---
title: "Design: 250 m head plant, 66 kV double circuit"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2071-250m-0-92mh"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2071-250m-0-92mh
site: BCT Engineering
---

# Design: 250 m head plant, 66 kV double circuit

Power Plant Design · Chapter 4: Electric System Design of a Power Plant

- Asked in: 2071 Magh, 16 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2071-250m-0-92mh
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-4

## Question

A typical hydro power plant has the following details

| Month | Discharge (m³/s) | Flow duration | Discharge (m³/s) |
|---|---|---|---|
| January | 4.9 | Qmax (Q0) | 211.00 m³/s |
| February | 3.86 | Q25% | 18.24 m³/s |
| March | 3.58 | Q45% | 6.80 m³/s |
| April | 4.55 | Q65% | 4.70 m³/s |
| May | 3.00 | Q85% | 3.10 m³/s |
| June | 12.77 | Q95% | 2.17 m³/s |
| July | 32.95 | Qmin | 0.64 m³/s |
| August | 40.26 |  |  |
| September | 31.23 |  |  |
| October | 15.05 |  |  |
| November | 7.38 |  |  |
| December | 4.85 |  |  |

| Turbine type | Head range (m) |
|---|---|
| Kaplan | 2 < H < 40 |
| Francis | 10 < H < 350 |
| Pelton | 50 < H < 1300 |

| Parameter | Value |
|---|---|
| Gross head | 250 m |
| Turbine efficiency | 91% |
| Generator efficiency | 97% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 21% |
| Transformer reactance | 9.15% |
| Transmission line inductance | 0.92 mH/km |
| Transmission line length | 35 km |
| Transmission voltage level | 66 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Double circuit |

Design a power plant so as to obtain the following parameters:
a) Electrical power output taking design discharge as Q45%
b) Select the numbers of units and installed capacity
c) Select the type of turbine
d) Draw the single line diagram showing generators, power transformers, station transformer and diesel generator for black start
e) Calculate the rating of generator circuit breaker to be used in your design
f) Suggest the type of bus bar to be used
(Assume suitable data if necessary and necessary graph is attached herewith)

## Short answer

Design discharge Q45% after 10% riparian release is 6.5 m³/s, net head 237.5 m; power delivered to 66 kV bus is 13.1 MW. Install two 6.7 MW Pelton turbines (4‑jet, 428.6 rpm), total 13.4 MW. Generator breakers rated 11 kV, 630 A, 25 kA; use single busbar with bus section breaker.

## Answer

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 $Q_{45\%}$

Driest month = May (3 m³/s), so riparian release $Q_r = 0.1 \times 3 = 0.3$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 6.8 - 0.3 = 6.500\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 250 - 12.50 = 237.50\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 6.500 \times 237.50 = 15.144\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 15.144 = 13.781\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 13.781 = 13.368\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 13.368 = 13.100\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 13.368 MW; power delivered to the 66 kV bus = **13.100 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: $Q = 6.500/2 = 3.250$ m³/s, generator output $= 13.368/2 = 6.684$ MW → rate each unit at **6.7 MW**.
- Generator rating: $S = 6.7/0.85 = 7.88$ MVA → **8 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **8 MVA, 11/66 kV, YNd11** (standard size matching the 8 MVA generator).

**Answer: installed capacity = 2 × 6.7 MW = 13.4 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 $P = 13.781/2 = 6890.6$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 932.4$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 89.0 |
| 750 | 8 | 66.8 |
| 600 | 10 | 53.4 |
| 500 | 12 | 44.5 |
| 428.6 | 14 | 38.2 |
| 375 | 16 | 33.4 |
| 333.3 | 18 | 29.7 |

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): $N_s = 38.2$. With **4 jets**, $N_{s,jet} = 38.2/\sqrt{4} = 19.1$, 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 $D = 60u/(\pi N) = 1.37$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.124$ m; $D/d = 11.0 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 89.0$) 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: 2 × vertical-axis 4-jet Pelton turbines, 428.6 rpm, about 6.89 MW each.**

#### d) Single line diagram

```text
 Line-1          Line-2  (66 kV, 35 km)
    |               |
  [CB]            [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
```

| Item | Rating |
|---|---|
| Generators G1–G2 | 8 MVA, 11 kV, 0.85 pf, 428.6 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T2 | 8 MVA each, 11/66 kV, YNd11, $X$ = 9.15% |
| Station transformer ST | 250 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 160 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 66 kV, double circuit, 35 km to grid substation |
| NGR | neutral grounding resistor (low-resistance earthing, one generator earthed at a time if preferred) |

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:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{8} = 2.6250\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{8} = 1.1438\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.92\times10^{-3} \times 35 = 10.116\ \Omega \text{ per circuit; two in parallel} = 5.058\ \Omega \\
x_l &= \frac{X_L \times 100}{66^2} = \frac{5.058}{43.56} = 0.1161\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\
x_{grid\ path} &= 0.1000 + 0.1161 = 0.2161\ \text{pu}
\end{aligned}
$$

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):

$$
\begin{aligned}
S_{gen} &= 2 \times \frac{100}{2.6250} = 76.2\ \text{MVA} \\
S_{grid} &= \frac{100}{x_t/2 + x_{grid\ path}} = \frac{100}{0.5719 + 0.2161} = 126.9\ \text{MVA} \\
S_{11kV} &= 76.2 + 126.9 = 203.1\ \text{MVA}
\end{aligned}
$$

$$
\begin{aligned}
I_{sc} &= \frac{203.1}{\sqrt{3} \times 11} = 10.66\ \text{kA}, \quad I_{make} = 2.55 \times 10.66 = 27.18\ \text{kA peak} \\
I_{FL} &= \frac{8}{\sqrt{3} \times 11} = 420\ \text{A}, \quad 1.25 I_{FL} = 525\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 525 A | 630 A |
| Breaking capacity | 203.1 MVA (10.66 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 27.18 kA peak | 63 kA peak |
| Short-time current | – | 25 kA, 3 s |
| Type | – | Vacuum (or SF6), generator-duty tested (IEEE C37.013) |

**Answer: GCB fault level = 203.1 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) 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.
- **66 kV:** single bus bar, sectionalised, with transformer bays and two line bays (double circuit).

Reason: for a 13.4 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With Qd = 6.5 m3/s and Hn = 237.5 m, which value is the power delivered to the 66 kV bus?
   A. 13.100 MW, after turbine, generator and transformer efficiencies
   B. 15.144 MW, the hydraulic power
   C. 13.781 MW, the shaft power
   D. 13.368 MW, the generator output only
   Answer: A. Phyd = 15.144 MW; x0.91 = 13.781; x0.97 = 13.368; x0.98 = 13.100 MW at the bus. (Key point: Power chain)

2. How was a 4-jet Pelton justified at 428.6 rpm?
   A. Ns = 38.2 is in the Kaplan range, so jets are added
   B. Ns = 38.2 for the machine; divided by sqrt(4) gives 19.1 per jet, inside the Pelton range
   C. Ns per jet = 38.2 x 4 = 152.8
   D. Jets are added to raise the head above 250 m
   Answer: B. For multi-jet Pelton Ns = Ns,jet x sqrt(z), so Ns,jet = 38.2/2 = 19.1, which is within about 10-30. (Key point: Pelton jets)

3. Why is the Pelton preferred over a 1000 rpm Francis here?
   A. Pelton has higher specific speed than Francis
   B. Pelton needs no nozzles or needles
   C. Flat efficiency from about 20% to 100% load and better tolerance of sediment
   D. Francis cannot be used above 200 m head
   Answer: C. Run-of-river flow drops far below design in dry months, and jets can be shut off to keep efficiency; Himalayan sediment is also tolerated better. (Key point: Why Pelton)

4. Why is the GCB rated for the full 11 kV bus fault level of 203.1 MVA?
   A. Only one generator can feed the fault
   B. The grid is ignored for GCBs
   C. The transformer reactance is added to the generator reactance
   D. All generators and the grid (through the step-up transformers) feed a fault on the common 11 kV bus
   Answer: D. S = generators (76.2 MVA) + grid through transformers (126.9 MVA) = 203.1 MVA, giving Isc = 10.66 kA. (Key point: GCB fault level)

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