fluid_mechanics
V 1 W 1 = W 2 U 676 Chapter 12 ■ Turbomachines ω r m Tangential Radial a V 1 V 2 U b F I G U R E 12.25 Ideal fluid velocities for a Pelton wheel turbine. Pelton wheel turbines operate most efficiently with a larger head and lower flowrates. V12.3 Pelton wheel lawn sprinkler As shown in Fig. 12.24, a high-speed jet of water strikes the Pelton wheel buckets and is deflected. The water enters and leaves the control volume surrounding the wheel as free jets 1atmospheric pressure2. In addition, a person riding on the bucket would note that the speed of the water does not change as it slides across the buckets 1assuming viscous effects are negligible2. That is, the magnitude of the relative velocity does not change, but its direction does. The change in direction of the velocity of the fluid jet causes a torque on the rotor, resulting in a power output from the turbine. Design of the optimum, complex shape of the buckets to obtain maximum power output is a very difficult matter. Ideally, the fluid enters and leaves the control volume shown in Fig. 12.25 with no radial component of velocity. 1In practice there often is a small but negligible radial component.2 In addition, the buckets would ideally turn the relative velocity vector through a 180° turn, but physical constraints dictate that b, the angle of the exit edge of the blade, is less than 180°. Thus, the fluid leaves with an axial component of velocity as shown in Fig. 12.26. The inlet and exit velocity triangles at the arithmetic mean radius, r m , are assumed to be as shown in Fig. 12.27. To calculate the torque and power, we must know the tangential components of the absolute velocities at the inlet and exit. 1Recall from the discussion in Section 12.3 that neither the radial nor the axial components of velocity enter into the torque or power equations.2 From Fig. 12.27 we see that V u1 V 1 W 1 U (12.48) a b Blade cross section W 1 = V 1 – U Tangential a W 2 = W 1 = V 1 – U β Axial b F I G U R E 12.26 Flow as viewed by an observer riding on the Pelton wheel—relative velocities. a b U W 1 V 2 F I G U R E 12.27 Inlet and exit β a b velocity triangles for a Pelton wheel turbine.
12.8 Turbines 677 ⎥T shaft ⎥ = mr ⋅ β max m V 1 (1– cos ) ⋅ ⎥ W ⎥ = 0.25mV ⋅ 2 shaft max 1 (1– cos β ) Actual power ⋅ Wshaft T shaft ⋅ –W shaft –T shaft U = 0.5V 1 max power Actual torque 0 0.2 V 1 0.4 V 1 0.6 V 1 0.8 V 1 1.0 V 1 U = ω r m F I G U R E 12.28 Typical theoretical and experimental power and torque for a Pelton wheel turbine as a function of bucket speed. and In Pelton wheel analyses, we assume the relative speed of the fluid is constant (no friction). V u2 W 2 cos b U (12.49) Thus, with the assumption that W 1 W 2 1i.e., the relative speed of the fluid does not change as it is deflected by the buckets2, we can combine Eqs. 12.48 and 12.49 to obtain V u2 V u1 1U V 1 211 cos b2 (12.50) This change in tangential component of velocity combined with the torque and power equations developed in Section 12.3 1i.e., Eqs. 12.2 and 12.42 gives T shaft m # r m 1U V 1 211 cos b2 where m # rQ is the mass flowrate through the turbine. Since U vR m , it follows that W # shaft T shaft v m # U1U V 1 211 cos b2 (12.51) These results are plotted in Fig. 12.28 along with typical experimental results. Note that 1i.e., the jet impacts the bucket2, and W # V 1 7 U shaft 6 0 1i.e., the turbine extracts power from the fluid2. Several interesting points can be noted from the above results. First, the power is a function of b. However, a typical value of b 165° 1rather than the optimum 180° 2 results in a relatively small 1less than 2%2 reduction in power since 1 cos 165° 1.966, compared to 1 cos 180° 2. Second, although the torque is maximum when the wheel is stopped 1U 02, there is no power under this condition—to extract power one needs force and motion. On the other hand, the power output is a maximum when U ƒ max power V 1 2 (12.52) This can be shown by using Eq. 12.51 and solving for U that gives dW # shaftdU 0. A bucket speed of one-half the speed of the fluid coming from the nozzle gives the maximum power. Third, the maximum speed occurs when T shaft 0 1i.e., the load is completely removed from the turbine, as would happen if the shaft connecting the turbine to the generator were to break and frictional torques were negligible2. For this case U vR V 1 , the turbine is “free wheeling,” and the water simply passes across the rotor without putting any force on the buckets. Although the actual flow through a Pelton wheel is considerably more complex than assumed in the above simplified analysis, reasonable results and trends are obtained by this simple application of the moment-of-momentum principle.
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V 1<br />
W 1 = W 2<br />
U<br />
676 Chapter 12 ■ Turbomachines<br />
ω<br />
r m<br />
Tangential<br />
Radial<br />
a<br />
V 1<br />
V 2<br />
U<br />
b<br />
F I G U R E 12.25 Ideal <strong>fluid</strong><br />
velocities for a Pelton wheel turbine.<br />
Pelton wheel turbines<br />
operate most<br />
efficiently with a<br />
larger head and<br />
lower flowrates.<br />
V12.3 Pelton wheel<br />
lawn sprinkler<br />
As shown in Fig. 12.24, a high-speed jet of water strikes the Pelton wheel buckets and is deflected.<br />
The water enters and leaves the control volume surrounding the wheel as free jets 1atmospheric<br />
pressure2. In addition, a person riding on the bucket would note that the speed of the water<br />
does not change as it slides across the buckets 1assuming viscous effects are negligible2. That is, the<br />
magnitude of the relative velocity does not change, but its direction does. The change in direction of<br />
the velocity of the <strong>fluid</strong> jet causes a torque on the rotor, resulting in a power output from the turbine.<br />
Design of the optimum, complex shape of the buckets to obtain maximum power output<br />
is a very difficult matter. Ideally, the <strong>fluid</strong> enters and leaves the control volume shown in Fig.<br />
12.25 with no radial component of velocity. 1In practice there often is a small but negligible radial<br />
component.2 In addition, the buckets would ideally turn the relative velocity vector through<br />
a 180° turn, but physical constraints dictate that b, the angle of the exit edge of the blade, is less<br />
than 180°. Thus, the <strong>fluid</strong> leaves with an axial component of velocity as shown in Fig. 12.26.<br />
The inlet and exit velocity triangles at the arithmetic mean radius, r m , are assumed to be as<br />
shown in Fig. 12.27. To calculate the torque and power, we must know the tangential components<br />
of the absolute velocities at the inlet and exit. 1Recall from the discussion in Section 12.3 that neither<br />
the radial nor the axial components of velocity enter into the torque or power equations.2 From<br />
Fig. 12.27 we see that<br />
V u1 V 1 W 1 U<br />
(12.48)<br />
a<br />
b<br />
Blade cross section<br />
W 1 = V 1 – U<br />
Tangential<br />
a<br />
W 2 = W 1 = V 1 – U<br />
β<br />
Axial<br />
b<br />
F I G U R E 12.26 Flow as viewed<br />
by an observer riding on the Pelton wheel—relative<br />
velocities.<br />
a<br />
b<br />
U<br />
W 1<br />
V 2<br />
F I G U R E 12.27 Inlet and exit<br />
β<br />
a<br />
b<br />
velocity triangles for a Pelton wheel turbine.