Turbine setting; three single-jet Pelton wheels

Setting of a turbine

Setting means fixing the elevation of the turbine (runner outlet or centre line) relative to the tailwater level. For a reaction turbine it is limited by cavitation, $H_s \le H_{atm} - H_v - \sigma_c H$. For a Pelton wheel the wheel centre is set above the maximum flood level in the tailrace (clearance 1 to 2 m) so that the runner never dips in water.

Data

Net head at nozzle $H = 360$ m, $C_v = 0.98$, $\eta = 0.85$, $N = 560$ rpm, $N_s = 15.3$ (for each single-jet wheel), $L = 1200$ m, $h_f = 18$ m, $f = 0.02$ (Darcy).

(i) Total power developed

Power of one wheel from specific speed (kW, m, rpm):

$$P = \left(\frac{N_s H^{5/4}}{N}\right)^2 = \left(\frac{15.3 \times 360^{1.25}}{560}\right)^2 = 1835.5\ \text{kW}$$ $$P_{total} = 3 \times 1835.5 = 5506.6\ \text{kW}$$

(ii) Discharge

$$Q = \frac{P_{total}}{\eta \gamma H} = \frac{5506.6}{0.85 \times 9.81 \times 360} = 1.834\ \text{m}^3/\text{s}$$

Per wheel (one jet): $q = 0.6115\ \text{m}^3/\text{s}$.

(iii) Jet diameter and pipe diameter

Jet velocity $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 360} = 82.36$ m/s.

$$d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 0.6115}{\pi \times 82.36}} = 0.0972\ \text{m}$$

Pipeline: $h_f = \dfrac{fLv^2}{2gD}$ with $v = \dfrac{4Q}{\pi D^2}$:

$$D^5 = \frac{16 f L Q^2}{\pi^2 \, 2g \, h_f} = \frac{16 \times 0.02 \times 1200 \times 1.834^2}{\pi^2 \times 2 \times 9.81 \times 18}$$ $$D = 0.820\ \text{m},\quad v = 3.47\ \text{m/s}$$

Answer: total power $= 5507$ kW; $Q = 1.834$ m³/s; jet diameter $= 97$ mm; pipe diameter $= 0.82$ m.