NEVER MISS A FORMULA SERIES FORMULA BOOK POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ Unit - I Generating Power Stations 1) Water power p = 0.736 75 iv. QπH kW Q → Discharge : π3 /sec H → Water head : m π = Overall efficiency v. π√ππ‘ ππ → Specific speed in metric units N → Speed of turbine in rpm ππ‘ = Output in metric h.p H → effective head in metres 3) Base load High Peak load 4) Low Fuel cost Low Typical annual load factor Type of Plant 65-75 Nuclear, thermal 5-10 Gas, hydro, pump storage High For thermal power station: i. ππ‘βπππππ = iii. Overall efficiency πππ£πππππ = π»πππ‘ πππ’ππ£πππππ‘ ππ πππππ‘πππππ π/π π»πππ‘ ππ πππππ’π π‘πππ ππ ππππ iv. Overall efficiency = Thermal efficiency × electrical efficiency v. Energy output = Coal consumption × calorific value = coal consumption × 6500 K.cal πππ₯πππ’π ππππππ πΆππππππ‘ππ ππππ Hydro power station: i. Metric output = π΄π£πππππ ππππ ππππ ππππ πππ» × π 75 (H.P) 1 H.P = 75 kg - m/sec ππ’π ππ πππππ£πππ’ππ πππ₯πππ’π ππππππ πππ₯πππ’π ππππππ ππ ππππ’π ii. Metric output in watt = πππ» × π 75 × 735.5 iii. Volume of water = Catchment (Diversity factor > 1) 73 73 00 77 34 π»πππ‘ πππ’ππ£πππππ‘ ππ πππβ−ππππππ¦ π‘ππππ πππ‘π‘ππ π‘π π‘π’πππππ π»πππ‘ ππ ππππ πππππ’π π‘πππ ii. ππ‘βπππππ = πππππππ × ππ‘π’πππππ (Demand factor < 1) iii. Diversity factor= πππ₯πππ’π ππππ 5) 6) ii. Load factor = × ππππ factor Utilization factor = π ππ‘ππ πππππ‘ πππππππ‘π¦ Operational factors: i. Demand factors = πππ₯πππ’π ππππ πππππ‘ πππππππ‘π¦ vi. Q → Quantity of water flow π3 /sec g → Acceleration due to gravity = 9.81m/π ππ 2 H → Water head, metre π → density of water Capital cost π΄π£πππππ ππππ’πππ ππππ π ππ‘ππ πππππ‘ πππππππ‘π¦ Capacity factor = Load factor × Utilization factor Tidal power output P = QπgH Watts Type of Load Capacity factor = Capacity factor = Specific speed ππ = π»1.25 2) Peak diversity factor πππ₯πππ’π ππππππ ππ π ππππ π’πππ ππππ’π = π·πππππ ππ π‘βπ ππππ π’πππ ππππ’π ππ‘ π‘βπ π‘πππ ππ ππππ ππππππ 1 available area × Annual per annum Rainfall POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ iv. Electric energy = weight × head generated × overall π 7) Gas turbine power plant: i. Engine efficiency ππππππππ = πππ£ππ πππ ππππ‘ ii. Heat produced by = coal consumption π·π = per day × calorific fuel per day 9 √(π·π . π. π)3 = 3√(π·π . π)2 iii. For four conductor arrangement value Unit - II Transmission and distribution 1) String efficiency ππππ‘πππ πππππ π π‘βπ π€βπππ π π‘ππππ = π × ππππ‘πππ πππππ π π‘βπ π’πππ‘ ππππππππ‘ π‘π ππππ πππππ’ππ‘πππ ) 16 π·π = √(π·π . π. π. π√2) 4 = 1.09 × √π·π . π3 6) n→number of insulator discs in the string 2) Capacitance canculation: i. Capacitance of Two wire line –q b +q a Inductance of a single phase two wire line D π· π L = 4× 10−7 ππ ( 1 ) H/m 3) πΆππ = Inductance of composite Conductor lines ii. Line to neutral capacitance −7 π· π πππ( ) ππΉ/ππ πΆππ =πΆππ =2πΆππ πΊππ· ππ (πΊππ ) iii. Capacitance of 3π line with ο equilateral spacing πΆπ = Inductance of 3Ο line with equivalent spacing 0.02412 π· π log( ) ππΉ/ππ ο asymmetrical spacing πΆπ = π· πΏπ = 2× 10−7 ππ (π1 ) H/m D = Distance between any two phases 5) 0.01206 Where, π ′ = 0.7788r L = 2× 10 4) 4 ππΉ/ππ Inductance of Bundled conductors line] 0.02412 log( π·ππ ) π π·ππ = 3√π·12 π·23 π·31 2 i. For two conductor arrangement D12 D23 4 π·π = √(π·π . π)2 = π·π . π 1 ii. For three conductor arrangement 73 73 00 77 34 2 D31 3 POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ iv. Capacitance of Bundled 1 π ππ =[ ][ ] 0 1 πΌπ 0.02412 Conductor lines πΆπ = π·ππ ππΉ/ππ log( ) So A = 1, B = Z, C = 0, D = 1 √ππ v. For a two conductor bundle D = √ππ 10) For a three conductor bundle D= √ππ2 condition ππ = Receiving end voltage under full load 4 condition D = 1.09 √ππ3 Ground line (method of images) capacitance 11) πΆππ = 2 πΆππ = Regulation = πΌπ π πππ ππ ± πΌπ ππ ππππ ππ + → For lag PF for 1Ο line 0.02412 π· log( ′) π - → For lead PF ππΉ/ππ i. Max voltage regulation ππ = π π π = π‘ππ−1 π π· 2π» π ′ = r √1 + ( )2 ii. Zero regulation π ππ + π = 8.a) Performance of Tr.line Type of Tr.line Classifi-cation based on f × l Short line <4000Hz km Medium line 4000 < fl 12000Hz km long line fl=12000 Hz km < 12) For nominal T-circuit π [ π ] πΌπ l >240km Effect of capacitance Short line 0-20 kv neglected Medium line 20-100 kv capacitance is lumped and constant long line >100 kv capacitance is uniformly distributed 1+ ππ 2 ππ + 4) 1+ =[ π(1 ππ ) 2 ππ ] 2 π 1+ ππ] 2 π [ π] πΌπ π [ π] πΌπ Long Tr.line π [ π ] πΌπ πΆππ βπΎπ = [1 π ππβπΎπ π π ππ π ππβπΎπ cos βπΎπ π ] [ π] πΌπ πΎ = √π§π¦ πΎ = ∝ + jπ½ π½ = π√πΏπ l→length of Tr line ∝→ attention constant Neper/sec π½ → phase constant in rad/km For short transmission lines 73 73 00 77 34 π(1 + π [ π ] πΌπ 8.b) π΄ =[ πΆ ππ 2 80km<l< 240km 13) π [ π ] πΌπ 1+ =[ π For nominal π-circuit Based on operating voltage when Medium length Tr.line l < 80 km Type of Tr.line occur 2 Classification based on length of line (i.e) power line i.e f = 50Hz 9) π ′ π − ππ ππ π ′ π = Receiving end voltage under no load 3 For a four conductor bundle 7) Voltage Regulation = π΅ ππ ][ ] π· πΌπ Sin πΎπ = √π¦π§ [1 + 3 π¦π§ 3! + (π¦π§)2 5! + …] POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ π¦ (π¦π§)2 17) Surge Impedance Cos πΎπ = 1 + π§ + +… 2! 14) 4! π§ 2π Velocity of wave propagation ππ = f π 15) π +ππ€πΏ ππ = √π¦ = √πΊ+ππ€πΆ Wave length (π) π= π½ π΅ ππ = √πΆ = √πππ . ππ π Equivalent π-ckt using ABCD parameters For loss less transmission line, R=0, G=0 πΏ ππ = √πΆ Note: ππ = 400 β¦ for transmission line ππ = 40 β¦ for cable π′π′ π§ π ππ 2 [ π ] = ′ ′ ′ ′ [ ] π π π π πΌπ πΌπ π ′ (1 + ) 1+ 4 2 ] [ 18) 1+ SIL = πΎπ tanh(πΎπ/2) (πΎπ/2) Y’ = π§ tan h 2 = y π 19) Equivalent T-ckt using ABCD parameter i) π¦"π§" 1+ 2 π [ π ] = [ πΌπ π¦" π§" 2 cos βπΎπ−1 ) sin hπΎπ = π§π ( 1 = (πΎπ)2 ππ MW Tr.line connected in series π π΄ [ π ] = [ 1 πΌπ πΆ1 π΅1 π΄2 ][ π·1 πΆ2 π§ tan β(πΎπ/2) πΎπ/2 iii) π΅2 ππ ][ ] π·2 πΌπ Tr.line connected in cascade π π΄ π΄ + π΅1 πΆ2 [ π] = [ 1 2 πΌπ πΆ1 π΄2 + π·1 πΆ2 =2 Y” = π§π sin hπΎπ = π¦ 16) ii) π§"π¦" ) ππ 4 ][ ] π¦"π§" πΌπ + 2 π§"(1 + 1 ππ ππ ππ ο loading <SIL→PF will be leading ο Loading>SIL→Pf will be lagging ο Loading = SIL →PF will be unity π ππβπΎπ πΎπ π§′ = π§π sin hπΎπ = z 2 Surge impendence loading π΄1 π΅1 + π΅1 π·2 ππ ][ ] πΆ1 π΅2 + π·1 π·2 πΌπ Tr.line connected in parallel sin πΎπ πΎπ ππ = |ππ |<πΏ, ππ = |ππ |<0 D = A = |A|<πΌ, B = |B| <π½ ππ = |ππ ||ππ | |π΅| 73 73 00 77 34 <(π½ − πΏ) – |π΄||ππ |2 |π΅| < (π½ − πΌ) 4 POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ 2π§ ο Refraction coefficient = π§ +π§ πΏ ο Reflection coefficient = 21) π ππΏ −ππΆ ππΏ +ππΆ Parameter of single core cables; i) Insulation Resistance π π Rins =2πln( π ) ohms/metre A= π΄1 π΅2 +π΄2 π΅1 π΅1 +π΅2 π΅ π΅ B=π΅ 1+π΅2 1 2 π· −π· C = (πΆ1 + πΆ2 ) + π΅1+π΅2 1 D= 20) 2 π·1 π΅2 +π·2 π΅1 π΅1 +π΅2 Where resistance Reflection and refraction of waves Line terminated by impedance (z) Rins→Insulation π →Resistivity material of insulating R→inside radius of sheath r→conductor radius ππ‘ = ii)Insulation 2π§ ππΏ +ππΆ Resistance ∝ 1 πππππ‘β ππ πππππ ππ ππ‘ = Transmitted (or) Refracted voltage Rins∝ 1 π π −π ππ = ππΏ +ππΆ ππ πΏ iii) Capacitance between core and sheath πΆ ππ = Reflected voltage ππ‘ = 2 ππΏ +ππΆ 2π∈ ∈ 0 π c = ππ(π /π) F/m ππ π iv) Potential V = 2ππ ππ(π /π) ππ = Incident voltage ππ‘ = transmitted current r→radius of conductor ππ = reflected current R→Inner radius of sheath πΌπ = incident current v) Gradient is maximum at the π −π surface of conductor ππ = ππΆ+ππΏ πΌπ πΏ (or) πΆ ππππ₯ = π −π − (ππΏ +ππΆ ) πΌπ πΏ πΆ 73 73 00 77 34 5 π π ππ π π POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ π ππ 1 1 1 ππ0 = e = 2.718 π [ππ ] = [1 π 2 π ] [ππ1 ] 1 π π 2 ππ2 ππ vi) Gradient is minimum at the inner ππ0 1 1 [ππ1 ] = 3 [1 ππ2 1 radius of conductor ππππ = π π ππ π π 7) Unit – III Symmetrical Components and faults 1) 1 ππ π 2 ] [ππ ] π ππ 1 π π2 Unsymmetrical faults: i) Single line to ground fault Per unit value= π΄ππ‘π’ππ π£πππ’π ππ π πππ π’πππ‘π π΅ππ π (ππ)πππππππππ π£πππ’π ππ π πππ π’πππ‘π 2) Base Current = π΅ππ π πΎππ΄ √3 ×π΅ππ π πΎπ A Base impedence = πΏπππ π‘π πππ’π‘πππ π£πππ’π ππ πππ π π£πππ‘πππ π΅ππ π ππ’πππππ‘ ο πΌπ 0 = πΌπ 1 = πΌπ 2 = 2 = 3) π΅ππ π πΎπ Ω π΅ππ π πππ΄ πΌ Per unit reactance πππ’ = ππππ‘π’ππ ππππ π 1ππ +π2ππ +π0ππ 1ππ +π2ππ +π0ππ +3π§π ο πΌπΉ = πΌπ =3πΌπ 0 = 3πΌπ 1 π 3π √ πΏ 3πππ΄π π1ππ +π2ππ +π0ππ ο Actual fault current πΌπΏ = 2 ο Short circuit MVA = π΅ππ ππππ΄ πππ’ old× [π΅ππ π πΎπ πππ ] × [ π΅ππ ππππ΄πππ€ ] πππ€ 5) 3πΈπ πΌ ο πΌπΉ = πΌπ = π (πππ΄)πππ π (πΎπ)2 πππ π πππ’ new = π΅ππ π πΎπ πΈπ πΌ ο πΌπ 1 = π πππ π πππ’ = ππππ‘π’ππ 4) ο ππ 1+ππ 2+ππ 0 = 0 ο If πΌπ = 3πΌπ 0 πΌπ 3 ο ππ = πΌπ ππ =3πΌπ 0 ππ πππ short circuit KVA = ii) Line to line fault: 100 Rated (or) Base KVa× %π 6) Symmetrical components ππ = ππ1 +ππ2 +ππ0 ππ = ππ1 +ππ2 +ππ0 ππ = ππ1 +ππ2 +ππ0 ππ1 = πΎ 2 ππ1 ππ2 = πΎ ππ2 ππ0 = ππ0 ππ1 = Kππ1 ππ2 = πΎ 2 ππ2 ππ0 = ππ0 Where K = 1∠120° ο πΌπΉ = πΌπ¦ =-πΌπ ο πΌπ = 0 In matrix form 73 73 00 77 34 6 POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ ο πΌπ¦ + πΌπ΅ = 0 ο πΌπ 2 = πΌπ 1 , ππ 1 = ππ 2 Unit – IV Power System Stability πΈπ 1 +π 1ππ 2ππ +ππΉ ο πΌπ 1 = π πΈπ 1 ο πΌπΉ = √3πΌπ 1 πΌπΉ = √3 π 1ππ +π2ππ +ππΉ 1) Synchronous Generator connected to infinite bus ππ = 0 (zero seq not present) iii) Double line to ground fault S= |πΈ||π| |π| β90° − π - |π|2 β90° |π| Real power output P= |πΈ||π| |π| Sin πΏ P ∝ Sin S Max Real power output P = ππππ₯ / πΏ = 90° ππππ₯ = ο πΌπΉ = πΌπ¦ +πΌπ΅ , πΌπ = 0 ο ππ¦ = ππ΅ = 0 ο ππ 0 = ππ 1 = ππ 2 = ο πΌπ 1 = πΈπ 1 π1ππ + ππ 3 |π| P = ππππ₯ Sin πΏ * πΌπ = 0 πΏ < 90° → machine is stable π2ππ π0ππ π2ππ +π0ππ ο πΌπ 2 = -πΌπ 1 π |πΈ||π| S > 90° → machine is unstable π0ππ 2ππ +π0ππ 2) ο πΌπ = 3πΌπ 0 = πΌπ ο ππ = πΌπ ππ = 3πΌπ 0 ππ 8) ο ο ο ο Syn Gen connected to load by Tr.line ππ = π΄ππ + π΅πΌπ Symmetrical fault πΌπ = πΆππ + π·πΌπ i) LLL fault Polar form of ABCD πΌπ = πΌπ = πΌπ¦ = πΌπ΅ πΌπ + πΌπ¦ + πΌπ΅ = 0 ππ = ππ¦ = ππ΅ πΌπ = πΌπ = πΌπ 1 , ππ 1 = 0 A = |A| β ∝ π B = |B| β π½ → Imp angle = πππ−1 π C = |C| βπΎ → admittance angle = 90° πΈπ πΌ 1ππ +ππΉ ο πΌπΉ = π ο Short circuit MVA = D = |D| β ∝ πππ΄π πππ ο ππ = πΌπ ππ = 0 Power at Receiving end: ππ = ii) LLLG fault |ππ ||ππ | |π΅| βπ½ − πΏ − |π΄| |π΅| |ππ |2 βπ½−∝ ο πΌπΉ = πΌπ + πΌπ + πΌπ΅ = 0 πΈ ο πΌπΉ = πΌπ = πΌπ 1 = π π 1 p.u 1ππ 73 73 00 77 34 7 POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ |B| = |Z| = |X|, π½ = π = 90° ππππ₯ = Real power o/p: ππ = |ππ ||ππ | |π΅| |π΄| Cos (π½ − πΏ) |π΅| |ππ |2 Cos βπ½−∝ ππππ₯ = Max Real power output: 3) iii) Inertia constant |π΄| πΏ πΉ (H) = Ratio: i) For short Tr. Line H= A = 1.0 = D ππ½ πππ΄ (or) sec ππ» M = 180π MJ – sec/elect deg ππππ₯ = |ππ ||ππ | |π§| - 1.0 |π§| Where, |ππ |2 cos π π → angular velocity (rad/sec) =0 ∝ → angular acceleration (rad/π ππ 2) X = √3R M →angular momentum πΏ = π½ = 60° H → Inertia constant (sec) ii) For long Tr. Line S → Rating of syn m/c πππ > ππ , |A| < 1.0, ∝ ≠ 0 Pa → accelerating power π π I →moment of inertia (kg - π2 ) = 2πππΏ ×π ππ/π πβ¦/ππ = π β¦/ππ 7) Swing equation of two machines Capacitor Compensation in the line: π πππ = ππΆ p.u πΏ πΏ = π½ = πππ−1 ( 5) π ππ» C=0 4) 1 2 πΌπ 2 πππππ‘ππ ππππππ¦ π π‘ππππ ππ π‘βπ π π¦π π/π π ππ‘πππ ππ π π¦π π/π iv) M = ππ MJ – sec/elect rad B = Z, ∝ = 0° πππππ₯ ππ₯ (ππ = ππ = π) ii) M = Iπ - |π΅| |ππ |2 Cos (π½− ∝) Calculation of |π|2 |π| i) Pa = M∝ πΏ = π½ = πππ−1(X / R) |ππ ||ππ | |π΅| |π| 6) Transient stability: ππ = ππππ₯ = SSSL (Steady State Stability Limit) ππππ₯ = |ππ ||ππ | = ππππ i) If two m/c’s are swing together ππΏ − ππΆ ) π πππ = π1 + π2 π»ππ = π»1 + π»2 Approximate power transfer equation ii) If two m/c’s are do not swing together For loss less line π1π2 πππ = π |A| = 1,0, ∝= 0 73 73 00 77 34 π2 πΏ ππ‘ 2 1 8 + π2 π»1π»2 π»ππ = π» 1 + π»2 POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ 8) Fault clearing angle: 4) Incremental fuel rate π(ππππ’π‘) ππΉ = π(ππ’π‘ππ’π‘) = ππ πΏπ = πππ −1 Where F → Fuel input (Btu/hr) π (πΏ − πΏ )+ π cos πΏ −ππ2 cos πΏ0 [ π π π ππππ − πππ ] 3 2 P → Power output (W) elect deg ππ πΏ0 = π π ππ−1 ( π ) ππ1 πΏπ = 180 - ο Incremental efficiency = ππΉ π π ππ−1 ( π )elett ππ3 deg 5) Where, πΏ0 = initial angle Penalty factor πΏπ = πππΏ πππ πΏπ = max swinging angle 1 1− πππΏ πππ → Incremental transmission loss at plant n. ππ1 = max mech power before fault Unit - VI Protective Relays ππ2 = max mech power during fault ππ3 = max mech power after fault 1) Time Multiplier Settings (TMS) TMS = Unit - V Economic Load Dispatch π‘ππππ’ππππ π‘(πππ=1) π‘ππππ’ππππ = Required time of operation π‘(πππ=1) = Time of operation when TMS =1 1) KVA loading on generator 2) =√ππ 2 + ππ 2 Plug setting Multiplier (PSM) πΉππ’ππ‘ ππ’πππππ‘ PSM=ππ’πππππ‘ π ππ‘π‘πππ×πΆπ π ππππππππ¦ πππ‘ππ ππ’πππππ‘ ×πΆ.π πππ‘πππ ππ = Generator active power 3) ππ = Generator reactive power 2) ππππ π’π ππ’πππππ‘ = π ππ‘ππ π ππππππππ¦ ππ’πππππ‘ × 100% ππ 2 + ππ 2 ≤ πΆπ 2 ππ πΆ.π πΆπ → pre specified value 3) Current setting 4) πππππ ≤ ππ ≤ πππππ₯ Torque equation T = πΎ1 πΌ 2 +πΎ2 π 2 +πΎ3 VI cos(π-π) + k πππππ , πππππ₯ → min and max active power generation of a source Where I→RMS value of current V→ RMS value of voltage πππππ ≤ ππ ≤ πππππ₯ π →angle b/w v and I πππππ , πππππ₯ → min and max reactive power generation of a source π → maximum torque angle K→Restraining Torque 73 73 00 77 34 9 POWER SYSTEMS MADURA COACHING #62, Rakame Complex, T.P.K Road, Madurai-625011 ______________________________________________________________________________ 6) Resistance Switching πΎ1, πΎ2 , πΎ3 →Relay constant 5) i) Over current relay T = πΎ1 πΌ 2 -K ii) Directional relay = T = πΎ3 VI cos(π-π) – k iii) Impedance relay T = πΎ1 πΌ 2 -πΎ2 π 2 1 For relay operation z = π πΌ 1 f = 2π √πΏπΆ − πΎ <√ πΎ1 2 7) 1 4π 2 πΆ 2 Transient Oscillation iv) Reactance relay If Resistance connected across circuit Breaker T = πΎ1 πΌ 2 -πΎ3 VI sinπ; π = 90° π πΎ For relay operation πΌ sinπ = X < πΎ1 v) Mho relay T = πΎ3 VI cos(π-π) –πΎ2 π 2 π 1 πΏ 1 πΏ ο R = 2 √πΆ → No oscillation 3 ο R > 2 √πΆ → Oscillation πΎ For relay operation πΌ = Z< πΎ3 cos(π-π) 2 Unit - VII Circuit Breakers 1) Restriking voltage V = E (1 – cos 2) Maximum value of 1 )t) √πΏπΆ = 2× πΈππππ Restriking voltage πΈππππ = peak value of system voltage 3) Natural frequency of oscillation ππ = 4) 1 √πΏπΆ Rate of rise of restriking voltage RRRV = πΈππ sinππ π‘ 5) Maximum value of RRRV = πΈππππ π€π 73 73 00 77 34 10 POWER SYSTEMS