Design: 220 m head plant, optimal voltage

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).

Optimal transmission voltage and number of circuits

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

$$\begin{aligned} V &= 5.5\sqrt{\frac{L}{1.6} + \frac{P}{100}} = 5.5\sqrt{\frac{25}{1.6} + \frac{11800}{100}} = 63.6\ \text{kV} \end{aligned}$$

The nearest standard voltage is 66 kV. Line current at 66 kV: $I = \dfrac{11.8/0.85}{\sqrt3 \times 66} = 121$ A, which a single ACSR Dog or Wolf circuit carries easily. Choose 66 kV, single circuit: an 11.8 MW plant does not justify the extra cost of a second circuit, and 33 kV would give high losses and voltage drop over 25 km.

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

Driest month = May (3.22678 m³/s), so riparian release $Q_r = 0.1 \times 3.22678 = 0.3227$ m³/s.

$$\begin{aligned} Q_d &= Q_{45\%} - Q_r = 6.87999 - 0.3227 = 6.557\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 220 - 11.00 = 209.00\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 6.557 \times 209.00 = 13.444\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 13.444 = 12.100\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 12.100 = 11.737\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 11.737 = 11.502\ \text{MW} \end{aligned}$$

Answer: generator output = 11.737 MW; power delivered to the 66 kV bus = 11.502 MW.

b) Number of units (turbine type in part c)

• 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: $Q = 6.557/2 = 3.279$ m³/s, generator output $= 11.737/2 = 5.868$ MW → rate each unit at 5.9 MW.
• Generator rating: $S = 5.9/0.85 = 6.94$ MVA → 7 MVA, 11 kV, 0.85 pf, 50 Hz.
• Step-up transformer, one per unit: 8 MVA, 11/66 kV, YNd11 (next standard size above 7 MVA).

Answer: installed capacity = 2 × 5.9 MW = 11.8 MW.

c) Type of turbine

$H_n = 209.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 12.100/2 = 6050.0$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 794.7$:

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 $N = 375$ rpm (16 poles): $N_s = 36.7$. With 4 jets, $N_{s,jet} = 36.7/\sqrt{4} = 18.4$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 62.8$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.47$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.129$ m; $D/d = 11.4 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 97.9$) 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: 2 × vertical-axis 4-jet Pelton turbines, 375 rpm, about 6.05 MW each.

d) Single line diagram

 Line to grid  (66 kV, 25 km)
    |
  [CB]
    |
 ===+=====+=============+=== 66 kV bus
          |             |
        [CB]          [CB]
          |             |
        (T1)          (T2)
          |             |
        [CB]          [CB]
          |             |
 =========+====[BS]=====+=========+=== 11 kV bus
          |             |         |
        [GCB]         [GCB]     [CB]
          |             |         |
        (G1)          (G2)      (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) Rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$\begin{aligned} x_g &= 0.21 \times \frac{100}{7} = 3.0000\ \text{pu}, \quad x_t = 0.09 \times \frac{100}{8} = 1.1250\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 25 = 7.461\ \Omega \\ x_l &= \frac{X_L \times 100}{66^2} = \frac{7.461}{43.56} = 0.1713\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\ x_{grid\ path} &= 0.1000 + 0.1713 = 0.2713\ \text{pu} \end{aligned}$$

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):

$$\begin{aligned} S_{gen} &= 2 \times \frac{100}{3.0000} = 66.7\ \text{MVA} \\ S_{grid} &= \frac{100}{x_t/2 + x_{grid\ path}} = \frac{100}{0.5625 + 0.2713} = 119.9\ \text{MVA} \\ S_{11kV} &= 66.7 + 119.9 = 186.6\ \text{MVA} \end{aligned}$$ $$\begin{aligned} I_{sc} &= \frac{186.6}{\sqrt{3} \times 11} = 9.79\ \text{kA}, \quad I_{make} = 2.55 \times 9.79 = 24.97\ \text{kA peak} \\ I_{FL} &= \frac{7}{\sqrt{3} \times 11} = 367\ \text{A}, \quad 1.25 I_{FL} = 459\ \text{A} \end{aligned}$$

Answer: GCB fault level = 186.6 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.