---
title: "Design: 589 m head plant at Q40%"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2079-589m-q40"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2079-589m-q40
site: BCT Engineering
---

# Design: 589 m head plant at Q40%

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

- Asked in: 2079 Shrawan, 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-2079-589m-q40
- 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) |
|---|---|---|---|
| Baishakh | 6.26 | Qmax (Q0) | 157.00 m³/s |
| Jestha | 12.45 | Q25% | 18.24 m³/s |
| Ashad | 32.12 | Q40% | 12.85 m³/s |
| Shrawan | 49.08 | Q65% | 10.7 m³/s |
| Bhadra | 44.47 | Q85% | 6.40 m³/s |
| Ashoj | 23.88 | Q95% | 4.91 m³/s |
| Kartik | 11.57 | Qmin | 3.35 m³/s |
| Mangsir | 7.2 |  |  |
| Poush | 5.06 |  |  |
| Magh | 4.1 |  |  |
| Falgun | 3.69 |  |  |
| Chaitra | 4.25 |  |  |

| 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 | 589 m |
| Turbine efficiency | 91% |
| Generator efficiency | 97% |
| Transformer efficiency | 99% |
| Head loss | 4% |
| Riparian release | 10% of driest flow |
| Generator sub-transient reactance | 18% |
| Transformer reactance | 7.15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 12 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 Q40%. [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

Design discharge Q40% after riparian release yields net head 565.44 m, generator output 61.111 MW, delivered 60.500 MW. Install three 20.4 MW Pelton units (61.2 MW total) with unit transformer scheme and GCB rated 24 MVA.

## Answer

Assumptions: riparian release = 10% of the driest monthly mean flow; net head = gross head minus 4% 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 $Q_{40\%}$

Driest month = Falgun (3.69 m³/s), so riparian release $Q_r = 0.1 \times 3.69 = 0.369$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{40\%} - Q_r = 12.85 - 0.369 = 12.481\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.04) = 589 - 23.56 = 565.44\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 12.481 \times 565.44 = 69.232\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 69.232 = 63.001\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 63.001 = 61.111\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 61.111 = 60.500\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 61.111 MW; power delivered to the 132 kV bus = **60.500 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: $Q = 12.481/3 = 4.160$ m³/s, generator output $= 61.111/3 = 20.370$ MW → rate each unit at **20.4 MW**.
- Generator rating: $S = 20.4/0.85 = 24.00$ MVA → **24 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **25 MVA, 11/132 kV, YNd11** (next standard size above 24 MVA).

**Answer: installed capacity = 3 × 20.4 MW = 61.2 MW.**

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

$H_n = 565.44$ m is above the Francis limit (350 m), so only a **Pelton** turbine is possible; specific speed fixes the speed and number of jets.

Shaft power per unit $P = 63.001/3 = 21000.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 2757.3$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 52.6 |
| 750 | 8 | 39.4 |
| 600 | 10 | 31.5 |
| 500 | 12 | 26.3 |
| 428.6 | 14 | 22.5 |
| 375 | 16 | 19.7 |
| 333.3 | 18 | 17.5 |

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 $N = 750$ rpm (8 poles): $N_s = 39.4$. With **4 jets**, $N_{s,jet} = 39.4/\sqrt{4} = 19.7$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 103.2$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.21$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.113$ m; $D/d = 10.7 > 10$, acceptable.

The Pelton also keeps high efficiency at part load and handles sediment well.

**Answer: 3 × vertical-axis 4-jet Pelton turbines, 750 rpm, about 21.00 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 (24 MVA each, about 1260 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, 12 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 | 24 MVA, 11 kV, 0.85 pf, 750 rpm, $X''_d$ = 18% |
| Step-up transformers T1–T3 | 25 MVA each, 11/132 kV, YNd11, $X$ = 7.15% |
| 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, 12 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:

$$
\begin{aligned}
x_g &= 0.18 \times \frac{100}{24} = 0.7500\ \text{pu}, \quad x_t = 0.0715 \times \frac{100}{25} = 0.2860\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 12 = 3.581\ \Omega \text{ per circuit; two in parallel} = 1.791\ \Omega \\
x_l &= \frac{X_L \times 100}{132^2} = \frac{1.791}{174.24} = 0.0103\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\
x_{grid\ path} &= 0.0500 + 0.0103 = 0.0603\ \text{pu}
\end{aligned}
$$

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).

$$
\begin{aligned}
S_a &= \frac{100}{x_g} = \frac{100}{0.7500} = 133.3\ \text{MVA} \\
X_{ext} &= x_{grid\ path} \parallel (x_g + x_t)/2 = 0.0603 \parallel 0.5180 = 0.0540\ \text{pu} \\
S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.2860 + 0.0540} = 294.1\ \text{MVA}
\end{aligned}
$$

Design fault level for the GCB = larger value = **294.1 MVA**.

$$
\begin{aligned}
I_{sc} &= \frac{294.1}{\sqrt{3} \times 11} = 15.44\ \text{kA}, \quad I_{make} = 2.55 \times 15.44 = 39.37\ \text{kA peak} \\
I_{FL} &= \frac{24}{\sqrt{3} \times 11} = 1260\ \text{A}, \quad 1.25 I_{FL} = 1575\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 1575 A | 1600 A |
| Breaking capacity | 294.1 MVA (15.44 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 39.37 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 = 294.1 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. How is the electrical output at Q40% found?
   A. Qd = 12.85 - 0.369 = 12.481 m3/s; Hn = 589 x 0.96 = 565.44 m; then rho g Qd Hn and the three efficiencies
   B. Qd = 12.85 m3/s and Hn = 589 m with no deductions
   C. Qd = 12.85 + 0.369 m3/s with a 4% head gain
   D. Qd = 18.24 - 0.369 m3/s from Q25%
   Answer: A. Riparian release (10% of Falgun's 3.69) is deducted and 4% head loss applies, giving 60.5 MW to the bus. (Key point: Power output at Q40%)

2. Why can only a Pelton turbine be used here?
   A. Net head is below 40 m
   B. Net head 565 m is above the Francis limit of 350 m
   C. Discharge is too low for Francis at any head
   D. Pelton is the only turbine with a draft tube
   Answer: B. The Francis range ends at 350 m, so only Pelton (50-1300 m) covers 565 m. (Key point: Turbine type by head)

3. Why are 3 units of about 20.4 MW chosen?
   A. Three units are required by the double-circuit line
   B. Three units give a lower head loss
   C. Two units would be very large for hill-road transport; three follow seasonal flow and one outage loses only a third
   D. Generator rating must be 24 MVA only with three units
   Answer: C. Three units ease transport, suit flow variation and limit the effect of one unit being out. (Key point: Units)

4. Why is the unit scheme chosen with a GCB for each generator?
   A. It uses fewer transformers than a common bus scheme
   B. It avoids the need for a station transformer
   C. It removes the need for earthing
   D. It keeps the 11 kV fault level and switchgear small, and a transformer fault loses only one unit
   Answer: D. Each generator has its own transformer, so fault levels at 11 kV are low and faults are isolated to one unit. (Key point: Unit scheme)

## Close variants set in other exams

- [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×

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