Design: 330 m head plant, 132 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 2,000 MVA at the 132 kV grid substation at the far end of the line (an assumed typical value, since grid data is not given). The line is taken as a single circuit (not stated).

(i) Installed capacity at $Q_{45\%}$

Driest month = March (4.55 m³/s), so riparian release $Q_r = 0.1 \times 4.55 = 0.455$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 9.82 - 0.455 = 9.365\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 330 - 16.50 = 313.50\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 9.365 \times 313.50 = 28.801\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 28.801 = 25.921\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 25.921 = 25.144\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 25.144 = 24.892\ \text{MW} \end{aligned}$$

Answer: generator output = 25.144 MW; power delivered to the 132 kV bus = 24.892 MW.

(ii) Number of units

• 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: $Q = 9.365/2 = 4.683$ m³/s, generator output $= 25.144/2 = 12.572$ MW → rate each unit at 12.6 MW.
• Generator rating: $S = 12.6/0.85 = 14.82$ MVA → 15 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 16 MVA, 11/132 kV, YNd11 (next standard size above 15 MVA).

Answer: installed capacity = 2 × 12.6 MW = 25.2 MW.

(iii) Type of turbine

$H_n = 313.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 25.921/2 = 12960.7$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1319.2$:

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 $N = 428.6$ rpm (14 poles): $N_s = 37.0$. With 4 jets, $N_{s,jet} = 37.0/\sqrt{4} = 18.5$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 76.9$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.58$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.139$ m; $D/d = 11.3 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 86.3$) 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, 428.6 rpm, about 12.96 MW each.

(iv) Generator–transformer scheme

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

Reasons: the units are large (15 MVA each, about 787 A at 11 kV), the transmission voltage is 132 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.

(v) Single line diagram

 Line to grid  (132 kV, 20 km)
    |
  [CB]
    |
 ===+====+================+=== 132 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.

Bus bar used in the SLD

• 132 kV switchyard: double main bus bar with a bus coupler. Each bay (transformer or line) can be put on either bus, so one bus can be maintained without shutting down the plant and a bus fault affects only the circuits on that bus. This is the NEA practice at 132 kV and above.
• 11 kV: no common bus (unit scheme); each generator connects to its transformer through the GCB.

(vi) Rating of the transmission line side circuit breaker (HVCB)

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{15} = 1.4000\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{16} = 0.5719\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 20 = 5.969\ \Omega \\ x_l &= \frac{X_L \times 100}{132^2} = \frac{5.969}{174.24} = 0.0343\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\ x_{grid\ path} &= 0.0500 + 0.0343 = 0.0843\ \text{pu} \end{aligned}$$

For a fault on the 132 kV bus, the plant and the grid feed in parallel:

$$\begin{aligned} X_{plant} &= \frac{x_g + x_t}{2} = \frac{1.4000 + 0.5719}{2} = 0.9859\ \text{pu} \\ S_{HV} &= \frac{100}{0.9859} + \frac{100}{0.0843} = 101.4 + 1186.8 = 1288.3\ \text{MVA} \\ I_{sc} &= \frac{1288.3}{\sqrt{3} \times 132} = 5.63\ \text{kA}, \quad I_{make} = 2.55 \times 5.63 = 14.37\ \text{kA peak} \\ I_{FL} &= \frac{2 \times 16}{\sqrt{3} \times 132} = 140\ \text{A}, \quad 1.25 I_{FL} = 175\ \text{A} \end{aligned}$$

Answer: HV bus fault level = 1288.3 MVA (5.63 kA); select a 145 kV, 1250 A, 31.5 kA SF6 circuit breaker. Of this, the plant itself contributes only 101.4 MVA; the rest comes from the grid, so the result depends on the assumed grid fault level.