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Number of Kaplan units (340 m³/s, 27.5 m)

Short answer

Total power is 80,716.7 kW; one unit delivers 44,754 kW, so two units are needed. Each Kaplan turbine output is 40,358 kW (40.4 MW) at 166.7 rpm.

In hydropower project the available river discharge is 340 m³/s under the net head of 27.5 m. If the speed of the turbine is to be 166.7 rpm and overall efficiency is 88%. Determine the number of unit required if Kaplan turbine of specific speed 560 rpm in SI unit is selected.

Given: Q=340Q = 340 m³/s, H=27.5H = 27.5 m, N=166.7N = 166.7 rpm, overall efficiency ηo=0.88\eta_o = 0.88, specific speed Ns=560N_s = 560 (SI: rpm, kW, m).

Formulas:

P=ηo ρgQH,Ns=NPH5/4P = \eta_o\,\rho g Q H, \qquad N_s = \frac{N\sqrt{P}}{H^{5/4}}

1. Total power developed

Ptotal=0.88×9.81×340×27.5=80,716.7 kW\begin{aligned} P_{total} &= 0.88 \times 9.81 \times 340 \times 27.5 \\ &= 80{,}716.7\ \text{kW} \end{aligned}

2. Power per unit from specific speed

H5/4=27.51.25=62.97P=NsH5/4N=560×62.97166.7=211.55Punit=211.552=44,754 kW\begin{aligned} H^{5/4} &= 27.5^{1.25} = 62.97 \\ \sqrt{P} &= \frac{N_s H^{5/4}}{N} = \frac{560 \times 62.97}{166.7} = 211.55 \\ P_{unit} &= 211.55^2 = 44{,}754\ \text{kW} \end{aligned}

3. Number of units

n=80,716.744,754=1.80n = \frac{80{,}716.7}{44{,}754} = 1.80

Adopt 2 units.

4. Output of each unit and check

Peach=80,716.72=40,358 kW≈40.4 MWP_{each} = \frac{80{,}716.7}{2} = 40{,}358\ \text{kW} \approx 40.4\ \text{MW}

Actual specific speed =166.740,35862.97=532<560= \dfrac{166.7\sqrt{40{,}358}}{62.97} = 532 < 560 — acceptable for a Kaplan turbine.

Answer: number of units required =2= 2 Kaplan turbines, each developing about 40,358 kW (40.4 MW).

✓ Answer checked by Claude Opus 5.5 (high effort)

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