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Number of Francis/Kaplan units (300 m³/s, 30 m)

Short answer

Total power = 0.88×9.81×300×30 = 77695 kW. With Ns ≤ 267, max power per Francis unit 12646 kW, requiring 7 units. With Ns ≤ 650, Kaplan unit power 74948 kW, requiring 2 units.

In a hydropower project the available river discharge is 300 m³/s and the net head is 30 m. If the speed of the turbine is to be 166.7 rpm and the overall efficiency is 88%, determine the number of units required for the turbine cases given below. (i) Francis turbines with specific speed not exceeding 267 rpm. (ii) Kaplan turbines with specific speed not exceeding 650 rpm.

Data

Q=300 m3/sQ = 300\ \text{m}^3/\text{s}, H=30H = 30 m, N=166.7N = 166.7 rpm, ηo=0.88\eta_o = 0.88.

Total power

Ptotal=ηγQH=0.88×9.81×300×30=77,695 kWP_{total} = \eta\gamma Q H = 0.88 \times 9.81 \times 300 \times 30 = 77{,}695\ \text{kW}

(i) Francis, Ns≤267N_s \le 267

Maximum power per unit:

Pu=(NsH5/4N)2=(267×301.25166.7)2=12,646 kWP_u = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{267 \times 30^{1.25}}{166.7}\right)^2 = 12{,}646\ \text{kW} n=77,69512,646=6.14⇒7 unitsn = \frac{77{,}695}{12{,}646} = 6.14 \Rightarrow 7\ \text{units}

(each 11,099 kW, Ns=250N_s = 250)

(ii) Kaplan, Ns≤650N_s \le 650

Pu=(650×301.25166.7)2=74,948 kWP_u = \left(\frac{650 \times 30^{1.25}}{166.7}\right)^2 = 74{,}948\ \text{kW} n=77,69574,948=1.04⇒2 unitsn = \frac{77{,}695}{74{,}948} = 1.04 \Rightarrow 2\ \text{units}

(each 38,848 kW, Ns=468N_s = 468)

Answer: 7 Francis units, or 2 Kaplan units.

✓ Answer checked by Claude Opus 5.5 (high effort)

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