BKCHAP15-2011 - Chapter 15 Electric Utility...

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Unformatted text preview: Chapter 15 Electric Utility Integration & Value Assessment 15.1 Introduction This chapter deals with the ways of determining the value of renewable and advanced energy sources to the electric utility. Various concepts are discussed in the outline form, and a technical paper by this author is attached to show an example. 15.2 Electric Power System Aspects A. Capacity Factor  ­ Measure of performance averaged over a certain period of time. C. F. = Actual Energy Produced Average Power xT = Rated Energy Rated Power xT Used for comparing different designs C.F. x Rated Power x Amount of time = Energy Produced. B. Energy Credit Savings in energy cost due to the presence of renewable energy sources. Usually after the fact  ­ energy credit Dual metering = Power In (+)  ­ Power Out x ( F )(-) = Net bill (F ← utility buyback factor) Standard form of arrangement for any renewable energy generator or co ­generator connected to the power supply system. C. Capacity Credit (capacity displacement) More controversial  ­ No industry ­wide standard for giving credit to co ­generators. 1 7 -1 Use probability functions to determine the expected output of the generator over the year/month. Use the negative load concept Reference load: Capacity (Ref. Reliability) Modified load: Capacitynew (Ref. Reliability) Capacity credit = Capold  ­ ­ Capnew Use conventional generators to meet the residual load. Number of conventional generators depends on the reliability levels chosen. Compare base case and modified load case difference in generating capacity  ­ capacity credit After the fact capacity factor 100 kW unit produces 60kW average for 10 hours/day. 12 hour min. duty ⇒ 60 X10 = 50 kW cap. credit 12 Determine the number of hours the WTG produced and the power produced. Use some average power value to determine the capacity credit. D. System Stability and Reliability  ­ Implications of change in operating conditions. Loss of Load probability (LOLP) LDC, ELDC E. Operating Reserve 1. Capacity Reserve: 20 per cent. 2. Spinning Reserve: equivalent to the largest unit 1 7 -2 15.3 Types of Generators Generators _______________________________I_______________________________ D.C. A.C. ______________I______________ I I Synchronous Asychronous (induction) D.C. Generator Field, Armature, Commutator, Brushes Field (stationary): Electromagnetic or permanent magnet poles No. of poles depend on the size of the m/c. Armature (Rotating): Armature conductors carry induced current. Commutator: Armature conductors are connected to the commutator. Brushes: Used to carry the current away # of Brushes = # of poles. Synchronous Generator Both stationary field and rotating field Stationary field generators are very similar to D.C. generators. Capacity < 5 kVA Larger sync. generators have rotating field. RPM = 120 f P Supply D.C. to the rotor using slip rings and brushes. 1 7 -3 Salient pole and cylindrical rotors Lower Speed and larger no. of poles (hydro) Higher speed, 2 or 4 poles Asynchronous Generators (Induction Generator) 3 phase induction motor mechanically driven at a higher than synchronous speed. Motor Armature on the stator Field on the rotor Electromagnetic power, (transferred to rotor) When S < O P < O ⇒ Power transferred from rotor Reactive power supplied by the line. Frequency and voltage is a function of line quantities. Service Factor "Multiplier which, applied to the rated output, indicates a permissible loading which may be carried continuously under the conditions specified for that service factor". (Usually 1.15) 15.4 RMS horsepower ∑ ( hp)2 ( time) RMS HP = Sq root of [ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ Running Time + (Standstill time/k) k ← accounts for poorer ventilation at standstill k = 3, for open motor 1 7 -4 Example 15.1 Following is a duty cycle for a motor load. Time represents duration. Compute the necessary RMS horsepower rating. Use k=1. Sec 10 10 6 20 14 HP 50 100 150 120 0 (50) 210 + (100) 210 + (150) 2 6 + (120) 2 20 + (0) 214 = 95.7 hp RMS HP = sq rt [ 10 + 10 + 6 + 20 + (14 /1) Use 100 hp motor. ! Example 15.2 Following is a duty cycle for a motor load. Compute the necessary RMS horsepower rating. Time, sec 0 5 36 39 55 80 Output, hp 160 1100 1470 400 160 160 1st period 1100 + 160 = 630 2 HP = sq.rt[ [5(630) 2 + 31(1285) 2 + 3(935) 2 + (280) 2 + 25(160) 2 ]1/2 1/2 [80] = 849 hp ! Use 850 hp motor Example 15.3 Following is a duty cycle for a motor load. Compute the necessary RMS horsepower rating. (Cumulative) Time, sec 0 40 160 180 260 280 HP 0 3 3 1.5 1.5 0 1 7 -5 ! (a) time in seconds ( 40)(1.5) 2 + 120( 3) 2 + 20(2.25) 2 + 60(1.5) 2 + 20(.75) 2 HP = 260 + 20 /1 Use k = 1 ∴ HP = 2.26 hp. Choose 2.5 hp motor (b) When time is in minutes HP = 3 hp Example 15.4 Following is a duty cycle for a motor load. Compute the necessary RMS horsepower rating. RMS horsepower Time, sec 0 10 20 30 40 50 60 HP 0 4 6 8 10 0 0 K = 2 HP RMS = = = 5.78 hp Therefore, use a 6 hp motor. In the following pages certain operation and planning functions of electric power system are discussed. 1 7 -6 ...
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This note was uploaded on 03/16/2012 for the course ECE 5374G taught by Professor Srahman during the Spring '12 term at Virginia Tech.

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