Design: 200 m head plant, bus MVA levels

Take riparian release as 10% of the driest monthly mean flow, net head = gross head minus 5% loss, and two generating units. The grid's fault contribution is neglected because grid data is not given (only the plant's generators feed a fault).

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

Driest month = May (3.22 m³/s), so riparian release $= 0.1 \times 3.22 = 0.322$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 6.82 - 0.322 = 6.498\ \text{m}^3/\text{s} \\ H_n &= H_g - 0.05H_g = 200 - 10.0 = 190.0\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 1000 \times 9.81 \times 6.498 \times 190.0 = 12.112\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.90 \times 12.112 = 10.900\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 10.900 = 10.573\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 10.573 = 10.362\ \text{MW} \end{aligned}$$

Answer: generator output = 10.573 MW; power delivered to the 33 kV bus = 10.362 MW.

b) Number of units and installed capacity

• Choose 2 units. The plant runs on $Q_{45\%}$, so in the dry season (flow below about $Q_{65\%}$) one unit can still run near its best efficiency; 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: $Q = 6.498/2 = 3.249$ m³/s, generator output $= 10.573/2 = 5.287$ MW.
• Generator rating: 5.3 MW at 0.85 pf, i.e. $S = 5.3/0.85 = 6.24$ MVA → 6.25 MVA, 11 kV, 50 Hz.
• Step-up transformer: one per unit, 6.3 MVA, 11/33 kV, YNd11, 9.15% (next standard size above 6.25 MVA).

Answer: installed capacity = 2 × 5.3 MW = 10.6 MW.

c) Single line diagram

     Line-1 33 kV        Line-2 33 kV
     (25 km)             (25 km)
         |                   |
        [CB]                [CB]
         |                   |
 33 kV ==+=======[BS]========+== 
         |                   |
        [CB]                [CB]
         |                   |
        (T1)                (T2)
     11/33 kV 6.3 MVA     11/33 kV 6.3 MVA
         |                   |
        [CB]                [CB]
         |                   |
 11 kV ==+=====[BS]======+===+===+==
         |               |       |
       [GCB]           [GCB]    [CB]
         |               |       |
        (G1)            (G2)    (ST) 11/0.4 kV
      6.25 MVA       6.25 MVA       250 kVA
                                 |
               0.4 kV aux bus ===+===+==
                                     |
                                   [ACB]
                                     |
                                   (DG) 200 kVA

BS = bus section breaker; ST = station (auxiliary) transformer; DG = diesel generator for black start. With the grid dead, the DG energises the 0.4 kV auxiliary bus (governor oil pumps, cooling water, excitation field flashing, lighting, battery chargers), so a unit can be started without outside supply.

d) Selection of turbine

$H_n = 190.0$ m lies in both the Francis (10–350 m) and Pelton (50–1300 m) ranges, so specific speed decides. Per-unit shaft power $P = 10.900/2 = 5450$ kW; take $N = 1000$ rpm (6-pole, 50 Hz synchronous speed).

$$\begin{aligned} N_s &= \frac{N\sqrt{P}}{H_n^{5/4}} = \frac{1000 \times 73.83}{705.4} = 104.7\ \text{(metric, kW)} \end{aligned}$$

A single-jet Pelton suits $N_s \approx 10$–$35$ (up to about 60–70 with multiple jets); a Francis suits $N_s \approx 60$–$300$. $N_s = 104.7$ falls in the slow-to-medium Francis range, and a Francis has good efficiency at this discharge per unit (about 3.249 m³/s).

Answer: 2 × vertical-axis Francis turbines, 1000 rpm.

e) Type of bus bar

• 11 kV: single bus bar with a bus-section breaker. Each half carries one unit, so a bus fault or maintenance takes out only one unit.
• 33 kV switchyard: single bus bar, sectionalised, with one bay for each circuit of the double-circuit line. This is simple and cheap for a 10.6 MW, 2-unit plant. A double bus or main-and-transfer bus costs more and is normally used only in larger plants or grid nodes.

f) MVA (fault) level of the LV, MV and HV buses

Take LV = 0.4 kV auxiliary bus, MV = 11 kV generator bus, HV = 33 kV switchyard bus. Base = 10 MVA. Station transformer assumed 250 kVA, 4.5%.

$$\begin{aligned} x_g &= 0.21 \times \frac{10}{6.25} = 0.3360\ \text{pu}, \quad x_t = 0.0915 \times \frac{10}{6.3} = 0.1452\ \text{pu} \\ x_{st} &= 0.045 \times \frac{10}{0.25} = 1.8\ \text{pu} \end{aligned}$$

Line check (remote end): $X_L = 2\pi \times 50 \times 0.95\times10^{-3} \times 25 = 7.461\ \Omega$ per circuit, 3.731 Ω for both circuits in parallel. $Z_b = 33^2/10 = 108.90\ \Omega$, so $x_L = 0.0343$ pu. Fault level at the far end of the line is $10/0.2749 = 36.4$ MVA (0.636 kA).

Answer: LV ≈ 5.08 MVA, MV ≈ 59.5 MVA, HV ≈ 41.6 MVA.

g) Rating of circuit breakers

Making current = $2.55 \times I_{sc}$ (peak, IEC 62271-100). Rated current ≥ 1.25 × full-load current.

Answer: GCB: 12 kV, 630 A, breaking capacity ≥ 59.5 MVA (3.12 kA); 33 kV CB: 36 kV, 630 A, breaking capacity ≥ 41.6 MVA (0.727 kA). Standard 25 kA breakers are chosen because they cost little more and leave margin for the grid's fault contribution and future units.