---
title: "Design: 95 m head plant at Q65%"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2079-95m-q65"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2079-95m-q65
site: BCT Engineering
---

# Design: 95 m head plant at Q65%

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

- Asked in: 2079 Chaitra, 1+2+3+2+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-2079-95m-q65
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-4

## Question

A typical hydropower plant has the following details:

| Month | Discharge (m³/s) | Flow duration | Discharge (m³/s) |
|---|---|---|---|
| January | 3.30 | Qmax (Q0) | 110.00 m³/s |
| February | 2.20 | Q25% | 16.42 m³/s |
| March | 1.40 | Q45% | 6.82 m³/s |
| April | 1.00 | Q65% | 3.7 m³/s |
| May | 3.50 | Q85% | 2.10 m³/s |
| June | 6.00 | Q95% | 1.17 m³/s |
| July | 14.00 | Qmin | 0.64 m³/s |
| August | 35.00 |  |  |
| September | 24.00 |  |  |
| October | 12.00 |  |  |
| November | 7.50 |  |  |
| December | 5.00 |  |  |

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

| Parameter | Value |
|---|---|
| Gross head | 95 m |
| Turbine efficiency | 90% |
| Generator efficiency | 97% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 20% |
| Transformer reactance | 11% |
| Transmission line inductance | 0.77 mH/km |
| Transmission line length | 80 km |
| Transmission voltage level | 66 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Single circuit |

Design a hydro power plant so as to obtain following parameters
a) Electrical power output taking design discharge as Q65%. [1]
b) Select the numbers of unit and installed capacity. [2]
c) Select the type of turbine. [3]
d) Select the type of bus bar to be used. [2]
e) Draw the single line diagram showing number of units, generator, power transformer, station transformer and diesel generator for black start. [4]
f) Calculate the rating of generator circuit breaker and high voltage circuit breaker in your design. [4]

## Short answer

Electrical output 2.73 MW; installed 2.8 MW (2×1.4 MW Francis, 1000 rpm). Single bus at 11 kV and 66 kV. Generator breaker: 11 kV, 630 A, 40 kA; HV breaker: 66 kV, 630 A, 20 kA.

## 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 $Q_{65\%}$

Driest month = April (1 m³/s), so riparian release $Q_r = 0.1 \times 1 = 0.1$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{65\%} - Q_r = 3.7 - 0.1 = 3.600\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 95 - 4.75 = 90.25\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 3.600 \times 90.25 = 3.187\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 3.187 = 2.869\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 2.869 = 2.782\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 2.782 = 2.727\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 2.782 MW; power delivered to the 66 kV bus = **2.727 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: $Q = 3.600/2 = 1.800$ m³/s, generator output $= 2.782/2 = 1.391$ MW → rate each unit at **1.4 MW**.
- Generator rating: $S = 1.4/0.85 = 1.65$ MVA → **2 MVA, 11 kV, 0.85 pf, 50 Hz**.
- One step-up transformer for both units: **4 MVA, 11/66 kV, YNd11** (standard size covering 2 × 1.65 MVA = 3.3 MVA).

**Answer: installed capacity = 2 × 1.4 MW = 2.8 MW.**

#### c) Type of turbine

$H_n = 90.25$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 2.869/2 = 1434.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 278.2$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 136.1 |
| 750 | 8 | 102.1 |
| 600 | 10 | 81.7 |
| 500 | 12 | 68.1 |
| 428.6 | 14 | 58.4 |
| 375 | 16 | 51.1 |
| 333.3 | 18 | 45.4 |

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 $N = 1000$ rpm (6 poles): $N_s = 136.1$, 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, 1000 rpm, about 1.43 MW each.**

#### d) Type of bus bar

- **11 kV:** single bus bar; the two small generators and the transformer feeder are connected to it through VCB panels.
- **66 kV:** single bus bar with one transformer bay and one line bay.

Reason: the plant is only 2.8 MW with one line, so a simple single bus is the most economical; an outage of the bus means loss of a small block of power only, and maintenance can be planned for the dry season.

#### e) Single line diagram

```text
 Line to grid  (66 kV, 80 km)
    |
  [CB]
    |
 ===+======+=== 66 kV bus
           |
         [CB]
           |
         (T1)
           |
         [CB]
           |
 ===+======+======+=========+=== 11 kV bus
    |             |         |
  [GCB]         [GCB]     [CB]
    |             |         |
  (G1)          (G2)      (ST)
                            |
 ===========================+===+ 0.4 kV
                                |
                              [ACB]
                                |
                              (DG) black start
```

| Item | Rating |
|---|---|
| Generators G1–G2 | 2 MVA, 11 kV, 0.85 pf, 1000 rpm, $X''_d$ = 20% |
| Step-up transformer T1 | 4 MVA, 11/66 kV, YNd11, $X$ = 11% |
| Station transformer ST | 100 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 62.5 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 66 kV, single circuit, 80 km to grid substation |
| NGR | neutral grounding resistor (low-resistance earthing, one generator earthed at a time if preferred) |

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.

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

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.2 \times \frac{100}{2} = 10.0000\ \text{pu}, \quad x_t = 0.11 \times \frac{100}{4} = 2.7500\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.77\times10^{-3} \times 80 = 19.352\ \Omega \\
x_l &= \frac{X_L \times 100}{66^2} = \frac{19.352}{43.56} = 0.4443\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\
x_{grid\ path} &= 0.1000 + 0.4443 = 0.5443\ \text{pu}
\end{aligned}
$$

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):

$$
\begin{aligned}
S_{gen} &= 2 \times \frac{100}{10.0000} = 20.0\ \text{MVA} \\
S_{grid} &= \frac{100}{x_t + x_{grid\ path}} = \frac{100}{2.7500 + 0.5443} = 30.4\ \text{MVA} \\
S_{11kV} &= 20.0 + 30.4 = 50.4\ \text{MVA}
\end{aligned}
$$

$$
\begin{aligned}
I_{sc} &= \frac{50.4}{\sqrt{3} \times 11} = 2.64\ \text{kA}, \quad I_{make} = 2.55 \times 2.64 = 6.74\ \text{kA peak} \\
I_{FL} &= \frac{2}{\sqrt{3} \times 11} = 105\ \text{A}, \quad 1.25 I_{FL} = 131\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 131 A | 630 A |
| Breaking capacity | 50.4 MVA (2.64 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 6.74 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 = 50.4 MVA; select a 12 kV, 630 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.

#### f) (contd.) Rating of HV circuit breaker

For a fault on the 66 kV bus, the plant and the grid feed in parallel:

$$
\begin{aligned}
X_{plant} &= \frac{x_g}{2} + \frac{x_t}{1} = 5.0000 + 2.7500 = 7.7500\ \text{pu} \\
S_{HV} &= \frac{100}{7.7500} + \frac{100}{0.5443} = 12.9 + 183.7 = 196.6\ \text{MVA} \\
I_{sc} &= \frac{196.6}{\sqrt{3} \times 66} = 1.72\ \text{kA}, \quad I_{make} = 2.55 \times 1.72 = 4.39\ \text{kA peak} \\
I_{FL} &= \frac{1 \times 4}{\sqrt{3} \times 66} = 35\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 66 kV system | 72.5 kV |
| Rated normal current | ≥ 44 A | 630 A |
| Breaking capacity | 196.6 MVA (1.72 kA) | 25 kA |
| Making capacity | 4.39 kA peak | 63 kA peak |
| BIL | – | 325 kVp |
| Type | – | SF6, outdoor, spring mechanism, 3-pole |

**Answer: HV bus fault level = 196.6 MVA (1.72 kA); select a 72.5 kV, 630 A, 25 kA SF6 circuit breaker.** Of this, the plant itself contributes only 12.9 MVA; the rest comes from the grid, so the result depends on the assumed grid fault level.

## Check yourself

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

1. What is the design discharge used for the power output?
   A. 3.600 m3/s: Q65% = 3.7 minus riparian release 0.1 (10% of April's 1.0 m3/s)
   B. 3.7 m3/s with no deduction
   C. 3.6 m3/s because 10% of Q65% is 0.37
   D. 6.82 m3/s from Q45%
   Answer: A. The driest month is April at 1.0 m3/s, so 0.1 m3/s is released and Qd = 3.7 - 0.1 = 3.6 m3/s. (Key point: Design discharge at Q65%)

2. Why is a Francis turbine selected at 1000 rpm instead of Pelton?
   A. Head of 90 m is above the Francis limit
   B. Ns = 136.1 is in the medium Francis range; a Pelton would need many jets and Kaplan is excluded above 40 m head
   C. Pelton cannot run at 1000 rpm
   D. Ns of 136.1 lies in the Kaplan range
   Answer: B. Ns of 136.1 is far above the 10-30 per jet Pelton range and below Kaplan range, so Francis fits. (Key point: Turbine choice)

3. Why is a single bus bar used at 11 kV and 66 kV?
   A. Single bus is the most reliable arrangement
   B. Double bus cannot be used at 66 kV
   C. The plant is small (2.8 MW) with one line, so a simple single bus is most economical and outages can be planned in the dry season
   D. A single bus is needed for black start
   Answer: C. For a 2.8 MW plant, a bus outage loses little power, so the cheapest arrangement is accepted. (Key point: Bus bar)

4. How is the GCB breaking duty found in this design where both units feed a common 11 kV bus?
   A. Fault MVA of one generator only (10 MVA)
   B. Only the grid contribution (30.4 MVA)
   C. Full-load current times 1.25
   D. Sum of both generators' fault MVA plus grid contribution through the transformer: 20 + 30.4 = 50.4 MVA
   Answer: D. The bus fault level includes both generators (2 x 100/10) and the grid via the step-up transformer, giving 50.4 MVA or 2.64 kA. (Key point: GCB fault level)

## Close variants set in other exams

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