---
title: "Design: 230 m head plant, 55 km line"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2074-230m-55km"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2074-230m-55km
site: BCT Engineering
---

# Design: 230 m head plant, 55 km line

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

- Asked in: 2074 Magh, 16 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-2074-230m-55km
- 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 | 4.62 | Qmax (Q0) | 210.00 m³/s |
| February | 3.80 | Q25% | 18.24 m³/s |
| March | 3.55 | Q45% | 7.82 m³/s |
| April | 4.55 | Q65% | 4.7 m³/s |
| May | 3.5 | Q85% | 3.10 m³/s |
| June | 12.70 | Q95% | 2.17 m³/s |
| July | 32.98 | Qmin | 0.64 m³/s |
| August | 40.26 |  |  |
| September | 31.23 |  |  |
| October | 15.00 |  |  |
| November | 7.37 |  |  |
| December | 4.80 |  |  |

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

| Parameter | Value |
|---|---|
| Gross head | 230 m |
| Turbine efficiency | 90% |
| Generator efficiency | 96% |
| Transformer efficiency | 99% |
| 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 | 55 km |
| Transmission voltage level | 33 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Double 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.
c) Select the type of turbine.
d) Draw the single line diagram showing number of units, generator, power transformer, station transformer and diesel generator for black start.
e) Calculate the rating of the generator circuit breaker (GCB) to be used in your design.
(Assume suitable data if necessary and necessary graph is attached herewith)
[Charts attached: turbine selection chart of discharge against head with Kaplan, Francis and Pelton envelopes; specific speed against net head chart; Pelton turbine efficiency curves for 1 to 4 injectors]

## Short answer

Generator output 13.834 MW, delivered 13.696 MW. Install 2×7 MW (14 MW total). Two vertical-axis 4-jet Pelton turbines, 375 rpm. Single-line diagram has two units, 11/33 kV transformers, double-circuit line; GCB rated 8.5 MVA, 11 kV.

## 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).

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

Driest month = May (3.5 m³/s), so riparian release $Q_r = 0.1 \times 3.5 = 0.35$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 7.82 - 0.35 = 7.470\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.05) = 230 - 11.50 = 218.50\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 7.470 \times 218.50 = 16.012\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.9 \times 16.012 = 14.411\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.96 \times 14.411 = 13.834\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 13.834 = 13.696\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 13.834 MW; power delivered to the 33 kV bus = **13.696 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: $Q = 7.470/2 = 3.735$ m³/s, generator output $= 13.834/2 = 6.917$ MW → rate each unit at **7 MW**.
- Generator rating: $S = 7/0.85 = 8.24$ MVA → **8.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **10 MVA, 11/33 kV, YNd11** (next standard size above 8.5 MVA).

**Answer: installed capacity = 2 × 7 MW = 14 MW.**

#### c) Type of turbine

$H_n = 218.50$ m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit $P = 14.411/2 = 7205.3$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 840.1$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 101.0 |
| 750 | 8 | 75.8 |
| 600 | 10 | 60.6 |
| 500 | 12 | 50.5 |
| 428.6 | 14 | 43.3 |
| 375 | 16 | 37.9 |
| 333.3 | 18 | 33.7 |

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): $N_s = 37.9$. With **4 jets**, $N_{s,jet} = 37.9/\sqrt{4} = 18.9$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 64.2$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.50$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.136$ m; $D/d = 11.0 > 10$, acceptable.

A Francis at 1000 rpm ($N_s = 101.0$) 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 7.21 MW each.**

#### d) Single line diagram

```text
 Line-1          Line-2  (33 kV, 55 km)
    |               |
  [CB]            [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 | 8.5 MVA, 11 kV, 0.85 pf, 375 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T2 | 10 MVA each, 11/33 kV, YNd11, $X$ = 9.15% |
| 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, double circuit, 55 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.

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

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{8.5} = 2.4706\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{10} = 0.9150\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 55 = 16.415\ \Omega \text{ per circuit; two in parallel} = 8.207\ \Omega \\
x_l &= \frac{X_L \times 100}{33^2} = \frac{8.207}{10.89} = 0.7537\ \text{pu}, \quad x_{grid} = \frac{100}{500} = 0.2000\ \text{pu} \\
x_{grid\ path} &= 0.2000 + 0.7537 = 0.9537\ \text{pu}
\end{aligned}
$$

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):

$$
\begin{aligned}
S_{gen} &= 2 \times \frac{100}{2.4706} = 81.0\ \text{MVA} \\
S_{grid} &= \frac{100}{x_t/2 + x_{grid\ path}} = \frac{100}{0.4575 + 0.9537} = 70.9\ \text{MVA} \\
S_{11kV} &= 81.0 + 70.9 = 151.8\ \text{MVA}
\end{aligned}
$$

$$
\begin{aligned}
I_{sc} &= \frac{151.8}{\sqrt{3} \times 11} = 7.97\ \text{kA}, \quad I_{make} = 2.55 \times 7.97 = 20.32\ \text{kA peak} \\
I_{FL} &= \frac{8.5}{\sqrt{3} \times 11} = 446\ \text{A}, \quad 1.25 I_{FL} = 558\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 558 A | 630 A |
| Breaking capacity | 151.8 MVA (7.97 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 20.32 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 = 151.8 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.

## Check yourself

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

1. How is the design discharge obtained from Q45% in this design?
   A. Add 10% of the driest monthly flow to Q45%
   B. Subtract 10% of the driest monthly mean flow (May, 3.5 m3/s) from Q45%
   C. Subtract 10% of Q45% itself
   D. Use Q45% directly with no deduction
   Answer: B. The riparian release is 10% of the driest monthly mean flow, 0.35 m3/s, which must stay in the river, so Qd = 7.82 - 0.35 = 7.47 m3/s. (Key point: Design discharge with riparian release)

2. After computing the hydraulic power from rho g Qd Hn, how is the power delivered to the 33 kV bus found?
   A. Multiply by turbine, generator and transformer efficiencies in turn (0.90, 0.96, 0.99)
   B. Multiply only by the turbine efficiency
   C. Add the three efficiencies and multiply once
   D. Multiply by the generator efficiency only, since the transformer is lossless
   Answer: A. Power passes through turbine, generator and transformer in series, so each efficiency multiplies: 16.012 x 0.9 x 0.96 x 0.99 = 13.696 MW. (Key point: Efficiency chain for electrical output)

3. Hn = 218.5 m falls in both the Francis and Pelton ranges. Why is a 4-jet Pelton chosen?
   A. Kaplan is the only turbine allowed above 50 m head
   B. Its specific speed per jet (about 18.9) lies in the Pelton range, and its efficiency stays flat at part load and it tolerates sediment
   C. Its specific speed is above 300, which only Pelton can handle
   D. Pelton turbines need no specific speed check
   Answer: B. Ns = 37.9 for 375 rpm gives 18.9 per jet with 4 jets, inside 10-30. Flat part-load efficiency and sediment tolerance suit a run-of-river Himalayan plant. (Key point: Turbine selection by specific speed)

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