the line
Go further
  • Add zone 2 at 120 pct with a 300 ms delay and check overlapping with the next line
  • Model a fault resistance of 5 ohm: the point moves right and can leave the circle
executed on MentDB

All-day efficiency of a distribution transformer

A 630 kVA distribution transformer (no-load losses 1.0 kW, load losses 6.5 kW) feeds a residential area with a known hourly load profile at power factor 0.9. The script computes the energy delivered, the iron and copper losses over 24 h, the all-day efficiency and the load of maximum efficiency.

Pipeline
01 · energyEnergy deliveredstat sum→02 · copperPk sum (S/Sn)^2stat sum_sq→03 · averageAverage load factormath avg→04 · optimumMax efficiency at sqrt(P0/Pk)math sqrt
Script
admin
#1. Hourly load profile in kVA from 0 h to 23 h;
-> "[S]" "[180, 160, 150, 145, 150, 190, 290, 380, 360, 330, 320, 340, 370, 350, 330, 340, 390, 480, 560, 590, 540, 450, 330, 230]";
-> "[Sn]" 630; -> "[P0]" 1.0; -> "[Pk]" 6.5; -> "[pf]" 0.9;
#2. Energy delivered and losses over 24 h in kWh;
-> "[E]" (* [pf] (stat sum [S]));
-> "[Wfe]" (* 24 [P0]);
-> "[Wcu]" (/ (* [Pk] (stat sum_sq [S])) (* [Sn] [Sn]));
#3. All-day efficiency, average load factor and load of maximum efficiency;
-> "[eta]" (/ [E] (+ [E] (+ [Wfe] [Wcu])));
concat "energy " (math round [E] 1) " kWh | iron " [Wfe] " kWh, copper " (math round [Wcu] 2) " kWh | all-day efficiency " (math round (* 100 [eta]) 3) " pct | mean load " (math round (* 100 (/ (math avg [S]) [Sn])) 1) " pct, max efficiency at " (math round (* 100 (math sqrt (/ [P0] [Pk]))) 1) " pct = " (math round (* [Sn] (math sqrt (/ [P0] [Pk]))) 0) " kVA"
Result
mentdb
energy 7159.5 kWh | iron 24 kWh, copper 49.61 kWh | all-day efficiency 98.982 pct | mean load 52.6 pct, max efficiency at 39.2 pct = 247 kVA
Go further
  • Compare with a 400 kVA amorphous core unit (P0 = 0.25 kW) over 30 years of losses
  • Add the hot-spot temperature from the IEC 60076-7 thermal model

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