A pelton wheel is to be designed for a head of 60m when running at 200 r.p.m. The pelton wheel develops 95.6475 KW shaft power. The velocity of the buckets = 0.45 times the velocity of the jet, overall efficiency = 0.85 and coefficient of the velocity is equal to 0.98.

Given :

1. Head, H = 60m

2. Speed, N = 200 r.p.m

3. Shaft power = S.P = 95.6475 KW.

4. Velocity of buckets, u = 0.45 x velocity of jet

5. Overall efficiency, no = 0.85

6. Co-efficient of velocity, $C_v$ = 0.98

Step no 1. Velocity of jet.

$v_1 = $C_v$ \times \sqrt{294}$

$= 0.98 \times \sqrt{2 \times 9.81 \times 60}$

$= 3362 m/s$

$\therefore$ Bucket velocity, $u = u_1 = u_2 = 0.45 \times 33.62$

= 15.13 m/s

But $u = \frac{ \pi DN}{60}$, [where, D = Dia of wheel]

$15.13 = \frac{\pi \times D \times 200}{60}$

$\therefore$ $D = \frac{60 \times 15.13}{\pi \times 200}$

[ D = 1.44m]

Step no (2) Diameter of jet (d)

overall efficiency, no = 0.85

But $n_o = \frac{S.P}{W.P}$

= $\frac{95.6475}{\frac{W.P}{1000}}$

= $\frac{95.6475 \times 1000}{p \times g \times Q \times H}$

= $\frac{95.6475 \times 1000}{1000 \times 9.81 \times Q \times 60}$

Q = $\frac{95.6475 \times 1000}{n_o \times 1000 \times 9.81 \times 60}$

= $\frac{95.6475 \times 1000}{0.85 \times 1000 \times 9.81 \times 60}$

$= 0.1912 m^3/sec$

But, Q = Area pf jet x Velocity of jet

$\therefore$ $ 0.1912 = \frac{\pi }{4} \times d^2 \times v_1$

$d = \sqrt{\frac{4 \times 0.1912}{\pi \times 33.62}}$

= 0.085m

[ d = 85mm]

Step no (3) Size of buckets.

width of buckets = 5 x d

= 5 x 85

= 425 mm

Depth of buckets = 1.2 x d

= 1.2 x 85

= 102 mm.

Step no (4) Number of buckets on the wheel is given by,

$z = 15 + \frac{D}{2d}$

$= 15 + \frac{1.44}{2 \times 0.85}$

= 15 + 8.5

[ Z = 23.5 say 24]