---
title: "Design: 260 m head plant, generator lead"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2078-260m-generator-lead"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2078-260m-generator-lead
site: BCT Engineering
---

# Design: 260 m head plant, generator lead

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

- Asked in: 2078 Kartik, 3+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-2078-260m-generator-lead
- 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) |
|---|---|---|---|
| January | 5.5 | Qmax (Q0) | 207.00 m³/s |
| February | 4.6 | Q25% | 28.24 m³/s |
| March | 3.58 | Q45% | 8.87 m³/s |
| April | 4.55 | Q65% | 5.7 m³/s |
| May | 4.8 | Q85% | 4.10 m³/s |
| June | 12.77 | Q95% | 2.91 m³/s |
| July | 39.95 | Qmin | 1.05 m³/s |
| August | 50.26 |  |  |
| September | 41.23 |  |  |
| October | 23.05 |  |  |
| November | 9.38 |  |  |
| December | 6.35 |  |  |

| 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 | 260 m |
| Turbine efficiency | 90% |
| Generator efficiency | 97% |
| Transformer efficiency | 98% |
| Head loss | 5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 21% |
| Transformer reactance | 9.15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 30 km |
| Transmission voltage level | 66 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 Q45%. [3]
b) Select the numbers of units and installed capacity. [3]
c) Choose the appropriate type of Generator lead. [2]
d) Make single line diagram for the project choosing your generator transformer scheme and bus bar system. [4]
e) Find Rating of generator circuit breaker of the project. [4]

## Short answer

Electrical power output at Q45% is 17.646 MW. Two 9.1 MW units give 18.2 MW installed capacity. Generator lead is 11 kV XLPE single-core copper cables in cable trench/tray, rated for 577 A full-load current.

## 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 = March (3.58 m³/s), so riparian release $Q_r = 0.1 \times 3.58 = 0.358$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 8.87 - 0.358 = 8.512\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 260 - 13.00 = 247.00\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 8.512 \times 247.00 = 20.625\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 20.625 = 18.563\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 18.563 = 18.006\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 18.006 = 17.646\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 18.006 MW; power delivered to the 66 kV bus = **17.646 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: $Q = 8.512/2 = 4.256$ m³/s, generator output $= 18.006/2 = 9.003$ MW → rate each unit at **9.1 MW**.
- Generator rating: $S = 9.1/0.85 = 10.71$ MVA → **11 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **12.5 MVA, 11/66 kV, YNd11** (next standard size above 11 MVA).

**Answer: installed capacity = 2 × 9.1 MW = 18.2 MW.**

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

$H_n = 247.00$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 18.563/2 = 9281.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 979.2$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 98.4 |
| 750 | 8 | 73.8 |
| 600 | 10 | 59.0 |
| 500 | 12 | 49.2 |
| 428.6 | 14 | 42.2 |
| 375 | 16 | 36.9 |
| 333.3 | 18 | 32.8 |

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 $N = 375$ rpm (16 poles): $N_s = 36.9$. With **4 jets**, $N_{s,jet} = 36.9/\sqrt{4} = 18.4$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 68.2$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.60$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.141$ m; $D/d = 11.3 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 98.4$) is also possible, but the Pelton is preferred because (i) its efficiency stays nearly flat from about 20% to 100% load (jets can be switched off), which suits a run-of-river plant whose flow falls far below the design flow in the dry season, and (ii) it tolerates the heavy sediment of Himalayan rivers better, and worn buckets and needles are easier to repair.

**Answer: 2 × vertical-axis 4-jet Pelton turbines, 375 rpm, about 9.28 MW each.**

#### c) Type of generator lead

Generator full-load current: $I = \dfrac{11 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 577$ A.

Choose **11 kV XLPE single-core copper cables** (generator lead in cable trench/tray) or a non-segregated phase bus duct.

- At about 577 A, two runs per phase of 1C 240 mm² Cu XLPE cable (about 400–450 A per cable laid in trefoil after derating) carry the current with margin.
- Cables are cheap, flexible to route inside the powerhouse, and easy to terminate at the GCB panel and transformer.
- A bus duct (NSPBD/SPBD) becomes economic only for higher currents (above about 1000–1500 A); IPB only for very large units.

#### d) 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: each unit is 11 MVA (about 577 A at 11 kV), the transmission voltage is 66 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) (contd.) Single line diagram

```text
 Line-1          Line-2  (66 kV, 30 km)
    |               |
  [CB]            [CB]
    |               |
 ===+====+==========+=====+=== 66 kV bus
         |                |
       [CB]             [CB]
         |                |
       (T1)             (T2)
         |                |
  ST1 ---+                +--- ST2
         |                |
       [GCB]            [GCB]
         |                |
       (G1)             (G2)
         |                |
       [NGT]            [NGT]

 ST1, ST2 (11/0.4 kV)
    |               |
 ===+====[BS]=======+=== 0.4 kV aux bus
              |
            [ACB]
              |
            (DG)  diesel set, black start
```

| Item | Rating |
|---|---|
| Generators G1–G2 | 11 MVA, 11 kV, 0.85 pf, 375 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T2 | 12.5 MVA each, 11/66 kV, YNd11, $X$ = 9.15% |
| Station transformer ST1, ST2 | 315 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 200 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 66 kV, double circuit, 30 km to grid substation |
| NGT | neutral grounding transformer with secondary resistor (high-resistance earthing) |

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

- **11 kV:** no common bus (unit scheme); each generator connects to its own step-up transformer through its GCB, and the station transformers are tapped off between the GCB and the transformer.
- **66 kV:** single bus bar, sectionalised by a bus-section breaker, with the two transformer bays on different sections and two line bays (double circuit), so a bus fault or maintenance takes out only part of the plant.

Reason: for an 18.2 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.

#### e) Rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{11} = 1.9091\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{12.5} = 0.7320\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 30 = 8.954\ \Omega \text{ per circuit; two in parallel} = 4.477\ \Omega \\
x_l &= \frac{X_L \times 100}{66^2} = \frac{4.477}{43.56} = 0.1028\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\
x_{grid\ path} &= 0.1000 + 0.1028 = 0.2028\ \text{pu}
\end{aligned}
$$

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).

$$
\begin{aligned}
S_a &= \frac{100}{x_g} = \frac{100}{1.9091} = 52.4\ \text{MVA} \\
X_{ext} &= x_{grid\ path} \parallel (x_g + x_t) = 0.2028 \parallel 2.6411 = 0.1883\ \text{pu} \\
S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.7320 + 0.1883} = 108.7\ \text{MVA}
\end{aligned}
$$

Design fault level for the GCB = larger value = **108.7 MVA**.

$$
\begin{aligned}
I_{sc} &= \frac{108.7}{\sqrt{3} \times 11} = 5.70\ \text{kA}, \quad I_{make} = 2.55 \times 5.70 = 14.54\ \text{kA peak} \\
I_{FL} &= \frac{11}{\sqrt{3} \times 11} = 577\ \text{A}, \quad 1.25 I_{FL} = 722\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 722 A | 800 A |
| Breaking capacity | 108.7 MVA (5.70 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 14.54 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 = 108.7 MVA; select a 12 kV, 800 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. Which set of steps gives the 17.646 MW delivered to the 66 kV bus?
   A. Qd = 8.87 - 0.358, Hn = 247 m, then multiply by turbine, generator and transformer efficiencies
   B. Only turbine efficiency applied to 9.81 x Qd x 260
   C. Qd = 8.87 - 0.887 with no head loss
   D. Qd = 8.87, Hn = 260 m, then efficiencies
   Answer: A. Release is 10% of the driest March flow (0.358), net head is 95% of 260 m, and the three efficiencies are applied in series. (Key point: Output at Q45%)

2. Why are 11 kV XLPE cables chosen as the generator lead here?
   A. Cables have no current-carrying limit so one run is enough
   B. Bus ducts cannot be used at 11 kV
   C. Full-load current is only about 577 A, and cables are cheap, flexible and easy to terminate
   D. Cables are needed because current is above 3000 A
   Answer: C. At about 577 A two cable runs per phase suffice; bus ducts only become economic above roughly 1000-1500 A. (Key point: Generator lead type)

3. How is the 577 A generator current obtained?
   A. I = S/V = 11 MVA / 11 kV
   B. I = P/(3 x V) using 18 MW
   C. I = S x sqrt(3) / V
   D. I = S/(sqrt(3) x V) = 11 MVA / (sqrt(3) x 11 kV)
   Answer: D. Three-phase line current is MVA divided by sqrt(3) times line voltage, giving 11e6/(1.732 x 11e3) = 577 A. (Key point: Generator current calculation)

4. Why is a 4-jet Pelton preferred even though Francis at 1000 rpm is possible?
   A. Pelton has higher efficiency than any Francis at full load only
   B. Pelton needs no maintenance of buckets
   C. Its efficiency is nearly flat from about 20% to 100% load and it tolerates sediment better
   D. Francis cannot operate at 247 m head
   Answer: C. A run-of-river plant often runs at part flow and carries sediment, which suits the Pelton's flat efficiency curve and repairable buckets. (Key point: Turbine choice for 247 m head)

## Close variants set in other exams

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