---
title: "Design: 120 m head plant, 33 kV"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2079-120m-33kv"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2079-120m-33kv
site: BCT Engineering
---

# Design: 120 m head plant, 33 kV

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

- Asked in: 2079 Jestha, 16 marks
- Video: [Design of single line diagram | power plant design | jitendra sah @smartstudy11](https://www.youtube.com/watch?v=_m_D_ErjyJU) — 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-2079-120m-33kv
- 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) |
|---|---|---|---|
| Jan | 11.95 | Qmax (Q0) | 674.19 m³/s |
| Feb | 9.94 | Q25% | 48.90 m³/s |
| Mar | 9.20 | Q45% | 16.75 m³/s |
| Apr | 11.93 | Q65% | 9.34 m³/s |
| May | 8.48 | Q85% | 6.21 m³/s |
| Jun | 42.28 | Q95% | 4.38 m³/s |
| July | 120.98 | Qmin | 1.40 m³/s |
| Aug | 143.00 |  |  |
| Sep | 110.27 |  |  |
| Oct | 50.95 |  |  |
| Nov | 24.66 |  |  |
| Dec | 16.17 |  |  |

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

| Parameter | Value |
|---|---|
| Gross head | 120 m |
| Turbine efficiency | 90% |
| Generator efficiency | 97% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 15% |
| Transformer reactance | 12% |
| Transmission line inductance | 0.96 mH/km |
| Transmission line length | 20 km |
| Transmission voltage level | 33 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Single circuit |

Design a power plant so as to obtain the following parameters
i) Electrical power output taking design discharge as Q45%
ii) Select the numbers of units and installed capacity with suitable reasons
iii) Draw the single line diagram showing numbers of units, generators, power transformer, station transformer and diesel generator for black start
iv) Select the type of turbine
v) Select the type of busbar with suitable reasons
vi) Calculate the MVA level of LV bus, MV bus and HV bus
vii) Calculate the rating of CB to be used in your design.
(Assume suitable data if required)

## Short answer

Electrical output 15.215 MW, two 7.8 MW Francis units (15.6 MW), unit transformers, single busbar with bus section. LV 0.4 kV (250 kVA), MV 11 kV fault level ~126 MVA, HV 33 kV; generator CB rated 9.5 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 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).

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

Driest month = May (8.48 m³/s), so riparian release $Q_r = 0.1 \times 8.48 = 0.848$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 16.75 - 0.848 = 15.902\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 120 - 6.00 = 114.00\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 15.902 \times 114.00 = 17.784\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 17.784 = 16.005\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 16.005 = 15.525\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 15.525 = 15.215\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 15.525 MW; power delivered to the 33 kV bus = **15.215 MW**.

#### ii) 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: $Q = 15.902/2 = 7.951$ m³/s, generator output $= 15.525/2 = 7.763$ MW → rate each unit at **7.8 MW**.
- Generator rating: $S = 7.8/0.85 = 9.18$ MVA → **9.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **10 MVA, 11/33 kV, YNd11** (next standard size above 9.5 MVA).

**Answer: installed capacity = 2 × 7.8 MW = 15.6 MW.**

#### iii) Single line diagram

```text
 Line to grid  (33 kV, 20 km)
    |
  [CB]
    |
 ===+=====+=============+=== 33 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 | 9.5 MVA, 11 kV, 0.85 pf, 600 rpm, $X''_d$ = 15% |
| Step-up transformers T1–T2 | 10 MVA each, 11/33 kV, YNd11, $X$ = 12% |
| 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 | 33 kV, single circuit, 20 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.

#### iv) Type of turbine

$H_n = 114.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 16.005/2 = 8002.7$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 372.5$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 240.2 |
| 750 | 8 | 180.1 |
| 600 | 10 | 144.1 |
| 500 | 12 | 120.1 |
| 428.6 | 14 | 102.9 |
| 375 | 16 | 90.1 |
| 333.3 | 18 | 80.0 |

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 $N = 600$ rpm (10 poles): $N_s = 144.1$, which is in the medium Francis range. A Pelton would need many jets (single-jet $N_s$ far above 30), and Kaplan is ruled out because the head is above 40 m.

**Answer: 2 × vertical-axis Francis turbines, 600 rpm, about 8.00 MW each.**

#### v) 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 one line bay.

Reason: for a 15.6 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.

#### vi) MVA level of LV, MV and HV buses

LV = 0.4 kV auxiliary bus, MV = 11 kV generator bus, HV = 33 kV switchyard bus. Station transformer 250 kVA, 4.5%.

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.15 \times \frac{100}{9.5} = 1.5789\ \text{pu}, \quad x_t = 0.12 \times \frac{100}{10} = 1.2000\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.96\times10^{-3} \times 20 = 6.032\ \Omega \\
x_l &= \frac{X_L \times 100}{33^2} = \frac{6.032}{10.89} = 0.5539\ \text{pu}, \quad x_{grid} = \frac{100}{500} = 0.2000\ \text{pu} \\
x_{grid\ path} &= 0.2000 + 0.5539 = 0.7539\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
x_{st} &= 0.045 \times \frac{100}{0.25} = 18.000\ \text{pu} \\
S_{MV} &= 2 \times \frac{100}{x_g} + \frac{100}{x_t/2 + x_{grid\ path}} = 126.7 + 73.9 = 200.5\ \text{MVA} \\
S_{HV} &= \frac{100}{x_g/2 + x_t/2} + \frac{100}{x_{grid\ path}} = 72.0 + 132.6 = 204.6\ \text{MVA} \\
X_{MV} &= \frac{100}{200.5} = 0.4987\ \text{pu}, \quad S_{LV} = \frac{100}{0.4987 + 18.000} = 5.41\ \text{MVA}
\end{aligned}
$$

| Bus | Fault MVA | $I_{sc}$ (kA) |
|---|---|---|
| LV (0.4 kV) | 5.41 | 7.80 |
| MV (11 kV) | 200.5 | 10.52 |
| HV (33 kV) | 204.6 | 3.58 |

**Answer: LV ≈ 5.41 MVA, MV ≈ 200.5 MVA, HV ≈ 204.6 MVA.**

#### vii) Rating of circuit breakers

Breaking current from the bus fault levels above; making current = 2.55 × $I_{sc}$ (IEC 62271-100); rated current ≥ 1.25 × full-load current.

| Breaker | Full-load current | Required breaking | Selected |
|---|---|---|---|
| GCB / 11 kV CBs | 499 A per generator | 200.5 MVA, 10.52 kA (make 26.8 kA) | 12 kV, 630 A, 25 kA VCB |
| HV CB (33 kV) | 350 A total | 204.6 MVA, 3.58 kA (make 9.1 kA) | 36 kV, 630 A, 16 kA SF6 |
| LV ACB (0.4 kV) | 361 A (station tr.) | 5.41 MVA, 7.80 kA (make 13.3 kA*) | 415 V, 630 A, 25 kA ACB |

*At 0.4 kV the making current uses the IEC 60947-2 factor n = 1.7 for this fault current.

**Answer: 11 kV GCB: 12 kV, 630 A, 25 kA; 33 kV CB: 36 kV, 630 A, 16 kA; 0.4 kV ACB: 630 A, 25 kA.**

## Check yourself

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

1. How is the design discharge for the Q45% design obtained in this answer?
   A. Q45% minus 10% of the driest monthly mean flow (16.75 - 0.848)
   B. Q45% minus 10% of Q45% itself
   C. The average of all twelve monthly flows
   D. Q45% plus the driest monthly flow
   Answer: A. The riparian release is taken as 10% of the driest month (May, 8.48 m3/s), giving 0.848 m3/s, which is subtracted from Q45%. (Key point: Riparian release and design discharge)

2. Which sequence gives the power delivered to the 33 kV bus from hydraulic power?
   A. Add the three efficiencies and multiply
   B. Multiply by generator and transformer efficiencies but not the turbine's
   C. Multiply by turbine efficiency only
   D. Multiply by turbine, generator and transformer efficiencies in turn (0.9 x 0.97 x 0.98)
   Answer: D. Losses occur in series, so hydraulic power 17.784 MW is reduced to 16.005, 15.525 and finally 15.215 MW. (Key point: Power chain to the 33 kV bus)

3. Why does the answer choose 2 units for this run-of-river plant?
   A. A single unit would always be cheaper and more reliable
   B. Two units are required by the Francis head range
   C. Two units double the energy available from the river
   D. One unit can run near full load on about half the design flow in the dry months, and one can be maintained while the other runs
   Answer: D. Two units keep efficiency high at low winter flow and allow maintenance without a full shutdown, while keeping cost low. (Key point: Reason for two units)

4. Net head is 114 m. Why is a Francis turbine chosen and Kaplan excluded?
   A. Kaplan has a higher specific speed requirement of 10-30
   B. Head is above Kaplan's 40 m limit and the specific speed (144 at 600 rpm) is in the Francis range
   C. Head is below 50 m so Pelton is impossible and Kaplan is too slow
   D. Francis is the only turbine that works above 100 m
   Answer: B. 114 m exceeds the Kaplan range, and Ns = 144.1 at 600 rpm fits medium Francis, whereas a Pelton would need many jets. (Key point: Turbine selection)

## Close variants set in other exams

- [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×

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