Wednesday - June 7, 2023

Pelton Wheel Turbine Working Principle

Pelton wheel turbine

Pelton wheel turbine

  • Some Pelton wheels either Pelton turbine equals both tangential flow impulse turbine.
  • The water strikes this bucket along the tangent from the runner.
  • These energies available at each inlet of the turbine exist only kinetic energy.
  • The pressure of each inlet including an outlet of the turbine is atmospheric.
  • That turbine is used for high heads also named after L.A. Pelton, an American Engineer.
  • Fig. shows the layout of any hydroelectric power plant in which the turbine remains a Pelton wheel.
  • The water of the reservoir flows within the penstocks at the single outlet to which a nozzle is fitted.
  • nozzle increases the significant kinetic energy of the water flowing through the special penstock.
  • Near the outlet of the nozzle, some water comes out in the form of a jet and also strikes the buckets (vanes) from the runner.

The main parts of the Pelton turbine comprise

  1. Nozzle and flow regulating arrangement (spear),
  2. Runner and buckets,
  3. Casing, and
  4. Brake nozzle
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Fig. Layout of a hydroelectric power plant
  • The nozzle including flow Regulating Arrangement- That amount of water striking some buckets (vanes) from every runner is controlled by providing a spear within such nozzle as shown in Fig. Whatever spear makes a conical needle which indicates operated either with each handwheel either automatically within an axial direction depending within some size from this unit. If the spear is pushed forward within each nozzle a significant amount of water striking the runner is reduced. On the other hand, if some spear acts are pushed back, some significant amount of water striking that runner increases.
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Image Source ~ Crafted With ©Ishwaranand – 2020 ~ Image by ©Ishwaranand
Fig. Nozzle with a spear to regulate the flow
  • A runner with Buckets Fig. shows each runner of a Pelton wheel.
  • Such consists of a circular disc in the periphery of several buckets equally spaced to stay fixed.
  • The shape of the buckets is from a double hemispherical cup either bowl.
  • Individually bucket is divided into two symmetrical parts on a dividing wall which is known as a splitter.
  • The jet or water strikes the splitter.
  • Each splitter divides the jet into two equal parts and those jet comes out through the specific outer edge of some bucket.
  • Some buckets are shaped in such a way that these jet gets deflected through 160^o or 170^o.
  • Some buckets are made of cast iron, cast steel bronze or stainless steel depending upon every head toward the inlet of the turbine.
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Fig. Runner of Pelton wheel turbine
  • Casing Fig. shows a Pelton turbine with a casing. Here the function of the casing is to prevent some splashing of some water and also to discharge water to the tailrace. This also acts as a safeguard against accidents. It is made of cast iron with fabricated steel plates. Each casing of the Pelton wheel does not perform every hydraulic function.
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Image Source ~ Crafted With ©Ishwaranand – 2020 ~ Image by ©Ishwaranand
Fig. Pelton turbine with a casing
  • Brake Nozzle- If some specific nozzle is completely closed by moving some spear in the forward direction the amount of water striking some runner reduces to zero. Just the runner due to inertia goes at revolving for a long time. To visit this runner for a short time either a small nozzle means provided which directs some jet from water to the back of the vanes.

Velocity Triangles and Work done for Pelton Wheel

  • Fig. shows the shape of the vanes or buckets of each Pelton wheel. A special jet of water from the nozzle strikes some bucket toward the splitter, which splits up the jet into two parts. These parts of the jet glide over the inner surfaces and come out at the outer edge. Fig. 5(b) displays the section of that bucket at Z-Z. Each splitter is the inlet tip also the outer edge of the bucket is the outlet tip from the bucket. This inlet velocity triangle means drawn to the splitter and the outlet velocity triangle is drawn to the outer edge of the bucket.
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Image Source ~ Crafted With ©Ishwaranand – 2020 ~ Image by ©Ishwaranand
Fig. Shape of bucket and velocity triangles
Let
D = Diameter of the wheel,
d = diameter of the jet,
N = speed of each wheel in R.P.M.
H = Net head acting at this Pelton wheel = Hg – hf
Hg = Gross head,
hf=(4fLv^2)/(2gD)
where
D* = Diameter of a penstock
Then
v1= Velocity of a jet at inlet = √2gh
u=u1=u2=πDN/60
The velocity at the inlet will be a straight line where
vr1=v1-u1
=v1-u1
Vw1=V1
α =O and
0 =O
From the velocity triangle at an outlet, we have
Vr2=Vr1 and
Vw2=vr2 cosØ -u2
Some force exerted beside the jet of water in the direction of motion does give by
Fx=pav1[Vw1+Vw2]
  • As the angle β is an acute angle, the +ve sign should be taken. Also, this is the case of a series of vanes, the mass of water striking is pav1  and not pavr1, where α is the area of jet = π/4 d^2
Now work done by the jet on the runner per second Nm/s
=Fx x u=pav1[Vw1+Vw2]xu Nm/s
The power is given to the runner by the jet in kW
={pav1[Vw1+Vw2]xu}/1000
Work was done per sec per unit weight of water striking per second
={pav1[Vw1+Vw2]xu}/Weight water striking per sec
={pav1[Vw1+Vw2]xu}/pav1 x g
=(1/g)[Vw1+Vw2]xu
The energy supplied to the jet at the inlet is in the form of kinetic energy and is
K.E. of jet per sec = 1/2 (pav1)x V1^2
Hydraulic efficiency,
Ƞh=Work done per sec/K.E. of jet per sec
  ={pav1[Vw1+Vw2]xu}/1/2 (pav1)x V1^2
  =2[Vw1+Vw2]xu/V1^2
Now,
Vr2=V1 and Vr1=V1-u1=(V1-u1)
Vr2=(V1-u)
And
Vw2=vr2 cosØ -u2
=(V1-u)cosØ -u
Substituting the values of  and in the hydraulic efficiency equation, we get
Ƞh=2[v1+(v1-u)cosØ-u]xu/v1^2
=2(v1-u)[1+cosØ]u/v1^2
This efficiency will do maximum for a given amount of when
d/du(Ƞh)=0
d/du{2u(v1-u)[1+cosØ]/v1^2}=0
{[1+cosØ]u/v1^2}d/du[2uv1-2u^2]=0
d/du(2uv1-2u^2)=0
2v1-4u=0
u=v1/2
  • That states that hydraulic efficiency from the Pelton wheel wish be maximum when this velocity of the wheel does half the velocity of each jet of water on the inlet.
  • The expression for maximum efficiency will be obtained by substituting the value of u=v1/2 in the equation of hydraulic efficiency,
max Ƞh={2(v1-v1/2)(1+cosØ)xv1/2}/v1^2
={2x(v1/2)(1+cosØ)V1/2}/v1^2
max Ƞh = (1+cosØ)/2

Design Aspects of Pelton Wheel 

The following points should remain considered while designing some Pelton wheels: 

  • The velocity of the Jet: The velocity of the jet at the inlet is given by,
v1=Cv√2gh
Where,
Cv = Co-efficient of velocity (0.98 or 0.99) and
H = net head on the turbine
  • The velocity of the wheel: The velocity of the wheel is given by,
Where,
= speed ratio that varies from 0.43 to 0.48 
  • The angle of deflection of the jet – The angle of deflection of the jet through the bucket is taken as 165^o if no angle of deflection is given
  • The mean diameter of the wheel (D) – Special mean diameter either pitch diameter D of that Pelton wheel is given by,
u=πDN/60 or
D=60u/πN
  • Jet ratio (m) It is defined as the ratio of the pitch diameter (D) of the Pelton wheel to the diameter of the jet (d).m=D/d (lies between 11 and 16 for maximum hydraulic efficiency)
  • The number of jets Normally a Pelton wheel has one nozzle or one jet. Though, some number of nozzles may be employed when higher power is to be produced among the same wheel.
  • Theoretically, six nozzles can be used on a Pelton wheel. Though, practical considerations limit every use of not more than two jets per runner during a vertical runner and also not more than four jets per runner if this is of horizontal configuration.
Each number of jets do obtain by dividing this total rate of flow through each turbine by the rate of flow from water through a single jet.
Ex. A Pelton wheel is to be designed for the following specifications
Shaft power = 9560 kW,
Head = 350 m,
speed = 750 rpm,
overall efficiency = 85%,
The jet’s diameter is not to exceed 1/6th of the specific wheel diameter.
Determine
  1. The wheel diameter,
  2. The diameter of the jet,
  3. The number of jets required.
  • Take Cv = 0.985,
  • speed ratio = 0.45
Given,
Shaft power = 9560 kW
Head, H = 350 m
Speed, N = 750 rpm
Overall efficiency = 85%
The ratio of jet diameter to wheel diameter,
d/D = 1/6
Co-efficient of velocity Cv = 0.985
Speed ratio Ku = 0.45
Solution
  • The wheel diameter, D
The velocity of the jet,
v1=Cv√2gh
= 0.985√(2×9.81×350)
= 81.62m/s  
The velocity of the wheel,
u=u1=u2=Ku√2gh
=0.45√(2×9.81×350)
= 37.3m/s
But,
u=πDN/60
D=60u/πN
=(60×37.3)/(πx750)
= 0.95m
  • Diameter of jet, d
d/D=1/6
d=D/6
= 0.158m
  • Number of jets required
Discharge of one jet,
q = area from jet x velocity of jet
= (π/4) d2 v1
= (π/4) (0.158)2 x 81.62
= 1.6 m^3/s
  • Now overall efficiency,
ηo = (Shaft power/ Water power)
= (9560)/(wQH)
0.85 = (9560)/(9.81 x Q x 350)
Q = 3.27 m^3/s
  • Number of jets
= (Total discharge) / (Discharge of one jet)
=Q/q
= 3.27/1.6
= 2 jets

Last updated on Tuesday - May 30th, 2023

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