Design: 150 m head plant, bus MVA levels

Assumptions: riparian release = 10% of the driest monthly mean flow; net head = gross head minus 5% loss; $\rho g = 9.81$ kN/m³; generator power factor 0.85; base = 100 MVA. For fault levels the grid is taken as a source of 1,000 MVA at the 66 kV grid substation at the far end of the line (an assumed typical value, since grid data is not given).

a. Electrical power output at $Q_{45\%}$

Driest month = May (4.35 m³/s), so riparian release $Q_r = 0.1 \times 4.35 = 0.435$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 11.17 - 0.435 = 10.735\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 150 - 7.50 = 142.50\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 10.735 \times 142.50 = 15.007\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 15.007 = 13.506\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 13.506 = 13.101\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 13.101 = 12.839\ \text{MW} \end{aligned}$$

Answer: generator output = 13.101 MW; power delivered to the 66 kV bus = 12.839 MW.

b. Number of units and installed capacity

• Choose 2 units. With a run-of-river flow that falls well below the design flow in the dry months, one unit can run near full load on half the design flow, so efficiency stays high in winter.
• One unit can be maintained while the other runs, and two units keep the number of machines (and cost) low for a plant of this size.
• Per unit: $Q = 10.735/2 = 5.367$ m³/s, generator output $= 13.101/2 = 6.550$ MW → rate each unit at 6.6 MW.
• Generator rating: $S = 6.6/0.85 = 7.76$ MVA → 8 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 8 MVA, 11/66 kV, YNd11 (standard size matching the 8 MVA generator).

Answer: installed capacity = 2 × 6.6 MW = 13.2 MW.

c. Single line diagram

 Line to grid  (66 kV, 35 km)
    |
  [CB]
    |
 ===+=====+=============+=== 66 kV bus
          |             |
        [CB]          [CB]
          |             |
        (T1)          (T2)
          |             |
        [CB]          [CB]
          |             |
 =========+====[BS]=====+=========+=== 11 kV bus
          |             |         |
        [GCB]         [GCB]     [CB]
          |             |         |
        (G1)          (G2)      (ST)
                                  |
 =================================+===+ 0.4 kV
                                      |
                                    [ACB]
                                      |
                                    (DG) black start

BS = bus section breaker, GCB = generator circuit breaker, ACB = air circuit breaker. Black start: with the grid dead, the DG energises the 0.4 kV auxiliary bus (governor oil pumps, cooling water, excitation field flashing, lighting and battery chargers), one unit is started and runs in island mode, and it then energises the step-up transformer and the line.

d. Type of turbine

$H_n = 142.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 13.506/2 = 6753.0$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 492.3$:

Usual ranges: Pelton about 10–30 per jet (multi-jet $N_s = N_{s,jet}\sqrt{z}$), Francis about 60–300, Kaplan 300–1000.

Choose $N = 750$ rpm (8 poles): $N_s = 125.2$, which is in the medium Francis range. A Pelton would need many jets (single-jet $N_s$ far above 30), and Kaplan is ruled out because the head is above 40 m.

Answer: 2 × vertical-axis Francis turbines, 750 rpm, about 6.75 MW each.

e. Type of bus bar

• 11 kV: single bus bar with a bus-section breaker, units split between the two sections, so a bus fault or maintenance takes out only part of the plant.
• 66 kV: single bus bar, sectionalised, with transformer bays and one line bay.

Reason: for a 13.2 MW plant a sectionalised single bus gives reasonable reliability at low cost; a double bus or main-and-transfer bus would cost much more and is normally used only for larger plants or grid substations.

f. MVA level of LV, MV and HV buses

LV = 0.4 kV auxiliary bus, MV = 11 kV generator bus, HV = 66 kV switchyard bus. Station transformer 200 kVA, 4.5%.

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.15 \times \frac{100}{8} = 1.8750\ \text{pu}, \quad x_t = 0.12 \times \frac{100}{8} = 1.5000\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.97\times10^{-3} \times 35 = 10.666\ \Omega \\ x_l &= \frac{X_L \times 100}{66^2} = \frac{10.666}{43.56} = 0.2449\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\ x_{grid\ path} &= 0.1000 + 0.2449 = 0.3449\ \text{pu} \end{aligned}$$ $$\begin{aligned} x_{st} &= 0.045 \times \frac{100}{0.2} = 22.500\ \text{pu} \\ S_{MV} &= 2 \times \frac{100}{x_g} + \frac{100}{x_t/2 + x_{grid\ path}} = 106.7 + 91.3 = 198.0\ \text{MVA} \\ S_{HV} &= \frac{100}{x_g/2 + x_t/2} + \frac{100}{x_{grid\ path}} = 59.3 + 290.0 = 349.2\ \text{MVA} \\ X_{MV} &= \frac{100}{198.0} = 0.5050\ \text{pu}, \quad S_{LV} = \frac{100}{0.5050 + 22.500} = 4.35\ \text{MVA} \end{aligned}$$

Answer: LV ≈ 4.35 MVA, MV ≈ 198.0 MVA, HV ≈ 349.2 MVA.

g. Rating of circuit breakers

Breaking current from the bus fault levels above; making current = 2.55 × $I_{sc}$ (IEC 62271-100); rated current ≥ 1.25 × full-load current.

*At 0.4 kV the making current uses the IEC 60947-2 factor n = 1.7 for this fault current.

Answer: 11 kV GCB: 12 kV, 630 A, 25 kA; 66 kV CB: 72.5 kV, 630 A, 25 kA; 0.4 kV ACB: 400 A, 25 kA.