Pelton wheel
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Pelton wheels are among the most efficient types of water turbines. It was invented by Lester Allan Pelton (1829-1908) in the 1870s, and is an impulse machine, meaning that it uses Newton's second law to extract energy from a jet of fluid. It should be noted that the original one piece cast impulse water turbine was invented by Samuel Knight in Sutter Creek, Ca in the California Mother Lode Gold Mining Region.[citation needed] Pelton modified this design to create his more efficient design. Knight Foundry is the last water powered foundry known to exist in the United States and is still operated using Knight impulse turbines. It is used for high heads and low discharge.
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[edit] Function
The water flows along the tangent to the path of the runner. Nozzles direct forceful streams of water against a series of spoon-shaped buckets mounted around the edge of a wheel. As water flows into the bucket, the direction of the water velocity changes in order to follow the contour of the bucket. When the water-jet contacts the bucket, the water exerts pressure on the bucket and the water is decelerated as it does a "u-turn" and flows out the other side of the bucket at low velocity. In the process, the water's momentum is transferred to the turbine. This "impulse" does work on the turbine. For maximum power and efficiency, the turbine system is designed such that the water-jet velocity is twice the velocity of the bucket. A very small percentage of the water's original kinetic energy will still remain in the water; however, this allows the bucket to be emptied at the same rate it is filled, (see conservation of mass), thus allowing the water flow to continue uninterrupted. Often two buckets are mounted side-by-side, thus splitting the water jet in half (see photo). This balances the side-load forces on the wheel, and helps to ensure smooth, efficient momentum transfer of the fluid jet to the turbine wheel.
Since 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.
[edit] Applications
Pelton wheels are the preferred turbine for hydro-power, when the available water source has relatively high hydraulic head at low flow rates. 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 up to 200 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 utilize household plumbing fixtures for water delivery. These small units are recommended for use with thirty metres 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 metres to 1,800 metres, although there is no theoretical limit.
The Pelton wheel is most efficient in high head applications (see the "Design Rules" section). Thus, more power can be extracted from a water source with high-pressure and low-flow than from a source with low-pressure and high-flow, even though the two flows theoretically contain the same power. Also a comparable amount of pipe material is required for each of the two sources, one requiring a long thin pipe, and the other a short wide pipe.
[edit] Design Rules
For a given turbine application, if one knows the water head, desired wheel speed and output power, then the following formula can indicate the appropriate type of turbine.
[edit] Imperial Units
The "specific speed" is defined as ns = n√(P)/h5/4
[edit] Metric Units
The "specific speed" is ns = 0.2626 n√(P)/h5/4
Well-designed efficient machines typically use the following values: Impulse turbines have the lowest ns values, typically ranging from 1 to 10, a Pelton wheel is typically around 4, Francis turbines fall in the range of 10 to 100, while Kaplan turbines are at least 100 or more. [1] The formula suggests that the Pelton turbine is most suitable for applications with relatively high hydraulic head, due to the 5/4 exponent being greater than unity, and given the low characteristic specific speed of the Pelton.
[edit] Turbine Physics and Derivation
[edit] 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). [1]
[edit] 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. All of this means that 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
[edit] 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.
[edit] Torque
By newton's second and third laws, the force F imposed by the jet on the runner is equal but opposite to the impulse or 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 a straight between these two points [(0, pQDVi ) and (Vi, 0)].[1]
[edit] 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. [1]
[edit] Efficiency
The 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.[1] Apparently, this basic theory does not suggest that efficiency will vary with hydraulic head, and further theory is required to show this.
[edit] Examples and Design Data
A working Pelton wheel was used to generate electricity in Southern California. The system had the following specifications. Pitch diameter, 162" (2.06 m); operating speed, 250 rpm (26.18 rad/s); head, 2200' (670.6 m). The theoretical jet velocity Vi = √(2gh), is calculated to be 114.6 m/s, and the wheel edge speed u = 53.86 m/s. Since u ~ Vi /2, this data is consistent with the theoretical model. The ratio of the runner velocity u to the ideal jet velocity √(2gh) is usually denoted φ. As the theoretical model suggests, for a Pelton wheel working at maximum efficiency, φ is about 1/2. This wheel is estimated to have produced about 60,000 HP (45 MW) on a flow of about 7 m3/s. [1]
[edit] 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. [1]
[edit] Additional physics information
The power potential is the product of the water head and the volume flow rate. Power can be expressed as Power = Force * velocity (where Power is measured in watts, Force is in newtons, and velocity is measured in metres per second). In the instance of fluid, force is typically reframed as the product of pressure difference and cross-sectional area, (F=P*A). Also, the product of cross-sectional area and average velocity, equals the volume flow rate. Thus the flow-power can be rewritten as P=kp(V/t) (where k is a constant representing the efficiency, p is the pressure difference, and V/t is the volume flow rate, or the volume of fluid flow per unit time). So the power, P, is directly proportional to both the pressure difference, and the flow rate.
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