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. A pelton wheel is working under a gross-head of 600 m. The water is supplied through penstock of diameter 1.25 m and length 10 km from the reservoir to the pelton wheel turbine. The diameter of the

. A pelton wheel is working under a gross-head of 600 m. The water is supplied through penstock of diameter 1.25 m and length 10 km from the reservoir to the pelton wheel turbine. The diameter of the jet is 200 mm and gets deflected through an angle of 160◦ . The relative velocity at the outlet is reduced by 20%. If U = 0.4V1, f = 0.0065 and ηm = 82%, determine: 1. Runner power 2. shaft power 3. hydraulic efficiency ηh 4. overall efficiency ηo Problem (4). A pelton wheel turbine wi

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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,

pelton

We know that V1 = $\sqrt{2gH}$

V1 = $\sqrt{2×9.81×600}$

V1 = 108.49 m/s

U= 0.4V1 = 43.4m/s

$Vω_1=V_1=108.4g$

$Vɛ_1=V_1-u=65.0g$

∴ $Vɛ_2$= 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% }$

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