Pelton wheel design (310 m, 5 m³/s)

Step 1: Power

$$P = \eta\gamma Q H = 0.90 \times 9.81 \times 5 \times 310 = 13{,}685\ \text{kW}$$

Step 2: Jet velocity and bucket speed

Assuming a nozzle velocity coefficient $C_v = 0.98$ (not given):

$V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 310} = 76.43$ m/s

Speed ratio $K_u = u/\sqrt{2gH} = 0.46$ (usual range 0.43 to 0.47) with $\sqrt{2gH} = 77.99$ m/s: $u = 0.46 \times 77.99 = 35.87$ m/s

Step 3: Number of jets and speed

For a single jet, the empirical (Siervo and Lugaresi) specific speed is $N_s \approx 85.49/H^{0.243} = 21.2$. For $z$ jets the wheel specific speed is $N_s\sqrt{z}$. Take $z = 2$ jets (horizontal shaft; 50 Hz assumed):

$$N = \frac{N_s\sqrt{z}\,H^{5/4}}{\sqrt{P}} = 333.5\ \text{rpm}$$

Synchronous speed for 50 Hz: $N = 6000/p$. Choose $p = 18$ poles, $N = 333.3$ rpm.

Step 4: Wheel (pitch circle) diameter

$$D = \frac{60u}{\pi N} = \frac{60 \times 35.87}{\pi \times 333.3} = 2.056\ \text{m}$$

Step 5: Jet diameter

Discharge per jet $q = 5/2 = 2.5\ \text{m}^3/\text{s}$:

$$d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 2.5}{\pi \times 76.43}} = 0.2041\ \text{m}$$

Jet ratio $m = D/d = 10.07$ (acceptable range 10 to 20).

Step 6: Bucket details

• Number of buckets $Z = 15 + \dfrac{D}{2d} = 15 + 5.04 \approx 20$
• Bucket width $B \approx 3.5d = 0.71$ m; depth about $1.0d$; length about $2.5d$.

Step 7: Specific speed of the wheel

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{333.3\sqrt{13{,}685}}{310^{1.25}} = 30.0$$

(about 21.2 per jet)

Answer: $P = 13.68$ MW; 2 jets; $N = 333.3$ rpm; $D = 2.06$ m; $d = 0.204$ m; specific speed of the turbine $N_s \approx 30$ (kW, m, rpm), i.e. 21 per jet.