Design: 260 m head plant, generator lead

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 = March (3.58 m³/s), so riparian release $Q_r = 0.1 \times 3.58 = 0.358$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 8.87 - 0.358 = 8.512\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 260 - 13.00 = 247.00\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 8.512 \times 247.00 = 20.625\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 20.625 = 18.563\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 18.563 = 18.006\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 18.006 = 17.646\ \text{MW} \end{aligned}$$

Answer: generator output = 18.006 MW; power delivered to the 66 kV bus = 17.646 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: $Q = 8.512/2 = 4.256$ m³/s, generator output $= 18.006/2 = 9.003$ MW → rate each unit at 9.1 MW.
• Generator rating: $S = 9.1/0.85 = 10.71$ MVA → 11 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 12.5 MVA, 11/66 kV, YNd11 (next standard size above 11 MVA).

Answer: installed capacity = 2 × 9.1 MW = 18.2 MW.

Type of turbine (for unit speed)

$H_n = 247.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 18.563/2 = 9281.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 979.2$:

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 $N = 375$ rpm (16 poles): $N_s = 36.9$. With 4 jets, $N_{s,jet} = 36.9/\sqrt{4} = 18.4$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 68.2$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.60$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.141$ m; $D/d = 11.3 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 98.4$) 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, 375 rpm, about 9.28 MW each.

c) Type of generator lead

Generator full-load current: $I = \dfrac{11 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 577$ A.

Choose 11 kV XLPE single-core copper cables (generator lead in cable trench/tray) or a non-segregated phase bus duct.

• At about 577 A, two runs per phase of 1C 240 mm² Cu XLPE cable (about 400–450 A per cable laid in trefoil after derating) carry the current with margin.
• Cables are cheap, flexible to route inside the powerhouse, and easy to terminate at the GCB panel and transformer.
• A bus duct (NSPBD/SPBD) becomes economic only for higher currents (above about 1000–1500 A); IPB only for very large units.

d) Generator–transformer scheme

Choose the unit scheme (one generator connected to its own step-up transformer, with a generator circuit breaker between them).

Reasons: each unit is 11 MVA (about 577 A at 11 kV), the transmission voltage is 66 kV, and the unit scheme keeps the 11 kV fault level and generator-voltage switchgear small. The GCB lets the unit be synchronised and tripped at 11 kV while the station transformer, tapped between the GCB and the step-up transformer, stays fed from the grid (back-feed) when the unit is stopped.

d) (contd.) Single line diagram

 Line-1          Line-2  (66 kV, 30 km)
    |               |
  [CB]            [CB]
    |               |
 ===+====+==========+=====+=== 66 kV bus
         |                |
       [CB]             [CB]
         |                |
       (T1)             (T2)
         |                |
  ST1 ---+                +--- ST2
         |                |
       [GCB]            [GCB]
         |                |
       (G1)             (G2)
         |                |
       [NGT]            [NGT]

 ST1, ST2 (11/0.4 kV)
    |               |
 ===+====[BS]=======+=== 0.4 kV aux bus
              |
            [ACB]
              |
            (DG)  diesel set, 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) (contd.) Bus bar system

• 11 kV: no common bus (unit scheme); each generator connects to its own step-up transformer through its GCB, and the station transformers are tapped off between the GCB and the transformer.
• 66 kV: single bus bar, sectionalised by a bus-section breaker, with the two transformer bays on different sections and two line bays (double circuit), so a bus fault or maintenance takes out only part of the plant.

Reason: for an 18.2 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.

e) Rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{11} = 1.9091\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{12.5} = 0.7320\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 30 = 8.954\ \Omega \text{ per circuit; two in parallel} = 4.477\ \Omega \\ x_l &= \frac{X_L \times 100}{66^2} = \frac{4.477}{43.56} = 0.1028\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\ x_{grid\ path} &= 0.1000 + 0.1028 = 0.2028\ \text{pu} \end{aligned}$$

A unit-connected GCB sees the larger of two currents: (a) a fault on the transformer side fed by its own generator, (b) a fault on the generator side fed by the system through the unit transformer (grid plus the other units).

$$\begin{aligned} S_a &= \frac{100}{x_g} = \frac{100}{1.9091} = 52.4\ \text{MVA} \\ X_{ext} &= x_{grid\ path} \parallel (x_g + x_t) = 0.2028 \parallel 2.6411 = 0.1883\ \text{pu} \\ S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.7320 + 0.1883} = 108.7\ \text{MVA} \end{aligned}$$

Design fault level for the GCB = larger value = 108.7 MVA.

$$\begin{aligned} I_{sc} &= \frac{108.7}{\sqrt{3} \times 11} = 5.70\ \text{kA}, \quad I_{make} = 2.55 \times 5.70 = 14.54\ \text{kA peak} \\ I_{FL} &= \frac{11}{\sqrt{3} \times 11} = 577\ \text{A}, \quad 1.25 I_{FL} = 722\ \text{A} \end{aligned}$$

Answer: GCB fault level = 108.7 MVA; select a 12 kV, 800 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.