* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Line Diagrams And The Per Unit System
Mechanical-electrical analogies wikipedia , lookup
Power over Ethernet wikipedia , lookup
Pulse-width modulation wikipedia , lookup
Opto-isolator wikipedia , lookup
Variable-frequency drive wikipedia , lookup
Power factor wikipedia , lookup
Current source wikipedia , lookup
Power inverter wikipedia , lookup
Audio power wikipedia , lookup
Electrification wikipedia , lookup
Ground (electricity) wikipedia , lookup
Electric power system wikipedia , lookup
Power MOSFET wikipedia , lookup
Nominal impedance wikipedia , lookup
Surge protector wikipedia , lookup
Amtrak's 25 Hz traction power system wikipedia , lookup
Electrical substation wikipedia , lookup
Transformer wikipedia , lookup
Stray voltage wikipedia , lookup
Earthing system wikipedia , lookup
Power electronics wikipedia , lookup
Buck converter wikipedia , lookup
History of electric power transmission wikipedia , lookup
Voltage optimisation wikipedia , lookup
Distribution management system wikipedia , lookup
Power engineering wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Three-phase electric power wikipedia , lookup
Line Diagrams And The Per Unit System Line Diagrams • In power engineering, one-line diagram or single-line diagram is a simplified notation for representing a three-phase power system. Electrical elements such as circuit breaker, transformer, capacitors, bus bars, and conductors are shown by standized schematic symbols. Instead of representing each of three phases with a separate line or terminal, only one conductor is represented. It is a form of block diagram graphically depicting the paths for power flow between entities of the system. Elements on the diagram do not represent the physical size or location of the equipment. An important point to note is that the diagram often represents the application, i.e. if load flow is being depicted then protective devices may not be shown Region - 3 2 Region - 1 Region 2 2 1 T2 L1 T2 1 2 L1 = impedance R1 = 5Ω X1 = 20Ω G1 Rating; 30 MVA 13.8KV R = 0.1 pu X = 1.0 pu T1 Rating; 35 MVA 13.2/115KV R = 0.01 pu X = 0.10 pu T2 Rating; 30 MVA 120/12.5KV R = 0.01 pu X = 0.08pu M1 Rating; 20 MVA 12.5KV R = 0.1 pu X = 1.1 pu M2 Rating; 10 MVA 12.5KV R = 0.1 pu X = 1.1 pu The Per-Unit System • In the analysis of power network, instead of using actual values of quantities it is usual to express them as fractions of reference quantities, such as rated or full load values. These fractions are called per unit ‘p.u.’ the ‘p.u.’ value of any quantity is defined as : 𝐴𝑛𝑦 𝑣𝑎𝑙𝑢𝑒 (𝑖𝑛 𝑎𝑛𝑦 𝑢𝑛𝑖𝑡) 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑝. 𝑢 = 𝐵𝑎𝑠𝑒 𝑜𝑟 𝑅𝑒𝑓𝑟𝑒𝑒𝑛𝑐𝑒 𝑉𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑢𝑛𝑖𝑡 The ‘p.u.’ value can also be expressed as a percentage of the base or reference value. The per-unit system is widely used in the power system industry to express values of voltages, currents, powers, and impedances of various power equipment. It is mainly used for transformers and AC machines. The advantages of using the ‘p.u.’ i. The apparatus considered may vary widely in size, losses and volt drops will also vary considerably. For apparatus of the same general type the ‘p.u.’ voltage drops and losses are in the same order regardless of size. ii. By choice of the appropriate voltage bases the solution of networks containing several transformer is facilitated. iii. When values are expressed in ‘p.u.’ the comparison of electrical quantities with their “normal” values is straightforward. iv. Calculations are simplified because quantities expressed as per-unit are the same regardless of the voltage level. Base Apparent Power ‘ SB’ The base apparent power, or base volt-amperes, represents the following equation : 𝑆𝐵 = 3 × 𝑉𝐵 × 𝐼 𝐵 where, ‘VB’ = Line Voltage in a 3 phase System ‘IB’ = Line/Phase Current in a 3 phase System Using the above equation for base apparent power it is possible to derive the base current and base voltage; 𝐼𝐵 = 𝑆𝐵 3×𝑉𝐵 Once the base current is defined then the base impedance is given by; • 𝑍𝐵 = 𝑉𝐵 3 𝐼𝐵 • 𝑍𝐵 = • 𝑍𝐵 = • 𝑍𝐵 = 𝑉𝐵 3 𝑆𝐵 3×𝑉𝐵 𝑉𝐵 3 𝑆𝐵 3×𝑉𝐵 𝑉𝐵2 𝑆𝐵 Hence, the per-unit impedance of an electrical item of equipment in the power system is given by: 𝑍𝑝𝑢 = 𝑧Ω 𝑍𝐵 𝑍𝑝𝑢 = 𝑍Ω 𝑉𝐵2 𝑆𝐵 𝑍𝑝𝑢 = 𝑍Ω ×𝑆𝐵 𝑉𝐵2 In load flow analysis, where the base apparent power may be different to that of the equipment rating, it is necessary to convert the per-unit impedance referenced to the ‘old base’ to the ‘new base’ per unit impedance, using the base apparent power selected for the whole power system. This conversion is achieved using the equation: 𝑍𝑝. 𝑢. 𝑛𝑒𝑤 = 𝑍𝑝. 𝑢. 𝑔𝑖𝑣𝑒𝑛 𝑉𝑔𝑖𝑣𝑒𝑛 𝑉𝑛𝑒𝑤 2 𝑆𝑛𝑒𝑤 𝑆𝑔𝑖𝑣𝑒𝑛 In some instances the per-unit impedance referenced to the ‘old base’ Is given as a percentage. When this is the case it is simple to convert the given percentage to its per-unit equivalent using ‘Zpu = Z%/100%. Transformer PU In an ideal transformer the following equation apply : |𝑉1| 𝑉2 = & 𝐼2 = 𝑛|𝐼1| 𝑛 Where n is the turn’s ration of the transformer 𝑛 = 𝑁1 𝑁2 Similarly |𝑉𝑏1| 𝑉𝑏2 = 𝑛 |𝑆𝑏| |𝑆𝑏| 𝐼𝑏2 = = 𝑛. = 𝑛|𝐼𝑏1| |𝑉𝑏2| |𝑉𝑏1| Dividing the voltage relation of the ideal transformer by Vb2 and substituting Vb2 = Vb1/n results in : |𝑉2| |𝑉1| |𝑉1| = = |𝑉𝑏2| 𝑛|𝑉𝑏2| |𝑉𝑏1| Similarly it can be shown that dividing the current relationship of the ideal transformer by the base current Ib2 and substituting Ib2 = nIb1 results in : |𝐼2| |𝐼1| |𝐼1| = 𝑛. = |𝐼𝑏2| |𝐼𝑏2| |𝐼𝑏1| Exercise 1 : A certain power element in a series circuit as impedance of 4+ J7Ω, if the output of the element is 5..773KV at a current of 150 at -36.9 degrees, determine the input voltage using conventional circuit theory, by phase or diagram and by using p.u. method. (assume Sbase = 106). Answer Vs.p.u = 1.19 + j1.19/V Region 3 Region 1 Region 2 2 2 1 G1 T2 L1 T2 1 L1 impedance R=5Ώ X = 20Ώ T1 Rating; 35 MVA 13.2/115KV R = 0.01 pu X = 0.10 pu T2 Rating; 30 MVA 120/12.5KV R = 0.01 pu X = 0.08pu 2 M1 Rating; 20 MVA 12.5KV R = 0.1 pu X = 1.1 pu M2 Rating; 10 MVA 12.5KV R = 0.1 pu X = 1.1 pu