---
title: "Design: 245 m head plant, standby generator"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2073-245m-standby"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2073-245m-standby
site: BCT Engineering
---

# Design: 245 m head plant, standby generator

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

- Asked in: 2073 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-2073-245m-standby
- 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% | 7.20 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 | 245 m |
| Turbine efficiency | 91% |
| Generator efficiency | 98% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 21% |
| Transformer reactance | 9.15% |
| Transmission line inductance | 0.9 mH/km |
| Transmission line length | 50 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 a stand by generator for black start
e) Calculate the rating of generator circuit breaker to be used in your design. (Assume suitable data if necessary and necessary graph is attached herewith)

## Short answer

Generator output 14.050 MW, delivered 13.769 MW. Install 2×7.1 MW (14.2 MW total). Two vertical-axis 4-jet Pelton turbines, 428.6 rpm. Single-line diagram includes diesel standby generator; GCB rated 8.5 MVA, 11 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 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.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 7.2 - 0.3 = 6.900\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 245 - 12.25 = 232.75\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 6.900 \times 232.75 = 15.755\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 15.755 = 14.337\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.98 \times 14.337 = 14.050\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 14.050 = 13.769\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 14.050 MW; power delivered to the 66 kV bus = **13.769 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: $Q = 6.900/2 = 3.450$ m³/s, generator output $= 14.050/2 = 7.025$ MW → rate each unit at **7.1 MW**.
- Generator rating: $S = 7.1/0.85 = 8.35$ MVA → **8.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **10 MVA, 11/66 kV, YNd11** (next standard size above 8.5 MVA).

**Answer: installed capacity = 2 × 7.1 MW = 14.2 MW.**

#### c) Type of turbine

$H_n = 232.75$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 14.337/2 = 7168.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 909.1$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 93.1 |
| 750 | 8 | 69.8 |
| 600 | 10 | 55.9 |
| 500 | 12 | 46.6 |
| 428.6 | 14 | 39.9 |
| 375 | 16 | 34.9 |
| 333.3 | 18 | 31.0 |

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): $N_s = 39.9$. With **4 jets**, $N_{s,jet} = 39.9/\sqrt{4} = 20.0$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 66.2$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.36$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.129$ m; $D/d = 10.5 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 93.1$) 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 7.17 MW each.**

#### d) Single line diagram (standby diesel generator for black start)

```text
 Line-1          Line-2  (66 kV, 50 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 | 8.5 MVA, 11 kV, 0.85 pf, 428.6 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T2 | 10 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, 50 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:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{8.5} = 2.4706\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{10} = 0.9150\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.9\times10^{-3} \times 50 = 14.137\ \Omega \text{ per circuit; two in parallel} = 7.069\ \Omega \\
x_l &= \frac{X_L \times 100}{66^2} = \frac{7.069}{43.56} = 0.1623\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\
x_{grid\ path} &= 0.1000 + 0.1623 = 0.2623\ \text{pu}
\end{aligned}
$$

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):

$$
\begin{aligned}
S_{gen} &= 2 \times \frac{100}{2.4706} = 81.0\ \text{MVA} \\
S_{grid} &= \frac{100}{x_t/2 + x_{grid\ path}} = \frac{100}{0.4575 + 0.2623} = 138.9\ \text{MVA} \\
S_{11kV} &= 81.0 + 138.9 = 219.9\ \text{MVA}
\end{aligned}
$$

$$
\begin{aligned}
I_{sc} &= \frac{219.9}{\sqrt{3} \times 11} = 11.54\ \text{kA}, \quad I_{make} = 2.55 \times 11.54 = 29.43\ \text{kA peak} \\
I_{FL} &= \frac{8.5}{\sqrt{3} \times 11} = 446\ \text{A}, \quad 1.25 I_{FL} = 558\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 558 A | 630 A |
| Breaking capacity | 219.9 MVA (11.54 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 29.43 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 = 219.9 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.

## Check yourself

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

1. Q45% = 7.2 m3/s and the driest month is 3 m3/s. What is the design discharge?
   A. 7.2 m3/s
   B. 6.9 m3/s
   C. 7.5 m3/s
   D. 4.2 m3/s
   Answer: B. Release = 10% of 3 = 0.3 m3/s, so Qd = 7.2 - 0.3 = 6.9 m3/s. (Key point: Design discharge)

2. Each unit gives about 7.1 MW at 0.85 power factor. How is the generator MVA rating obtained?
   A. S = 7.1 x 0.85 = 6.0 MVA
   B. S = 7.1 / 0.85 = 8.35 MVA, rounded up to 8.5 MVA
   C. S = 7.1 MVA directly
   D. S = 7.1 / 0.98 MVA
   Answer: B. Apparent power is real power divided by power factor, then rounded up to a standard size. (Key point: Generator rating)

3. At 428.6 rpm the specific speed is 39.9, so with 4 jets the value per jet is 20. What does this show?
   A. Kaplan is required
   B. A Francis is the only option
   C. The Pelton range (about 10-30 per jet) is satisfied, so a 4-jet Pelton is suitable
   D. The head is below the Pelton limit
   Answer: C. Multi-jet Ns = Ns,jet x sqrt(z); 39.9/2 = 20 lies in the Pelton range. (Key point: Turbine type by specific speed)

4. What is the role of the diesel generator in the single line diagram?
   A. It supplies the grid during normal operation
   B. It replaces the step-up transformer
   C. It feeds the main 11 kV bus permanently
   D. It energises the 0.4 kV auxiliary bus so a unit can be started with the grid dead
   Answer: D. The DG powers essential auxiliaries (governor oil, cooling, excitation, lighting, chargers) so one unit can start and then energise the transformer and line. (Key point: Black start)

## Close variants set in other exams

- [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×
- [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: 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: 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: 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: 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: 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: 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: 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: 120 m head plant, bus MVA levels](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2073-120m-66kv) — Power Plant Design, chapter 4, asked 8×
- [Design: 150 m head plant, bus MVA levels](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2072-150m-bus-mva) — Power Plant Design, chapter 4, asked 8×
- [Design: 300 m head plant, GCB rating](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2081-300m-gcb) — Power Plant Design, chapter 4, asked 11×

## Related in Power Plant Design

- [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×

## Related in other subjects

- [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
- [Micro hydro net head, power and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Amicro-hydro-power-output-numerical) — Energy Resources, chapter 5
- [Flow duration curve and power at mean flow](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-capacity-fdc-flow-table-79) — Hydropower Engineering, chapter 3