---
title: "Design: 250 m head plant at Q25%"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2072-250m-q25"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2072-250m-q25
site: BCT Engineering
---

# Design: 250 m head plant at Q25%

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

- Asked in: 2072 Magh, 32 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-2072-250m-q25
- 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 | 4.62 | Qmax (Q0) | 210.00 m³/s |
| February | 3.80 | Q25% | 18.24 m³/s |
| March | 3.55 | Q45% | 7.82 m³/s |
| April | 4.55 | Q65% | 4.7 m³/s |
| May | 3.5 | Q85% | 3.10 m³/s |
| June | 12.70 | Q95% | 2.17 m³/s |
| July | 32.98 | Qmin | 0.64 m³/s |
| August | 40.26 |  |  |
| September | 31.23 |  |  |
| October | 15.00 |  |  |
| November | 7.37 |  |  |
| December | 4.80 |  |  |

| 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 | 90% |
| Generator efficiency | 97% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 21% |
| Transformer reactance | 9.15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 55 km |
| Transmission voltage level | 33 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 Q25%.
b) Select the numbers of units and installed capacity.
c) Select the type of turbine.
d) Draw the single line diagram showing number of units, generator, power transformer, station transformer and diesel generator for black start.
e) Suggest the type of bus bar to be used
f) Calculate the rating of the generator circuit breaker (GCB) to be used in your design
(Assume suitable data if necessary and necessary graph is attached herewith)

## Short answer

Design discharge Q25% minus riparian release = 17.89 m³/s, net head = 237.5 m. Station output 35.660 MW. Three units ×12.2 MW, installed 36.6 MW. 4‑jet Pelton turbines, 333.3 rpm. SLD: 3 generators, 3 transformers, ST, DG, double‑circuit 33 kV line, single bus with section breaker. Busbar type single bus. GCB rating based on generator MVA and fault level.

## 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 500 MVA at the 33 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 $Q_{25\%}$

Driest month = May (3.5 m³/s), so riparian release $Q_r = 0.1 \times 3.5 = 0.35$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{25\%} - Q_r = 18.24 - 0.35 = 17.890\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 250 - 12.50 = 237.50\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 17.890 \times 237.50 = 41.681\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 41.681 = 37.513\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 37.513 = 36.388\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 36.388 = 35.660\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 36.388 MW; power delivered to the 33 kV bus = **35.660 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: $Q = 17.890/3 = 5.963$ m³/s, generator output $= 36.388/3 = 12.129$ MW → rate each unit at **12.2 MW**.
- Generator rating: $S = 12.2/0.85 = 14.35$ MVA → **14.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **16 MVA, 11/33 kV, YNd11** (next standard size above 14.5 MVA).

**Answer: installed capacity = 3 × 12.2 MW = 36.6 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 $P = 37.513/3 = 12504.4$ 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 | 119.9 |
| 750 | 8 | 90.0 |
| 600 | 10 | 72.0 |
| 500 | 12 | 60.0 |
| 428.6 | 14 | 51.4 |
| 375 | 16 | 45.0 |
| 333.3 | 18 | 40.0 |

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 $N = 333.3$ rpm (18 poles): $N_s = 40.0$. With **4 jets**, $N_{s,jet} = 40.0/\sqrt{4} = 20.0$, 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 $D = 60u/(\pi N) = 1.76$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.168$ m; $D/d = 10.5 > 10$, acceptable.

A Francis at 750 rpm ($N_s = 90.0$) 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: 3 × vertical-axis 4-jet Pelton turbines, 333.3 rpm, about 12.50 MW each.**

#### d) Single line diagram

```text
 Line-1          Line-2  (33 kV, 55 km)
    |               |
  [CB]            [CB]
    |               |
 ===+===+===========+===========+=== 33 kV bus
        |           |           |
      [CB]        [CB]        [CB]
        |           |           |
      (T1)        (T2)        (T3)
        |           |           |
      [CB]        [CB]        [CB]
        |           |           |
 =======+===[BS]====+===========+=========+=== 11 kV bus
        |           |           |         |
      [GCB]       [GCB]       [GCB]     [CB]
        |           |           |         |
      (G1)        (G2)        (G3)      (ST)
                                          |
 =========================================+===+ 0.4 kV
                                              |
                                            [ACB]
                                              |
                                            (DG) black start
```

| Item | Rating |
|---|---|
| Generators G1–G3 | 14.5 MVA, 11 kV, 0.85 pf, 333.3 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T3 | 16 MVA each, 11/33 kV, YNd11, $X$ = 9.15% |
| Station transformer ST | 630 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 400 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 33 kV, double circuit, 55 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) 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.
- **33 kV:** single bus bar, sectionalised, with transformer bays and two line bays (double circuit).

Reason: for a 36.6 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.

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

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{14.5} = 1.4483\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{16} = 0.5719\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 55 = 16.415\ \Omega \text{ per circuit; two in parallel} = 8.207\ \Omega \\
x_l &= \frac{X_L \times 100}{33^2} = \frac{8.207}{10.89} = 0.7537\ \text{pu}, \quad x_{grid} = \frac{100}{500} = 0.2000\ \text{pu} \\
x_{grid\ path} &= 0.2000 + 0.7537 = 0.9537\ \text{pu}
\end{aligned}
$$

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):

$$
\begin{aligned}
S_{gen} &= 3 \times \frac{100}{1.4483} = 207.1\ \text{MVA} \\
S_{grid} &= \frac{100}{x_t/3 + x_{grid\ path}} = \frac{100}{0.1906 + 0.9537} = 87.4\ \text{MVA} \\
S_{11kV} &= 207.1 + 87.4 = 294.5\ \text{MVA}
\end{aligned}
$$

$$
\begin{aligned}
I_{sc} &= \frac{294.5}{\sqrt{3} \times 11} = 15.46\ \text{kA}, \quad I_{make} = 2.55 \times 15.46 = 39.42\ \text{kA peak} \\
I_{FL} &= \frac{14.5}{\sqrt{3} \times 11} = 761\ \text{A}, \quad 1.25 I_{FL} = 951\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 951 A | 1250 A |
| Breaking capacity | 294.5 MVA (15.46 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 39.42 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 = 294.5 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. With Q25% = 18.24 m3/s and riparian release 0.35 m3/s, what is Qd?
   A. 18.24 m3/s
   B. 17.89 m3/s
   C. 18.59 m3/s
   D. 7.82 m3/s
   Answer: B. The release is deducted: Qd = 18.24 - 0.35 = 17.89 m3/s. (Key point: Design discharge at Q25%)

2. Why are 3 units chosen for this ~36 MW plant?
   A. Three units give three times the efficiency
   B. Transformers come only in threes
   C. Two large machines are hard to transport on hill roads, and three units follow the seasonal flow and lose only a third of the output in an outage
   D. The riparian release requires three units
   Answer: C. Smaller machines are easier to transport, units are loaded according to flow, and a single unit outage removes only one third of the output. (Key point: Number of units)

3. At 333.3 rpm with 4 jets the Ns per jet is 20. Which turbine does this support?
   A. Kaplan
   B. Francis at 333 rpm
   C. Propeller
   D. Pelton, with 4 jets per unit
   Answer: D. Ns of 40 divided by sqrt(4) gives 20, inside the Pelton range of about 10-30 per jet. (Key point: Turbine type)

4. What bus arrangement is suggested for 11 kV and 33 kV?
   A. Sectionalised single bus bars with a bus-section breaker
   B. Double bus bar at both levels
   C. No bus, direct connection
   D. Ring bus at 33 kV only
   Answer: A. A sectionalised single bus gives reasonable reliability at low cost for this plant; double or main-and-transfer buses cost more. (Key point: Bus bar type)

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