---
title: "Design: 150 m head plant, bus MVA levels"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2072-150m-bus-mva"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2072-150m-bus-mva
site: BCT Engineering
---

# Design: 150 m head plant, bus MVA levels

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

- Asked in: 2072 Asoj, 32 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-2072-150m-bus-mva
- 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 | 150 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 | 0.97 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 = 142.5 m. Station output 12.839 MW. Two units ×6.6 MW, installed13.2 MW. 8 MVA generators,8 MVA transformers. Francis turbine. SLD: 2 generators,2 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 study.

## 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.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 11.17 - 0.435 = 10.735\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 150 - 7.50 = 142.50\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 10.735 \times 142.50 = 15.007\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 15.007 = 13.506\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 13.506 = 13.101\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 13.101 = 12.839\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 13.101 MW; power delivered to the 66 kV bus = **12.839 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 $= 13.101/2 = 6.550$ MW → rate each unit at **6.6 MW**.
- Generator rating: $S = 6.6/0.85 = 7.76$ MVA → **8 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **8 MVA, 11/66 kV, YNd11** (standard size matching the 8 MVA generator).

**Answer: installed capacity = 2 × 6.6 MW = 13.2 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 | 8 MVA, 11 kV, 0.85 pf, 750 rpm, $X''_d$ = 15% |
| Step-up transformers T1–T2 | 8 MVA each, 11/66 kV, YNd11, $X$ = 12% |
| Station transformer ST | 200 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 125 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

$H_n = 142.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 13.506/2 = 6753.0$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 492.3$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 166.9 |
| 750 | 8 | 125.2 |
| 600 | 10 | 100.1 |
| 500 | 12 | 83.5 |
| 428.6 | 14 | 71.5 |
| 375 | 16 | 62.6 |
| 333.3 | 18 | 55.6 |

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): $N_s = 125.2$, 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 6.75 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 13.2 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 200 kVA, 4.5%.

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.15 \times \frac{100}{8} = 1.8750\ \text{pu}, \quad x_t = 0.12 \times \frac{100}{8} = 1.5000\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.97\times10^{-3} \times 35 = 10.666\ \Omega \\
x_l &= \frac{X_L \times 100}{66^2} = \frac{10.666}{43.56} = 0.2449\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\
x_{grid\ path} &= 0.1000 + 0.2449 = 0.3449\ \text{pu}
\end{aligned}
$$

$$
\begin{aligned}
x_{st} &= 0.045 \times \frac{100}{0.2} = 22.500\ \text{pu} \\
S_{MV} &= 2 \times \frac{100}{x_g} + \frac{100}{x_t/2 + x_{grid\ path}} = 106.7 + 91.3 = 198.0\ \text{MVA} \\
S_{HV} &= \frac{100}{x_g/2 + x_t/2} + \frac{100}{x_{grid\ path}} = 59.3 + 290.0 = 349.2\ \text{MVA} \\
X_{MV} &= \frac{100}{198.0} = 0.5050\ \text{pu}, \quad S_{LV} = \frac{100}{0.5050 + 22.500} = 4.35\ \text{MVA}
\end{aligned}
$$

| Bus | Fault MVA | $I_{sc}$ (kA) |
|---|---|---|
| LV (0.4 kV) | 4.35 | 6.27 |
| MV (11 kV) | 198.0 | 10.39 |
| HV (66 kV) | 349.2 | 3.06 |

**Answer: LV ≈ 4.35 MVA, MV ≈ 198.0 MVA, HV ≈ 349.2 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 | 420 A per generator | 198.0 MVA, 10.39 kA (make 26.5 kA) | 12 kV, 630 A, 25 kA VCB |
| HV CB (66 kV) | 140 A total | 349.2 MVA, 3.06 kA (make 7.8 kA) | 72.5 kV, 630 A, 25 kA SF6 |
| LV ACB (0.4 kV) | 289 A (station tr.) | 4.35 MVA, 6.27 kA (make 10.7 kA*) | 415 V, 400 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; 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. How is the design discharge obtained when the plant is designed at Q45%?
   A. Q45% minus the riparian release (10% of the driest monthly mean flow)
   B. Q45% plus the riparian release
   C. Q45% minus 10% of Q45%
   D. Q45% multiplied by the turbine efficiency
   Answer: A. The riparian release (10% of the driest month, May = 0.435 m3/s) must stay in the river, so Qd = 11.17 - 0.435 = 10.735 m3/s. (Key point: Design discharge)

2. Which sequence correctly gives the power delivered to the 66 kV bus?
   A. Hydraulic power x turbine eff x generator eff only
   B. Hydraulic power x turbine eff x generator eff x transformer eff
   C. Hydraulic power / (turbine eff x generator eff x transformer eff)
   D. Hydraulic power x turbine eff + generator eff + transformer eff
   Answer: B. Each stage multiplies the power by its efficiency: 15.007 x 0.9 x 0.97 x 0.98 = 12.839 MW. (Key point: Power chain)

3. Why is a Francis turbine chosen at Hn = 142.5 m?
   A. Kaplan turbines work best above 100 m head
   B. Pelton is impossible because the head is above 50 m
   C. Specific speed at 750 rpm is about 125, in the medium Francis range, and Kaplan is excluded above 40 m head
   D. Francis is always cheaper than Pelton at any head
   Answer: C. Head fits both Francis and Pelton ranges, so Ns decides; Ns = 125.2 suits Francis, and Pelton would need many jets. (Key point: Turbine selection)

4. Why is a sectionalised single bus with a bus-section breaker selected for this 13.2 MW plant?
   A. A double bus is not allowed at 11 kV
   B. It gives the highest possible reliability regardless of cost
   C. It removes the need for generator circuit breakers
   D. It gives reasonable reliability at low cost; a bus fault or maintenance affects only part of the plant
   Answer: D. A double or main-and-transfer bus costs much more and is used for larger plants or grid substations. (Key point: Busbar type)

## Close variants set in other exams

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