Design: 360 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).

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

Driest month = March (5.58 m³/s), so riparian release $Q_r = 0.1 \times 5.58 = 0.558$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 18.87 - 0.558 = 18.312\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 360 - 18.00 = 342.00\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 18.312 \times 342.00 = 61.437\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 61.437 = 55.293\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 55.293 = 53.635\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 53.635 = 52.562\ \text{MW} \end{aligned}$$

Answer: generator output = 53.635 MW; power delivered to the 132 kV bus = 52.562 MW.

b) Number of units and installed capacity

• 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: $Q = 18.312/3 = 6.104$ m³/s, generator output $= 53.635/3 = 17.878$ MW → rate each unit at 17.9 MW.
• Generator rating: $S = 17.9/0.85 = 21.06$ MVA → 21.5 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 25 MVA, 11/132 kV, YNd11 (next standard size above 21.5 MVA).

Answer: installed capacity = 3 × 17.9 MW = 53.7 MW.

Type of turbine (for unit speed)

$H_n = 342.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 55.293/3 = 18431.1$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1470.7$:

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): $N_s = 39.6$. With 4 jets, $N_{s,jet} = 39.6/\sqrt{4} = 19.8$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 80.3$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.65$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.156$ m; $D/d = 10.6 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 92.3$) 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: 3 × vertical-axis 4-jet Pelton turbines, 428.6 rpm, about 18.43 MW each.

c) 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 (21.5 MVA each, about 1128 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

 Line-1          Line-2  (132 kV, 22 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.

d) (contd.) Bus bar system

• 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.

e) MVA rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.19 \times \frac{100}{21.5} = 0.8837\ \text{pu}, \quad x_t = 0.0815 \times \frac{100}{25} = 0.3260\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 22 = 6.566\ \Omega \text{ per circuit; two in parallel} = 3.283\ \Omega \\ x_l &= \frac{X_L \times 100}{132^2} = \frac{3.283}{174.24} = 0.0188\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\ x_{grid\ path} &= 0.0500 + 0.0188 = 0.0688\ \text{pu} \end{aligned}$$

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).

$$\begin{aligned} S_a &= \frac{100}{x_g} = \frac{100}{0.8837} = 113.2\ \text{MVA} \\ X_{ext} &= x_{grid\ path} \parallel (x_g + x_t)/2 = 0.0688 \parallel 0.6049 = 0.0618\ \text{pu} \\ S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.3260 + 0.0618} = 257.9\ \text{MVA} \end{aligned}$$

Design fault level for the GCB = larger value = 257.9 MVA.

$$\begin{aligned} I_{sc} &= \frac{257.9}{\sqrt{3} \times 11} = 13.53\ \text{kA}, \quad I_{make} = 2.55 \times 13.53 = 34.51\ \text{kA peak} \\ I_{FL} &= \frac{21.5}{\sqrt{3} \times 11} = 1128\ \text{A}, \quad 1.25 I_{FL} = 1411\ \text{A} \end{aligned}$$

Answer: GCB fault level = 257.9 MVA; select a 12 kV, 1600 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.