---
title: "Design: 360 m head plant, 132 kV"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2078-360m-132kv"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2078-360m-132kv
site: BCT Engineering
---

# Design: 360 m head plant, 132 kV

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

- Asked in: 2078 Chaitra, 3+2+3+4+4 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-2078-360m-132kv
- 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 | 7.50 | Qmax (Q0) | 207.00 m³/s |
| February | 6.60 | Q25% | 28.24 m³/s |
| March | 5.58 | Q45% | 18.87 m³/s |
| April | 6.55 | Q65% | 13.7 m³/s |
| May | 8.8 | Q85% | 8.90 m³/s |
| June | 17.77 | Q95% | 7.91 m³/s |
| July | 56.95 | Qmin | 5.05 m³/s |
| August | 134.26 |  |  |
| September | 103.27 |  |  |
| October | 29.05 |  |  |
| November | 19.38 |  |  |
| December | 12.35 |  |  |

| 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 | 360 m |
| Turbine efficiency | 90% |
| Generator efficiency | 97% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 19% |
| Transformer reactance | 8.15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 22 km |
| Transmission voltage level | 132 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Double circuit |

(Please make your own smart assumption if needed beside given data above)
Design a power plant project so as to obtain following
a) Electrical power output taking design discharge as Q45%. [3]
b) Select the numbers of units and installed capacity. [2]
c) Recommend the appropriate type of Generator-transformer scheme. [3]
d) Make single line diagram for the project choosing your generator transformer scheme and bus bar system. [4]
e) Find MVA Rating of a generator circuit breaker of the project. [4]

## Short answer

At Q45%, net head 342 m gives 52.562 MW delivered. Install three 17.9 MW Pelton units (53.7 MW) with unit transformers; generator breaker rating 21.5 MVA.

## 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).

#### a) Electrical power output at $Q_{45\%}$

Driest month = March (5.58 m³/s), so riparian release $Q_r = 0.1 \times 5.58 = 0.558$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 18.87 - 0.558 = 18.312\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 360 - 18.00 = 342.00\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 18.312 \times 342.00 = 61.437\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 61.437 = 55.293\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 55.293 = 53.635\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 53.635 = 52.562\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 53.635 MW; power delivered to the 132 kV bus = **52.562 MW**.

#### b) Number of units and installed capacity

- 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: $Q = 18.312/3 = 6.104$ m³/s, generator output $= 53.635/3 = 17.878$ MW → rate each unit at **17.9 MW**.
- Generator rating: $S = 17.9/0.85 = 21.06$ MVA → **21.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **25 MVA, 11/132 kV, YNd11** (next standard size above 21.5 MVA).

**Answer: installed capacity = 3 × 17.9 MW = 53.7 MW.**

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

$H_n = 342.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 55.293/3 = 18431.1$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1470.7$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 92.3 |
| 750 | 8 | 69.2 |
| 600 | 10 | 55.4 |
| 500 | 12 | 46.2 |
| 428.6 | 14 | 39.6 |
| 375 | 16 | 34.6 |
| 333.3 | 18 | 30.8 |

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): $N_s = 39.6$. With **4 jets**, $N_{s,jet} = 39.6/\sqrt{4} = 19.8$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 80.3$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.65$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.156$ m; $D/d = 10.6 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 92.3$) 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, 428.6 rpm, about 18.43 MW each.**

#### c) Generator–transformer scheme

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 (21.5 MVA each, about 1128 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

```text
 Line-1          Line-2  (132 kV, 22 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 | 21.5 MVA, 11 kV, 0.85 pf, 428.6 rpm, $X''_d$ = 19% |
| Step-up transformers T1–T3 | 25 MVA each, 11/132 kV, YNd11, $X$ = 8.15% |
| 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, 22 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.

#### 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) MVA rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.19 \times \frac{100}{21.5} = 0.8837\ \text{pu}, \quad x_t = 0.0815 \times \frac{100}{25} = 0.3260\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 22 = 6.566\ \Omega \text{ per circuit; two in parallel} = 3.283\ \Omega \\
x_l &= \frac{X_L \times 100}{132^2} = \frac{3.283}{174.24} = 0.0188\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\
x_{grid\ path} &= 0.0500 + 0.0188 = 0.0688\ \text{pu}
\end{aligned}
$$

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).

$$
\begin{aligned}
S_a &= \frac{100}{x_g} = \frac{100}{0.8837} = 113.2\ \text{MVA} \\
X_{ext} &= x_{grid\ path} \parallel (x_g + x_t)/2 = 0.0688 \parallel 0.6049 = 0.0618\ \text{pu} \\
S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.3260 + 0.0618} = 257.9\ \text{MVA}
\end{aligned}
$$

Design fault level for the GCB = larger value = **257.9 MVA**.

$$
\begin{aligned}
I_{sc} &= \frac{257.9}{\sqrt{3} \times 11} = 13.53\ \text{kA}, \quad I_{make} = 2.55 \times 13.53 = 34.51\ \text{kA peak} \\
I_{FL} &= \frac{21.5}{\sqrt{3} \times 11} = 1128\ \text{A}, \quad 1.25 I_{FL} = 1411\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 1411 A | 1600 A |
| Breaking capacity | 257.9 MVA (13.53 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 34.51 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 = 257.9 MVA; select a 12 kV, 1600 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.

## Check yourself

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

1. For the 360 m plant, what are the design discharge and net head used?
   A. Qd = 18.87 m3/s and Hn = 360 m
   B. Qd = 18.312 m3/s (Q45% minus 0.558) and Hn = 342 m
   C. Qd = 18.312 m3/s and Hn = 360 m
   D. Qd = 19.428 m3/s and Hn = 342 m
   Answer: B. Riparian release is 10% of March flow (5.58), so 0.558 is subtracted, and 5% head loss reduces 360 m to 342 m. (Key point: Design discharge and net head)

2. Why are 3 units chosen rather than 2?
   A. Three units reduce the number of transformers needed
   B. The 132 kV level requires an odd number of units
   C. Two units would be very large and hard to transport, and three follow the seasonal flow change and lose only one third on an outage
   D. Three units give a higher total efficiency at every flow
   Answer: C. Transport of large machines on hill roads, flexibility with seasonal flow and smaller loss on outage justify three units. (Key point: Number of units)

3. What is the main benefit of the unit scheme (each generator with its own transformer and GCB)?
   A. A large 11 kV bus allowing power sharing
   B. Low 11 kV fault level and only one unit lost for a transformer fault
   C. Fewer transformers and lower cost
   D. No need for a generator circuit breaker
   Answer: B. Each machine is isolated, so fault level at 11 kV stays small and a transformer fault affects only one unit, though it needs more transformers. (Key point: Unit scheme advantages)

4. At 428.6 rpm the Ns is 39.6, so how is a Pelton justified?
   A. A single jet gives Ns of 39.6, which is in the Pelton range
   B. Ns is divided by 4 jets without a square root, giving 9.9
   C. With 4 jets, Ns per jet = 39.6/sqrt(4) = 19.8, within the Pelton range of about 10-30
   D. Pelton is justified because Ns is above 300
   Answer: C. Multi-jet Pelton uses Ns = Ns,jet x sqrt(z), so 4 jets give 19.8 per jet, and it also has flat part-load efficiency and sediment tolerance. (Key point: Pelton versus Francis choice)

## 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: 550 m head plant, transformer quotation](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2077-550m-quotation) — Power Plant Design, chapter 4, asked 3×
- [Design: 589 m head plant at Q40%](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2079-589m-q40) — Power Plant Design, chapter 4, asked 3×
- [Design: 310 m head plant, HV breaker](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2080-310m-hvcb) — Power Plant Design, chapter 4, asked 3×

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