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Design: 250 m head plant, 66 kV double circuit

Short answer

Design discharge Q45% after 10% riparian release is 6.5 m³/s, net head 237.5 m; power delivered to 66 kV bus is 13.1 MW. Install two 6.7 MW Pelton turbines (4‑jet, 428.6 rpm), total 13.4 MW. Generator breakers rated 11 kV, 630 A, 25 kA; use single busbar with bus section breaker.

A typical hydro power plant has the following details

MonthDischarge (m³/s)Flow durationDischarge (m³/s)
January4.9Qmax (Q0)211.00 m³/s
February3.86Q25%18.24 m³/s
March3.58Q45%6.80 m³/s
April4.55Q65%4.70 m³/s
May3.00Q85%3.10 m³/s
June12.77Q95%2.17 m³/s
July32.95Qmin0.64 m³/s
August40.26
September31.23
October15.05
November7.38
December4.85
Turbine typeHead range (m)
Kaplan2 < H < 40
Francis10 < H < 350
Pelton50 < H < 1300
ParameterValue
Gross head250 m
Turbine efficiency91%
Generator efficiency97%
Transformer efficiency98%
Head loss5%
Riparian release10%
Generator sub-transient reactance21%
Transformer reactance9.15%
Transmission line inductance0.92 mH/km
Transmission line length35 km
Transmission voltage level66 kV
Generator voltage level11 kV
Transmission circuitDouble 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 generators, power transformers, station transformer and diesel generator for black start e) Calculate the rating of generator circuit breaker to be used in your design f) Suggest the type of bus bar to be used (Assume suitable data if necessary and necessary graph is attached herewith)

Assumptions: riparian release = 10% of the driest monthly mean flow; net head = gross head minus 5% loss; ρg=9.81\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 Q45%Q_{45\%}

Driest month = May (3 m³/s), so riparian release Qr=0.1×3=0.3Q_r = 0.1 \times 3 = 0.3 m³/s.

Qd=Q45%−Qr=6.8−0.3=6.500 m3/sHn=Hg(1−0.05)=250−12.50=237.50 mPhyd=ρgQdHn=9.81×6.500×237.50=15.144 MWPshaft=ηtPhyd=0.91×15.144=13.781 MWPgen=ηgPshaft=0.97×13.781=13.368 MWPout=ηtrPgen=0.98×13.368=13.100 MW\begin{aligned} Q_d &= Q_{45\%} - Q_r = 6.8 - 0.3 = 6.500\ \text{m}^3/\text{s} \\ H_n &= H_g(1 - 0.05) = 250 - 12.50 = 237.50\ \text{m} \\ P_{hyd} &= \rho g Q_d H_n = 9.81 \times 6.500 \times 237.50 = 15.144\ \text{MW} \\ P_{shaft} &= \eta_t P_{hyd} = 0.91 \times 15.144 = 13.781\ \text{MW} \\ P_{gen} &= \eta_g P_{shaft} = 0.97 \times 13.781 = 13.368\ \text{MW} \\ P_{out} &= \eta_{tr} P_{gen} = 0.98 \times 13.368 = 13.100\ \text{MW} \end{aligned}

Answer: generator output = 13.368 MW; power delivered to the 66 kV bus = 13.100 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=6.500/2=3.250Q = 6.500/2 = 3.250 m³/s, generator output =13.368/2=6.684= 13.368/2 = 6.684 MW → rate each unit at 6.7 MW.
  • Generator rating: S=6.7/0.85=7.88S = 6.7/0.85 = 7.88 MVA → 8 MVA, 11 kV, 0.85 pf, 50 Hz.
  • Step-up transformer, one per unit: 8 MVA, 11/66 kV, YNd11 (standard size matching the 8 MVA generator).

Answer: installed capacity = 2 × 6.7 MW = 13.4 MW.

c) Type of turbine

Hn=237.50H_n = 237.50 m lies in the Francis (10–350 m) and Pelton (50–1300 m) ranges, so the specific speed decides.

Shaft power per unit P=13.781/2=6890.6P = 13.781/2 = 6890.6 kW. With N=120f/pN = 120f/p (synchronous speeds) and Ns=NP/Hn5/4N_s = N\sqrt{P}/H_n^{5/4}, where Hn5/4=932.4H_n^{5/4} = 932.4:

NN (rpm)PolesNsN_s (metric, kW)
1000689.0
750866.8
6001053.4
5001244.5
428.61438.2
3751633.4
333.31829.7

Usual ranges: Pelton about 10–30 per jet (multi-jet Ns=Ns,jetzN_s = N_{s,jet}\sqrt{z}), Francis about 60–300, Kaplan 300–1000.

Choose N=428.6N = 428.6 rpm (14 poles): Ns=38.2N_s = 38.2. With 4 jets, Ns,jet=38.2/4=19.1N_{s,jet} = 38.2/\sqrt{4} = 19.1, inside the Pelton range.

Check of jet ratio: jet velocity Vj=0.982gHn=66.9V_j = 0.98\sqrt{2gH_n} = 66.9 m/s; bucket speed u=0.46Vju = 0.46V_j; runner diameter D=60u/(πN)=1.37D = 60u/(\pi N) = 1.37 m; jet diameter d=4(Q/z)/(πVj)=0.124d = \sqrt{4(Q/z)/(\pi V_j)} = 0.124 m; D/d=11.0>10D/d = 11.0 > 10, acceptable.

A Francis at 1000 rpm (Ns=89.0N_s = 89.0) 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, 428.6 rpm, about 6.89 MW each.

d) Single line diagram

 Line-1          Line-2  (66 kV, 35 km)
    |               |
  [CB]            [CB]
    |               |
 ===+=====+=========+===+=== 66 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
ItemRating
Generators G1–G28 MVA, 11 kV, 0.85 pf, 428.6 rpm, Xd′′X''_d = 21%
Step-up transformers T1–T28 MVA each, 11/66 kV, YNd11, XX = 9.15%
Station transformer ST250 kVA, 11/0.4 kV, Dyn11, 4.5% (aux load about 1.5% of capacity)
Diesel generator DG160 kVA, 0.4 kV (essential auxiliaries for black start)
Line66 kV, double circuit, 35 km to grid substation
NGRneutral 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

Per-unit reactances on 100 MVA base:

xg=0.21×1008=2.6250 pu,xt=0.0915×1008=1.1438 puXL=2πfLl=2π×50×0.92×10−3×35=10.116 Ω per circuit; two in parallel=5.058 Ωxl=XL×100662=5.05843.56=0.1161 pu,xgrid=1001000=0.1000 puxgrid path=0.1000+0.1161=0.2161 pu\begin{aligned} x_g &= 0.21 \times \frac{100}{8} = 2.6250\ \text{pu}, \quad x_t = 0.0915 \times \frac{100}{8} = 1.1438\ \text{pu} \\ X_L &= 2\pi f L l = 2\pi \times 50 \times 0.92\times10^{-3} \times 35 = 10.116\ \Omega \text{ per circuit; two in parallel} = 5.058\ \Omega \\ x_l &= \frac{X_L \times 100}{66^2} = \frac{5.058}{43.56} = 0.1161\ \text{pu}, \quad x_{grid} = \frac{100}{1000} = 0.1000\ \text{pu} \\ x_{grid\ path} &= 0.1000 + 0.1161 = 0.2161\ \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):

Sgen=2×1002.6250=76.2 MVASgrid=100xt/2+xgrid path=1000.5719+0.2161=126.9 MVAS11kV=76.2+126.9=203.1 MVA\begin{aligned} S_{gen} &= 2 \times \frac{100}{2.6250} = 76.2\ \text{MVA} \\ S_{grid} &= \frac{100}{x_t/2 + x_{grid\ path}} = \frac{100}{0.5719 + 0.2161} = 126.9\ \text{MVA} \\ S_{11kV} &= 76.2 + 126.9 = 203.1\ \text{MVA} \end{aligned} Isc=203.13×11=10.66 kA,Imake=2.55×10.66=27.18 kA peakIFL=83×11=420 A,1.25IFL=525 A\begin{aligned} I_{sc} &= \frac{203.1}{\sqrt{3} \times 11} = 10.66\ \text{kA}, \quad I_{make} = 2.55 \times 10.66 = 27.18\ \text{kA peak} \\ I_{FL} &= \frac{8}{\sqrt{3} \times 11} = 420\ \text{A}, \quad 1.25 I_{FL} = 525\ \text{A} \end{aligned}
ParameterRequiredSelected
Rated voltage11 kV system12 kV
Rated normal current≥ 525 A630 A
Breaking capacity203.1 MVA (10.66 kA)25 kA (520 MVA at 12 kV)
Making capacity27.18 kA peak63 kA peak
Short-time current–25 kA, 3 s
Type–Vacuum (or SF6), generator-duty tested (IEEE C37.013)

Answer: GCB fault level = 203.1 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.

f) Type of bus bar

  • 11 kV: single bus bar with a bus-section breaker, units split between the two sections, so a bus fault or maintenance takes out only part of the plant.
  • 66 kV: single bus bar, sectionalised, with transformer bays and two line bays (double circuit).

Reason: for a 13.4 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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