* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Short-circuit current rating (SCCR) of industrial control panels
Induction motor wikipedia , lookup
Flexible electronics wikipedia , lookup
Electrical ballast wikipedia , lookup
Ground loop (electricity) wikipedia , lookup
Electric machine wikipedia , lookup
Immunity-aware programming wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Electric power system wikipedia , lookup
Mercury-arc valve wikipedia , lookup
Opto-isolator wikipedia , lookup
Transformer wikipedia , lookup
Stray voltage wikipedia , lookup
Buck converter wikipedia , lookup
Three-phase electric power wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Current source wikipedia , lookup
Integrated circuit wikipedia , lookup
Distribution management system wikipedia , lookup
History of electric power transmission wikipedia , lookup
Mains electricity wikipedia , lookup
Power engineering wikipedia , lookup
Surface-mount technology wikipedia , lookup
Transformer types wikipedia , lookup
Protective relay wikipedia , lookup
Ground (electricity) wikipedia , lookup
Surge protector wikipedia , lookup
Rectiverter wikipedia , lookup
Electrical substation wikipedia , lookup
Network analysis (electrical circuits) wikipedia , lookup
Alternating current wikipedia , lookup
Circuit breaker wikipedia , lookup
Residual-current device wikipedia , lookup
Fault tolerance wikipedia , lookup
Calculation of SCCR Ch 6,7: Faults and Fault Calculations 1 Short-circuit current rating (SCCR) of industrial control panels Introduction Article 409 on Industrial Control Panels was added to the NEC in its 2005 edition. This Article requires all Industrial Control Panels to be marked with a short-circuit current rating. The short-circuit current rating (SCCR) requirements for UL 508A came into force in April 2006. These changes impact control panel builders, OEMs and end users in a number of different ways: ● The correct choice of power circuit components of a control panel Specification of preferred device manufacturers Design and marking of panels Correct installation and modification of control panels 2 http://www.easypower.com/videos/intro-to-arc-flash-video.php http://www.easypower.com/videos/device-coordination-intro-video.php 3 As follows, the NEC defines two acceptable methods for labeling an industrial control panel’s short circuit current rating: 1. Short-circuit current rating of the listed and labeled assembly 2. Short-circuit current rating established utilizing an approved method Thus, the two options for panel designers and builders are: 1. Using the SCCR of a listed and labeled assembly, which requires testing the individual panel design and then recording the test result for each panel design. Note: The variety of possible combinations is usually very high, and this option therefore requires a lengthy test procedure. In addition, testing is a timeconsuming and costly undertaking, since it is “destructive” (testing is conducted until the product fails). After testing, the results must be recorded in the panel builder’s file. 2. Utilizing an approved method. This translates to applying the method described in UL 508A Second Ed., Dec. 2013, Supplement SB. 4 Note: The latest edition of the NEC (NEC Ed. 2014) refers to an “Informational Note in Article 409,” “Industrial Control Panels,” which addresses the short circuit current ratings of the standard UL508A Supplement SB, as an example for an approved calculation method. For many years, the UL508A Supplement SB has been the approved method for the calculation and determination of the short circuit current rating for industrial control panels, and one could proceed without further tests. It is important to note that the calculation method and the rules in the UL508A were changed in the latest edition, which was reissued in December 2013. The calculation method and all the latest changes will be explained step-by-step below. 5 The NFPA 70: National Electrical Code includes Article 409 on the Construction of Industrial Control Panels operating at 600 volts or less. Section 409.110 requires a short-circuit current rating (SCCR) to be marked on all industrial control panels. This rating must be based on the rating of a listed and labeled assembly or on another approved method for determining the rating. It also includes a fine print note (FPN) reference to UL 508A Supplement SB as an example of an approved method for determining the SCCR that may be marked on the panel. 6 UL508A is the safety standard for industrial control panels. NEC Article 409 references UL508A Supplement SB as an approved method for determining the SCCR of an industrial control panel. The specific method is outlined in Section SB4. Paragraph SB3.2.1 states that the primary short-circuit protective device for the control circuit is also included in the calculation for the SCCR for the power circuit. Therefore, the SCCR of the overcurrent protective devices (except for supplementary protectors recognized according to UL1077 or sets of supplemental fuses recognized according to UL248-13) are included in calculation of the SCCR of the control panel. Control circuit components on the load side of these devices are not included in calculation of the SCCR. 7 SB4.2 – Determining the short-circuit current ratings (SCCR) of individual power circuit components Determining the short circuit current ratings (SCCR) of individual power circuit components using three possible methods. Based on device markings or component instruction sheets Based on assumed short-circuit current ratings Based on device markings or component instruction sheets 8 9 10 11 12 13 14 15 16 17 18 19 NEC 2014 – Section 409.110 Marking An industrial control panel shall be marked with the following information that is plainly visible after installation: Short-circuit current rating of the industrial control panel based on one of the following: a. Short-circuit current rating of a listed and labeled assembly b. Short-circuit current rating established utilizing an approved method Informational Note: UL 508A, Standard for Industrial Control Panels, Supplement SB, is an example of an approved method. Exception to (4): Short-circuit current rating markings are not required for industrial control panels containing only control circuit components. 20 II. Overview of the UL508A Supplement SB for Calculating the SCCR for Industrial Control Panels Calculating the overall SCCR of an industrial control panel involves three essential steps: 1. Establishing the short circuit current ratings of the individual, relevant power circuit components 2. Applying current limiting components to modify the SCCR within a portion of a circuit in the panel 3. Determining the overall SCCR of the industrial control panel Step 1: Establishing the short circuit current ratings of the individual, relevant power circuit components. 21 Which components are relevant? All power circuit components, including the disconnect switches, branch circuit protective devices, branch circuit fuse holders, load controllers, motor overload relays, terminal blocks, and bus bars Exempt from this rule are the following components: • Power transformers • Reactors • Current transformers • Dry-type capacitors • Resistors • Varistors • Voltmeters • “S” contactor of a wye-delta motor controller Enclosed air conditioners, multi-motor, and combination load equipment are also exempt from having a SCCR rating, as long as one of the following conditions is met: • The equipment is cord-and-attachment-plug connected. • The equipment is protected by a branch circuit protective device with a rated current of no more than 60 A. 22 Note: The primary short circuit protective device for the control circuit is also included in the calculation of the SCCR for the power circuit. Therefore, the SCCR of the overcurrent protective device (i.e., a supplementary protector or set of fuses) used on the primary side of a control power transformer is included in the determination of the SCCR for the control panel. Control circuit components downstream from these devices would not be included in the calculation of the SCCR. Where can the SCCR ratings of the individual power circuit components be found? Option 1: Capture this information from the device markings or component instruction sheets. Most Siemens power control and circuit protection components include a standard short circuit rating on the front or side label. 23 24 25 Option 3: High-capacity SCCR is based on testing a combination of components per UL508 Supplement SB, so follow those guidelines. Within the UL508A Supplement SB, the application of previously investigated and tested assemblies from a supplier of equipment can be utilized as described in the manufacturer’s procedures to determine the SCCR. Manufacturers of low-voltage protection, distribution, and control equipment may perform tests according to UL508 “Standard for Industrial Equipment,” which allows obtaining “high-capacity short circuit current ratings.” These high capacity short circuit current ratings exceed the standard short circuit current ratings, in most cases. Note: The High Capacity Short Circuit Current Ratings for Siemens components can be found at: www.usa.siemens.com/sccr All the ratings for Siemens components are conveniently summarized in Excel spreadsheets. 26 Important note for tested combinations: If the specified protective device is a Class CC, G, J, L, RK1, RK5, or T fuse, a fuse of a different class may be used, provided that the peak let-through current (Ip) and the peak let-through energy (I2t) is equal to or lower than the specified fuse. The peak let-through currents and peak let-through energy shall be taken from Table SB4.2 in the UL508A standard. If the specified protective device is a “current limiting” circuit breaker, a different current limiting circuit breaker may be used, provided that the peak let-through current and the peak let-through energy is equal to or lower than the specified circuit breaker. The values shall be taken from the data sheets that are provided by the manufacturer of the circuit breaker (see example). If the specified protective device is a “non-current limiting” overcurrent protective device, a current limiting device is able to be used at the same high fault rating, as long as the interrupting rating of the current limiting overcurrent device is equal to or greater than the specified overcurrent device. A graphic explanation of Step 1 for the determination of the SCCR for individual power circuit components is available in Figure 2. 27 Step 2: Applying current limiting components to modify the SCCR within a portion of a circuit in the panel Which components are considered to be “current limiting”? UL508A allows the use of one of the following components to limit the available fault current to components downstream of the current limiting device: 1. Power transformers with an isolated secondary winding 2. Circuit breakers that are marked as “current limiting” 3. Fuses of Class CC, G, J, L, RK1, RK5, CF or T Note: The current limiting component shall be installed in the Feeder Circuit! A graphic explanation of the feeder circuit and the branch circuit can be seen in Figure 1. The following information details how to effectively apply the current limiting components in the feeder circuit. Three scenarios are detailed. Option 1: Use of power transformer with an isolated secondary winding, installed in the feeder circuit 28 In general, the SCCR on the line side of the transformer shall be the interrupting rating of the overcurrent protection device on the primary side of the power transformer, provided that the short circuit current of all the components and overcurrent protective devices is equal to or higher than the available short circuit current on the secondary side of the transformer. How can the available secondary short circuit current of a transformer be determined? Method A: Calculation with Formulas Single-phase Transformers Transformer Full-Load Current (IFL) = (Transformer kVA × 1000) / Voltage* Short Circuit Current (ISC line-to-line) = ((Transformer Full Load Current (IFL )) / Transformer Impedance (Z) Three-phase Transformers Transformer Full-Load Current (IFL) = (Transformer kVA × 1000) / (Voltage** × 1.732) Short Circuit Current (ISC line-to-line-to-line) = ((Transformer Full Load Current (IFL )) / Transformer Impedance (Z) 29 30 31 Option 2: Use of circuit breaker marked as “current limiting” installed in the feeder circuit In General: The SCCR on the line side of feeder circuit breaker shall be the interrupting rating of the breaker if the following two conditions are fulfilled: 1. The components on the load side of the circuit breaker have a SCCR equal to or higher than the peak let-through current of the feeder circuit breaker. 2. The branch protection devices have an interrupting rating equal or higher than the interrupting rating of the circuit breaker in the feeder circuit. If condition 1. is not fulfilled, the lowest SCCR any component on the load side of the circuit breaker shall be the SCCR for the entire circuit on the line side of the feeder circuit breaker. If condition 2. is not fulfilled, the interrupting rating of the branch circuit protective device shall be the SCCR of the entire circuit on the line side of the feeder circuit breaker. Note: The peak let-through values of the Circuit Breaker need to be provided by the Circuit Breaker manufacturer. 32 Example: Siemens ED Circuit Breaker, CED 6 33 Option 3: Use of current limiting fuses installed in the feeder circuit Fuses of the Class CC, G, J, L, RK1, RK5, or T are current limiting. The SCCR on the line side of fuse in the feeder circuit shall be the interrupting rating of the fuse if the following two conditions are fulfilled: 1. The components on the load side of the fuse have a SCCR equal to or higher than the peak let-through current of the fuse in the feeder circuit. 2. The branch protection devices have an interrupting rating equal to or higher than the interrupting rating of the fuse in the feeder circuit. If condition 1. is not fulfilled, the lowest SCCR of any component on the load side of the circuit breaker shall be the SCCR for the entire circuit on the line side of the feeder circuit breaker. If condition 2. is not fulfilled, the interrupting rating of the branch circuit protective device shall be the SCCR of the entire circuit on the line side of the feeder circuit fuse. Note: The peak let-through values of the Fuse shall be taken of the table SB 4.2 in the UL508A Standard for Industrial Control Panels! 34 Extract of the table SB 4.2 UL508A , Second Edition 35 II. Overview of the UL508A Supplement SB for Calculating the SCCR for Industrial Control Panels Step 3: Determining the overall SCCR of the industrial control panel To do so, establish the overall rating for the industrial control panel, which cannot exceed the rating of the lowest rated component or circuit, including • the modified rating determined in Step 2 (applying current limiting components to modify the SCCR within a portion of a circuit in the panel) above, and • the overcurrent protection device on the primary side of the control circuit. 36 Figure 1: Graphic explanation of branch circuit and feeder circuit 37 Figure 2: Graphic explanation of Step 1: Determination of the SCCR for individual power circuit components 38 Figure 3.1: Graphic explanation / example of Step 2: Use of current limiting transformers in the feeder circuit 39 Figure 3.2: Graphic explanation / example of Step 2: Use of current limiting fuses in the feeder circuit 40 Figure 3.3: Graphic explanation / example of Step 2: Use of circuit breaker marked as “current limiting” in the feeder circuit 41 42 43 44 45 46 47 V. Glossary Available Fault Current R.M.S. value of the current that would flow if the supply conductors to the circuit are short-circuited by a conductor of negligible impedance located as near as practicable to the supply terminals of the industrial control panel. The available fault current at the point of the supply to the machine shall not be greater than the short circuit current rating marked on the industrial control panel nameplate. Branch Circuit The conductors and components following the last overcurrent protective device protecting a load. Current-Limiting Overcurrent Protective Device A device that, when interrupting currents in its current limiting range, reduces the current flowing in the faulted circuit to a magnitude substantially less than that obtainable in the same circuit if the device were replaced with a solid conductor having comparable impedance. Feeder Circuit The conductors and circuitry on the supply side of the branch circuit overcurrent protective device. High Fault / Capacity Short Circuit Current Rating A marked short circuit current rating of a motor controller that is greater than the standard fault short circuit current rating.. 48 Industrial Control Panel An assembly of two or more components • In the power circuit, such as motor controllers, overload relays, fused disconnect switches, and circuit breakers • In the control circuit, such as pushbuttons, signal lamps, selector switches, time-delay switches/relays, switches, control relays • In a combination of two circuits These components are mounted in an enclosure or panel with the associated wiring and terminals. The industrial control panel does not include the controlled equipment Interrupting Rating (aka Available Interrupting Capacity – A.I.C.) The highest current at rated voltage that a device is identified to interrupt under standard test conditions. Overcurrent Protection A device designed to open a circuit when the current through it exceeds a predetermined value. The ampere rating of the device is selected for a circuit to terminate a condition where the current exceeds the rating of conductors and equipment due to overloads, short circuits, and faults to ground. 49 Peak Let-Through Current - IP The highest instantaneous current passed by the over-current protection device during the interruption of the current SCCR - Short Circuit Current Rating A prospective symmetrical fault current at a nominal voltage to which an apparatus or system is able to be connected without sustaining damage exceeding defined acceptance criteria. Standard Fault Short Circuit Current Rating Short circuit current rating of a motor controller as specified in Table SB4.1. UL508A, Second Edition. 50 51 52 53 In this case, the worst case available fault current level at the transformer’s secondary is 53,704 amps. Upon inspecting Machine 1, it is determined by the equipment label that Machine 1’s control panel is rated for 42kA. However, Machine 2 is older equipment and its equipment label does not list an assembly SCCR for the control panel. Using the worst case available fault current level, Machine 1 is not compliant (42kA < 54kA). The facility decides to pursue a more precise available fault current calculation at Machine 1. There are three 500MCM cables per phase running approximately 500 feet between the substation transformer and Machine 1. Using Eaton’s FC2 fault current calculator, they determine the calculated available fault current to be 27,782 amps. Based on this calculation, the equipment SCCR of Machine 1 is adequate for the available fault current at its location in the electrical distribution system (42kA > 27.782kA). There are several options to resolve Machine 2’s unknown equipment SCCR issue. One is to assume the minimum 5kA equipment SCCR on the control panel and determine a more precise fault current level calculation for the point in the electrical distribution system where Machine 2 is located. In this case, a copper bus runs approximately 100 feet to Machine 2. 54 Using Eaton’s FC2 calculator, the available fault current is 48,817 amps. This does not resolve the issue as the assumed 5kA default equipment SCCR is less than the available fault current (5kA < 48.817kA). Machine 2 is older and will be replaced in a few years. Management considers a reworking and recertification of its control panel to be a significant investment for a piece of equipment that will soon be replaced. Machine 2 has a relatively small load, so a decision is made to investigate lowering the available fault current level below 5kA by installing an isolation transformer ahead of Machine 2. It is determined that a 15kVA transformer is properly sized to support the Machine 2’s load, and a calculation of the worst case available fault current on the isolation transformer’s secondary resulted in an available fault current level of 1023 amps, which is below the panel’s minimum assembly SCCR (1.023kA < 5kA). To remedy the 5kA default equipment SCCR assumed for Machine 2, the facility hired an electrical contractor to install a 15kVA transformer and associated circuit protection, and the issue of the unmarked equipment SCCR on Machine 2’s control panel was resolved. 55 Now that the existing equipment SCCR has been resolved, the facility determines its equipment SCCR specifications for any new equipment. The substation transformer is fairly new and expected to be in service for some time. However, some portions of the electrical distribution system have aged, and it is expected some portions may be replaced with busway to lower energy losses and system impedance. Thus the facility management determines to standardize on a minimum 55kA equipment SCCR for any new equipment purchases. This standardized 55kA equipment SCCR will provide flexibility and accommodate any utility changes or electrical distribution system upgrades while ensuring the required short-circuit event protection for personnel. Facility management wants to sustain their equipment SCCR plan it has established for the existing and new equipment. They elect to post available fault current labels at key points in the electrical distribution system along with the minimum acceptable equipment SCCR for new or relocated equipment. These labels also include warnings to maintenance personnel and contractors that they are not to make any changes or additions to the electrical system without prior approval from facility management. They follow up these events with annual training for all personnel to advise them 56 of these changes, associated risks, and their respective responsibilities. Ch 6: Fault Calculations 57 The operation of a power system departs from normal after the occurrence of a fault. Faults give rise to abnormal operating conditions-usually excessive currents and voltages at certain points on the system-which are guarded against with various types of protective equipment. 6.1 TYPES OF FAULTS Various types of short-circuit faults that can occur on a transmission line are depicted in Fig. 6-1; the frequency of occurrence decreases from part (a) to part (f). Although the balanced three-phase short circuit in Fig. 6-1(d) is relatively uncommon, it is the most severe fault and therefore determines the rating of the line-protecting circuit breaker. A fault study includes the following: 1. Determination of the maximum and minimum three-phase short-circuit currents 2. Determination of unsymmetrical fault currents, as in single ground, line-to-line, and open-circuit faults 3. Determination of the ratings of required circuit breakers 4. Investigation of schemes of protective relaying 58 5. Determination of voltage levels at strategic points during a fault The short-circuit faults depicted in Fig. 6-1 are called shunt faults; open circuits, which may be caused by broken conductors, for instance, are categorized as series faults. 59 6.2 SYMMETRICAL FAULTS A balanced three-phase short circuit [Fig. 6-1(d)] is an example of a symmetrical fault. Balanced three-phase fault calculations can be carried out on a per-phase basis, so that only single-phase equivalent circuits need be used in the analysis. Invariably, the circuit constants are expressed in per-unit terms, and all calculations are made on a per-unit basis. In shortcircuit calculations, we often evaluate the short-circuit MVA (megavoltamperes), which is equal to 3 Vl If where Vl is the nominal line voltage in kilovolts, and If is the fault current in kiloamperes. 60 An example of a three-phase symmetrical fault is a sudden short at the terminals of a synchronous generator. The symmetrical trace of a shortcircuited stator-current wave is shown in Fig. 6-2. The wave, whose envelope is shown in Fig. 6-3, may be divided into three periods or time regimes: the subtransient period, lasting only for the first few cycles, during which the current decrement is very rapid; the transient period, covering a relatively longer time during which the current decrement is more moderate; and finally the steady-state period. The difference Δi' (in Fig. 6-3) between the transient envelope and the steady-state amplitude is plotted on a logarithmic scale as a function of time in Fig. 6-4, along with the difference Ai" between the subtransient envelope and an extrapolation of the transient envelope. Both plots closely approximate straight lines, illustrating the essentially exponential nature of the decrement. 61 62 63 The currents during these three regimes are limited primarily by various reactances of the synchronous machine (we neglect the armature resistance, which is relatively small). These currents and reactances are defined by the following equations, provided the alternator was operating at no load before the occurrence of a three-phase fault at its terminals: 64 where lEgl is the no-load voltage of the generator, the currents are rms currents, and O, a, b, and c are shown in Fig. 6-2. The machine reactances Xs, X’d, and X’’d are known as the direct-axis synchronous reactance, directaxis transient reactance, and direct-axis subtransient reactance, respectively. The currents I, i', and i" are known as the steady-state, transient, and subtransient currents. From (6.1) through (6.3) it follows that the fault currents in a synchronous generator can be calculated when the machine reactances are known. 65 Suppose now that a generator is loaded when a fault occurs. Figure 6-5(a) shows the corresponding equivalent circuit with the fault to occur at point P. The current flowing before the fault occurs is IL, the voltage at the fault is Vf, and the terminal voltage of the generator is Vt. When a three-phase fault occurs at P, the circuit shown in Fig. 6-5(b) becomes the appropriate equivalent circuit (with switch S closed). Here a voltage E’’g in series with X’’d supplies the steady-state current IL when switch S is open, and supplies the current to the short circuit through X’’d and Zext when switch S is closed. If we can determine E’’g, we can find this current through X’’d ,which will be i". With switch S open, we have 66 which defines E’’g, the subtransient internal voltage. Similarly, for the transient internal voltage we have Clearly E’’g and E’g are dependent on the value of the load before the fault occurs. 67 6.3 UNSYMMETRICAL FAULTS AND SYMMETRICAL COMPONENTS Unsymmetrical faults such as line-to-line and line-to-ground faults (which occur more frequently than three-phase short circuits) can be analyzed on a perphase basis. For such faults the method of symmetrical components is used. This method is based on the fact that a set of three-phase unbalanced phasors can be resolved into three sets of symmetrical components, which are termed the positive-sequence, negative-sequence, and zero-sequence components. The phasors of the set of positive-sequence components have a counterclockwise phase rotation (or phase sequence) abc, the negativesequence components have the reverse phase sequence acb; and the zerosequence components are all in phase with each other. 68 Short Circuit Fault Calculations A power system functions normally until after the occurrence of a fault in the system. The good news is fault events can be minimized or avoided through diligent electrical design, accurate record keeping information on equipment/devices/motors, proper installation, and use of agency-certified equipment. There are three major sources of fault current: an electric utility power system, a generator, and a motor. Short circuit faults are called shunt faults. An open-circuit condition is known as a series fault. Any phase/circuit to ground condition is called a ground fault. Among all faults, a balanced 3phase short circuit is the most critical and serious. However, it is one of the least likely of faults to occur. The elements in a power distribution system that limit or impede the fault current value include: cables, transformers, and reactors. The NEC requires protection to personnel and electrical systems against damage during short circuit conditions. Generally, circuit breaker ratings are determined for the worst-case fault situation. 69 You typically perform short circuit calculations when working with complex and interactive power distribution systems; however, manual calculations can be used for more simplified systems. A short circuit calculation determines the amount of current that can flow at certain points in the distribution system. An electrical device or piece of equipment can then be selected for appropriate rating (withstand or interrupting rating) based on these calculations. Let's work through a few simple examples to show how you can quickly and easily calculate fault currents. We'll first use the concept of admittance to calculate the fault current in the system shown in Fig. 1. Admittance is a measure of how easily a circuit or device will allow a current to flow. It is the inverse of impedance, which is defined as a measure of opposition to current. 70 71 Step 1: Because the electric utility can't provide us with fault information at this particular site, we assume it to be infinite. We calculate the maximum amount of power that the transformer (XFMR) will allow to flow to its secondary (i.e., load side) using this formula. XFMR let through power = XFMR kVA rating ÷ [% Impedance ÷ 100] = 5,000kVA ÷ [5 ÷ 100] = 100,000kVA Step 2: Now we can calculate the cable let-through power. This is defined as the amount of power that the cable would let through from an infinite source to the load side of the XFMR. The formula we use for this step is as follows: Cable let through power = [1,000 x (kV phase-phase)2] ÷ [cable impedance (ohms) per phase, per 1,000 ft x total distance (ft)] = [1,000 x (12kV)2] ÷ {0.15 ÷ 1,000 ft} x 100,000 ft = 9,600kVA let through 72 Step 3: Next, we calculate the total let-through fault power by using the following formula: Net fault power = 1 ÷ [(1 ÷ XFMR let through power) + (1 ÷ cable let through power)] = 1 ÷ [(1 ÷ 100,000kVA) + (1 ÷ 9,600kVA)] = 8,759kVA Step 4: Now we can find the fault current using the following formula: Fault current = net fault power ÷ (secondary XFMR voltage rating x √3) = 8,759kVA ÷ (12kV x √3) = 421A 73 Let's work through another sample calculation where the electric utility fault power level is known. The given value is 50,000kVA. In this situation, we assume cable length to be minimal and therefore neglect its effect since the impedance is minimal. Step 1: XFMR let through power = 2,000kVA ÷ [5 ÷ 100] = 40,000kVA Step 2: Net fault power = 1 ÷ [(1 ÷ 40,000kVA) + (1 ÷ 50,000kVA)] = 22,222kVA Step 3: Fault current = 22,222kVA ÷ (0.48kV x √3) = 26,729A 74 75 For our third and final example calculation, we'll work a problem where generator data is available to us. When working with generators, we introduce a new value known as sub-transient reactance, which is referred to as X"d. This value is typically shown on the generator nameplate or can be obtained directly from the manufacturer. Step 1: The short circuit kVA available at the generator is calculated using the following formula: Generator fault power (MVA) = generator MVA rating ÷ X"d = 800 MVA ÷ 0.17 = 4,706 MVA Step 2: XFMR let through power = 1,000 MVA ÷ [10 ÷ 100] = 10,000 MVA Step 3: Net fault power = 1 ÷ [(1 ÷ 10,000 MVA) + (1 ÷ 4,706 MVA)] = 3,200 MVA 76 As you can see, simplified short circuit fault calculations can be performed using a basic understanding of sources of fault power and current and impedance values that impede the short circuit power flow. 77 Electrical networks, machines and equipment are often subjected to various types of faults while they are in operation. When a fault occurs, the characteristic values (such as impedance) of the machines may change from existing values to different values till the fault is cleared. There may be lot of probabilities of faults to appear in the power system network, including lighting, wind, tree falling on lines, apparatus failure, etc. A fault in an electric power system can be defined as any abnormal condition of the system that involves the electrical failure of the equipment, such as , transformers, generators, busbars, etc. The fault inception also involves in insulation failures and conducting path failures which results short circuit and open circuit of conductors. 78 Under normal or safe operating conditions, the electric equipment in a power system network operate at normal voltage and current ratings. Once the fault takes place in a circuit or device, voltage and current values deviates from their nominal ranges. The faults in power system causes over current, under voltage, unbalance of the phases, reversed power and high voltage surges. This results in the interruption of the normal operation of the network, failure of equipment, electrical fires, etc. Usually power system networks are protected with switchgear protection equipment such as circuit breakers and relays in order to limit the loss of service due to the electrical failures. 79 Symmetrical and Unsymmetrical Faults As discussed above that faults are mainly classified into open and short circuit faults and again these can be symmetrical or unsymmetrical faults. Symmetrical Faults A symmetrical fault gives rise to symmetrical fault currents that are displaced with 1200 each other. Symmetrical fault is also called as balanced fault. This fault occurs when all the three phases are simultaneously short circuited. These faults rarely occur in practice as compared with unsymmetrical faults. Two kinds of symmetrical faults include line to line to line (L-L-L) and line to line to line to ground (L-L-L-G) as shown in figure below. 80 Protection Devices against Faults When the fault occurs in any part of the system, it must be cleared in a very short period in order to avoid greater damage to equipment and personnel and also to avoid interruption of power to the customers. The fault clearing system uses various protection devices such as relays and circuit breakers to detect and clear the fault. Some of these fault clearing or faults limiting devices are given below. Fuse It opens the circuit whenever a fault exists in the system. It consists of a thin copper wire enclosed in a glass or a casing with two metallic contacts. The high fault current rises the temperature of the wire and hence it melts. A fuse necessitates the manual replacement of wire each time when it blows. Circuit Breaker It is the most common protection device that can make or break the circuit either manually or through remote control under normal operating conditions. 81 Circuit Breakers 82 Protective Relays These are the fault detecting devices. These devices detect the fault and initiate the operation of the circuit breaker so as to isolate the faulty circuit. A relay consists of a magnetic coil and contacts (NC and NO). The fault current energizes the coil and this causes to produce the field, thereby the contacts get operated. 83 Ch 7: General Methods for Network Calculations 84 General Methods for Network Calculations In this chapter we develop general solution methods that are amenable to the computer solution of power system network problems. We begin from the basic network theorems. 7.1 SOURCE TRANSFORMATIONS The voltage source of Fig. 7-1(a) may be transformed to the current source of Fig. 7-1(b) and vice versa, provided that 85 7.2 BUS ADMITTANCE MATRIX The four-bus system that corresponds to the one-line diagram of Fig. 72(a) may be represented by the network of Fig. 7-2(b). In terms of the node voltages V1, 112, V3 and V4 and the given admittances, Kirchhoff's current law yields 86 Rearranging these equations and rewriting them in matrix form, we obtain 87 88 89 Each admittance Yii (i = 1, 2, 3, 4) is called the self-admittance (or driving-point admittance) of node i and is equal to the algebraic sum of all the admittances terminating on the node. Each off-diagonal term Yik (i, k = i, 2, 3, 4) is called the mutual admittance (or transfer admittance) between nodes i and k and is equal to the negative of the sum of all admittances connected directly between those nodes. Further, Yik = Yki. For a general network with N nodes, therefore, Kirchhoff's current law in terms of node voltages may be written as 90 is called the bus admittance matrix, and V and I are the N-element node voltage matrix and node current matrix, respectively. In (7.6), the first subscript on each Y indicates the node at which the current is being expressed, and the second subscript indicates the node whose voltage is responsible for a particular component of the current. Further, the admittances along the diagonal are the self-admittances, and the off-diagonal admittances are the mutual admittances. It follows from (7.5) and (7.6) that the current entering a node k is given by 91 A Short Circuit analysis is used to determine the magnitude of short circuit current the system is capable of producing and compares that magnitude with the interrupting rating of the overcurrent protective devices (OCPD). Since the interrupting ratings are based by the standards, the methods used in conducting a short circuit analysis must conform to the procedures which the standard making organizations specify for this purpose. In the United States, the America National Standards Institute (ANSI) publishes both the standards for equipment and the application guides, which describes the calculation methods. Short circuit currents impose the most serious general hazard to power distribution system components and are the prime concerns in developing and applying protection systems. Fortunately, short circuit currents are relatively easy to calculate. The application of three or four fundamental concepts of circuit analysis will derive the basic nature of short circuit currents. These concepts will be stated and utilized in a step-by step development. 92 The three phase bolted short circuit currents are the basic reference quantities in a system study. In all cases, knowledge of the three phase bolted fault value is wanted and needs to be singled out for independent treatment. This will set the pattern to be used in other cases. A device that interrupts short circuit current, is a device connected into an electric circuit to provide protection against excessive damage when a short circuit occurs. It provides this protection by automatically interrupting the large value of current flow, so the device should be rated to interrupt and stop the flow of fault current without damage to the overcurrent protection device. The OCPD will also provide automatic interruption of overload currents. 93 Listed here are reference values that will be needed in the calculation of fault current. Impedance Values for Three phase transformers 94 95 TRANSFORMER FAULT CURRENT Calculating the Short Circuit Current when there is a Transformer in the circuit. Every transformer has “ %” impedance value stamped on the nameplate. Why is it stamped? It is stamped because it is a tested value after the transformer has been manufactured. The test is as follows: A voltmeter is connected to the primary of the transformer and the secondary 3-Phase windings are bolted together with an ampere meter to read the value of current flowing in the 3Phase bolted fault on the secondary. The voltage is brought up in steps until the secondary full load current is reached on the ampere meter connected on the transformer secondary. So what does this mean for a 1000KVA 13.8KV – 480Y/277V. First you will need to know the transformer Full Load Amps Full Load Ampere = KVA / 1.73 x L-L KV FLA = 1000 / 1.732 x 0.48 FLA = 1,202.85 The 1000KVA 480V secondary full load ampere is 1,202A. 96 When the secondary ampere meter reads 1,202A and the primary Voltage Meter reads 793.5V. The percent of impedance value is 793.5 / 13800 = 0.0575. Therefore: % Z = 0.0575 x 100 = 5.75% This shows that if there was a 3-Phase Bolted fault on the secondary of the transformer then the maximum fault current that could flow through the transformer would be the ratio of 100 / 5.75 times the FLA of the transformer, or 17.39 x the FLA = 20,903A. Based on the infinite source method at the primary of the transformer. A quick calculation for the Maximum Fault Current at the transformer secondary terminals is FC = FLA / %PU Z FC = 1202 / 0.0575 = 20,904A. 97 This quick calculation can help you determine the fault current on the secondary of a transformer for the purpose of selecting the correct overcurrent protective devices that can interrupt the available fault current. The main breaker that is to be installed in the circuit on the secondary of the transformer has to have a KA Interrupting Rating greater then 21,000A. Be aware that feeder breakers should include the estimated motor contribution too. If the actual connected motors are not known, then assume the contribution to be 4 x FLA of the transformer. Therefore, in this case the feeders would be sized at 20,904 + (4 x 1202) = 25,712 Amps 98 GENERATOR FAULT CURRENT Generator fault current differs from a Transformer. Below, we will walk through a 1000KVA example. 800KW 0.8% PF 1000KVA 480V 1,202FLA KVA = KW / PF KVA = 800 / .8 KVA = 1000 FLA = KVA / 1.732 x L-L Volts FLA = 1000 / 1.732 x 0.48 FLA = 1,202 (As listed in the table for generator subtransient X” values is 0.16) FC = FLA / X” FC = 1202 / 0.16 FC = 7,513A So, the fault current of a 1000KVA Generator is a lot less then a 1000KVA transformer. The reason is the impedance value at the transformer and Generator reactance values are very different. Transformer 5.75% vs. a Generator 16% 99 SYSTEM FAULT CURRENT Below is a quick way to get a MVA calculated value. The MVA method is fast and simple as compared to the per unit or ohmic methods. There is no need to convert to an MVA base or worry about voltage levels. This is a useful method to obtain an estimated value of fault current. The elements have to be converted to an MVA value and then the circuit is converted to admittance values. 100 Utility MVA at the Primary of the Transformer MVAsc = 500MVA Transformer Data 13.8KV - 480Y/277V 1000KVA Transformer Z = 5.75% MVA Value 1000KVA / 1000 = 1 MVA MVA Value = 1MVA / Zpu = 1MVA / .0575 = 17.39 MVA Use the admittance method to calculate Fault Current 1 / Utility MVA + 1 / Trans MVA = 1 / MVAsc 1 / 500 + 1 / 17.39 = 1 / MVAsc 0.002 + 0.06 = 1/ MVAsc MVAsc = 1 / (0.002 + 0.06) MVAsc = 16.129 FC at 480V = MVAsc / (1.73 x 0.48) FC = 16.129 / 0.8304 FC = 19.423KA FC = 19, 423 A 101 The 480V Fault Current Value at the secondary of the 1000KVA transformer based on an Infinite Utility Source at the Primary of the transformer as calculated in the Transformer Fault Current section in this article is 20,904A. The 480V Fault Current Value at the secondary of the 1000KVA transformer based on a 500MVA Utility Source at the Primary of the transformer as calculated in the System Fault Current section in this article is 19,432A. The 480V Fault Current Value at the secondary of the 1000KVA transformer based on a 250MVA Utility Source at the Primary of the transformer the calculated value is 18,790A. 102 When the cable and its length is added to the circuit the fault current in a 480V system will decrease to a smaller value. To add cable into your calculation use the formula. Cable MVA Value MVAsc = KV2/ Z cable. Use the cable X & R values to calculate the Z value then add to the Admittance calculation as shown in this article. The conclusion is that you need to know the fault current value in a system to select and install the correct Overcurrent Protective Devices (OCPD). The available FC will be reduced as shown in the calculations when the fault current value at the primary of the transformer is reduced. If the infinite method is applied when calculating fault current and 4 x FLA is added for motor contributions, then the fault current value that is obtained will be very conservative. This means the calculated value in reality will never be reached, so you reduce any potential overcurrent protection device failures due to fault current. 103