written 3.0 years ago by
DevarenjiniP
• 3.2k

•
modified 3.0 years ago

Given data:
gross head (H)= 600m,
Dp = diameter of penstock: 1.25m,
diameter of jet (Dj) = 200mm
angle of deflection (δ) = 160°
$Vɛ_2$=relative velocity at outlet
∴ $Vɛ_2$= 0.8$Vɛ_1$, U = 0.4 $V_1$
$n_m$=82.1
The velocity diagłam of pelton wheel can be drawn as below,
We know that V1 = $\sqrt{2gH}$
V1 = $\sqrt{2×9.81×600}$
V1 = 108.49 m/s
U= 0.4V1 = 43.4m/s
$Vω_1=V_1=108.4g$
$Vɛ_1=V_1u=65.0g$
∴ $Vɛ_2$= 0.8$Vɛ_1$ = 52.072 m/s
now in Δ ABC
$cosϕ = \frac {U +Vω_2}{Vɛ_2}$
$cosϕ = \frac {43.4 +Vω_2}{52.072}$
∴ $Vω_2=5.53m/s$
Now $Q=\frac{π}{4}×{d_j^2} × V_1 $
$\frac{π}{4}×{0.2^2} × 108.4 g $
$Q=3.4 m^3/s$
1.Runner power (Rp)
=m($Vω_1 + Vω_2.$). U
=δQ($Vω_1 + Vω_2.$). U
=$10^3$ x 3.4 x (108.49 + 5.53) x 43.4
$ \bbox[yellow]
{
Runner \ power= 16824.791 kJ/sec
}
$
2.Shaft power
$n_m =\frac {shaft power} {runner power}$
0.82 = $\frac {Shaft Power}{ 16824.791}$
$ \bbox[yellow]
{
Shaft \ power = 13796.32 KJ/sec
}
$
3. Hydraulic efficiency
$η_h$= $\frac{RP }{SgQH }$=$\frac {Runner power}{SgQH}$
$\frac {16824.791×10^3} {10^3 × 9.8× 3.4× 600}$
$ \bbox[yellow]
{
Hydraulic \ efficiency = 0.84= 84%
}
$
4. overall efficiency = hydraulic efficiency × mechanical efficiency
= 0.84 x 0.82
$ \bbox[yellow]
{
overall \ efficiency = 68.93%
}$