A pelton wheel is to be designed for the following specifications-

Shaft power = 11,772 kw,

Head = 380 meters,

Speed =750 r.p.m ,

Overall efficiency = 86%,

Jet diameter is not to exceed one sixth of the wheel diameter.

Determine:

1] The wheel diameter.

2] The number of jet required.

3] Diameter of the jet.

Take $kv_1 = 0.985$ and $ku_1 = 0.45$

Given-

1. Shaft power, S.P = 11,772 kw.

2. Head, H = 380 m

3. Speed, N = 750 r.p.m

4. Overall efficiency $n_o$ = 86% or 0.86.

   Ratio of jet dia to wheel dia.

   $= \frac{d}{D} = \frac{1}{6}$

5. Co-efficient of velocity, $kv_1 = cv = 0.985$

6. Speed ratio, $ku_1 = 0.45$

7. velocity of jet, $v_1 = cv\sqrt{29H}$

   $= 0.985\sqrt{2 \times 9.81 \times 380}$

   = 85.05 m/s

8. The velocity of wheel,

   $u = u_1 = u_2$

   $= speed ratio \times \sqrt{29H}$

   $= 0.45 \times \sqrt{2 \times 9.81 \times 380}$

   = 38.85 m/s

   But $u = \frac{\pi DN}{60}$

   $38.85 = \frac{\pi DN}{60}$

   $\therefore$ $D = \frac{60 \times 38.85}{\pi \times N}$

   $= \frac{60 \times 38.85}{\pi \times 750}$

   $\therefore$ $D = 0.989m$

   But $\frac{d}{D} = \frac{1}{6}$

9. Dia of jet, $d = \frac{1}{6} \times D$

   $= \frac{0.989}{6}$

   d = 0.165m

   Discharge of on jet,

   Q = Area of jet x Velocity of jet

   $= \frac{\pi }{4} (d^2) \times v_1$

   $= \frac{\pi }{4} (0.165)^2 \times 85.05 m^3/s$

   q = 1.818 $m^3/s$

   Now, $n_o = \frac{S.P}{W.P}$

   $= \frac{11773}{\frac{pg \times q \times H}{1000}}$

   $0.86 = \frac{11772 \times 1000}{1000 \times 9.81 \times Q \times 380}$

   $\therefore$ $Q = \frac{11772 \times 1000}{1000 \times 9.81 \times 380 \times 0.86}$

   $\therefore$ $Q = 3.672 m^3/s$

   $\therefore$ Number of jets = $\frac{Q}{q}$

   $= \frac{3.672}{1.818}$

   = 2 jets.