Design: 310 m head plant, HV breaker

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).

Transmission voltage and circuits (assumed as required)

Use Still's formula for the economical voltage (L in km, P in kW):

$$\begin{aligned} V &= 5.5\sqrt{\frac{L}{1.6} + \frac{P}{100}} = 5.5\sqrt{\frac{30}{1.6} + \frac{56700}{100}} = 133.1\ \text{kV} \end{aligned}$$

Choose 132 kV, the nearest standard voltage, with a double-circuit line so that the 56.7 MW output is not lost when one circuit trips (N-1 security).

a) Installed capacity at $Q_{45\%}$

Driest month = March (6.58 m³/s), so riparian release $Q_r = 0.1 \times 6.58 = 0.658$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 22.87 - 0.658 = 22.212\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 310 - 15.50 = 294.50\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 22.212 \times 294.50 = 64.171\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 64.171 = 58.396\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 58.396 = 56.644\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 56.644 = 56.078\ \text{MW} \end{aligned}$$

Answer: generator output = 56.644 MW; power delivered to the 132 kV bus = 56.078 MW.

a) (contd.) Number of units

• Choose 3 units. A plant of this size would need very large single machines with only two units, which are hard to transport on Nepal's hill roads; three units also let the plant follow the large seasonal change in flow (one or two units in the dry months, all three in the wet months).
• Loss of one unit (forced outage or maintenance) removes only one third of the output.
• Per unit: $Q = 22.212/3 = 7.404$ m³/s, generator output $= 56.644/3 = 18.881$ MW → rate each unit at 18.9 MW.
• Generator rating: $S = 18.9/0.85 = 22.24$ MVA → 22.5 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 25 MVA, 11/132 kV, YNd11 (next standard size above 22.5 MVA).

Answer: installed capacity = 3 × 18.9 MW = 56.7 MW.

Type of turbine (for unit speed)

$H_n = 294.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 58.396/3 = 19465.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1220.0$:

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 $N = 333.3$ rpm (18 poles): $N_s = 38.1$. With 4 jets, $N_{s,jet} = 38.1/\sqrt{4} = 19.1$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 74.5$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.96$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.178$ m; $D/d = 11.0 > 10$, acceptable.

A Francis at 750 rpm ($N_s = 85.8$) 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: 3 × vertical-axis 4-jet Pelton turbines, 333.3 rpm, about 19.47 MW each.

b) Type of generator lead

Generator full-load current: $I = \dfrac{22.5 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 1181$ A.

Choose a segregated phase bus duct (SPBD) from the generator terminals to the GCB and the step-up transformer LV side.

• About 1181 A is too high for a practical number of parallel cables (several runs per phase, poor current sharing, many terminations).
• An isolated phase bus duct (IPB) is used for very large machines (above about 3000–5000 A, i.e. roughly 60–100 MVA or more at 11 kV); it is not needed here.
• SPBD has each phase in its own compartment of an earthed metal enclosure, so a phase-to-phase fault is very unlikely; it is also compact, self-cooled and low-maintenance.
• Rating: 12 kV, 1600 A continuous, short-time withstand ≥ the GCB rating (e.g. 25 kA for 1 s), aluminium bars, with tap-off for the station transformer, surge capacitors and PTs.

c) Generator–transformer scheme and reasons

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 (22.5 MVA each, about 1181 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.

d) Single line diagram with earthing, excitation and bus bar

 Line-1          Line-2  (132 kV, 30 km)
    |               |
  [CB]            [CB]
    |               |
 ===+===+===========+===========+=== 132 kV bus
        |           |           |
      [CB]        [CB]        [CB]
        |           |           |
      (T1)        (T2)        (T3)
        |           |           |
 ST1 ---+           +           +--- ST2
        |           |           |
      [GCB]       [GCB]       [GCB]
        |           |           |
      (G1)        (G2)        (G3)
        |           |           |
      [NGT]       [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.

Choices shown on the SLD: earthing – each generator neutral through a neutral grounding transformer with secondary resistor (high-resistance earthing, fault current about 10 A), the HV star of each GSU solidly earthed, and all equipment bonded to the powerhouse/switchyard earth mat; excitation – static (thyristor) excitation fed from an excitation transformer at each generator terminal, with field flashing from the 110 V DC battery (fast response for a large grid-connected unit); bus bar – 132 kV double main bus with bus coupler, two line bays for the double-circuit line.

e) Rating of high voltage circuit breaker

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{22.5} = 0.9333\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{25} = 0.3660\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 30 = 8.954\ \Omega \text{ per circuit; two in parallel} = 4.477\ \Omega \\ x_l &= \frac{X_L \times 100}{132^2} = \frac{4.477}{174.24} = 0.0257\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\ x_{grid\ path} &= 0.0500 + 0.0257 = 0.0757\ \text{pu} \end{aligned}$$

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

$$\begin{aligned} X_{plant} &= \frac{x_g + x_t}{3} = \frac{0.9333 + 0.3660}{3} = 0.4331\ \text{pu} \\ S_{HV} &= \frac{100}{0.4331} + \frac{100}{0.0757} = 230.9 + 1321.1 = 1552.0\ \text{MVA} \\ I_{sc} &= \frac{1552.0}{\sqrt{3} \times 132} = 6.79\ \text{kA}, \quad I_{make} = 2.55 \times 6.79 = 17.31\ \text{kA peak} \\ I_{FL} &= \frac{3 \times 25}{\sqrt{3} \times 132} = 328\ \text{A}, \quad 1.25 I_{FL} = 410\ \text{A} \end{aligned}$$

Answer: HV bus fault level = 1552.0 MVA (6.79 kA); select a 145 kV, 1250 A, 31.5 kA SF6 circuit breaker. Of this, the plant itself contributes only 230.9 MVA; the rest comes from the grid, so the result depends on the assumed grid fault level.