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Chapter 9—Wind Power Plants 9–35 4. Vulcan Materials uses rectangular copper conductors 6 inches by 12 inches in cross section and a total circuit length of 400 ft to carry 15,000 A of direct current to a set of electrolysis cells. What is the power loss in 400 ft of this conductor, assuming the conductor temperature is 20 o C? Use the techniques in Appendix C. 5. A wind turbine rated at 200 kVA, 480 V, three-phase, is located 500 ft from its stepup transformer and circuit breaker. Assume that the turbine operates at full power 10 percent of the time, half power 40 percent of the time, and is off the remaining 50 percent of the time. (a) What is the minimum wire size of aluminum underground cable, according to the National Electrical Code? (b) What is the voltage drop in the conductors, assuming a power factor of 0.9? (c) What is the yearly energy loss in the conductors between turbine and transformer. 6. An underground 12.47 kV loop carries 200 A in a three-conductor, single-circuit configuration. Full current flows 20 percent of the time and zero current the other 80 percent. Energy losses cost 5 cents/kWh, escalating at 3 percent per year. The wire is purchased on a 20 year note at 9 percent interest. Should you buy 1/0 AWG aluminum wire at $860/1000 ft or 4/0 AWG aluminum wire at $1500/1000 ft? 7. A wind turbine rated at 250 kVA, 480 V, three-phase, is connected by aluminum underground cable to a circuit breaker/starter and transformer located 700 feet away. (a) What size wire should be used, according to the NEC? (b) What is the rated current of this wire? (c) What NEMA size of circuit breaker/starter should be used? (d) Estimate the line-to-line voltage drop in volts for this installation when operating at rated kVA and 0.9 power factor lag. 8. An underground 12.47 kV loop is being designed to carry 10 MVA. (a) Assuming aluminum wire, is it more economical to use one circuit or two? Include trenching costs. (b) Is there any advantage to increasing the voltage from 12.47 kV to 13.8 kV? Explain. Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 9—Wind Power Plants 9–36 APPENDIX A: CONVERSION FACTORS To convert from to multiply by acre ft 2 43,560 atmosphere pascal 1.01325 × 10 5 barrel (oil, 42 gal) m 3 0.15899 Btu (mean) joule(J) 1056 Btu/h watt(W) 0.293 calorie (mean) joule(J) 4.190 cm Hg (0 o C) pascal 1.333 × 10 3 degree Celsius kelvin(K) T K = To C + 273.15 degree Fahrenheit degree Celsius To C =(To F − 32)/1.8 degree Fahrenheit kelvin(K) T K =(To F + 459.67)/1.8 foot meter 0.3048 ft 3 gallon (US liquid) 7.4805 gallon (US liquid) m 3 0.003785 hp watt(W) 745.7 in. Hg (32 o F) pascal 3.386 × 10 3 in. of H 2 O(39 o F) pascal 2.491 × 10 2 kelvin degree Celsius To C = T K − 273.15 knot mi/h 1.151 knot km/h 1.852 knot m/s 0.5144 kWh Btu 3410 kWh joule(J) 3.6 × 10 6 liter m 3 0.001 m/s mi/h 2.237 m/s knot 1.944 m 3 gallon (US liquid) 264.17 m 3 /s cfs 35.315 m 3 /s gal/min 15,850 mile (statute) meter 1609 mile (statute) feet 5280 mile (nautical) meter 1852 mi/h feet/s 1.467 mi/h km/h 1.609 mi/h knot 0.8690 mi/h m/s 0.447 N·m pound-feet 0.7376 N·m pound-inch 8.8507 psi pascal 6.895 × 10 3 Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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