---
title: "Design: 300 m head plant, GCB rating"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2081-300m-gcb"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2081-300m-gcb
site: BCT Engineering
---

# Design: 300 m head plant, GCB rating

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

- Asked in: 2081 Chaitra, 3+3+3+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-2081-300m-gcb
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-4

## Question

A typical hydropower plant has the following details:

| Month | Discharge (m³/s) | Flow duration | Discharge (m³/s) |
|---|---|---|---|
| January | 6.90 | Qmax (Q0) | 218.00 m³/s |
| February | 5.86 | Q25% | 28.24 m³/s |
| March | 4.58 | Q45% | 9.80 m³/s |
| April | 3.55 | Q65% | 6.7 m³/s |
| May | 5.10 | Q85% | 4.1 m³/s |
| June | 17.70 | Q95% | 2.9 m³/s |
| July | 72.95 | Qmin | 1.64 m³/s |
| August | 120.26 |  |  |
| September | 98.23 |  |  |
| October | 25.05 |  |  |
| November | 8.38 |  |  |
| December | 7.85 |  |  |

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

| Parameter | Value |
|---|---|
| Gross head | 300 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 | 0.92 mH/km |
| Transmission line length | 20 km |
| Generator voltage level | 11 kV |

Design a power plant so to obtain following parameters:
(i) Installed capacity in MW taking design discharge as Q45% [3]
(ii) Select numbers of units [3]
(iii) Select the type of turbine [3]
(iv) Choose suitable generator transformer scheme. [3]
(v) Draw single line diagram showing number of units, generator, power transformer, station transformer and diesel generator for black start. [4]
(vi) Calculate the rating of the generator circuit breaker (GCB) to be used in your design. [4]
[Charts attached: turbine head-range table; turbine application chart for higher head (head against flow with Pelton, Turgo and Francis envelopes); chart for lower head (Pelton, Turgo, Crossflow, Francis and Kaplan envelopes); specific speed against net head chart]

## Short answer

Design discharge Q45% minus riparian release yields 9.445 m³/s, net head 285 m. Generator output is 23.309 MW. Install two 11.7 MW units, each with 14 MVA, 11 kV generator. Turbine: 4-jet Pelton, 428.6 rpm. Unit scheme with GCB; transmission voltage 132 kV.

## 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 (not given)

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

$$
\begin{aligned}
V &= 5.5\sqrt{\frac{L}{1.6} + \frac{P}{100}} = 5.5\sqrt{\frac{20}{1.6} + \frac{23400}{100}} = 86.4\ \text{kV}
\end{aligned}
$$

Still's value of 86.4 kV lies between 66 kV and 132 kV; choose **132 kV, single circuit**, the NEA grid voltage, which also leaves room for future projects in the area.

#### (i) Installed capacity in MW at $Q_{45\%}$

Driest month = April (3.55 m³/s), so riparian release $Q_r = 0.1 \times 3.55 = 0.355$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 9.8 - 0.355 = 9.445\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 300 - 15.00 = 285.00\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 9.445 \times 285.00 = 26.407\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 26.407 = 24.030\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 24.030 = 23.309\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 23.309 = 23.076\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 23.309 MW; power delivered to the 132 kV bus = **23.076 MW**.

#### (ii) Number of units

- 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: $Q = 9.445/2 = 4.723$ m³/s, generator output $= 23.309/2 = 11.655$ MW → rate each unit at **11.7 MW**.
- Generator rating: $S = 11.7/0.85 = 13.76$ MVA → **14 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **16 MVA, 11/132 kV, YNd11** (next standard size above 14 MVA).

**Answer: installed capacity = 2 × 11.7 MW = 23.4 MW.**

#### (iii) Type of turbine

$H_n = 285.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 24.030/2 = 12015.1$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 1171.0$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 93.6 |
| 750 | 8 | 70.2 |
| 600 | 10 | 56.2 |
| 500 | 12 | 46.8 |
| 428.6 | 14 | 40.1 |
| 375 | 16 | 35.1 |
| 333.3 | 18 | 31.2 |

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 $N = 428.6$ rpm (14 poles): $N_s = 40.1$. With **4 jets**, $N_{s,jet} = 40.1/\sqrt{4} = 20.1$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 73.3$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.50$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.143$ m; $D/d = 10.5 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 93.6$) 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 12.02 MW each.**

#### (iv) 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 (14 MVA each, about 735 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.

#### (v) Single line diagram

```text
 Line to grid  (132 kV, 20 km)
    |
  [CB]
    |
 ===+====+================+=== 132 kV bus
         |                |
       [CB]             [CB]
         |                |
       (T1)             (T2)
         |                |
  ST1 ---+                +--- ST2
         |                |
       [GCB]            [GCB]
         |                |
       (G1)             (G2)
         |                |
       [NGT]            [NGT]

 ST1, ST2 (11/0.4 kV)
    |               |
 ===+====[BS]=======+=== 0.4 kV aux bus
              |
            [ACB]
              |
            (DG)  diesel set, black start
```

| Item | Rating |
|---|---|
| Generators G1–G2 | 14 MVA, 11 kV, 0.85 pf, 428.6 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T2 | 16 MVA each, 11/132 kV, YNd11, $X$ = 9.15% |
| Station transformer ST1, ST2 | 400 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 250 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 132 kV, single circuit, 20 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.

#### (vi) Rating of generator circuit breaker (GCB)

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{14} = 1.5000\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{16} = 0.5719\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.92\times10^{-3} \times 20 = 5.781\ \Omega \\
x_l &= \frac{X_L \times 100}{132^2} = \frac{5.781}{174.24} = 0.0332\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\
x_{grid\ path} &= 0.0500 + 0.0332 = 0.0832\ \text{pu}
\end{aligned}
$$

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).

$$
\begin{aligned}
S_a &= \frac{100}{x_g} = \frac{100}{1.5000} = 66.7\ \text{MVA} \\
X_{ext} &= x_{grid\ path} \parallel (x_g + x_t) = 0.0832 \parallel 2.0719 = 0.0800\ \text{pu} \\
S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.5719 + 0.0800} = 153.4\ \text{MVA}
\end{aligned}
$$

Design fault level for the GCB = larger value = **153.4 MVA**.

$$
\begin{aligned}
I_{sc} &= \frac{153.4}{\sqrt{3} \times 11} = 8.05\ \text{kA}, \quad I_{make} = 2.55 \times 8.05 = 20.53\ \text{kA peak} \\
I_{FL} &= \frac{14}{\sqrt{3} \times 11} = 735\ \text{A}, \quad 1.25 I_{FL} = 919\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 919 A | 1250 A |
| Breaking capacity | 153.4 MVA (8.05 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 20.53 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 = 153.4 MVA; select a 12 kV, 1250 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. How is the design discharge obtained at Q45% in this design?
   A. Q45% minus the riparian release (10% of the driest monthly flow)
   B. Q45% plus the riparian release
   C. Mean of all monthly flows minus Q45%
   D. Q45% multiplied by 0.9 for the efficiency
   Answer: A. The riparian release (0.355 m3/s, 10% of April's 3.55 m3/s) must stay in the river, so Qd = 9.8 - 0.355 = 9.445 m3/s. (Key point: Design discharge)

2. Which sequence correctly gives the power delivered to the 132 kV bus?
   A. rho g Qd Hg only, with no efficiencies
   B. rho g Qd Hn, then multiply by turbine, generator and transformer efficiencies in turn
   C. rho g Qd Hn divided by each efficiency
   D. Hydraulic power times the transformer efficiency only
   Answer: B. Hydraulic power 26.407 MW is multiplied by 0.91, 0.97 and 0.99 to give 24.03, 23.309 and 23.076 MW. (Key point: Power chain)

3. Why is net head 285 m rather than the gross head of 300 m?
   A. The turbine efficiency reduces the head
   B. The riparian release reduces the head
   C. 5% of the gross head is lost in the water conductor system
   D. Tail water level adds 15 m
   Answer: C. Head loss of 5% (15 m) is subtracted: Hn = 300 x 0.95 = 285 m. (Key point: Net head)

4. Both Francis and Pelton cover 285 m, so why is a 4-jet Pelton chosen?
   A. Francis cannot operate above 100 m
   B. Pelton always has higher peak efficiency than any other turbine
   C. Pelton needs no draft tube so it is cheaper in every case
   D. Its efficiency stays flat at part load and it tolerates sediment, suiting run-of-river with low dry-season flow
   Answer: D. Pelton jets can be switched off to keep efficiency high at low flow, and it handles Himalayan sediment better. (Key point: Turbine selection)

## Close variants set in other exams

- [Design: 230 m head plant, 55 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2074-230m-55km) — Power Plant Design, chapter 4, asked 12×
- [Design: 330 m head plant, 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2082-330m-q45) — Power Plant Design, chapter 4, asked 12×
- [Design: 95 m head plant at Q65%](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2079-95m-q65) — Power Plant Design, chapter 4, asked 12×
- [Design: 230 m head plant, 45 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2071-230m-45km) — Power Plant Design, chapter 4, asked 12×
- [Design: 250 m head plant, 35 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2074-250m-35km) — Power Plant Design, chapter 4, asked 12×
- [Design: 250 m head plant at Q25%](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2072-250m-q25) — Power Plant Design, chapter 4, asked 14×
- [Design: 280 m head plant, 66 kV](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2075-280m-66kv) — Power Plant Design, chapter 4, asked 12×
- [Design: 200 m head plant, bus MVA levels](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2070-200m-bus-mva) — Power Plant Design, chapter 4, asked 14×
- [Design: 230 m head plant, 20 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2070-230m-20km) — Power Plant Design, chapter 4, asked 14×
- [Design: 245 m head plant, standby generator](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2073-245m-standby) — Power Plant Design, chapter 4, asked 14×

## Related in Power Plant Design

- [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×
- [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: 250 m head plant, 66 kV double circuit](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2071-250m-0-92mh) — Power Plant Design, chapter 4, asked 13×

## Related in other subjects

- [Francis design (90 m, 40 m³/s, 86%)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-design-90m-40cumec) — Hydropower Engineering, chapter 7
- [Load duration curve; diesel plant; ROR flow duration](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aload-duration-diesel-import-fdc) — Hydropower Engineering, chapter 3
- [Flow duration curve; firm and secondary energy](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afdc-firm-secondary-energy-69) — Hydropower Engineering, chapter 3