2072 Kartik · 8 marks

Load sharing: 600 MVA (6%) and 500 MVA (4%)

Short answer

Steady-state frequency deviation is –0.24 Hz (59.76 Hz). Unit 1 generates 540 MW and Unit 2 generates 450 MW.

Two generating units have following ratings and droop: Unit 1: 600MVA, R1 = 6%; Unit 2: 500MVA, R2 = 4%. The units are operating in parallel sharing a load of 900 MW at the normal frequency of 60Hz. Unit 1 supplies 500 MW and unit 2 supplies 400MW. If the load is increased by 90MW calculate the steady state frequency deviation and new generation on each unit.

Both units drop along their droop lines to a common frequency; the 90 MW is shared in proportion to $P_r/R$ . MVA ratings are used as the MW base.

Method: a unit with droop $R$ (pu on its own rating $P_r$ ) and rated frequency $f_0$ changes its output by

\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f

where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$ .

Data: Unit 1: 600 MVA, $R_1 = 0.06$ , 500 MW; Unit 2: 500 MVA, $R_2 = 0.04$ , 400 MW; $f_0 = 60$ Hz; $\Delta P_L = 90$ MW.

\begin{aligned} K_1 &= \frac{600}{0.06 \times 60} = 166.67\ \text{MW/Hz} \\ K_2 &= \frac{500}{0.04 \times 60} = 208.33\ \text{MW/Hz} \\ \Delta f &= -\frac{90}{166.67 + 208.33} = -\frac{90}{375} = -0.24\ \text{Hz} \end{aligned}

In per unit: $\Delta f = -0.24/60 = -0.004$ pu. New frequency $= 59.76$ Hz.

\begin{aligned} \Delta P_1 &= 166.67 \times 0.24 = 40\ \text{MW} \\ \Delta P_2 &= 208.33 \times 0.24 = 50\ \text{MW} \end{aligned}

Per-unit cross-check (base 1000 MVA): $R_1 = 0.06 \times 1000/600 = 0.1$ , $R_2 = 0.04 \times 1000/500 = 0.08$ ; $\Delta f = -0.09/(10 + 12.5) = -0.004$ pu ✓.

Unit	Initial (MW)	Increase (MW)	New (MW)
1	500	40	540
2	400	50	450

Answer: Steady-state frequency deviation = −0.24 Hz (−0.004 pu), i.e. 59.76 Hz; Unit 1 = 540 MW, Unit 2 = 450 MW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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