Design: 250 m head plant at Q25%

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 500 MVA at the 33 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 $Q_{25\%}$

Driest month = May (3.5 m³/s), so riparian release $Q_r = 0.1 \times 3.5 = 0.35$ m³/s.

$$\begin{aligned} Q_d &= Q_{25\%} - Q_r = 18.24 - 0.35 = 17.890\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 250 - 12.50 = 237.50\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 17.890 \times 237.50 = 41.681\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 41.681 = 37.513\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 37.513 = 36.388\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 36.388 = 35.660\ \text{MW} \end{aligned}$$

Answer: generator output = 36.388 MW; power delivered to the 33 kV bus = 35.660 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: $Q = 17.890/3 = 5.963$ m³/s, generator output $= 36.388/3 = 12.129$ MW → rate each unit at 12.2 MW.
• Generator rating: $S = 12.2/0.85 = 14.35$ MVA → 14.5 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 16 MVA, 11/33 kV, YNd11 (next standard size above 14.5 MVA).

Answer: installed capacity = 3 × 12.2 MW = 36.6 MW.

c) Type of turbine

$H_n = 237.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 37.513/3 = 12504.4$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 932.4$:

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): $N_s = 40.0$. With 4 jets, $N_{s,jet} = 40.0/\sqrt{4} = 20.0$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 66.9$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.76$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.168$ m; $D/d = 10.5 > 10$, acceptable.

A Francis at 750 rpm ($N_s = 90.0$) 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 12.50 MW each.

d) Single line diagram

 Line-1          Line-2  (33 kV, 55 km)
    |               |
  [CB]            [CB]
    |               |
 ===+===+===========+===========+=== 33 kV bus
        |           |           |
      [CB]        [CB]        [CB]
        |           |           |
      (T1)        (T2)        (T3)
        |           |           |
      [CB]        [CB]        [CB]
        |           |           |
 =======+===[BS]====+===========+=========+=== 11 kV bus
        |           |           |         |
      [GCB]       [GCB]       [GCB]     [CB]
        |           |           |         |
      (G1)        (G2)        (G3)      (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) Type of bus bar

• 11 kV: single bus bar with a bus-section breaker, units split between the two sections, so a bus fault or maintenance takes out only part of the plant.
• 33 kV: single bus bar, sectionalised, with transformer bays and two line bays (double circuit).

Reason: for a 36.6 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.

f) Rating of generator circuit breaker (GCB)

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{14.5} = 1.4483\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{16} = 0.5719\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 55 = 16.415\ \Omega \text{ per circuit; two in parallel} = 8.207\ \Omega \\ x_l &= \frac{X_L \times 100}{33^2} = \frac{8.207}{10.89} = 0.7537\ \text{pu}, \quad x_{grid} = \frac{100}{500} = 0.2000\ \text{pu} \\ x_{grid\ path} &= 0.2000 + 0.7537 = 0.9537\ \text{pu} \end{aligned}$$

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):

$$\begin{aligned} S_{gen} &= 3 \times \frac{100}{1.4483} = 207.1\ \text{MVA} \\ S_{grid} &= \frac{100}{x_t/3 + x_{grid\ path}} = \frac{100}{0.1906 + 0.9537} = 87.4\ \text{MVA} \\ S_{11kV} &= 207.1 + 87.4 = 294.5\ \text{MVA} \end{aligned}$$ $$\begin{aligned} I_{sc} &= \frac{294.5}{\sqrt{3} \times 11} = 15.46\ \text{kA}, \quad I_{make} = 2.55 \times 15.46 = 39.42\ \text{kA peak} \\ I_{FL} &= \frac{14.5}{\sqrt{3} \times 11} = 761\ \text{A}, \quad 1.25 I_{FL} = 951\ \text{A} \end{aligned}$$

Answer: GCB fault level = 294.5 MVA; select a 12 kV, 1250 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.