## Pelton Wheel Failure Modes

El g where hT is the turbine head. This simplifies to hr

since p1 = p2 and z1 = z2. Note that the impulse turbine obtains its energy from a reduction in the velocity head. The largest turbine head possible (and therefore the largest power) occurs when all of the kinetic energy available is extracted by the turbine, giving

This is consistent with the maximum power condition represented by Fig. E12.7&.

As indicated by Eq. 1, if the exit absolute velocity is not in the plane of the rotor (i.e., b < 180°), there is a reduction in the power available (by a factor of 1 - cos b). This is also supported by the energy equation, Eq. 2, as follows. For b < 180° the inlet and exit velocity triangles are as shown in Fig. E12.7d. Regardless of the bucket speed, U, it is not possible to reduce the value of V2 to zero—there is always a component in the axial direction. Thus, according to Eq. 2, the turbine cannot extract the entire velocity head; the exiting fluid has some kinetic energy left in it.

Fluid jets

Control volume

Control volume

shaft

Fluid jets

■ FIGURE 12.29 A multinozzle, non-Pelton wheel impulse turbine commonly used with air as the working fluid.

shaft

Dentist drill turbines are usually of the impulse class.

A second type of impulse turbine that is widely used (most often with gas as the working fluid) is indicated in Fig. 12.29. A circumferential series of fluid jets strikes the rotating blades which, as with the Pelton wheel, alter both the direction and magnitude of the absolute velocity. As with the Pelton wheel, the inlet and exit pressures (i.e., on either side of the rotor) are equal, and the magnitude of the relative velocity is unchanged as the fluid slides across the blades (if frictional effects are negligible).

Typical inlet and exit velocity triangles (absolute, relative, and blade velocities) are shown in Fig. 12.30. As discussed in Section 12.2, in order for the absolute velocity of the fluid to be changed as indicated during its passage across the blade, the blade must push on the fluid in the direction opposite of the blade motion. Hence, the fluid pushes on the blade in the direction of the blade's motion—the fluid does work on the blade (a turbine).

velocity triangles for the impulse turbine shown in Fig. 12.29.

XAMPLE 12.8

An air turbine used to drive the high-speed drill used by your dentist is shown in Fig. E12.8a. Air exiting from the upstream nozzle holes forces the turbine blades to move in the direction shown. Estimate the shaft energy per unit mass of air flowing through the turbine under the following conditions. The turbine rotor speed is 300,000 rpm, the tangential component of velocity out of the nozzle is twice the blade speed, and the tangential component of the absolute velocity out of the rotor is zero.

### Solution.

We use the fixed, nondeforming control volume that includes the turbine rotor and the fluid in the rotor blade passages at an instant of time (see Fig. E12.8b). The only torque acting on this control volume is the shaft torque. For simplicity we analyze this problem using an arithmetic mean radius, rm, where

A sketch of the velocity triangles at the rotor entrance and exit is shown in Fig. E12.8c. Application of Eq. 12.5 (a form of the moment-of-momentum equation) gives

where wshaft is shaft energy per unit of mass flowing through the turbine. From the problem

Reaction turbines are best suited for higher flowrate and lower head situations.

U = vrm = (300,000 rev/min)(1 min/60 s)(2p rad/rev)

= 394 ft/s is the mean-radius blade velocity. Thus, Eq. (1) becomes wshaft = -U1VU1 = -2U 2 = -2(394 ft/s)2 = -310,000 ft2/s2

= (-310,000 ft2/s2)(1 lb/slug • ft/s2) = -310,000 ft • lb/slug (Ans)

For each slug of air passing through the turbine there is 310,000 ft • lb of energy available at the shaft to drive the drill. However, because of fluid friction, the actual amount of energy given up by each slug of air will be greater than the amount available at the shaft. How much greater depends on the efficiency of the fluid-mechanical energy transfer between the fluid and the turbine blades. •

Recall that the shaft power, Wshaft is given by

Wshaft = mWshaft

Hence, to determine the power we need to know the mass flowrate, m, which depends on the size and number of the nozzles. Although the energy per unit mass is large (i.e., 310,000

ft • lb/slug), the flowrate is small, so the power is not "large." ^_

### 12.8.2 Reaction Turbines

As indicated in the previous section, impulse turbines are best suited (i.e., most efficient) for lower flowrate and higher head operations. Reaction turbines, on the other hand, are best suited for higher flowrate and lower head situations such as are often encountered in hydroelectric power plants associated with a dammed river, for example.

In a reaction turbine the working fluid completely fills the passageways through which it flows (unlike an impulse turbine, which contains one or more individual unconfined jets of fluid). The angular momentum, pressure, and velocity of the fluid decrease as it flows through the turbine rotor—the turbine rotor extracts energy from the fluid.

As with pumps, turbines are manufactured in a variety of configurations—radial-flow, mixed-flow, and axial-flow. Typical radial- and mixed-flow hydraulic turbines are called Francis turbines, named after James B. Francis, an American engineer. At very low heads the most efficient type of turbine is the axial-flow or propeller turbine. The Kaplan turbine, named after Victor Kaplan, a German professor, is an efficient axial-flow hydraulic turbine with adjustable blades. Cross sections of these different turbine types are shown in Fig. 12.31.

As shown in Fig. 12.31a, flow across the rotor blades of a radial-inflow turbine has a major component in the radial direction. Inlet guide vanes (which may be adjusted to allow optimum performance) direct the water into the rotor with a tangential component of velocity. The absolute velocity of the water leaving the rotor is essentially without tangential velocity. Hence, the rotor decreases the angular momentum of the fluid, the fluid exerts a torque on the rotor in the direction of rotation, and the rotor extracts energy from the fluid. The Euler turbomachine equation (Eq. 12.2) and the corresponding power equation (Eq. 12.4) are equally valid for this turbine as they are for the centrifugal pump discussed in Section 12.4.

As shown in Fig. 12.31b, for an axial-flow Kaplan turbine, the fluid flows through the inlet guide vanes and achieves a tangential velocity in a vortex (swirl) motion before it reaches the rotor. Flow across the rotor contains a major axial component. Both the inlet guide vanes

■ FIGURE 12.31 (a) Typical radial-flow Francis turbine, (b) typical axial-flow Kaplan turbine.

In a crude sense, pumps and turbines are the reverse flow versions of each other.

and the turbine blades can be adjusted by changing their setting angles to produce the best match (optimum output) for the specific operating conditions. For example, the operating head available may change from season to season and/or the flow rate through the rotor may vary.

Pumps and turbines are often thought of as the "inverse" of each other. Pumps add energy to the fluid; turbines remove energy. The propeller on an outboard motor (a pump) and the propeller on a Kaplan turbine are in some ways geometrically similar, but they perform opposite tasks. Similar comparisons can be made for centrifugal pumps and Francis turbines. In fact, some large turbomachines at hydroelectric power plants are designed to be run as turbines during high-power demand periods (i.e., during the day) and as pumps to re-supply the upstream reservoir from the downstream reservoir during low-demand times (i.e., at night). Thus, a pump type often has its corresponding turbine type. However, is it possible to have the "inverse" of a Pelton wheel turbine—an impulse pump?

As with pumps, incompressible flow turbine performance is often specified in terms of appropriate dimensionless parameters. The flow coefficient, Cq = Q/wD3, the head coefficient, CH = ghT/w2D2, and the power coefficient, Cp = Wshaft/pw3D5, are defined in the same way for pumps and turbines. On the other hand, turbine efficiency, h, is the inverse of pump efficiency. That is, the efficiency is the ratio of the shaft power output to the power available in the flowing fluid, or

PgQhT

For geometrically similar turbines and for negligible Reynolds number and surface roughness difference effects, the relationships between the dimensionless parameters are given functionally by that shown in Eqs. 12.29, 30, and 31. That is,

where the functions f 1, f2, and f 3 are dependent on the type of turbine involved. Also, for turbines the efficiency, h, is related to the other coefficients according to h = Cp/ChCq.

As indicated above, the design engineer has a variety of turbine types available for any given application. It is necessary to determine which type of turbine would best fit the job (i.e., be most efficient) before detailed design work is attempted. As with pumps, the use of a specific speed parameter can help provide this information. For hydraulic turbines, the rotor diameter D is eliminated between the flow coefficient and the power coefficient to obtain the power specific speed, N's, where

Ns _ ighTr

We use the more common, but not dimensionless, definition of specific speed

_ oXrpm)V^shaft (bhp) Nsd " [hT(ft)]5/4 (12'53)

That is, N'sd is calculated with angular velocity, v, in rpm; shaft power, Wshaft, in brake horsepower; and head, hT, in feet. Optimum turbine efficiency (for large turbines) as a function of specific speed is indicated in Fig. 12.32. Also shown are representative rotor and casing cross sections. Note that impulse turbines are best at low specific speeds; that is, when operating with large heads and small flowrate. The other extreme is axial-flow turbines, which are the

Specific speed may be used to approximate what kind of turbine geometry (axial to radial) would operate most efficiently.

Impulse turbines Reaction turbines

Impulse turbines Reaction turbines

10 20 40 60 80 100

10 20 40 60 80 100

■ FIGURE 12.32 Typical turbine cross sections and maximum efficiencies as a function of specific speed.

most efficient type if the head is low and if the flowrate is large. For intermediate values of specific speeds, radial- and mixed-flow turbines offer the best performance.

The data shown in Fig. 12.32 are meant only to provide a guide for turbine-type selection. The actual turbine efficiency for a given turbine depends very strongly on the detailed design of the turbine. Considerable analysis, testing, and experience are needed to produce an efficient turbine. However, the data of Fig. 12.32 are representative. Much additional information can be found in the literature (Ref. 1).

XAMPLE 12.9

A hydraulic turbine is to operate at an angular velocity of 6 rev/s, a flowrate of 10 ft3/s, and a head of 20 ft. What type of turbine should be selected? Explain.

^OLUTION.

The most efficient type of turbine to use can be obtained by calculating the specific speed, N'sd, and using the information of Fig. 12.32. To use the dimensional form of the specific speed indicated in Fig. 12.32 we must convert the given data into the appropriate units. For the rotor speed we get v = 6 rev/s x 60 s/min = 360 rpm

To estimate the shaft power, we assume all of the available head is converted into power and multiply this amount by an assumed efficiency (94%).

Wshaft -

Wshaft -Thus for this turbine,

(hTr

According to the information of Fig. 12.32,

A mixed-flow Francis turbine would probably give the highest efficiency and an assumed efficiency of 0.94 is appropriate.

What would happen if we wished to use a Pelton wheel for this application? Note that with only a 20-ft head, the maximum jet velocity, V1, obtainable (neglecting viscous effects) would be

As shown by Eq. 12.52, for maximum efficiency of a Pelton wheel the jet velocity is ideally two times the blade velocity. Thus, V1 = 2vR, or the wheel diameter, D = 2R, is

0.952 ft

To obtain a flowrate of Q be given by

10 ft3/s at a velocity of V1 = 35.9 ft/s, the jet diameter, d1, must v shaft p 1 Q = p d 1 V

 C 4Q d 1/2 4(10 ft3/s 2 " 1/2 . pV1 . p (35.9 ft/s). 0.596 ft A Pelton wheel with a diameter of D = 0.952 ft supplied with water through a nozzle of diameter d1 = 0.596 ft is not a practical design. Typically d1 << D (see Fig. 12.22). By using multiple jets it would be possible to reduce the jet diameter. However, even with 8 jets, the jet diameter would be 0.211 ft, which is still too large (relative to the wheel diameter) to be practical. Hence, the above calculations reinforce the results presented in Fig. 12.32—a Pelton wheel would not be practical for this application. If the flowrate were considerably smaller, the specific speed could be reduced to the range where a Pelton wheel would be the type to use (rather than a mixed-flow reaction turbine).
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### Responses

• Celendine
When and where we used pelton wheel turbine?
2 years ago