characteristic curves
of pelton wheel:-
Function:-
Nozzles direct forceful, high-speed streams of
water against a rotary series of spoon-shaped buckets, also known as impulse
blades, which are mounted around the circumferential rim of a drive wheel—also
called a runner (see photo, 'Old Pelton wheel..'). As the water jet impinges
upon the contoured bucket-blades, the direction of water velocity is changed to
follow the contours of the bucket. Water impulse energy exerts torque on the
bucket-and-wheel system, spinning the wheel; the water stream itself does a
"u-turn" and exits at the outer sides of the bucket, decelerated to a
low velocity. In the process, the water jet's momentum is transferred to the
wheel and thence to a turbine. Thus, "impulse" energy does work on
the turbine. For maximum power and efficiency, the wheel and turbine system is
designed such that the water jet velocity is twice the velocity of the rotating
buckets. A very small percentage of the water jet's original kinetic energy
will remain in the water, which causes the bucket to be emptied at the same
rate it is filled, (see conservation of mass) and thereby allows the
high-pressure input flow to continue uninterrupted and without waste of energy.
Typically two buckets are mounted side-by-side on the wheel, which permits
splitting the water jet into two equal streams (see photo). This balances the
side-load forces on the wheel and helps to ensure smooth, efficient transfer of
momentum of the fluid jet of water to the turbine wheel.
Because water and most liquids are nearly
incompressible, almost all of the available energy is extracted in the first
stage of the hydraulic turbine. Therefore, Pelton wheels have only one turbine
stage, unlike gas turbines that operate with compressible fluid.It is used for
generating electricity.
Applications:-
Pelton wheels are the preferred turbine for
hydro-power, when the available water source has relatively high hydraulic head
at low flow rates, where the Pelton wheel geometry is most suitable. Pelton
wheels are made in all sizes. There exist multi-ton Pelton wheels mounted on
vertical oil pad bearings in hydroelectric plants. The largest units can be
over 400 megawatts. The smallest Pelton wheels are only a few inches across,
and can be used to tap power from mountain streams having flows of a few
gallons per minute. Some of these systems use household plumbing fixtures for
water delivery. These small units are recommended for use with 30 metres (100
ft) or more of head, in order to generate significant power levels. Depending
on water flow and design, Pelton wheels operate best with heads from 15–1,800
metres (50–5,910 ft), although there is no theoretical limit.
Design
rules:-
The specific speed s parameter is independent of a
particular turbine's size.
Compared to other turbine designs, the relatively
low specific speed of the Pelton wheel, implies that the geometry is inherently
a "low gear" design. Thus it is most suitable to being fed by a hydro
source with a low ratio of flow to pressure, (meaning relatively low flow and/or
relatively high pressure).
The specific speed is the main criterion for
matching a specific hydro-electric site with the optimal turbine type. It also
allows a new turbine design to be scaled from an existing design of known
performance.
where:
n = Frequency of rotation (rpm)
P = Power (W)
H = Water head (m)
rho =
Density (kg/m3)
The formula implies that the Pelton turbine is
geared most suitably for applications with relatively high hydraulic head H,
due to the 5/4 exponent being greater than unity, and given the
characteristically low specific speed of the Pelton.
Turbine physics and derivation
Energy and initial jet velocity
In the ideal (frictionless) case, all of the
hydraulic potential energy (Ep = mgh) is converted into kinetic energy (Ek =
mv2/2) (see Bernoulli's principle). Equating these two equations and solving
for the initial jet velocity (Vi) indicates that the theoretical (maximum) jet
velocity is Vi = √(2gh)
. For simplicity, assume that all of the velocity vectors are parallel to each
other. Defining the velocity of the wheel runner as: (u), then as the jet
approaches the runner, the initial jet velocity relative to the runner is: (Vi
− u). The initial jet velocity of jet is Vi
Final
jet velocity:-
Assuming that the jet velocity is higher than the
runner velocity, if the water is not to become backed-up in runner, then due to
conservation of mass, the mass entering the runner must equal the mass leaving
the runner. The fluid is assumed to be incompressible (an accurate assumption
for most liquids). Also it is assumed that the cross-sectional area of the jet
is constant. The jet speed remains constant relative to the runner. So as the
jet recedes from the runner, the jet velocity relative to the runner is: −(Vi −
u) = −Vi + u. In the standard reference frame (relative to the earth), the
final velocity is then: Vf = (−Vi + u) + u = −Vi + 2u.
Optimal
wheel speed:-
We know that the ideal runner speed will cause all
of the kinetic energy in the jet to be transferred to the wheel. In this case
the final jet velocity must be zero. If we let −Vi + 2u = 0, then the optimal
runner speed will be u = Vi /2, or half the initial jet velocity.
Torque:-
By Newton's second and third laws, the force F
imposed by the jet on the runner is equal but opposite to the rate of momentum
change of the fluid, so:
F = −m( Vf − Vi) = −ρQ[(−Vi + 2u) − Vi] = −ρQ(−2Vi
+ 2u) = 2ρQ(Vi − u)
where (ρ) is the density and (Q) is the volume
rate of flow of fluid. If (D) is the wheel diameter, the torque on the runner
is: T = F(D/2) = ρQD(Vi − u). The torque is at a maximum when the runner is
stopped (i.e. when u = 0, T = ρQDVi ). When the speed of the runner is equal to
the initial jet velocity, the torque is zero (i.e. when u = Vi, then T = 0). On
a plot of torque versus runner speed, the torque curve is straight between
these two points, (0, pQDVi) and (Vi, 0).
Power:-
The power P = Fu = Tω, where ω is the angular
velocity of the wheel. Substituting for F, we have P = 2ρQ(Vi − u)u. To find
the runner speed at maximum power, take the derivative of P with respect to u
and set it equal to zero, [dP/du = 2ρQ(Vi − 2u)]. Maximum power occurs when u =
Vi /2. Pmax = ρQVi2/2.
Substituting the initial jet power Vi = √(2gh), this simplifies to Pmax = ρghQ. This quantity
exactly equals the kinetic power of the jet, so in this ideal case, the
efficiency is 100%, since all the energy in the jet is converted to shaft
output.
Efficiency:-
A wheel power divided by the initial jet power, is
the turbine efficiency, η = 4u(Vi − u)/Vi2. It is zero for u = 0 and for u =
Vi. As the equations indicate, when a real Pelton wheel is working close to
maximum efficiency, the fluid flows off the wheel with very little residual
velocity. In theory, the energy efficiency varies only with the efficiency of
the nozzle and wheel, and does not vary with hydraulic head. The term
"efficiency" can refer to: Hydraulic, Mechanical, Volumetric, Wheel,
or overall efficiency.
System
components:-
The conduit bringing high-pressure water to the
impulse wheel is called the penstock. Originally the penstock was the name of
the valve, but the term has been extended to include all of the fluid supply
hydraulics. Penstock is now used as a general term for a water passage and
control that is under pressure, whether it supplies an impulse turbine or not.
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