Question 9.4.3: A single jet Pelton turbine is required to drive a generator......
A single jet Pelton turbine is required to drive a generator to develop 10 MW. The available head at the nozzle is 760 m. Assuming electric generation efficiency 95%, Pelton wheel efficiency 87%, coefficient of velocity for nozzle 0.97, mean bucket velocity 0.46 of jet velocity, outlet angle of bucket 15^0 and the relative velocity of the water leaving the buckets 0.85 of that at inlet, find: (i) the flow in m³/s, (ii) the diameter of jet, (iii) force exerted by the jet on the bucket, (iv) the best synchronous speed for generation at 50 Hz and (v) the corresponding mean diameter if the ratio of the mean bucket circle diameter to the jet diameter is not to be less than 10.
Given:
\mathrm{P=10000 kW H=760 m \eta_gen =95 \eta _h =87 }
\mathrm{C_\nu =0.97 K_u=.46 \beta _2=15 K = 0.85}
m ≥ 10 f = 50 Hz
1. Here Pelton wheel efficiency is nothing but the hydraulic efficiency. To find Q, Eq. 9.4.2 may be used as all variables except Q are known. Otherwise also power generated will depend on discharge and hence this information should be used for determination of Q. Moreover, hydraulic power of turbine will be more than what is available at generator outlet because of mechanical losses, accounted for by mechanical efficiency, and generator losses, taken care of by generator efficiency. This gives,
\mathrm{P=w QH \times \eta _o/1000 kW} (9.4.2)
\mathrm{P = 10000/(η_{gen} × η_{mech}) = 9.81 × Q × 760} ⇒
Q = 1.62 m³/s
2. Now jet diameter can be obtained from Q = πd^2/4 C_1, where C_1 = Cν × \sqrt{2gH} This gives,
C_1 = 0.97 × √(2 × 9.81 × 760) = 118.45 and d = 0.132 m
3. Force exerted on bucket is given by ρ Q (C_{w1} – C_{w2}), for which velocity triangle, as shown in Fig. 9.4.3 are drawn. From given data:
\begin{matrix} \text{At inlet,} & C_{w1} = C_1 = 118.45, u = K_u × C_1 = 54.5, C_{r1} = C_1 – u = 64 m/s\\ \\ \text{and at outlet,} & C_{a2} = K × C_{r1} = 54.4, C_{w2} = u – C_{r2} \cos β_2 = 1.95\\ &\mathrm{a + ve sign means that α_2 < 90°}\end{matrix}∴ Force on bucket = 1000 × 1.62 × (118.45 – 1.95) = 188.8 kN
4. We mentioned earlier that turbine speed is chosen according to that of generator for which it depends on the number of poles. Assuming minimum jet ratio 10, wheel diameter = 10 × d = 1.32 m and for this diameter, rpm of turbine is N = 60u/ π D = 788. For generator, N_g = generator rpm = 120 f/p= 120 × 50/p = 6000/p, where p is number of poles, which will be always even number as poles are present in pair in a generator.
For its rpm nearest to that of turbine, p = 8 and N = 750. Therefore diameter of the wheel is to be changed to give desired bucket velocity. Thus D = 60 × 54.5/(π × 750) = 1.387 m
