Question 9.4.2: A Pelton wheel is to be designed for shaft power 9560 kW, he......
A Pelton wheel is to be designed for shaft power 9560 kW, head 350 m, speed 750 rpm, overall efficiency 0.85, and jet diameter not to exceed 1/6 of the wheel diameter. Determine: (i) wheel diameter, (ii) jet diameter, and (iii) Number of jets required. Take nozzle velocity coefficient Cv = 0.985 and speed ratio = 0.45
Given:
\mathrm{P = 9560 kW N = 750 rpm H = 350 m} \mathrm{η_o = 0.85 D/D ≤ 1/6 C_v = 0.985 K_u = 0.45}Steps:
1. Wheel diameter can be obtained if u is known since N is given. But C_1 = C_ν × 2gh = 81.62
∴ u = K_u × C_1 = 37.3 m/s and D = 60 u/(πN) = 0.95 m
2. Taking d/D = 1/6, we get d = 0.158 m
3. To find out number of jets, what we require is the total discharge and discharge of each nozzle. Discharge of each nozzle is = π d^2/4 × C_1 = 1.6 m^3/s. Further,
wQH/1000 = η_o × P ⇒ Q = .85 × 9560/(9.81^7 × 350) = 3.27 m^3/s
and number of jets required = 3.27/1.6 = 2
Now we have to recalculate diameter of nozzle so that each of the nozzle has discharge = 3.27/2 = 1.635 m³/s.
This gives: d=\sqrt{\frac{4 \times 1.635}{\pi \times 81.62} }=0.1597 m