--- title: "Design: 230 m head plant, 20 km line" subject: "Power Plant Design" chapter: 4 key: "semester_8/power_plant_design:design-2070-230m-20km" source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2070-230m-20km site: BCT Engineering --- # Design: 230 m head plant, 20 km line Power Plant Design · Chapter 4: Electric System Design of a Power Plant - Asked in: 2070 Bhadra, 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-2070-230m-20km - 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 | 4.9 | Qmax (Q0) | 207.00 m³/s | | February | 3.86 | Q25% | 18.24 m³/s | | March | 3.58 | Q45% | 6.87 m³/s | | April | 4.55 | Q65% | 4.7 m³/s | | May | 3.22 | Q85% | 3.10 m³/s | | June | 12.77 | Q95% | 2.17 m³/s | | July | 32.95 | Qmin | 0.64 m³/s | | August | 40.26 | | | | September | 31.23 | | | | October | 15.05 | | | | November | 7.38 | | | | December | 4.85 | | | | 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 | 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 | 20 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) Draw the single line diagram showing number of units, generator, power transformer, station transformer and diesel generator for black start d) Select the type of turbine e) Suggest the type of bus bar to be used f) Calculate the MVA level of LV bus, MV bus and HV bus. g) Calculate the rating of circuit breaker to be used in your design (Assume suitable data if required and the necessary graph is attached herewith) ## Short answer Assume riparian release 0.322 m³/s and net head 218.5 m; choose two units. Generator output is 12.253 MW and 33 kV bus output 12.008 MW; installed capacity is 2×6.2 MW = 12.4 MW with 7.5 MVA, 11 kV generators and 8 MVA, 11/33 kV transformers. Turbine is Francis (Ns=94.6). ## Answer Take riparian release as 10% of the driest monthly mean flow, net head = gross head minus 5% loss, and two generating units. The grid's fault contribution is neglected because grid data is not given (only the plant's generators feed a fault). #### a) Electrical power output at $Q_{45\%}$ Driest month = May (3.22 m³/s), so riparian release $= 0.1 \times 3.22 = 0.322$ m³/s. $$ \begin{aligned} Q_d &= Q_{45\%} - Q_r = 6.87 - 0.322 = 6.548\ \text{m}^3/\text{s} \\ H_n &= H_g - 0.05H_g = 230 - 11.5 = 218.5\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 1000 \times 9.81 \times 6.548 \times 218.5 = 14.036\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.90 \times 14.036 = 12.632\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 12.632 = 12.253\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 12.253 = 12.008\ \text{MW} \end{aligned} $$ **Answer:** generator output = 12.253 MW; power delivered to the 33 kV bus = **12.008 MW**. #### b) Number of units and installed capacity - Choose **2 units**. The plant runs on $Q_{45\%}$, so in the dry season (flow below about $Q_{65\%}$) one unit can still run near its best efficiency; 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: $Q = 6.548/2 = 3.274$ m³/s, generator output $= 12.253/2 = 6.127$ MW. - Generator rating: 6.2 MW at 0.85 pf, i.e. $S = 6.2/0.85 = 7.29$ MVA → **7.5 MVA, 11 kV, 50 Hz**. - Step-up transformer: one per unit, **8 MVA, 11/33 kV, YNd11, 9.15%** (next standard size above 7.5 MVA). **Answer: installed capacity = 2 × 6.2 MW = 12.4 MW.** #### c) Single line diagram ```text Line-1 33 kV Line-2 33 kV (20 km) (20 km) | | [CB] [CB] | | 33 kV ==+=======[BS]========+== | | [CB] [CB] | | (T1) (T2) 11/33 kV 8 MVA 11/33 kV 8 MVA | | [CB] [CB] | | 11 kV ==+=====[BS]======+===+===+== | | | [GCB] [GCB] [CB] | | | (G1) (G2) (ST) 11/0.4 kV 7.5 MVA 7.5 MVA 250 kVA | 0.4 kV aux bus ===+===+== | [ACB] | (DG) 200 kVA ``` BS = bus section breaker; ST = station (auxiliary) transformer; DG = diesel generator for black start. With the grid dead, the DG energises the 0.4 kV auxiliary bus (governor oil pumps, cooling water, excitation field flashing, lighting, battery chargers), so a unit can be started without outside supply. #### d) Selection of turbine $H_n = 218.5$ m lies in both the Francis (10–350 m) and Pelton (50–1300 m) ranges, so specific speed decides. Per-unit shaft power $P = 12.632/2 = 6316$ kW; take $N = 1000$ rpm (6-pole, 50 Hz synchronous speed). $$ \begin{aligned} N_s &= \frac{N\sqrt{P}}{H_n^{5/4}} = \frac{1000 \times 79.47}{840.1} = 94.6\ \text{(metric, kW)} \end{aligned} $$ A single-jet Pelton suits $N_s \approx 10$–$35$ (up to about 60–70 with multiple jets); a Francis suits $N_s \approx 60$–$300$. $N_s = 94.6$ falls in the slow-to-medium Francis range, and a Francis has good efficiency at this discharge per unit (about 3.274 m³/s). **Answer: 2 × vertical-axis Francis turbines, 1000 rpm.** #### e) Type of bus bar - **11 kV:** single bus bar with a bus-section breaker. Each half carries one unit, so a bus fault or maintenance takes out only one unit. - **33 kV switchyard:** single bus bar, sectionalised, with one bay for each circuit of the double-circuit line. This is simple and cheap for a 12.4 MW, 2-unit plant. A double bus or main-and-transfer bus costs more and is normally used only in larger plants or grid nodes. #### f) MVA (fault) level of the LV, MV and HV buses Take LV = 0.4 kV auxiliary bus, MV = 11 kV generator bus, HV = 33 kV switchyard bus. Base = 10 MVA. Station transformer assumed 250 kVA, 4.5%. $$ \begin{aligned} x_g &= 0.21 \times \frac{10}{7.5} = 0.2800\ \text{pu}, \quad x_t = 0.0915 \times \frac{10}{8} = 0.1144\ \text{pu} \\ x_{st} &= 0.045 \times \frac{10}{0.25} = 1.8\ \text{pu} \end{aligned} $$ | Fault location | Equivalent $X$ (pu) | Fault MVA $=10/X$ | $I_{sc}$ (kA) | |---|---|---|---| | MV bus (11 kV) | $x_g/2 = 0.1400$ | 71.4 | 3.75 | | HV bus (33 kV) | $(x_g+x_t)/2 = 0.1972$ | 50.7 | 0.887 | | LV bus (0.4 kV) | $0.1400 + 1.8 = 1.940$ | 5.15 | 7.44 | Line check (remote end): $X_L = 2\pi \times 50 \times 0.95\times10^{-3} \times 20 = 5.969\ \Omega$ per circuit, 2.985 Ω for both circuits in parallel. $Z_b = 33^2/10 = 108.90\ \Omega$, so $x_L = 0.0274$ pu. Fault level at the far end of the line is $10/0.2246 = 44.5$ MVA (0.779 kA). **Answer: LV ≈ 5.15 MVA, MV ≈ 71.4 MVA, HV ≈ 50.7 MVA.** #### g) Rating of circuit breakers Making current = $2.55 \times I_{sc}$ (peak, IEC 62271-100). Rated current ≥ 1.25 × full-load current. | Item | Generator CB (11 kV) | HV line/transformer CB (33 kV) | |---|---|---| | Full-load current | $7.5/(\sqrt3 \times 11) = 394$ A | 140 A per transformer, 280 A total | | Required breaking | 3.75 kA (71.4 MVA) | 0.887 kA (50.7 MVA) | | Required making | 9.56 kA peak | 2.26 kA peak | | Selected rating | 12 kV, 630 A, 25 kA, VCB | 36 kV, 630 A, 25 kA, SF6 | **Answer:** GCB: 12 kV, 630 A, breaking capacity ≥ 71.4 MVA (3.75 kA); 33 kV CB: 36 kV, 630 A, breaking capacity ≥ 50.7 MVA (0.887 kA). Standard 25 kA breakers are chosen because they cost little more and leave margin for the grid's fault contribution and future units. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Generator output per unit is 6.127 MW, rated 6.2 MW at 0.85 pf. What follows? A. S = 6.2 MVA, so a 6.3 MVA generator B. S = 7.29 MVA, so a 7.5 MVA transformer C. S = 5.3 MVA, so a 6 MVA generator D. S = 7.29 MVA, so a 7.5 MVA generator and an 8 MVA step-up transformer Answer: D. S = P/pf = 6.2/0.85 = 7.29 MVA; generator 7.5 MVA, transformer the next standard size 8 MVA. (Key point: Generator and transformer rating) 2. What is the role of the diesel generator for black start? A. Energise the 0.4 kV auxiliary bus (oil pumps, cooling, excitation, lighting) so a unit can start with the grid dead B. Supply the whole 33 kV line directly C. Replace the step-up transformer D. Run in parallel with the generators at all times Answer: A. With no grid, the DG powers the essential auxiliaries so one unit can start and then energise the transformer and line. (Key point: Black start) 3. The MV bus fault is 71.4 MVA (3.75 kA). How is the GCB rated? A. Breaking 9.56 kA and making 3.75 kA B. Breaking at least 3.75 kA and making 2.55 x Isc = 9.56 kA peak C. Rated only on full-load current of 394 A D. Breaking 0.887 kA, the HV value Answer: B. The CB breaking capacity must exceed the fault level; making current is 2.55 x Isc and rated current at least 1.25 x full-load. 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