---
title: "Design: 310 m head plant, HV breaker"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2080-310m-hvcb"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2080-310m-hvcb
site: BCT Engineering
---

# Design: 310 m head plant, HV breaker

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

- Asked in: 2080 Chaitra, 3+2+3+4+4 marks
- Video: [power plant design | calculation of fault current and fault MVA @smartstudy11](https://www.youtube.com/watch?v=lwieP4cT3WA) — jitendra sah
- 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-2080-310m-hvcb
- 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 | 13.5 | Qmax (Q0) | 149.00 m³/s |
| February | 8.6 | Q25% | 72.5 m³/s |
| March | 6.58 | Q45% | 22.87 m³/s |
| April | 7.5 | Q65% | 12.7 m³/s |
| May | 12.8 | Q85% | 8.70 m³/s |
| June | 25.7 | Q95% | 7.1 m³/s |
| July | 72.5 | Qmin | 5.05 m³/s |
| August | 101.2 |  |  |
| September | 81.3 |  |  |
| October | 27.05 |  |  |
| November | 21.38 |  |  |
| December | 14.35 |  |  |

| Turbine type | Head range (m) |
|---|---|
| Kaplan and Propeller | 2 < H < 40 |
| Francis | 10 < H < 350 |
| Pelton | 50 < H < 1300 |
| Michell-Banki | 3 < H < 250 |
| Turgo | 50 < H < 250 |

| Parameter | Value |
|---|---|
| Gross head | 310 m |
| Turbine efficiency | 91% |
| Generator efficiency | 97% |
| Transformer efficiency | 99% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 21% |
| Transformer reactance | 9.15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 30 km |
| Transmission voltage level | As required |
| Generator voltage level | 11 kV |
| Transmission circuit | Choose appropriately as required (single or double) |

(Please make your own smart assumption if needed beside given data above)
Design a power plant project so as to obtain following,
a) Installed capacity with design discharge as Q45% and appropriate number of units. [3]
b) Choose the appropriate type of Generator lead. [2]
c) Choose your generator transformer scheme mentioning the reasons. [3]
d) Make single line diagram for the project choosing earthing system, excitation system and bus bar system. [4]
e) Find Rating of high voltage circuit breaker of the project. [4]
[Charts attached: turbine head-range table; turbine application chart of head (m) against flow (m³/s) with Pelton, Turgo and Francis envelopes and power lines from 20 kW to 20 MW; low-head chart with Pelton, Turgo, Crossflow, Francis and Kaplan envelopes; specific speed against net head chart for Bulbo, Helice, Kaplan, Francis, Crossflow and Pelton; turbine efficiency against Q/Q0 curves for Full Kaplan, Pelton, Francis, Crossflow and Fixed propeller]

## Short answer

Installed capacity 56.7 MW (3×18.9 MW) with Pelton turbines (4 jets). Design discharge 22.21 m³/s, net head 294.5 m. HV circuit breaker: 132 kV, 630 A, 20 kA based on short-circuit calculation from system and generator contributions.

## 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 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).

#### Transmission voltage and circuits (assumed as required)

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

$$
\begin{aligned}
V &= 5.5\sqrt{\frac{L}{1.6} + \frac{P}{100}} = 5.5\sqrt{\frac{30}{1.6} + \frac{56700}{100}} = 133.1\ \text{kV}
\end{aligned}
$$

Choose **132 kV**, the nearest standard voltage, with a **double-circuit** line so that the 56.7 MW output is not lost when one circuit trips (N-1 security).

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

Driest month = March (6.58 m³/s), so riparian release $Q_r = 0.1 \times 6.58 = 0.658$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 22.87 - 0.658 = 22.212\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 310 - 15.50 = 294.50\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 22.212 \times 294.50 = 64.171\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 64.171 = 58.396\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 58.396 = 56.644\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 56.644 = 56.078\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 56.644 MW; power delivered to the 132 kV bus = **56.078 MW**.

#### a) (contd.) 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: $Q = 22.212/3 = 7.404$ m³/s, generator output $= 56.644/3 = 18.881$ MW → rate each unit at **18.9 MW**.
- Generator rating: $S = 18.9/0.85 = 22.24$ MVA → **22.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **25 MVA, 11/132 kV, YNd11** (next standard size above 22.5 MVA).

**Answer: installed capacity = 3 × 18.9 MW = 56.7 MW.**

#### Type of turbine (for unit speed)

$H_n = 294.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 58.396/3 = 19465.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1220.0$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 114.4 |
| 750 | 8 | 85.8 |
| 600 | 10 | 68.6 |
| 500 | 12 | 57.2 |
| 428.6 | 14 | 49.0 |
| 375 | 16 | 42.9 |
| 333.3 | 18 | 38.1 |

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): $N_s = 38.1$. With **4 jets**, $N_{s,jet} = 38.1/\sqrt{4} = 19.1$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 74.5$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.96$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.178$ m; $D/d = 11.0 > 10$, acceptable.

A Francis at 750 rpm ($N_s = 85.8$) 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 19.47 MW each.**

#### b) Type of generator lead

Generator full-load current: $I = \dfrac{22.5 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 1181$ A.

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

- About 1181 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 3000–5000 A, i.e. roughly 60–100 MVA or more 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, 1600 A continuous, short-time withstand ≥ the GCB rating (e.g. 25 kA for 1 s), aluminium bars, with tap-off for the station transformer, surge capacitors and PTs.

#### c) Generator–transformer scheme and reasons

Choose the **unit scheme** (one generator connected to its own step-up transformer, with a generator circuit breaker between them).

| Point | Unit scheme (chosen) | Common bus / group scheme |
|---|---|---|
| Fault level at 11 kV | Low (one machine only) | High (all machines in parallel) |
| Effect of a transformer fault | Only one unit lost | Whole plant may trip |
| Maintenance outage | One unit at a time | Common transformer outage stops all |
| 11 kV switchgear | Only GCB and short leads | Large 11 kV bus and breakers |
| Losses / efficiency | No 11 kV bus losses | More 11 kV copper losses |
| Cost | More transformers | Fewer, larger transformers |

Reasons: the units are large (22.5 MVA each, about 1181 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) Single line diagram with earthing, excitation and bus bar

```text
 Line-1          Line-2  (132 kV, 30 km)
    |               |
  [CB]            [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
```

| Item | Rating |
|---|---|
| Generators G1–G3 | 22.5 MVA, 11 kV, 0.85 pf, 333.3 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T3 | 25 MVA each, 11/132 kV, YNd11, $X$ = 9.15% |
| Station transformer ST1, ST2 | 1000 kVA, 11/0.4 kV, Dyn11, 5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 625 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 132 kV, double circuit, 30 km to grid substation |
| NGT | neutral grounding transformer with secondary resistor (high-resistance earthing) |
| 132 kV bus | double main bus with bus coupler (drawn as one line) |

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.

Choices shown on the SLD: **earthing** – each generator neutral through a neutral grounding transformer with secondary resistor (high-resistance earthing, fault current about 10 A), the HV star of each GSU solidly earthed, and all equipment bonded to the powerhouse/switchyard earth mat; **excitation** – static (thyristor) excitation fed from an excitation transformer at each generator terminal, with field flashing from the 110 V DC battery (fast response for a large grid-connected unit); **bus bar** – 132 kV double main bus with bus coupler, two line bays for the double-circuit line.

#### e) Rating of high voltage circuit breaker

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{22.5} = 0.9333\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{25} = 0.3660\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 30 = 8.954\ \Omega \text{ per circuit; two in parallel} = 4.477\ \Omega \\
x_l &= \frac{X_L \times 100}{132^2} = \frac{4.477}{174.24} = 0.0257\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\
x_{grid\ path} &= 0.0500 + 0.0257 = 0.0757\ \text{pu}
\end{aligned}
$$

For a fault on the 132 kV bus, the plant and the grid feed in parallel:

$$
\begin{aligned}
X_{plant} &= \frac{x_g + x_t}{3} = \frac{0.9333 + 0.3660}{3} = 0.4331\ \text{pu} \\
S_{HV} &= \frac{100}{0.4331} + \frac{100}{0.0757} = 230.9 + 1321.1 = 1552.0\ \text{MVA} \\
I_{sc} &= \frac{1552.0}{\sqrt{3} \times 132} = 6.79\ \text{kA}, \quad I_{make} = 2.55 \times 6.79 = 17.31\ \text{kA peak} \\
I_{FL} &= \frac{3 \times 25}{\sqrt{3} \times 132} = 328\ \text{A}, \quad 1.25 I_{FL} = 410\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 132 kV system | 145 kV |
| Rated normal current | ≥ 410 A | 1250 A |
| Breaking capacity | 1552.0 MVA (6.79 kA) | 31.5 kA |
| Making capacity | 17.31 kA peak | 80 kA peak |
| BIL | – | 650 kVp |
| Type | – | SF6, outdoor, spring mechanism, 3-pole, single-pole tripping for auto-reclose |

**Answer: HV bus fault level = 1552.0 MVA (6.79 kA); select a 145 kV, 1250 A, 31.5 kA SF6 circuit breaker.** Of this, the plant itself contributes only 230.9 MVA; the rest comes from the grid, so the result depends on the assumed grid fault level.

## Check yourself

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

1. Why is a 132 kV double-circuit line chosen for 56.7 MW?
   A. Still's formula gives about 133 kV, and two circuits keep the output available if one circuit trips (N-1)
   B. Still's formula gives 66 kV and double circuit is cheaper
   C. Double circuit halves the generator voltage
   D. 132 kV is the only voltage allowed in Nepal
   Answer: A. Still's formula gives 133.1 kV, so 132 kV is the nearest standard; a double circuit gives N-1 security. (Key point: Transmission voltage)

2. Why are 3 units chosen instead of 2?
   A. Three units always cost less than two
   B. Two units would be very large and hard to transport on hill roads; three follow the seasonal flow and losing one removes only a third
   C. Three units give higher generator voltage
   D. Two units cannot be synchronised
   Answer: B. Smaller machines are easier to transport, and units can be added in the wet months while one outage costs only one third of output. (Key point: Number of units)

3. Why is a segregated phase bus duct (SPBD) chosen for the generator lead?
   A. Current is above 4000 A so isolated phase bus is not possible
   B. SPBD is the only duct that can be earthed
   C. Full-load current is about 1181 A, too high for cables, but not large enough to need isolated phase bus
   D. Cables are never used at 11 kV
   Answer: C. 1181 A needs many parallel cables, while IPB is only for very large machines; SPBD gives phase compartments and is compact. (Key point: Generator lead)

4. What is the main advantage of the unit scheme (generator with own transformer and GCB) over a common bus scheme?
   A. Fewer transformers are needed
   B. A common 11 kV bus is available for all units
   C. Transformer maintenance stops all units
   D. Low 11 kV fault level and loss of only one unit for a transformer fault
   Answer: D. Each machine feeds its own transformer, so fault level at 11 kV is low and a fault affects only one unit. (Key point: Unit scheme)

## Close variants set in other exams

- [Design: 260 m head plant, generator lead](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2078-260m-generator-lead) — Power Plant Design, chapter 4, asked 5×
- [Design: 360 m head plant, 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2078-360m-132kv) — Power Plant Design, chapter 4, asked 5×

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