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WIND ENERGY SYSTEMS - Cd3wd

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Chapter 7—Asynchronous Loads 7–20<br />

A =<br />

(20 hp)(746 W/hp)<br />

=98.8 m2<br />

(0.9)(0.647)(0.35)(9.05)<br />

3<br />

The rotor diameter for this area is 11.22 m. Since rotors are only available in integer meter lengths,<br />

you select the 11 m rotor. This reduces the area by about 4 percent, which in turn increases the wind<br />

speed necessary to get 20 hp by slightly over 1 percent or to 9.17 m/s, an amount which seems quite<br />

acceptable.<br />

The mechanical angular velocity of the rotor in a wind speed of 9.17 m/s and a tip speed ratio of<br />

8 would be, from Eq. 11,<br />

The turbine rotational speed in r/min is then<br />

ω m = u meλ<br />

= 9.17(8) =13.34 rad/s<br />

r m 5.5<br />

n = 30 (13.34) = 127.4 r/min<br />

π<br />

The pump rotational speed needs to be 1750 r/min at this operating point, so the gear box ratio should<br />

be 1750/127.4 = 13.74:1.<br />

You note from Fig. 7a that the maximum flow rate is 1200 gal/min for a NPSH of 25 ft. This<br />

corresponds to a pump speed of 1900 r/min and an input shaft power of 28 hp according to Fig. 7b.<br />

The wind speed required for this shaft power is, from Eq. 13 and using a 0.9 air density correction,<br />

[<br />

] 1/3<br />

28(746)<br />

u R =<br />

0.647(0.9)(0.35)(π/4)(11) 2 =10.25 m/s<br />

From Fig. 7a, you estimate by extrapolation that water will start to flow at an input shaft power<br />

of about 7.5 hp, which corresponds to a cut-in wind speed of 6.61 m/s. The proposed system will,<br />

therefore, pump water at wind speeds between 6.61 and 10.25 m/s. Higher wind speeds can be used<br />

if a blade pitching mechanism can restrict the shaft speed to less than 1900 r/min so that shaft power<br />

does not increase above 28 hp.<br />

The capacity factor can be determined from Eq. 14 as CF = 0.207. The rated power would be 28<br />

hp, so the average power is (0.207)(28) = 5.79 hp. The average power required by the electric motor<br />

driven pump is 20 hp for three hours averaged over a 24 hour day or 20(3/24) = 2.5 hp. The wind<br />

turbine will have to be shut down over half the time because all the required water has been pumped.<br />

You report to the city commissioner that the system should work satisfactorily if a good speed<br />

control system is used.<br />

We should emphasize that pump characteristics vary significantly with pump design. The<br />

curves in Fig. 7 are only valid for that particular pump and should not be considered a good<br />

representation for all centrifugal pumps. Another pump design may yield a much better load<br />

match for a variable speed wind turbine than the one illustrated. It may be necessary to<br />

Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001

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