Design: 95 m head plant at Q65%

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 1,000 MVA at the 66 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 $Q_{65\%}$

Driest month = April (1 m³/s), so riparian release $Q_r = 0.1 \times 1 = 0.1$ m³/s.

$$\begin{aligned} Q_d &= Q_{65\%} - Q_r = 3.7 - 0.1 = 3.600\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 95 - 4.75 = 90.25\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 3.600 \times 90.25 = 3.187\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 3.187 = 2.869\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 2.869 = 2.782\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 2.782 = 2.727\ \text{MW} \end{aligned}$$

Answer: generator output = 2.782 MW; power delivered to the 66 kV bus = 2.727 MW.

b) Number of units and installed capacity

• 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: $Q = 3.600/2 = 1.800$ m³/s, generator output $= 2.782/2 = 1.391$ MW → rate each unit at 1.4 MW.
• Generator rating: $S = 1.4/0.85 = 1.65$ MVA → 2 MVA, 11 kV, 0.85 pf, 50 Hz.
• One step-up transformer for both units: 4 MVA, 11/66 kV, YNd11 (standard size covering 2 × 1.65 MVA = 3.3 MVA).

Answer: installed capacity = 2 × 1.4 MW = 2.8 MW.

c) Type of turbine

$H_n = 90.25$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 2.869/2 = 1434.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 278.2$:

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 $N = 1000$ rpm (6 poles): $N_s = 136.1$, which is in the medium Francis range. A Pelton would need many jets (single-jet $N_s$ far above 30), and Kaplan is ruled out because the head is above 40 m.

Answer: 2 × vertical-axis Francis turbines, 1000 rpm, about 1.43 MW each.

d) Type of bus bar

• 11 kV: single bus bar; the two small generators and the transformer feeder are connected to it through VCB panels.
• 66 kV: single bus bar with one transformer bay and one line bay.

Reason: the plant is only 2.8 MW with one line, so a simple single bus is the most economical; an outage of the bus means loss of a small block of power only, and maintenance can be planned for the dry season.

e) Single line diagram

 Line to grid  (66 kV, 80 km)
    |
  [CB]
    |
 ===+======+=== 66 kV bus
           |
         [CB]
           |
         (T1)
           |
         [CB]
           |
 ===+======+======+=========+=== 11 kV bus
    |             |         |
  [GCB]         [GCB]     [CB]
    |             |         |
  (G1)          (G2)      (ST)
                            |
 ===========================+===+ 0.4 kV
                                |
                              [ACB]
                                |
                              (DG) black start

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.

f) Rating of generator circuit breaker (GCB)

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.2 \times \frac{100}{2} = 10.0000\ \text{pu}, \quad x_t = 0.11 \times \frac{100}{4} = 2.7500\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.77\times10^{-3} \times 80 = 19.352\ \Omega \\ x_l &= \frac{X_L \times 100}{66^2} = \frac{19.352}{43.56} = 0.4443\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\ x_{grid\ path} &= 0.1000 + 0.4443 = 0.5443\ \text{pu} \end{aligned}$$

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):

$$\begin{aligned} S_{gen} &= 2 \times \frac{100}{10.0000} = 20.0\ \text{MVA} \\ S_{grid} &= \frac{100}{x_t + x_{grid\ path}} = \frac{100}{2.7500 + 0.5443} = 30.4\ \text{MVA} \\ S_{11kV} &= 20.0 + 30.4 = 50.4\ \text{MVA} \end{aligned}$$ $$\begin{aligned} I_{sc} &= \frac{50.4}{\sqrt{3} \times 11} = 2.64\ \text{kA}, \quad I_{make} = 2.55 \times 2.64 = 6.74\ \text{kA peak} \\ I_{FL} &= \frac{2}{\sqrt{3} \times 11} = 105\ \text{A}, \quad 1.25 I_{FL} = 131\ \text{A} \end{aligned}$$

Answer: GCB fault level = 50.4 MVA; select a 12 kV, 630 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) (contd.) Rating of HV circuit breaker

For a fault on the 66 kV bus, the plant and the grid feed in parallel:

$$\begin{aligned} X_{plant} &= \frac{x_g}{2} + \frac{x_t}{1} = 5.0000 + 2.7500 = 7.7500\ \text{pu} \\ S_{HV} &= \frac{100}{7.7500} + \frac{100}{0.5443} = 12.9 + 183.7 = 196.6\ \text{MVA} \\ I_{sc} &= \frac{196.6}{\sqrt{3} \times 66} = 1.72\ \text{kA}, \quad I_{make} = 2.55 \times 1.72 = 4.39\ \text{kA peak} \\ I_{FL} &= \frac{1 \times 4}{\sqrt{3} \times 66} = 35\ \text{A} \end{aligned}$$

Answer: HV bus fault level = 196.6 MVA (1.72 kA); select a 72.5 kV, 630 A, 25 kA SF6 circuit breaker. Of this, the plant itself contributes only 12.9 MVA; the rest comes from the grid, so the result depends on the assumed grid fault level.