--- title: "Design: 200 m head plant, bus MVA levels" subject: "Power Plant Design" chapter: 4 key: "semester_8/power_plant_design:design-2070-200m-bus-mva" source: https://www.bctengineer.com/q/semester_8/power_plant_design/design-2070-200m-bus-mva site: BCT Engineering --- # Design: 200 m head plant, bus MVA levels Power Plant Design · Chapter 4: Electric System Design of a Power Plant - Asked in: 2070 Magh, 16 marks - Video: [power plant design | calculation of fault current and fault MVA @smartstudy11](https://www.youtube.com/watch?v=lwieP4cT3WA) — jitendra sah - 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-2070-200m-bus-mva - 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) | 207.00 m³/s | | February | 3.80 | Q25% | 18.24 m³/s | | March | 3.55 | Q45% | 6.82 m³/s | | April | 4.55 | Q65% | 4.7 m³/s | | May | 3.22 | 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 | 200 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 | 25 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 Design discharge Q45% after riparian release is 6.5 m³/s, net head 190 m; output 10.36 MW. Install two 5.3 MW Francis turbines, total 10.6 MW. Use single busbar with bus section breaker. ## 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. $$ \begin{aligned} Q_d &= Q_{45\%} - Q_r = 6.82 - 0.322 = 6.498\ \text{m}^3/\text{s} \\ H_n &= H_g - 0.05H_g = 200 - 10.0 = 190.0\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 1000 \times 9.81 \times 6.498 \times 190.0 = 12.112\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.90 \times 12.112 = 10.900\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 10.900 = 10.573\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 10.573 = 10.362\ \text{MW} \end{aligned} $$ **Answer:** generator output = 10.573 MW; power delivered to the 33 kV bus = **10.362 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: $Q = 6.498/2 = 3.249$ m³/s, generator output $= 10.573/2 = 5.287$ MW. - Generator rating: 5.3 MW at 0.85 pf, i.e. $S = 5.3/0.85 = 6.24$ MVA → **6.25 MVA, 11 kV, 50 Hz**. - Step-up transformer: one per unit, **6.3 MVA, 11/33 kV, YNd11, 9.15%** (next standard size above 6.25 MVA). **Answer: installed capacity = 2 × 5.3 MW = 10.6 MW.** #### c) Single line diagram ```text Line-1 33 kV Line-2 33 kV (25 km) (25 km) | | [CB] [CB] | | 33 kV ==+=======[BS]========+== | | [CB] [CB] | | (T1) (T2) 11/33 kV 6.3 MVA 11/33 kV 6.3 MVA | | [CB] [CB] | | 11 kV ==+=====[BS]======+===+===+== | | | [GCB] [GCB] [CB] | | | (G1) (G2) (ST) 11/0.4 kV 6.25 MVA 6.25 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 $H_n = 190.0$ m lies in both the Francis (10–350 m) and Pelton (50–1300 m) ranges, so specific speed decides. Per-unit shaft power $P = 10.900/2 = 5450$ kW; take $N = 1000$ rpm (6-pole, 50 Hz synchronous speed). $$ \begin{aligned} N_s &= \frac{N\sqrt{P}}{H_n^{5/4}} = \frac{1000 \times 73.83}{705.4} = 104.7\ \text{(metric, kW)} \end{aligned} $$ A single-jet Pelton suits $N_s \approx 10$–$35$ (up to about 60–70 with multiple jets); a Francis suits $N_s \approx 60$–$300$. $N_s = 104.7$ falls in the slow-to-medium Francis range, and a Francis has good efficiency at this discharge per unit (about 3.249 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 10.6 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%. $$ \begin{aligned} x_g &= 0.21 \times \frac{10}{6.25} = 0.3360\ \text{pu}, \quad x_t = 0.0915 \times \frac{10}{6.3} = 0.1452\ \text{pu} \\ x_{st} &= 0.045 \times \frac{10}{0.25} = 1.8\ \text{pu} \end{aligned} $$ | Fault location | Equivalent $X$ (pu) | Fault MVA $=10/X$ | $I_{sc}$ (kA) | |---|---|---|---| | MV bus (11 kV) | $x_g/2 = 0.1680$ | 59.5 | 3.12 | | HV bus (33 kV) | $(x_g+x_t)/2 = 0.2406$ | 41.6 | 0.727 | | LV bus (0.4 kV) | $0.1680 + 1.8 = 1.968$ | 5.08 | 7.33 | Line check (remote end): $X_L = 2\pi \times 50 \times 0.95\times10^{-3} \times 25 = 7.461\ \Omega$ per circuit, 3.731 Ω for both circuits in parallel. $Z_b = 33^2/10 = 108.90\ \Omega$, so $x_L = 0.0343$ pu. Fault level at the far end of the line is $10/0.2749 = 36.4$ MVA (0.636 kA). **Answer: LV ≈ 5.08 MVA, MV ≈ 59.5 MVA, HV ≈ 41.6 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 | $6.25/(\sqrt3 \times 11) = 328$ A | 110 A per transformer, 220 A total | | Required breaking | 3.12 kA (59.5 MVA) | 0.727 kA (41.6 MVA) | | Required making | 7.97 kA peak | 1.85 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 ≥ 59.5 MVA (3.12 kA); 33 kV CB: 36 kV, 630 A, breaking capacity ≥ 41.6 MVA (0.727 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. What is the design discharge when the driest month (May) is 3.22 m3/s and Q45% is 6.82 m3/s? A. 6.82 m3/s B. 3.60 m3/s C. 6.50 m3/s taken from 10% of Q45% D. 6.498 m3/s Answer: D. Riparian release is 0.322 m3/s (10% of May flow); Qd = 6.82 - 0.322 = 6.498 m3/s. (Key point: Design discharge) 2. Ns = 104.7 at 1000 rpm. Which turbine does this indicate? A. Francis, since it is within about 60-300 B. Pelton, since Ns is above 35 C. Kaplan, since head is below 350 m D. Pelton, since head is above 50 m Answer: A. Single-jet Pelton suits Ns of about 10-35, Francis 60-300; 104.7 is slow-to-medium Francis. (Key point: Turbine choice) 3. Why is the LV (0.4 kV) bus fault level only about 5 MVA, far lower than the MV bus? A. The LV bus is fed only by the diesel generator B. The station transformer's large per-unit reactance (1.8 pu) is in series with the MV source reactance C. The 33 kV line reactance limits it D. Generators are disconnected during LV faults Answer: B. LV X = 0.168 + 1.8 = 1.968 pu, so S = 10/1.968 = 5.08 MVA. (Key point: LV fault level) 4. Why is the HV (33 kV) bus fault MVA lower than the MV bus in this answer? A. The 33 kV bus has more generators B. Transformer reactance is subtracted for HV faults C. Grid contribution is neglected, so HV X includes the transformer reactance in series: (xg+xt)/2 D. The base MVA changes at 33 kV Answer: C. MV X = xg/2 = 0.168 gives 59.5 MVA; HV X = (xg+xt)/2 = 0.2406 gives 41.6 MVA. 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