written 5.0 years ago by |
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-
Shaft power, S.P = 11,772 kw.
Head, H = 380 m
Speed, N = 750 r.p.m
Overall efficiency $n_o$ = 86% or 0.86.
Ratio of jet dia to wheel dia.
$= \frac{d}{D} = \frac{1}{6}$
Co-efficient of velocity, $kv_1 = cv = 0.985$
Speed ratio, $ku_1 = 0.45$
velocity of jet, $v_1 = cv\sqrt{29H}$
$= 0.985\sqrt{2 \times 9.81 \times 380}$
= 85.05 m/s
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 $\frac{d}{D} = \frac{1}{6}$
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.