---
title: "Design: 550 m head plant, transformer quotation"
subject: "Power Plant Design"
chapter: 4
key: "semester_8/power_plant_design:design-2077-550m-quotation"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2077-550m-quotation
site: BCT Engineering
---

# Design: 550 m head plant, transformer quotation

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

- Asked in: 2077 Chaitra, 3+3+3+5+4+6 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-2077-550m-quotation
- 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 | 4.15 | Qmax (Q0) | 120.00 m³/s |
| February | 3.60 | Q25% | 43.24 m³/s |
| March | 3.44 | Q45% | 16.17 m³/s |
| April | 3.89 | Q65% | 5.99 m³/s |
| May | 5.54 | Q85% | 4.40 m³/s |
| June | 16.67 | Q95% | 3.71 m³/s |
| July | 43.36 | Qmin | 2.65 m³/s |
| August | 53.29 |  |  |
| September | 37.84 |  |  |
| October | 16.56 |  |  |
| November | 8.15 |  |  |
| December | 5.53 |  |  |

| Parameter | Value |
|---|---|
| Gross head | 550 m |
| Turbine efficiency | 91% |
| Generator efficiency | 97% |
| Transformer efficiency | 99% |
| Head loss | 4.5% |
| Riparian release | 10% |
| Generator sub-transient reactance | 21% |
| Transformer reactance | 9.15% |
| Transmission line inductance | 0.95 mH/km |
| Transmission line length | 25 km |
| Transmission voltage level | 132 kV |
| Generator voltage level | 11 kV |
| Transmission circuit | Single 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 Turbine Type. [3]
c) What kind of Generator lead do you choose and why? [3]
d) Make single line diagram for the project choosing your generator transformer scheme and bus bar system. [5]
e) Find Rating of generator circuit breaker of the project. [4]
f) Make a tentative quotation for purchasing suitable "Generator Transformer" as per the requirement of the project. [6]

## Short answer

Power output at Q45% is 71.261 MW. Installed capacity is 3 × 24 MW (72 MW) using 4-jet Pelton turbines. Generator lead is a segregated phase bus duct due to 1496 A current; step-up transformers are 31.5 MVA, 11/132 kV, YNd11.

## Answer

Assumptions: riparian release = 10% of the driest monthly mean flow; net head = gross head minus 4.5% 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 2,000 MVA at the 132 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 (3.44 m³/s), so riparian release $Q_r = 0.1 \times 3.44 = 0.344$ m³/s.

$$
\begin{aligned}
Q_d &= Q_{45\%} - Q_r = 16.17 - 0.344 = 15.826\ \text{m}^3/\text{s} \\
H_n &= H_g(1 - 0.045) = 550 - 24.75 = 525.25\ \text{m} \\
P_{hyd} &= \rho g Q_d H_n = 9.81 \times 15.826 \times 525.25 = 81.547\ \text{MW} \\
P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 81.547 = 74.207\ \text{MW} \\
P_{gen} &= \eta_g P_{shaft} = 0.97 \times 74.207 = 71.981\ \text{MW} \\
P_{out} &= \eta_{tr} P_{gen} = 0.99 \times 71.981 = 71.261\ \text{MW}
\end{aligned}
$$

**Answer:** generator output = 71.981 MW; power delivered to the 132 kV bus = **71.261 MW**.

#### b) Number of units

- Choose **3 units**. A plant of this size would need very large single machines with only two units, which are hard to transport on Nepal's hill roads; three units also let the plant follow the large seasonal change in flow (one or two units in the dry months, all three in the wet months).
- Loss of one unit (forced outage or maintenance) removes only one third of the output.
- Per unit: $Q = 15.826/3 = 5.275$ m³/s, generator output $= 71.981/3 = 23.994$ MW → rate each unit at **24 MW**.
- Generator rating: $S = 24/0.85 = 28.24$ MVA → **28.5 MVA, 11 kV, 0.85 pf, 50 Hz**.
- Step-up transformer, one per unit: **31.5 MVA, 11/132 kV, YNd11** (next standard size above 28.5 MVA).

**Answer: installed capacity = 3 × 24 MW = 72 MW.**

#### b) (contd.) Turbine type

$H_n = 525.25$ m is above the Francis limit (350 m), so only a **Pelton** turbine is possible; specific speed fixes the speed and number of jets.

Shaft power per unit $P = 74.207/3 = 24735.8$ kW. With $N = 120f/p$ (synchronous speeds) and $N_s = N\sqrt{P}/H_n^{5/4}$, where $H_n^{5/4} = 2514.5$:

| $N$ (rpm) | Poles | $N_s$ (metric, kW) |
|---|---|---|
| 1000 | 6 | 62.5 |
| 750 | 8 | 46.9 |
| 600 | 10 | 37.5 |
| 500 | 12 | 31.3 |
| 428.6 | 14 | 26.8 |
| 375 | 16 | 23.5 |
| 333.3 | 18 | 20.8 |

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 $N = 600$ rpm (10 poles): $N_s = 37.5$. With **4 jets**, $N_{s,jet} = 37.5/\sqrt{4} = 18.8$, inside the Pelton range.

Check of jet ratio: jet velocity $V_j = 0.98\sqrt{2gH_n} = 99.5$ m/s; bucket speed $u = 0.46V_j$; runner diameter $D = 60u/(\pi N) = 1.46$ m; jet diameter $d = \sqrt{4(Q/z)/(\pi V_j)} = 0.130$ m; $D/d = 11.2 > 10$, acceptable.

The Pelton also keeps high efficiency at part load and handles sediment well.

**Answer: 3 × vertical-axis 4-jet Pelton turbines, 600 rpm, about 24.74 MW each.**

#### c) Generator lead and reason

Generator full-load current: $I = \dfrac{28.5 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 1496$ A.

Choose a **segregated phase bus duct (SPBD)** from the generator terminals to the GCB and the step-up transformer LV side.

- About 1496 A is too high for a practical number of parallel cables (several runs per phase, poor current sharing, many terminations).
- An isolated phase bus duct (IPB) is used for very large machines (above about 3000–4000 A, i.e. more than about 60–75 MVA at 11 kV); it is not needed here.
- SPBD has each phase in its own compartment of an earthed metal enclosure, so a phase-to-phase fault is very unlikely; it is also compact, self-cooled and low-maintenance.
- Rating: 12 kV, 2000 A continuous, short-time withstand ≥ the GCB rating (25 kA for 1 s), aluminium bars, with tap-off for the station transformer, surge capacitors and PTs.

#### 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: the units are large (28.5 MVA each, about 1496 A at 11 kV), the transmission voltage is 132 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 to grid  (132 kV, 25 km)
    |
  [CB]
    |
 ===+===+===========+===========+=== 132 kV bus
        |           |           |
      [CB]        [CB]        [CB]
        |           |           |
      (T1)        (T2)        (T3)
        |           |           |
 ST1 ---+           +           +--- ST2
        |           |           |
      [GCB]       [GCB]       [GCB]
        |           |           |
      (G1)        (G2)        (G3)
        |           |           |
      [NGT]       [NGT]       [NGT]

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

| Item | Rating |
|---|---|
| Generators G1–G3 | 28.5 MVA, 11 kV, 0.85 pf, 600 rpm, $X''_d$ = 21% |
| Step-up transformers T1–T3 | 31.5 MVA each, 11/132 kV, YNd11, $X$ = 9.15% |
| Station transformer ST1, ST2 | 1250 kVA, 11/0.4 kV, Dyn11, 5% (aux load about 1.5% of capacity) |
| Diesel generator DG | 750 kVA, 0.4 kV (essential auxiliaries for black start) |
| Line | 132 kV, single circuit, 25 km to grid substation |
| NGT | neutral grounding transformer with secondary resistor (high-resistance earthing) |
| 132 kV bus | double main bus with bus coupler (drawn as one line) |

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

- **132 kV switchyard:** double main bus bar with a bus coupler. Each bay (transformer or line) can be put on either bus, so one bus can be maintained without shutting down the plant and a bus fault affects only the circuits on that bus. This is the NEA practice at 132 kV and above.
- **11 kV:** no common bus (unit scheme); each generator connects to its transformer through the GCB.

#### e) Rating of generator circuit breaker

Per-unit reactances on 100 MVA base:

$$
\begin{aligned}
x_g &= 0.21 \times \frac{100}{28.5} = 0.7368\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{31.5} = 0.2905\ \text{pu} \\
X_L &= 2\pi f L l = 2\pi \times 50 \times 0.95\times10^{-3} \times 25 = 7.461\ \Omega \\
x_l &= \frac{X_L \times 100}{132^2} = \frac{7.461}{174.24} = 0.0428\ \text{pu}, \quad x_{grid} = \frac{100}{2000} = 0.0500\ \text{pu} \\
x_{grid\ path} &= 0.0500 + 0.0428 = 0.0928\ \text{pu}
\end{aligned}
$$

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).

$$
\begin{aligned}
S_a &= \frac{100}{x_g} = \frac{100}{0.7368} = 135.7\ \text{MVA} \\
X_{ext} &= x_{grid\ path} \parallel (x_g + x_t)/2 = 0.0928 \parallel 0.5137 = 0.0786\ \text{pu} \\
S_b &= \frac{100}{x_t + X_{ext}} = \frac{100}{0.2905 + 0.0786} = 270.9\ \text{MVA}
\end{aligned}
$$

Design fault level for the GCB = larger value = **270.9 MVA**.

$$
\begin{aligned}
I_{sc} &= \frac{270.9}{\sqrt{3} \times 11} = 14.22\ \text{kA}, \quad I_{make} = 2.55 \times 14.22 = 36.26\ \text{kA peak} \\
I_{FL} &= \frac{28.5}{\sqrt{3} \times 11} = 1496\ \text{A}, \quad 1.25 I_{FL} = 1870\ \text{A}
\end{aligned}
$$

| Parameter | Required | Selected |
|---|---|---|
| Rated voltage | 11 kV system | 12 kV |
| Rated normal current | ≥ 1870 A | 2000 A |
| Breaking capacity | 270.9 MVA (14.22 kA) | 25 kA (520 MVA at 12 kV) |
| Making capacity | 36.26 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 = 270.9 MVA; select a 12 kV, 2000 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) Tentative quotation for the generator transformers

Requirement: 3 generator step-up transformers, one per unit (unit scheme), each above the generator rating of 28.5 MVA, so **31.5 MVA, 11/132 kV** units are quoted. The site is in a hilly region, so weight and size must suit hill-road transport.

| Item | Offered particulars |
|---|---|
| Type | 3-phase, oil-immersed, outdoor, two-winding step-up (GSU) |
| Rating | 31.5 MVA ONAN/ONAF (ONAF rating about 25% higher) |
| Voltage ratio | 11 kV / 132 kV, off-circuit taps ±2 × 2.5% on HV |
| Vector group | YNd11, HV neutral solidly earthed |
| Impedance | 9.15% (as used in the fault study) ± IEC tolerance |
| Insulation (HV) | BIL 650 kVp, 275 kV power frequency; altitude corrected |
| Guaranteed losses | No-load and load losses stated; capitalised in bid evaluation |
| Accessories | Buchholz, PRV, OTI/WTI, MOG, breather, bushing CTs for REF/differential, fans, marshalling box |
| Standard | IEC 60076 |

| S.N. | Description | Qty | Unit price (NPR) | Amount (NPR) |
|---|---|---|---|---|
| 1 | 31.5 MVA, 11/132 kV GSU transformer as above, with oil | 3 | 6,50,00,000 | 19,50,00,000 |
| 2 | Mandatory spares (bushings, gaskets, relays) | 1 lot | – | 25,00,000 |
| 3 | Transport to site, erection supervision, testing | 1 lot | – | 60,00,000 |
| | **Subtotal** | | | **20,35,00,000** |
| | VAT 13% | | | 2,64,55,000 |
| | **Grand total** | | | **22,99,55,000** |

Terms: delivery 8–10 months after LC to site; routine tests witnessed, type test reports supplied; warranty 24 months after commissioning; payment by LC; validity 120 days. Prices are indicative for budgeting only; actual prices come from competitive bids.

## Check yourself

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

1. Net head is 525.25 m. Why is only a Pelton turbine possible?
   A. The head is below the Kaplan limit
   B. Pelton is always cheaper than Francis
   C. The head is above the 350 m Francis limit
   D. Francis turbines only work with sediment-free water
   Answer: C. Francis is limited to 350 m, so above that only Pelton remains. (Key point: Turbine type from head)

2. What changes in this problem compared with the 5% head-loss problems?
   A. Riparian release is taken from Q45%
   B. Head loss is 4.5%, so Hn = 550 x 0.955 = 525.25 m, and efficiencies are 91%, 97%, 99%
   C. No head loss is applied
   D. Head loss is 5%, giving Hn = 522.5 m
   Answer: B. The stated 4.5% head loss and the given efficiencies are used, giving 71.261 MW at the 132 kV bus. (Key point: Design power calculation)

3. Why is a segregated phase bus duct chosen at about 1496 A?
   A. Cables are always preferred above 1000 A
   B. Too much current for a practical number of parallel cables, while isolated phase bus is only needed for very large machines
   C. Current is above 4000 A so IPB is required
   D. SPBD has all three phases in one open compartment
   Answer: B. At about 1496 A many cable runs would be needed; SPBD has each phase in its own compartment, making phase faults unlikely, and IPB is excessive. (Key point: Generator lead selection)

4. How is the 28.5 MVA generator rating found for each of the 3 units?
   A. 24 MW divided by sqrt(3)
   B. Generator output 24 MW divided by 0.85 pf = 28.24 MVA, rounded to 28.5 MVA
   C. 24 MW multiplied by 0.85
   D. 72 MW divided by 0.85
   Answer: B. Apparent power equals active power over power factor, rounded up to a standard size. (Key point: Generator rating)

## Close variants set in other exams

- [Design: 260 m head plant, generator lead](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2078-260m-generator-lead) — Power Plant Design, chapter 4, asked 5×
- [Design: 360 m head plant, 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2078-360m-132kv) — Power Plant Design, chapter 4, asked 5×

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