Question 8.4: The effect of motors is neglected in Example 8.3. Calculate ......

Effect of Motor Contribution at 13.8 kV Bus, Fault F2
For a fault at the 13.8 kV bus F2, an equivalent impedance of the motors through transformers and cables is calculated. The partial currents from medium- and low-voltage motors are calculated in Tables 8.10 and 8.11, respectively. The equivalent impedance of low-voltage motors of two identical groups, from Table 8.11, is 6.45 + j15.235 per unit.
The per unit impedance of transformer T3 from Table 7.13 is 0.639 + j3.780. Therefore, the lowvoltage motor impedance through transformer T3, seen from the 4.16 kV bus, is 7.089 + j19.015 per unit.
From Table 8.10, the equivalent impedance of medium-voltage motors is 0.58 + j4.55 pu. The equivalent impedances of low- and medium-voltage motors in parallel are
(7.089 + j19.015) || (0.58 + j4.55) = 0.637 + j3.707
To this, add the impedance of cable C1 and transformer T2 from Table 7.13, which gives 0.701 + j4.606 pu. This is the equivalent impedance as seen from the 13.8 kV bus. Thus, the initial short-circuit current from the motor contribution is 1.1/(0.701 + j4.606) per unit = |0.99| kA.
The effect of motors in this example can be ignored, and the above calculation of currents from motor contributions is not necessary. From Equation 8.33,
\frac{ΣP_{rM}}{ΣS_{rT}}≤ \frac{0.8}{|\frac{c100 ΣS_{r\overline{T}}}{S^{\prime \prime}_{kQ} – \ 0.3}|}      (8.33)
ΣP_{rm} = sum of the active powers of all medium- and low-voltage motors = 0.86 MW. Also, ΣS_{rT} = rated apparent power of the transformer = 7.5 MVA. The left-hand side of Equation 8.33 = 0.1147. Symmetrical short-circuit power at the point of connection, without effect of motors, is
S^{\prime \prime}_{kQ} = \sqrt{3} I^{\prime \prime}_{k}U_{n} = \sqrt{3} (56.5)(13.8) = 1350.7 \ MVA
The right-hand side of Equation 8.33 gives 2.571, and the identity in Equation 8.33 is satisfied. The effect of motors can be ignored for a fault at 13.8 kV.
If the calculation reveals that motor contributions should be considered, we have to modify i_{p} at the fault point. This requires calculation of χ, which is not straightforward. High-voltage motors have χ = 1.75 or 1.65, and low-voltage motors have χ = 1.65. For a combination load, χ = 1.7 can be used to calculate i_{p} approximately.
Three-phase short-circuit at F3. For a fault at F3, we will first calculate the motor contributions. The low-voltage motor impedance plus transformer T3 impedance is 7.089 + j19.015, as calculated above. The initial short-circuit contribution from the low-voltage motor contribution is 1.1/(7.089 + j19.015) = 0.019 – j0.051 pu or | I^{\prime \prime}_{k}| = 0.76 kA. Medium-voltage impedance, from Table 8.10, is 0.58 + j4.55 per unit. The medium-voltage motor contribution is 0.028- j0.239 pu or | I^{\prime \prime}_{k}| = 3.43 kA.
To calculate the generator and utility source contributions, the impedances Z_{G,PSU} are in || with (Z_{T,PSU} + Z_{Q}), i.e., 0.0074 + j0.1483 in || with 0.006074 + j0.16378. This gives 0.0034 + j0.0778 pu. Add transformer T2 impedance (0.06349 + j0.89776) and cable C1 impedance (0.00038 + j0.00101) from Table 7.13. This gives an equivalent impedance of 0.0673 + j0.976 per unit. Thus, the initial short-circuit current is 0.077 – j1.122 per unit or | I^{\prime \prime}_{k}| = 15.61 kA. The total initial symmetrical
current, considering low- and medium-voltage motor contributions, is 19.80 kA. To calculate i_{p}, χ must be calculated for the component currents.
For contribution through transformer T2, using Equation 8.4:
χ= 1.02 + 0.98e^{-3R/X}        (8.4)
χ_{AT} = 1.02 + 0.98ε^{-3(0.06895)}
= 1.82
As this is calculated from a meshed network, a safety factor of 1.15 is applicable, i.e., χ = 1.15 multiplied by 1.82 = 2.093. However, for high-voltage systems, χ is not > 2.0. This gives i_{pAT} = (2)( \sqrt{2} )(15.65) = 44.27. For medium-voltage motors, χ can be calculated from Table 8.10:
χ_{MV} =\frac{ i_{p}}{ \sqrt{2}  I^{\prime \prime}_{k}}= \frac{7.39}{ \sqrt{2}(3.06)} = 1.71
For low-voltage motors through transformer T3:
χ_{LVT} = 1.02 + 0.98e^{-3(6.97/19.02)} = 1.346
i_{pLV} = (1.346)( \sqrt{2} )(0.76) = 1.45 \ kA
Total peak current by summation = (44.27 +7.39 + 1.45) = 53.11 kA.
The breaking current is the summation of individual breaking currents:
Breaking current through transformers at 4.16 kV = | I^{\prime \prime}_{k}| = 15.65 kA
Breaking current medium-voltage motors from Table 8.10 = 1.78 kA
For low-voltage motors with I^{\prime \prime}_{KM}/I_{RM} = 6.6, μ = 0.78, q can be conservatively calculated for m ≤ 0.3 and p = 2. This gives q = 0.64. The component breaking current from low-voltage motors is therefore, 0.78 × 0.64 × 0.76 = 0.38 kA. Total symmetrical breaking current = 17.81 kA.
To calculate the asymmetrical breaking current, the dc components of the currents should be calculated:
The dc component of the low-voltage motor contribution is practically zero.
The dc component of the medium-voltage motors at contact parting time of 0.05 s = 0.5 kA.
The dc component of current through transformer T2 = 6.07 kA.
Total dc current at contact parting time = 6.57 kA; this gives asymmetrical breaking current of 18.98 kA.
The results are shown in Table 8.8.