Design: 550 m head plant, transformer quotation

Assumptions: riparian release = 10% of the driest monthly mean flow; net head = gross head minus 4.5% 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 (3.44 m³/s), so riparian release $Q_r = 0.1 \times 3.44 = 0.344$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 16.17 - 0.344 = 15.826\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.045) = 550 - 24.75 = 525.25\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 15.826 \times 525.25 = 81.547\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 81.547 = 74.207\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 74.207 = 71.981\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 71.981 = 71.261\ \text{MW} \end{aligned}$$

Answer: generator output = 71.981 MW; power delivered to the 132 kV bus = 71.261 MW.

b) 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: $Q = 15.826/3 = 5.275$ m³/s, generator output $= 71.981/3 = 23.994$ MW → rate each unit at 24 MW.
• Generator rating: $S = 24/0.85 = 28.24$ MVA → 28.5 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 31.5 MVA, 11/132 kV, YNd11 (next standard size above 28.5 MVA).

Answer: installed capacity = 3 × 24 MW = 72 MW.

b) (contd.) Turbine type

$H_n = 525.25$ m is above the Francis limit (350 m), so only a Pelton turbine is possible; specific speed fixes the speed and number of jets.

Shaft power per unit $P = 74.207/3 = 24735.8$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 2514.5$:

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 $N = 600$ rpm (10 poles): $N_s = 37.5$. With 4 jets, $N_{s,jet} = 37.5/\sqrt{4} = 18.8$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 99.5$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.46$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.130$ m; $D/d = 11.2 > 10$, acceptable.

The Pelton also keeps high efficiency at part load and handles sediment well.

Answer: 3 × vertical-axis 4-jet Pelton turbines, 600 rpm, about 24.74 MW each.

c) Generator lead and reason

Generator full-load current: $I = \dfrac{28.5 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 1496$ A.

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

• About 1496 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 3000–4000 A, i.e. more than about 60–75 MVA 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, 2000 A continuous, short-time withstand ≥ the GCB rating (25 kA for 1 s), aluminium bars, with tap-off for the station transformer, surge capacitors and PTs.

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: the units are large (28.5 MVA each, about 1496 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) (contd.) Single line diagram

 Line to grid  (132 kV, 25 km)
    |
  [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) Rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{28.5} = 0.7368\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{31.5} = 0.2905\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 25 = 7.461\ \Omega \\ x_l &= \frac{X_L \times 100}{132^2} = \frac{7.461}{174.24} = 0.0428\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\ x_{grid\ path} &= 0.0500 + 0.0428 = 0.0928\ \text{pu} \end{aligned}$$

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).

$$\begin{aligned} S_a &= \frac{100}{x_g} = \frac{100}{0.7368} = 135.7\ \text{MVA} \\ X_{ext} &= x_{grid\ path} \parallel (x_g + x_t)/2 = 0.0928 \parallel 0.5137 = 0.0786\ \text{pu} \\ S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.2905 + 0.0786} = 270.9\ \text{MVA} \end{aligned}$$

Design fault level for the GCB = larger value = 270.9 MVA.

$$\begin{aligned} I_{sc} &= \frac{270.9}{\sqrt{3} \times 11} = 14.22\ \text{kA}, \quad I_{make} = 2.55 \times 14.22 = 36.26\ \text{kA peak} \\ I_{FL} &= \frac{28.5}{\sqrt{3} \times 11} = 1496\ \text{A}, \quad 1.25 I_{FL} = 1870\ \text{A} \end{aligned}$$

Answer: GCB fault level = 270.9 MVA; select a 12 kV, 2000 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.

f) Tentative quotation for the generator transformers

Requirement: 3 generator step-up transformers, one per unit (unit scheme), each above the generator rating of 28.5 MVA, so 31.5 MVA, 11/132 kV units are quoted. The site is in a hilly region, so weight and size must suit hill-road transport.

Terms: delivery 8–10 months after LC to site; routine tests witnessed, type test reports supplied; warranty 24 months after commissioning; payment by LC; validity 120 days. Prices are indicative for budgeting only; actual prices come from competitive bids.