---
title: "Design: 120 m head plant, bus MVA levels"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2073-120m-66kv"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2073-120m-66kv
site: BCT Engineering
---

# Design: 120 m head plant, bus MVA levels

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

- Asked in: 2073 Bhadra, 16 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-2073-120m-66kv
- 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 | 6.13 | Qmax (Q0) | 449.46 m³/s |
| Feb | 5.10 | Q25% | 32.60 m³/s |
| Mar | 4.72 | Q45% | 11.17 m³/s |
| Apr | 6.12 | Q65% | 6.23 m³/s |
| May | 4.35 | Q85% | 4.14 m³/s |
| Jun | 21.68 | Q95% | 2.92 m³/s |
| July | 62.04 | Qmin | 0.93 m³/s |
| Aug | 73.37 |  |  |
| Sep | 56.55 |  |  |
| Oct | 26.13 |  |  |
| Nov | 12.65 |  |  |
| Dec | 8.24 |  |  |

| 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 | 20% |
| Transformer reactance | 15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 35 km |
| Transmission voltage level | 66 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Single 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 with suitable reasons
c. Draw the single line diagram showing numbers of units, generators, power transformer, station transformer and diesel generator for black start
d. Select the type of turbine
e. Select the type of busbar with suitable reasons
f. Calculate the MVA level of LV bus, MV bus and HV bus
g. Calculate the rating of CB to be used in your design.
(Assume suitable data if required and the necessary graph is attached herewith)

## Short answer

Design discharge Q45% minus riparian release = 10.735 m³/s, net head = 114 m. Generator output 10.481 MW, station output 10.271 MW. Two units ×5.3 MW, installed 10.6 MW. Francis turbine. SLD: 2 generators, 2 step‑up transformers, ST, DG; single 66 kV bus, 11 kV bus with section breaker. Busbar type single bus. LV, MV, HV bus MVA and CB ratings from fault calculations.

## 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 (4.35 m³/s), so riparian release $Q_r = 0.1 \times 4.35 = 0.435$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 11.17 - 0.435 = 10.735\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 120 - 6.00 = 114.00\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 10.735 \times 114.00 = 12.005\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 12.005 = 10.805\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 10.805 = 10.481\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 10.481 = 10.271\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 10.481 MW; power delivered to the 66 kV bus = **10.271 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: $Q = 10.735/2 = 5.367$ m³/s, generator output $= 10.481/2 = 5.240$ MW → rate each unit at **5.3 MW**.
- Generator rating: $S = 5.3/0.85 = 6.24$ MVA → **6.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **8 MVA, 11/66 kV, YNd11** (next standard size above 6.5 MVA).

**Answer: installed capacity = 2 × 5.3 MW = 10.6 MW.**

#### c. Single line diagram

```text
 Line to grid  (66 kV, 35 km)
    |
  [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 | 6.5 MVA, 11 kV, 0.85 pf, 750 rpm, $X''_d$ = 20% |
| Step-up transformers T1–T2 | 8 MVA each, 11/66 kV, YNd11, $X$ = 15% |
| Station transformer ST | 160 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 100 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 66 kV, single circuit, 35 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.

#### d. 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 $P = 10.805/2 = 5402.4$ 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 | 197.3 |
| 750 | 8 | 148.0 |
| 600 | 10 | 118.4 |
| 500 | 12 | 98.7 |
| 428.6 | 14 | 84.6 |
| 375 | 16 | 74.0 |
| 333.3 | 18 | 65.8 |

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): $N_s = 148.0$, 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, 750 rpm, about 5.40 MW each.**

#### 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.
- **66 kV:** single bus bar, sectionalised, with transformer bays and one line bay.

Reason: for a 10.6 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. MVA level of LV, MV and HV buses

LV = 0.4 kV auxiliary bus, MV = 11 kV generator bus, HV = 66 kV switchyard bus. Station transformer 160 kVA, 4.5%.

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.2 \times \frac{100}{6.5} = 3.0769\ \text{pu}, \quad x_t = 0.15 \times \frac{100}{8} = 1.8750\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 35 = 10.446\ \Omega \\
x_l &= \frac{X_L \times 100}{66^2} = \frac{10.446}{43.56} = 0.2398\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\
x_{grid\ path} &= 0.1000 + 0.2398 = 0.3398\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
x_{st} &= 0.045 \times \frac{100}{0.16} = 28.125\ \text{pu} \\
S_{MV} &= 2 \times \frac{100}{x_g} + \frac{100}{x_t/2 + x_{grid\ path}} = 65.0 + 78.3 = 143.3\ \text{MVA} \\
S_{HV} &= \frac{100}{x_g/2 + x_t/2} + \frac{100}{x_{grid\ path}} = 40.4 + 294.3 = 334.7\ \text{MVA} \\
X_{MV} &= \frac{100}{143.3} = 0.6979\ \text{pu}, \quad S_{LV} = \frac{100}{0.6979 + 28.125} = 3.47\ \text{MVA}
\end{aligned}
$$

| Bus | Fault MVA | $I_{sc}$ (kA) |
|---|---|---|
| LV (0.4 kV) | 3.47 | 5.01 |
| MV (11 kV) | 143.3 | 7.52 |
| HV (66 kV) | 334.7 | 2.93 |

**Answer: LV ≈ 3.47 MVA, MV ≈ 143.3 MVA, HV ≈ 334.7 MVA.**

#### g. 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 | 341 A per generator | 143.3 MVA, 7.52 kA (make 19.2 kA) | 12 kV, 630 A, 25 kA VCB |
| HV CB (66 kV) | 140 A total | 334.7 MVA, 2.93 kA (make 7.5 kA) | 72.5 kV, 630 A, 25 kA SF6 |
| LV ACB (0.4 kV) | 231 A (station tr.) | 3.47 MVA, 5.01 kA (make 7.5 kA*) | 415 V, 400 A, 25 kA ACB |

*At 0.4 kV the making current uses the IEC 60947-2 factor n = 1.5 for this fault current.

**Answer: 11 kV GCB: 12 kV, 630 A, 25 kA; 66 kV CB: 72.5 kV, 630 A, 25 kA; 0.4 kV ACB: 400 A, 25 kA.**

## Check yourself

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

1. Why is a Francis turbine chosen for Hn = 114 m at 750 rpm?
   A. Kaplan is best because the head is above 40 m
   B. Ns = 148 is in the medium Francis range, a Pelton would need many jets and Kaplan is ruled out above 40 m
   C. Pelton is best because it is cheapest
   D. Ns = 148 lies in the Pelton single-jet range
   Answer: B. Ns of 148 is typical Francis (about 60-300), far above the Pelton single-jet limit of about 30, and Kaplan only works up to 40 m. (Key point: Turbine type by head and Ns)

2. Q45% = 11.17 m3/s and the driest month (May) is 4.35 m3/s. What is Qd?
   A. 11.17 m3/s
   B. 11.605 m3/s
   C. 10.735 m3/s
   D. 6.82 m3/s
   Answer: C. Riparian release = 0.435 m3/s, so Qd = 11.17 - 0.435 = 10.735 m3/s. (Key point: Design discharge)

3. In this plant what are the LV, MV and HV buses?
   A. 11 kV, 66 kV and 0.4 kV
   B. 0.4 kV auxiliary, 11 kV generator and 66 kV switchyard
   C. 66 kV, 11 kV and 33 kV
   D. 0.4 kV, 33 kV and 66 kV
   Answer: B. LV is the 0.4 kV station auxiliary bus, MV the 11 kV generator bus, and HV the 66 kV transmission switchyard. (Key point: Meaning of LV, MV, HV buses)

4. Why is the LV bus fault MVA (3.47 MVA) far smaller than the MV bus (143.3 MVA)?
   A. The station transformer is small (160 kVA, 4.5%), so its per-unit reactance of 28.1 pu on 100 MVA limits the fault level
   B. LV buses have no source
   C. The voltage is higher on LV
   D. The diesel generator absorbs the fault
   Answer: A. The large series reactance of the small station transformer is added to the MV equivalent reactance, giving S = 100/(0.6979 + 28.125). (Key point: LV bus fault level)

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