Question 6.8: Using the manufacturer’s data in Table 6.1, calculate the ma......
Using the manufacturer’s data in Table 6.1, calculate the machine parameters in the d-q axes.
| TABLE 6.1 Generator Data |
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| Description | Symbol | Data | |
| Generator 112.1 MVA, 2-pole, 13.8 kV, 0.85 PF, 95.285 MW, 4690 A, 0.56 SCR, 235 field V, wye connected |
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| Per unit reactance data, direct axis | |||
| Saturated synchronous | X_{dv} | 1.949 | |
| Unsaturated synchronous | X_{d} | 1.949 | |
| Saturated transient | X_{dv}^{\prime} | 0.207 | |
| Unsaturated transient | X_{d}^{\prime} | 0.278 | |
| Saturated subtransient | X_{dv}^{\prime \prime} | 0.164 | |
| Unsaturated subtransient | X_{d}^{\prime \prime} | 0.193 | |
| Saturated negative sequence | X_{2v} | 0.137 | |
| Unsaturated negative sequence | X_{2I} | 0.185 | |
| Saturated zero sequence | X_{0v} | 0.092 | |
| Leakage reactance, overexcited | X_{0I} | 0.111 | |
| Leakage reactance, underexcited | X_{LM,OXE} X_{LM,UXE} |
0.164 0.164 |
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| Per unit reactance data, quadrature axis | |||
| Saturated synchronous | X_{qv} | 1.858 | |
| Unsaturated synchronous | X_{q} | 1.858 | |
| Unsaturated transient | X_{q}^{\prime} | 0.434 | |
| Saturated subtransient | X_{qv}^{\prime \prime} | 0.140 | |
| Unsaturated subtransient | X_{q}^{\prime \prime} | 0.192 | |
| Field time constant data, direct axis | |||
| Open circuit | T_{d0}^{\prime} | 5.615 | |
| Three-phase short-circuit transient | T_{d3}^{\prime} | 0.597 | |
| Line-to-line short-circuit transient | T_{d2}^{\prime} | 0.927 | |
| Line-to-neutral short-circuit transient | T_{d1}^{\prime} | 1.124 | |
| Short-circuit subtransient | T_{d}^{\prime \prime} | 0.015 | |
| Open circuit subtransient | T_{d0}^{\prime \prime} | 0.022 | |
| Field time constant data quadrature axis | |||
| Open circuit | T_{q0}^{\prime} | 0.451 | |
| Three-phase short-circuit transient | T_{q}^{\prime} | 0.451 | |
| Short-circuit subtransient | T_{q}^{\prime \prime} | 0.015 | |
| Open circuit subtransient | T_{q0}^{\prime \prime} | 0.046 | |
| Armature dc component time constant data | |||
| Three-phase short circuit | T_{a3} | 0.330 | |
| Line-to-line short circuit | T_{a2} | 0.330 | |
| Line-to-neutral short circuit | T_{a1} | 0.294 | |
Applying the equations in Section 6.9.1:
L_{ad} = L_{d} – l_{l} = 1.949 – 0.164 = 1.785 \ pu = kM_{F} = kM_{D} = M_{R}
L_{aq} = L_{q} – l_{l} = 1.858 – 0.164 = 1.694 \ pu = kM_{Q}
l_{f} = (1.758)(0.278 – 0.164)=(1.964 – 0.278) = 0.121 \ pu
L_{F} = 0.121 + 1.785 = 1.906 \ pu
L_{D} = 0.026 + 1.785 = 1.811 \ pu
l_{kq} = (1.694)(0.192 – 0.164)/(1.858 – 0.192) =0.028 \ pu
L_{Q} = 0.028 + 1.694 = 1.722 \ pu
T_{do}^{\prime} = 5.615s = 2116.85 \ rad
r_{F} = 1.906/2116.85 = 1.005 × 10^{-5} \ pu
r_{D} =\frac{ (1.811)(1.906) \ – \ 1.785^2}{(0.015)(377)(1.906)}(\frac{0.193}{0.278}) = 0.0131 \ pu
r_{Q} =(\frac{ 0.192}{1.858}) (\frac{ 1.722}{0.015 \ × \ 377 })= 0.031 \ pu
Note that in Table 6.1, as the data is in per unit, we consider X_{d} = L_{d}, \ X_{l} = l_{l}, etc. The per unit system for synchronous machines is not straightforward and variations in the literature exist. Reference [6] provides further reading.