Design: 280 m head plant, 66 kV

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 (3.2 m³/s), so riparian release $Q_r = 0.1 \times 3.2 = 0.32$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 7 - 0.32 = 6.680\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 280 - 14.00 = 266.00\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 6.680 \times 266.00 = 17.431\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 17.431 = 15.862\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 15.862 = 15.387\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 15.387 = 15.233\ \text{MW} \end{aligned}$$

Answer: generator output = 15.387 MW; power delivered to the 66 kV bus = 15.233 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: $Q = 6.680/2 = 3.340$ m³/s, generator output $= 15.387/2 = 7.693$ MW → rate each unit at 7.7 MW.
• Generator rating: $S = 7.7/0.85 = 9.06$ MVA → 9.5 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 10 MVA, 11/66 kV, YNd11 (next standard size above 9.5 MVA).

Answer: installed capacity = 2 × 7.7 MW = 15.4 MW.

c) Selection and recommendation of turbine

$H_n = 266.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 15.862/2 = 7931.2$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1074.2$:

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 $N = 500$ rpm (12 poles): $N_s = 41.5$. With 4 jets, $N_{s,jet} = 41.5/\sqrt{4} = 20.7$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 70.8$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.24$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.123$ m; $D/d = 10.2 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 82.9$) is also possible, but the Pelton is preferred because (i) its efficiency stays nearly flat from about 20% to 100% load (jets can be switched off), which suits a run-of-river plant whose flow falls far below the design flow in the dry season, and (ii) it tolerates the heavy sediment of Himalayan rivers better, and worn buckets and needles are easier to repair.

Answer: 2 × vertical-axis 4-jet Pelton turbines, 500 rpm, about 7.93 MW each.

From the efficiency versus $Q/Q_0$ curves: Pelton and full Kaplan stay above about 85% down to 20–30% flow, while Francis drops sharply below about 40–50% flow and a fixed propeller is the worst. Since this plant runs at part flow for much of the year, the flat Pelton curve supports the choice.

d) Single line diagram

 Line-1          Line-2  (66 kV, 30 km)
    |               |
  [CB]            [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.

e) Rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{9.5} = 2.2105\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{10} = 0.9150\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.92\times10^{-3} \times 30 = 8.671\ \Omega \text{ per circuit; two in parallel} = 4.335\ \Omega \\ x_l &= \frac{X_L \times 100}{66^2} = \frac{4.335}{43.56} = 0.0995\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\ x_{grid\ path} &= 0.1000 + 0.0995 = 0.1995\ \text{pu} \end{aligned}$$

All generators feed the common 11 kV bus, so the GCB is rated for the full 11 kV bus fault level (generators plus the grid through the step-up transformers):

$$\begin{aligned} S_{gen} &= 2 \times \frac{100}{2.2105} = 90.5\ \text{MVA} \\ S_{grid} &= \frac{100}{x_t/2 + x_{grid\ path}} = \frac{100}{0.4575 + 0.1995} = 152.2\ \text{MVA} \\ S_{11kV} &= 90.5 + 152.2 = 242.7\ \text{MVA} \end{aligned}$$ $$\begin{aligned} I_{sc} &= \frac{242.7}{\sqrt{3} \times 11} = 12.74\ \text{kA}, \quad I_{make} = 2.55 \times 12.74 = 32.48\ \text{kA peak} \\ I_{FL} &= \frac{9.5}{\sqrt{3} \times 11} = 499\ \text{A}, \quad 1.25 I_{FL} = 623\ \text{A} \end{aligned}$$

Answer: GCB fault level = 242.7 MVA; select a 12 kV, 630 A, 25 kA generator circuit breaker. The margin covers the DC component and delayed current zeros of generator faults; surge arresters and surge capacitors are fitted at the generator terminals when a VCB is used.