08/10/2012CHAPTER 3 Engr. Dr. Kok Boon Ching 2012@JEK/FKEE Introduction er 3 ¾ ¾ Voltage Ranges 42803 – Chapte ¾ Definition of Voltage Drop ¾ Cable Impedances ¾ Transformer Voltage Drop ¾ Voltage Drop Due to Motor Starting BEX 4 ¾ Symmetrical S l and dAAsymmetricall Faultl C Currents ¾ Equivalent System Impedance ¾ Short Circuit Analysis in Three-phase Systems ¾ Short Circuit Analysis in Single-phase Systems 2 1 08/10/2012 Voltage drop on electrical power distribution er 3 systems is mainly caused by cables, cables BEX 42803 – Chapte transformers, and motors. Voltage drop happens when load current (Ib) flows through a conductor or transformer having a finite impedance. Severe voltage g dropp will result in motor failures, dimming of lamps, and CPU shutdown. Voltage drop calculation is important to system designer for maintaining nominal voltage at servicing sides. 3 According to 17th Edition of IEE Wiring Regulations er 3 (BS7671: 2008, Table 12A), the voltage drop 42803 – Chapte between the origin of an installation and any load point should not be greater than the values in Table 12A expressed with respect to the value of the nominal voltage of the installation. Max. 100m only. Increase 0.005% per meter if beyond 100m. BEX 4 T bl 12 A – Voltage Table V lt d drop Lighting Other uses i. Low voltage installation supplied directly from a public low voltage 3% 5% distribution system ii. Low voltage installation supplied 6% 8% from private LV supply 4 2 08/10/2012 er 3 VD(R) BEX 42803 – Chapte VD(X) VSEND θ VREC jXSIA IARA IA 5 Approximation method: er 3 Vdrop ≈ I b × [R L × cosθ − X L × sinθ ] 42803 – Chapte RL = circuit resistance in Ohms XL = circuit reactance in Ohms Ib = design current/ line current BEX 4 θ = phase h angle l off line l current If VA = system voltage, Vdrop %Vdrop = × 100% VA 6 3 08/10/2012 Determine the percentage voltage drop along er 3 a 415V, 415V three-phase feeder, feeder 85ft in length, length BEX 42803 – Chapte consisting of one 400 THW (Thermoplastic Heat and Water Resistant Insulated Wire) copper conductor per phase. The current is 350A at 0.85 PF lagging. Assume steel conduit. 7 From the table of 600V cables, resistance = er 3 0 035Ω/1000ft reactance = 0.049Ω/1000ft. 0.035Ω/1000ft, 0 049Ω/1000ft 42803 – Chapte 0.035 RL = × 85 = 0.002975Ω 1000 0.049 XL = × 85 = 0.004165Ω 1000 θ = − cos −1(0.85) = −31.79° Vdrop = 350 A × [0.002975Ω × cos((−31.79°) − 0.004165Ω × sin(( −31.79°)] = 1.65V BEX 4 1.65V %Vdrop = × 100% = 0.69% 239.6V 8 4 If Ib = 120A.95. tabulated (mV/A/m) z × I b × l × cos θ Vdrop = volts 1000 For AC circuits using conductors of 25mm2 or greater greater. what is the percentage voltage drop? BEX 4 10 5 . and (mV/A/m)z = 1. length = 27m. [tabulated (mV/A/m)r × cos θ + tabulated (mV/A/m)x × sin θ ] × I b × l Vdrop = volts 1000 Note: Refer to Tabulated Table of Voltage Drop (17th IEE Regulations) 9 A 415V three-phase AC circuit is wired in a er 3 four-core armoured cable to BS5467 having 42803 – Chapte XLPE insulation and aluminum conductors of 35mm2 cross-sectional area. 08/10/2012 Tabulated mV/A/m values: er 3 tabulated (mV/A/m) z × I b × l Vdrop = volts BEX 42803 – Chapte d 1000 Taking account of load power factor (for AC circuits using conductors of 16mm2 or less). 95 ×120 × 27 Vdrop = V = 6. 08/10/2012 1.32 %Vdrop = ×100% = 2.64% 415 / 3 11 Three-phase voltage drop calculations is er 3 referred to as the “one-way” impedance one-way impedance.32V er 3 1000 BEX 42803 – Chapte 6. 42803 – Chapte The ohmic cable impedances: Resistance in Ohms/1000ft RL = × (Cable length in ft) 1000 Reactance in Ohms/1000ft XL = × (Cable length in ft) 1000 BEX 4 12 6 . 08/10/2012 Single-phasevoltage drop considers the load er 3 current flowing from the source to the load BEX 42803 – Chapte and back to the source. the ohmic cable impedances are calculated as: Resistance in Ohms/1000ft RL = 2× × (Cable length in ft) 1000 Reactance in Ohms/1000ft XL = 2 × × (Cable length in ft) 1000 13 Determine the voltage drop along a 240V. 42803 – Chapte consisting of #10 AWG THW copper conductor. Thus. Assume PVC conduit. BEX 4 14 7 . The load current is 13∠-25°A. er 3 length single-phase branch circuit 135ft in length. 324Ω × cos(−25°) − 0. Vdrop BEX 4 %Vdrop = × 100% VLS 16 8 .0135Ω × sin(−25°)] = 3.89V 3.89V %Vdrop = × 100% = 1. resistance = er 3 1.050Ω/1000ft. 0 050Ω/1000ft BEX 42803 – Chapte 1.0135Ω 1000 Vdrop = 13 A × [0.324Ω 1000 0.2Ω/1000ft. 1 2Ω/1000ft reactance = 0. 08/10/2012 From the table of 600V cables.62% 240 240V 15 The voltage drop through the transformer er 3 is using approximation method is.050 XL = 2× × 135 = 0. 42803 – Chapte Vdrop ≈ I LS × [R TR × cosθ − X TR × sinθ ] In percentage. 2 RL = 2 × × 135 = 0. 82 lagging power factor. by BEX 42803 – Chapte 1 ⎡ (%R)(Secondary line voltage)2 ⎤ R TR = ⎢ ⎥ 100 ⎢ Transformer voltampere rating ⎥ ⎣ ⎦ 1 ⎡ (%X)(Secondary line voltage)2 ⎤ X TR = ⎢ ⎥ 100 ⎢ Transformer voltampere rating ⎥ ⎣ ⎦ If given in %ZTR and X/R ratio. three-phase 42803 – Chapte transformer having an impedance of 4%. 3300-415Y V V. 08/10/2012 Three-phase transformer impedances er 3 reflected to the low-voltage side are given by. and an X/R ratio of 1. The transformer is operating at full load. BEX 4 18 9 .8. 0. ratio θ = tan -1(X/R) %R TR = %ZTR × cosθ %X TR = %ZTR × sinθ 17 Determine the voltage drop through a er 3 115kVA 115kVA. 61V %Vdrop = ×100% = 2. %RTR = 4% x cos(60. 08/10/2012 The impedance angle.61V 8.05242Ω 100 ⎣ 115kVA ⎦ 19 Full-load current of the transformer.92°)] = 8.07% 415V 415V BEX 4 20 10 .02905Ω × cos(−34.05242Ω × sin( −34. er 3 115kVA 42803 – Chapte I LS = ∠ − cos −1 (0.02905Ω 100 ⎣ 115kVA ⎦ 1 ⎡ (3.94% %XTR = 4% x sin(60.94%)(415V ) 2 ⎤ RTR = ⎢ ⎥ = 0.95° 95° BEX 42803 – Chapte Transformer %RTR and %XTR.50%)(415V ) 2 ⎤ X TR = ⎢ ⎥ = 0.95°) = 1. 1 ⎡ (1. Vdrop = 160 A × [0.8) (1 8) = 60 60.92°) − 0. er 3 θ = tan-1 (1.92° A 3 × 415V Voltage drop.82) = 160∠ − 34.95°) = 3.50% RTR and XTR. 415V. 2 common methods to determine voltage dropp due to the motor starting: g Constant Impedance Constant Current 21 A 50HP. starting BEX 42803 – Chapte The voltage drop is large when starting large motors applied to systems having a relatively high source impedance.01+j0. 08/10/2012 Voltage drop or voltage dips occurs due to er 3 large motor starting. Assume a locked-rotor power factor of 35% lagging. Calculate the percentage voltage drop during starting using (a) the constant BEX 4 i impedance d and d (b) the th constant t t currentt representations.02 ohms/phase. code letter G induction motor er 3 is to be started with full voltage applied 42803 – Chapte from a 415V/240V system whose equivalent impedance is 0. 22 11 . 315kVA I LR = = 438.3 kVA/HP x 50HP = 315 kVA The locked-rotor current.35) = 110. P = (315kVA) x (0.1913Ω ⎟ 3 ⎜⎝ 438.5122 ⎞ VM = 240∠0°⎜⎜ ⎟⎟ = 230.1913 + j 0.24°V ⎝ 0.232 ⎟⎠ Voltage at motor terminal. er 3 1 ⎛ 110.3 (take the er 3 worst case).91% 240V 24 12 .25 kW Q = (315kVA) x [sin(cos-1(0.35))] = 295.23 A 3 × 415V The h active andd reactive power during d starting.1kVAr 23 Locked-rotor R and X.1kVAr ⎞ X = × ⎜⎜ ⎟ = 0.02 ⎠ Voltage drop. 08/10/2012 (a) The locked-rotor kVA/HP is 6.62∠0.232 ⎠ 1 ⎛ 295.5122Ω 3 ⎝ 438. kVALR = 6.25kW ⎞ 42803 – Chapte R= ×⎜ ⎟ = 0. BEX 4 ⎛ 0.62V %Vdrop = × 100% = 3.5122 + 0.01 + j 0. The locked rotor kVA during locked-rotor BEX 42803 – Chapte starting.1913 + j 0. 240V − 230. 8V. %Vdrop = (9.23 A × [0. it is possible to have an approximation value of voltage drop due to motor starting by assuming that the voltage drop is in-phase with the source voltage. the Vdrop.01Ω × cos(−69. 08/10/2012 (b) Constant current. known 42803 – Chapte In this case.51°)] = 9. the locked-rotor power factor er 3 (cosθ) of the motor may not be known. For p previous Examplep 5.05% 240V 25 In some cases.51° 438 23∠ 69 51° BEX 42803 – Chapte Voltage drop (using approximation method). Vdrop = 438.23∠-69. BEX 4 Vdrop = I x Z = 438.02Ω × sin( −69.23A x |0.74V 9. The %Vdrop.74V %Vdrop = × 100% = 4.02| = 9.8V/240V) x 100% = 4. er 3 IM = 438.01+j0.51°) − 0.08% 26 13 . 5%)(240V ) 2 ⎤ X TR = ⎢ ⎥ = 0.13 RL = 2 × × (120 ft ) = 0. 415-240V 30kVA R=1.03456Ω 100 ⎣ 30kVA ⎦ 1 ⎡ (1.0312Ω 1000 0. 0.8% X=1.0101Ω 1000 28 14 .042 X L = 2× × (120 ft ) = 0. below The total load is BEX 42803 – Chapte [email protected]%)(240V ) 2 ⎤ RTR = ⎢ ⎥ = 0.0288Ω 100 ⎣ 30kVA ⎦ BEX 4 C bl Cable/Wire: 0.5% Service #3/0 AWG aluminum PVC conduit 120ft 40A Service entrance panel 8kVA.85 lagging power factor.85 PF lagging 27 Step1 – Determine all system impedances er 3 Transformer: 42803 – Chapte 1 ⎡ (1. 08/10/2012 Determine the voltage drop at the service er 3 panel for the system below. 79°) − 0. starting at source source.48V %Vdrop = × 100% = 0.79°)] = 1. 42803 – Chapte Transformer: Vdrop ≈ 33.0101Ω × sin( −31.62% 240V BEX 4 Cable/Wire: Vdrop ≈ 33.85)° = 33.06V %Vdrop = × 100% = 0.85 PF lagging.44% 240V 30 15 .06V 1.79°)] = 1.0312Ω × cos(−31.33∠ − 31.33 A × [0. 8kVA I= = 33. 08/10/2012 Step2 – Determine load supplied at the end er 3 of each segment or portion of the system. system BEX 42803 – Chapte The loading is 8kVA@0. Step 3 – Determine the load current magnitude and phase angle.33 A∠ − cos −1 (0.48V 1.0288Ω × sin( −31.79° A 240V 29 Step4 – Calculate the %Vdrop along each er 3 segment of the circuit circuit.79°) − 0.33 A × [0.03456Ω × cos(−31. er 3 starting at the source to the point of BEX 42803 – Chapte interest. 08/10/2012 Step 5 – Add the %Vdrop along each segment.44% Total :11.06% 06% 31 Determine the %Vdrop at the Main Distribution er 3 Panel (MDP) and at the end of the Service 42803 – Chapte Panel (SP) for the system shown below.62% Cable/Wire : 0. The total voltage drop at the panel is: Transformer : 0. BEX 4 32 16 . 5%.5%. 08/10/2012 Transformer TR1 800kVA 6600 – 1000V R = 1. X = 5% er 3 30ft two 400 kcmil copper/phase.049 XL = × (30 ft ) = 0.035 RL = × (30 ft ) = 0.00105Ω 1000 0.0625Ω 100 ⎣ 800kVA ⎦ BEX 4 C bl (400 Cable 00 kcmil): k l 0.9 lagging PF 1500A 400A 250A 35ft #8 AWG aluminum.85 lagging PF 33 Step1 – Determine all system impedances er 3 Transformer (TR1): 42803 – Chapte 1 ⎡ (1.5%)(1000V ) 2 ⎤ RTR1 = ⎢ ⎥ = 0. 0. X = 4% 12ft #12 AWG copper.00147Ω 1000 34 17 . steel conduit BEX 42803 – Chapte 2000A MDP 600kVA. 0. aluminum conduit 150A SP 100A 20kVA.01875Ω 100 ⎣ 800kVA ⎦ 1 ⎡ (5%)(1000V ) 2 ⎤ X TR1 = ⎢ ⎥ = 0. steel conduit Transformer TR2 40kVA 1000 – 415V R = 3. 002275Ω 1000 Transformer (TR2): 1 ⎡ (3.9 lagging PF.5%)(415V ) 2 ⎤ RTR 2 = ⎢ ⎥ = 0.0 42803 – Chapte RL = × (12 ft ) = 0. 08/10/2012 Cable (#8 AWG): er 3 1 .000648Ω 1000 Step2 – Determine load supplied at the end BEX 4 of each segment or portion of the system.85 lagging PF.054 XL = × (12 ft ) = 0.3 BEX 42803 – Chapte RL = × (35 ft ) = 0.1722Ω 100 ⎣ 40kVA ⎦ 35 Cable (#12 AWG): er 3 2 .065 XL = × (35 ft ) = 0.1507Ω 100 ⎣ 40kVA ⎦ 1 ⎡ (4%)(415V ) 2 ⎤ X TR 2 = ⎢ ⎥ = 0. 36 18 .0455Ω 1000 0. system Loading until MDP is [email protected]Ω 1000 0. Loading from MDP until SP is 20kVA@0. 84° A Cable (400 kcmil) 173.85)° A = 11.82∠ − 31. 20kVA I= = 27.79° A BEX 4 Cable (#12 AWG) 27.90)° A = 346.79° A 38 19 .82∠ − 31.85)° A = 27. 600kVA I= = 346.41∠ − 25.55∠ − 31. 20kVA I= = 11.21∠ − 25. 08/10/2012 Step 3 – Determine the load current er 3 magnitude and phase angle.41∠ − 25.79° A Transformer (TR2) 27.82∠ − 31.82∠ − cos −1 (0.79° A 3 × 415V 37 Summary of load currents: er 3 42803 – Chapte Component Load Current (A) Transformer (TR1) 346.79° A 3 × 1000V Through Transformer TR2.41∠ − cos −1 (0.84° A / conductor Cable (#8 AWG) 11. angle BEX 42803 – Chapte Through Transformer TR1.55∠ − cos −1 (0.84° A 3 × 1000V Through Cable (#8 AWG).55∠ − 31. 08% 577.79°) − 0.01875Ω × cos(−25. 08/10/2012 Step4 – Calculate the %Vdrop along each er 3 segment of the circuit circuit.09V BEX 4 6.21A × [0.82 A × [0.79°)] = 0.53% 1000V Cable (400 kcmil): Vdrop ≈ 173.1722Ω × sin(−31.84°) − 0.79°)] = 6.00105Ω × cos(−25.35V Transformer (TR2): Vdrop ≈ 27.002275Ω × sin( −31.0625Ω × sin( −25.27V %Vdrop = ×100% = 0. BEX 42803 – Chapte Transformer (TR1): Vdrop ≈ 346.84°)] = 0.28V %Vdrop = ×100% = 1.28V 15.47% 415V 40 20 .84°)] = 15.27V 0.79°) − 0.0455Ω × cos(−31.84°) − 0.09V %Vdrop = ×100% = 1.00147Ω × sin( −25.46V 0.1507Ω × cos(−31.35V 39 Cable (#8 AWG): er 3 42803 – Chapte Vdrop ≈ 11.55 A × [0. starting at source source.46V %Vdrop = ×100% = 0.05% 577.41A × [0. 42803 – Chapte The total voltage drop at service panel is: Transformer (TR1) : 1.79°) − 0.82 A × [0.05% BEX 4 Cable (#8 AWG) :00. er 3 interest starting at the source to the point of interest.58V 0.024Ω × cos(−31.53% Cable (400 kcmil) : 0.47% Cable (#12 AWG) : 0.08% 08% Transformer (TR2) : 1.24% Total : 3.79°)] = 0.6V 41 Step 5 – Add the %Vdrop along each segment. 08/10/2012 Cable (#12 AWG): er 3 BEX 42803 – Chapte Vdrop ≈ 27.58V %Vdrop = ×100% = 0.24% 239.37% 42 21 .000648Ω × sin( −31. The normal circuit path includes the phase and neutral conductors. 44 22 . It does not include equipment grounding conductors. Normal motor current varies from low values (under light loading) to medium values (under medium loading) to maximum values (under maximum loading). or load. BEX 4 Normal current flows only in the normal circuit path. 08/10/2012 er 3 BEX 42803 – Chapte Types of current Normal Overload Short-circuit Ground-fault current current current current 43 Normal. current may be defined as the er 3 current specifically designed to be drawn by a 42803 – Chapte load under normal operating conditions. The maximum value is limited by the maximum short-circuit current available on the system at the fault point. are only temporary. such as motor starting currents (or locked-rotor current). and thus is considered to be an abnormal current. 45 Short-circuit current is greater than locked- er 3 rotor current and may range upwards of 42803 – Chapte thousands of amperes. or low line voltage. 08/10/2012 Overload current is greater in magnitude than er 3 full load current and flows only in the normal BEX 42803 – Chapte circuit path. single-phasing. Some overload currents. Short-circuit current may y be further classified BEX 4 as bolted or arcing. Overload current is greater in magnitude than full-load amperes but less than locked-rotor amperes. Large amounts of short- circuit current will flow into a bolted fault than the arcing fault. It is commonly caused by overloaded equipment. 46 23 . 47 Synchronous generators . path BEX 42803 – Chapte Ground-fault current flow in the equipment grounding conductor for low-voltage systems. 08/10/2012 Ground-fault current consists of any current er 3 which flows outside the normal circuit path.short short-circuit circuit current decays BEX 4 very quickly. 48 24 . bolted.delivers short-circuit current into the fault until the motor completely stops Induction motors .when a short-circuit er 3 occurs downstream of a synchronous generator. Ground-fault current on low-voltage systems may be classified as leakage.Transformer impedances will also limit the amount of short-circuit current from the utility generators. In medium. ground- fault current may return to the source through the earth. or arcing. Synchronous motors .and high-voltage systems. Supply transformers . 42803 – Chapte it may continue to produce output voltage and current. 08/10/2012 er 3 BEX 42803 – Chapte Synchronous Generator Induction Motor Synchronous Motor 49 er 3 42803 – Chapte Totally Symmetrical Current BEX 4 Totally Asymmetrical Current Partially Asymmetrical Current 50 25 . The asymmetrical short circuit current is the actual current that flows during a fault condition. in the same circuit. 08/10/2012 “Symmetrical" and “Asymmetrical” are terms er 3 used to describe the symmetry of the short- BEX 42803 – Chapte circuit current waveform around the zero axis. If a short-circuit. It is applicable only for balanced three- phase power system and can be calculated as the total line-to-neutral voltage over the total impedances on the power system. the resulting short-circuit current will be totally symmetrical. occurs at the zero of the voltage waveform. 51 The symmetrical short circuit current consists only er 3 the pure AC component inside its sinusoidal 42803 – Chapte waveform. 52 26 . If a short-circuit occurs in an inductive reactive circuit at the peak of the voltage waveform. The amount of ‘DC offset’ or asymmetry depends on the point when the fault occurs. the resulting short-circuit current will be totally asymmetrical. It BEX 4 consists of DC and AC components that contribute to a certain amount of ‘DC offset’ in the waveform immediately after the initiation of the fault. Instantaneous peak short circuit current Asymmetrical short circuit current Symmetrical y short circuit current 53 er 3 jXL i(t) R 42803 – Chapte t=0s + Vm sin(ωt + θ ) Fault - Line-to-Neutral Equivalent Circuit BEX 4 [ i (t ) = 2 ⋅ I rms sin(ωt − θ Z ) + sin(θ Z ) ⋅ e − (ωR / X ) t ] Vm ⎛ X ⎞ I rms = θ Z = tan −1 ⎜ ⎟ 2 ⋅ ZS ZS = R2 + X 2 ⎝ R ⎠ 54 27 . 08/10/2012 Theinstantaneous peak short circuit current is er 3 the maximum peak instantaneous fault current BEX 42803 – Chapte on the asymmetrical short circuit current waveform. It is a function of X/R of the system. 47kV distribution er 3 0 4 + j1.cycle rms multiplying factor = rms Symmetrical short circuit current 55 The source impedance at a 12.1 2 = ( rms half .cycle factor) × I rms The rms half-cycle factor: T 1 2 T ∫0 i (t ) dt First half . (b) the maximum peak instantaneous value of fault current.4 j1 5 ohms per phase. and (c) the rms value of the half-cycle fault current if a balanced three-phase fault occurs. phase 42803 – Chapte Calculate (a) the rms fault current. BEX 4 56 28 .5 substation bus is 0. 08/10/2012 First half-cycle asymmetrical fault current: er 3 BEX 42803 – Chapte I rms . 47 kV BEX 42803 – Chapte VLN = = 7.5/0.2kV 3 The rms symmetrical fault current: 7.0892 – 1. the instantaneous peak factor is determine by interpolation: = (2.4 = 3.75 – 3.4 2 + 1.52 ) 57 (b) The system X/R ratio = 1.75 er 3 42803 – Chapte From table.0543)(4638A) = 9528A 58 29 .2kV I rms = = 4638 A (0.0) + 1.9495)(3.9495 = 2. 08/10/2012 (a) The line-to-neutral voltage: er 3 12.0543 BEX 4 The maximum peak instantaneous value of fault current is Ip = (2. table BEX 42803 – Chapte = (1.172 The rms half-cycle asymmetrical fault current is.0) + 1.1/2 = (1.75 – 3. Common system impedances – equivalent system. is Irms. 08/10/2012 (c) The rms half-cycle multiplying factor is er 3 determine by interpolation from table. need to be reflected to its low voltage BEX 4 60 30 . transformers.115)(3. All impedances placed before transformer g side.115 = 1.172)(4638A) = 5436A 59 To determine short circuit current. etc. the total er 3 impedances of the system to the fault point 42803 – Chapte must be established.191 – 1. cables. current the rms BEX 42803 – Chapte short circuit current at a particular fault point is calculated as: Line . 61 Determine the RMS symmetrical. Half-cycle factor is used to calculate the asymmetrical fault current. RMS er 3 asymmetrical and peak short circuit 42803 – Chapte magnitudes for a three-phase fault occurring at (a) F1 and (b) F2 for the power system shown in Figure below.neutral voltage I rms = Z total The X/R ratio is used to determine the instantaneous peak factor and half-cycle factor. BEX 4 62 31 .to . 08/10/2012 Since the three-phase fault condition results in er 3 a balanced set of short circuit current. 08/10/2012 Equivalent system 3-phase SC MVA = [email protected]°) = 0. X/R = 5 er 3 50ft three 400 kcmil copper/phase.57° Rsys = 0.1675Ω 65MVA Impedance angle. steel conduit 400A 250A F2 63 Equivalent system impedance.5 10ft 250 kcmil copper. X/R = 3 Transformer TR1 750kVA 3300 – 1100V Z = 5. θ = tan −1 (3) = 71. er 3 42803 – Chapte (3300V ) 2 Z sys = = 0.1675 × cos(71. X/R = 1.3kV. steel conduit Transformer TR2 75kVA 1100 – 415V Z = 1.57°) = 0.05295Ω BEX 4 X sys = 0.75%.1675 × sin(71.8%. steel conduit BEX 42803 – Chapte 1000A F1 1000A 400A 250A 10ft #4/0 AWG copper.1589Ω 64 32 . 01766Ω ⎝ 3300V ⎠ %R and %X of TR1.69° % RTR1 = 5.01823Ω 100 ⎣ 750kVA ⎦ 1 ⎡ (5. er 3 42803 – Chapte 1 ⎡ (1.13% % X TR1 = 5.00058Ω ⎝ 3 ⎠ ⎣ 1000 ⎦ R = 0.1589 × ⎜ ⎟ = 0.69°) = 1.005883Ω ⎝ 3300V ⎠ 2 ⎛ 1100V ⎞ X sys ( LS ) = 0.049 ⎤ X = 0. BEX 4 ⎛ 1 ⎞ ⎡ 0.75% × sin(78.09099Ω 100 ⎣ 750kVA ⎦ Rc and Xc of 400 kcmil cable (From Table). TR1 θ = tan −1 (5) = 78.049Ω / 1000 ft Xc = ⎜ ⎟ ⎢ ⎥ × 50 ft = 0.64% 65 RTR1 and XTR1 of TR1.035 ⎤ Rc = ⎜ ⎟ ⎢ ⎥ × 50 ft = 0.64%)(1100V ) 2 ⎤ X TR1 = ⎢ ⎥ = 0.035Ω / 1000 ft ⎛ 1 ⎞ ⎡ 0. TR1 BEX 42803 – Chapte 2 ⎛ 1100V ⎞ Rsys ( LS ) = 0.05295 × ⎜ ⎟ = 0.75% × cos(78.69°) = 5.13%)(1100V ) 2 ⎤ RTR1 = ⎢ ⎥ = 0. 08/10/2012 The equivalent system impedance referred to er 3 the low voltage side of TR1.00082Ω ⎝ 3 ⎠ ⎣ 1000 ⎦ 66 33 . 005883 j0. 42803 – Chapte 635.00058 + j0.0892 = 2.1325 2 1325 BEX 4 Peak instantaneous current. er 3 Equivalent system :00.024783 + j0.070A 68 34 .00082 Total : 0.01766 005883 + j0 01766 BEX 42803 – Chapte Transformer (TR1) : 0. = (2.42 67 The RMS symmetrical short circuit current at er 3 F1 F1.10947 Z sys = 0. Ip = (2.0) + 2.660 A 0.01832 + j0. 08/10/2012 (a) The total impedance to fault at F1.1122Ω [X / R]sys = 4.1924 – 2.09V I RMS = = 5.1325) x (5.42 – 4.1122Ω Instantaneous peak factor (interpolation).0892) x (4.09099 Cable (400 kcmil) : 0.660A) = 12. 0) 4 0) + 1.00063Ω ⎣ 1000 ⎦ R = 0. 08/10/2012 Half-cycle factor. TR2 2 ⎛ 415V ⎞ Rc = 0.191 1 191 BEX 42803 – Chapte = 1.00051 × ⎜ ⎟ = 0.051⎤ X = 0.051Ω / 1000 ft Xc = ⎢ ⎥ × 10 ft = 0.00009 Ω ⎝ 1100V ⎠ 2 ⎛ 415V ⎞ Xc = 0.263 (1 263 – 1.2212) x (5.912A 69 (b) Rc and Xc of #4/0 AWG cable (from table). IRMS/1/2 = (1. er 3 = (1.191) (4.00051Ω ⎣ 1000 ⎦ Rc and Xc (#4/0 AWG) referred to low voltage BEX 4 side of TR2.063Ω / 1000 ft ⎡ 0.00063 × ⎜ ⎟ = 0.2212 Half-cycle RMS asymmetrical current.00007 Ω ⎝ 1100V ⎠ 70 35 .063 ⎤ 42803 – Chapte Rc = ⎢ ⎥ × 10 ft = 0.42 1 191) x (4 42 – 4. er 3 ⎡ 0.660A) = 6. 054Ω / 1000 ft ⎡ 0.TR 2 = 0.02296Ω 100 ⎣ 75kVA ⎦ 1 ⎡ (1.31°) = 1.024783 × ⎜ ⎟ = 0.5) = 56.10947 × ⎜ ⎟ = 0.8% × sin(56.00054Ω ⎣ 1000 ⎦ R = 0.054 ⎤ 42803 – Chapte Rc = ⎢ ⎥ × 10 ft = 0.052Ω / 1000 ft Xc = ⎢ ⎥ × 10 ft = 0.TR 2 = 0. er 3 θ = tan t −1 (1.50% 1 ⎡ (1.01558Ω ⎝ 1100V ⎠ 72 36 . TR2 2 ⎛ 415V ⎞ Rsys .03445Ω 100 ⎣ 75kVA ⎦ 71 The Rc and Xc of 250 kcmil cable.052 ⎤ X = 0.00052Ω ⎣ 1000 ⎦ R and X up to point F1 reflected to the low BEX 4 voltage side of TR2.00% % X TR 2 = 1.50%)( )(415V ) 2 ⎤ X TR 2 = ⎢ ⎥ = 0.31°) = 1.8% × cos(56.31° BEX 42803 – Chapte % RTR 2 = 1.00%)(415V ) 2 ⎤ RTR 2 = ⎢ ⎥ = 0. er 3 ⎡ 0.00353Ω ⎝ 1100V ⎠ 2 ⎛ 415V ⎞ X sys . 08/10/2012 %R and %X of TR2. 172A) = 7. 42803 – Chapte 239. Ip = (1.194A 74 37 .5122 = 1.5122) x (1.00007 Transformer (TR2) : 0.87 – 1.00054 + j0.00009 + j0.05743Ω [X / R ]sys = 1.0) + 1.00353 0 00353 + j0 j0.03445 Cable 250 kcmil : 0.7560 – 1.02712 + j0.172 A 0.7243) x (4.01558 01558 BEX 42803 – Chapte Cable #4/0 AWG : 0.87 73 The RMS symmetrical short circuit current at er 3 F2 F2. 08/10/2012 The total impedance to F2. er 3 Equivalent up to F1 : 0.00052 Total : 0.05743Ω Instantaneous peak factor (interpolation).05062 Z sys = 0.02296 + j0. = (1.6V I RMS = = 4.7243 1 7243 BEX 4 Peak instantaneous current. the only possible fault er 3 240V is line-to-ground fault across 240V.87 1 002) x (1 87 – 1.002 1 002 BEX 42803 – Chapte = 1.042 (1 042 – 1. 42803 – Chapte The short circuit current is calculated as: Line .0368) x (4.to .326A 75 In single-phase system.0) 1 0) + 1.neutral voltage 240V I rms = = Z total Z total BEX 4 76 38 . 08/10/2012 Half-cycle factor.002) (1. IRMS/1/2 = (1. er 3 = (1.172A) = 4.0368 Half-cycle RMS asymmetrical current. 4 er 3 50kVA 42803 – Chapte 6600 – 415V R = 1. rms er 3 asymmetrical. 08/10/2012 Determine the rms symmetrical. 77 Equivalent system Three-phase: MVA = 65MVA@6. line-to-ground fault occurring at point F1 for the power system shown below.6kV. steel conduit F1 78 39 .9% X = 2.2% 240V Service 100ft #4/0 AWG AL. X/R = 2. asymmetrical and peak short circuit current BEX 42803 – Chapte magnitudes for a single-phase. Steel conduit BEX 4 240V 50ft #12 AWG copper. X/R = 3 Single-phase: ILG = 2kA. 4°) = 1.9053 × sin(67.1φ = (1 / 3)(0.1φ = 1.7322Ω X sys . er 3 BEX 42803 – Chapte 6600V / 3 6600V Z sys .07578Ω) = 0.7590Ω × ⎜ ⎟ = 0.4° Rsys . 42803 – Chapte 1 ⎡ (1.4°) = 0.9053 × cos(67. RTR .006978Ω ⎝ 6600V / 3 ⎠ 79 Transformer R and X referred to low voltage er 3 side side.02526Ω 80 40 .1φ = 1.7590Ω R fl i system R and Reflecting d X to the h 2240V.9053Ω 2000 A θ = tan −1 (2. 0 2 ⎛ 240V ⎞ .4) = 67.9%)(415V ) 2 ⎤ RTR = ⎢ ⎥ = 0.002905Ω ' Rsys ⎝ 6600V / 3 ⎠ 2 ⎛ 240V ⎞ X ' sys .02181Ω X TR .2%)(415V ) 2 ⎤ X TR = ⎢ ⎥ = 0.06544Ω 100 ⎣ 50kVA ⎦ 1 ⎡ (2.1φ = 0.1φ = = 1.07578Ω 100 ⎣ 50kVA ⎦ BEX 4 Transformer R and X for the half-windinghalf winding (or single-phase) condition.1φ = 1.7322Ω × ⎜ ⎟ = 0.06544Ω) = 0. 08/10/2012 The equivalent system impedance.1φ = (1 / 3)(0. 2Ω 1000 0.0054Ω 1000 81 Total impedances to F1. 08/10/2012 The Rc and Xc of the #4/0 AWG AL cable.02181 + j0. er 3 0.244715 02 + j0.02526 Cable #4/0 : 0.1 BEX 42803 – Chapte Rc = 2 × ×100 ft = 0.2 + j0.02 + j0. er 3 42803 – Chapte System impedance : 0.0102Ω 1000 The Rc and Xc of the #12 AWG copper cable.047838 0 0 838 Z total = 0.0541 Xc = 2 × × 50 ft = 0.0 Rc = 2 × × 50 ft = 0.002905 + j0. 2.051 Xc = 2 × ×100 ft = 0.0054 BEX 4 Totall : 0.2 82 41 .02Ω 1000 0.2493 X / R ratio = 0.006978 Transformer : 0.0102 Cable #12 : 0. 7A) = 1. Ip = 1.361.5A The rms first half-cycle asymmetrical current. Irms. The peak current.0(962.1/2 = 1.4142(962. er 3 240V BEX 42803 – Chapte I rms = = 962.7 A 0.0.2493Ω The instantaneous peak factor = 1.4142. 08/10/2012 The rms symmetrical short circuit current. and the half-cycle rms factor = 1.7A 83 42 .7A) = 962.
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