Chapter 6: Load and Short-Circuit Calculations
Overview
Interpret the importance of load and short-circuit calculations.
Evaluate design calculations and revise load calculation values.
Create a set of load calculations.
Evaluate the procedure to complete a set of short-circuit calculations.
Create a set of short-circuit calculations.
Properly select equipment based on short-circuit calculations.
Introduction
Performing load calculations and short-circuit calculations is one of the final steps in completing an electrical design. At this stage in the design process, loads from all the various components of the design served by panelboards or the main switchboard must be brought together for load calculations that determine the amperage size of the main electrical service distribution equipment that will serve the electrical requirements of the facility.
If a short circuit or ground fault should occur anywhere in the electrical system, extremely high amperage values could flow through the system. If electrical components in the system are not properly protected from these currents, severe damage and possible electrical explosions can occur, destroying electrical equipment and causing injury or death to those working around the equipment (see FIGURE 6-1). Electrical designers perform short-circuit calculations to select protective devices to prevent any severe damage caused by short circuits or fault conditions.
Figure 6-1: Properly performing load and short-circuit calculations protects personnel and equipment from dangerous explosions caused by short-circuit or fault conditions
Tip | When designers work with a service planner from the local utility, the load calculations will be provided to the planner, who works with the utility to design appropriate service entrance requirements. Accurate load calculations enable the serving utility to provide a service entrance plan that meets all the necessary requirements for approval. |
Performing Load Calculations
It is not sufficient to simply add the panel schedule totals and equipment served directly by the main switchboard to calculate the total system electrical demands. As designers calculate electrical loads throughout the process and enter them into panel schedules, they enter these loads at 100 percent of rated capacity. Load calculations are necessary to evaluate what the total load will be for any one time during operation to properly size the panelboards and the feeder conductors that supply them. However, at 100 percent of rated capacity, these loads do not realistically represent the total electrical demand on the system at any one time because it is highly unlikely all loads will be operating at capacity at once. When calculating the actual demand for panelboards serving these types of loads, designers calculate allowances to more accurately reflect the demand on the electrical system.
For panelboards that serve equipment and motorized loads, allowances must be added to the total load. For the total calculated load from the panel schedule, designers must add an increase factor to ensure that the electrical system is designed to carry the equipment full-load amperages and can accommodate any additional amperage caused by inrush currents associated with motor loads.
Through the process of load calculations, designers evaluate these factors to determine the realistic electrical demand imposed on the electrical system at any one time.
General Electrical Loads
To determine the actual full load of general purpose receptacles, the designer must consider relevant demand factors. The NEC defines a demand factor as the ratio of the maximum demand to the total connected load under consideration [Article 100]. In essence, demand factors are reduction allowances that can be applied to certain types of loads without high demand, such as the general purpose receptacle loads, to get a more realistic volt-amperage value. For example, the NEC permits general purpose receptacle loads to be computed at a value of 100 percent ofthe first 10 kVA of load and 50 percent of the remaining load [Table 220.44].
Tip | It is important to perform load calculations correctly. Underestimating a facility’s demand limits its electrical capacity and ability to operate. Overestimating a facility’s demand results in a larger than necessary distribution system with excessive equipment and labor costs. |
To calculate demand factor for a commercial facility with the following specifications:
General receptacle load of 25,000 VA
Calculate the first 10,000 VA at 100 percent:
Apply a demand factor of 50 percent to the remainder of load:
Add the totals to determine the total load after applying the allowable demand factors:
Answer: The adjusted total for this load is 17,500 VA.
Specialized Electrical Loads
To size distribution systems for specialized equipment loads, the distribution equipment and conductors must be capable of supporting the higher inrush currents associated with motor loads without overload or voltage drop. Therefore, specialized equipment loads are calculated at 125 percent of the rated load of the device. When motor loads are supplied through panelboards, the 125 percent allowance is calculated for the largest motor supplied. If the allowance is added to the largest motor, then this allowance is sufficient for any motor(s) of smaller load supplied by the panelboard.
Tip | Equipment should be sized based on the total load derived from the panel schedule. Demand factors are used only to determine a more accurate load that may be imposed on the main electrical service equipment and should not be applied when sizing branch-circuit panelboard or feeder conductors. |
Tip | Demand factors that reduce the calculated load cannot be applied to specialized equipment loads or continuous loads such as lighting loads. |
Lighting System Loads
Because of their long operating hours, lighting loads are considered continuous loads. When panelboards serve lighting loads, the designer must calculate the demand on the system at a value of 125 percent of the full load value [215.2(A)(1)].
Distribution System Loads
When distribution equipment directly serves motor loads without distribution through a panelboard, the designer must calculate the equipment demand at 125 percent of the highest-rated motor load plus the amperage for all the other motors being served.
Tip | When panelboards or distribution equipment serve both specialized equipment and lighting loads, the calculation must add both 125 percent of the lighting load and 125 percent of the specialized equipment load. |
To determine the load calculations when more than one motor is being served:
Determine full-load amperage of the motor loads in volt-amperes using NEC Table 430.248 for single-phase motors or NEC Table 430.250 for 3-phase motors:
10-hp, 3-phase, 230-volt motor = 28 A
15-hp, 3-phase, 230-volt motor = 42 A
25-hp, 3-phase, 230-volt motor = 68 A
Use Ohm’s law to calculate this value in volt-amperes: For a 3-phase motor: Voltage (E) × Amperage (I) × 1.73
230 V × 28A × 1.73= 11,141 VA
230 V × 42 A × 1.73= 16,712 VA
230 V × 68 A × 1.73 = 27,057 VA
Calculate 125 percent of the volt-amperes of the highest-rated motor in the group:
Add the volt-amperes of the other motors in the group at 100 percent:
Answer: The total demand for this load is 61,674 VA.
Additional 125 percent demand values for subsequent motor loads are not necessary. The 125 percent allowance for the highest-rated motor provides for any other smaller motors served. In this type of application, when any motors in the group are of the same amperage rating, the designer must calculate 125 percent demand factor for the highest-rated motor in the group and then add the remaining amperages. If all motors are the same amperage, the designer must calculate 125 percent for any motor in the group and add the remaining amperages.
Performing Short-Circuit Calculations
If a short circuit or fault condition were to flow in a system, extremely high, damaging amperages would flow through the system unless proper protection was in place. These amperages are not just limited to the main entry point of the electrical system within a facility. If a short circuit or fault occurs at a panelboard located away from the main electrical service, the amperages flow through the entire path from the point of service entry to the location of the fault or short circuit, causing extreme electrical stress to any equipment in the path such as panelboards, the feeders that serve them, and any branch-circuit protective device in a panelboard.
Designers perform a set of short-circuit calculations to determine these amperage values, and then select components for the electrical system with rated values that ensure the equipment is properly protected. If extreme amperages occur, they are limited to values within the rated, short-circuit limits of the electrical equipment. According to the NEC:
The overcurrent protective devices, the total impedance, the component short-circuit current ratings, and other characteristics of the circuit to be protected shall be selected and coordinated to permit the circuit-protective devices used to clear a fault to do so without extensive damage to the electrical components of the circuit [110.10].
Tip | Distribution equipment follows a rule similar to feeder conductors. To calculate ampacity, take 125 percent of the highest rated motor in the group plus 100 percent of the remaining motor amperages [430.24]. |
Short-circuit amperages in a circuit are caused by the high levels of current available if a short circuit occurs. These short-circuit values may be provided by the serving utility or calculated manually to determine the maximum short-circuit amperage available. Like any electrical circuit, one factor that determines the amount of amperage is the impedance (AC resistance) in the circuit. However, with short-circuit amperages, higher and more extreme values are possible. For a service entrance point in a facility, assuming an infinite value available, the amount of short-circuit amperage is limited only to the impedance in the circuit from a serving utility transformer and the total impedance in the service entrance conductors.
Once the total available short-circuit amperage is determined, then the designer evaluates each point in the system (e.g., panelboards, equipment) to determine how much of the total short-circuit amperage is available at each point, based on the impedance in the circuit.
Transformer Impedance
All transformers 25 kVA and larger must include the transformer impedance rating on the nameplate [450.11]. Impedance ratings are given in percentages. Typical ratings are approximately 3 percent to 6 percent, which means that if the terminals of a transformer were purposely shorted out (nearly zero impedance) and voltage was supplied to them, the transformer would reach its full-load amperage when the input voltage reaches 3 percent to 6 percent of the full voltage rating of the transformer. For example, if a transformer with a 3 percent impedance rating was rated at 208 volts, and the leads of the transformer shorted as a result of a line-to-line short circuit, the transformer would deliver its full-load current at approximately 6.24 volts (which is 3 percent of 208 volts). With the transformer already reaching full-load current value at 3 percent of rated voltage, more extreme currents would be delivered when the transformer reaches 100 percent of rated voltage.
When designing an electrical plan, designers must complete the short-circuit calculations for several points in the distribution path, including the following ones:
Utility transformer (typically provided by the serving utility)
Main service entrance point
Distribution panelboards
Motor circuits
Motor control centers
To determine the available short-circuit current for a transformer with the following specifications:
225 kVA, 208-volt, 3-phase transformer
Impedance rating of 3 percent
Determine the full-load current of the transformer using Ohm’s law:
Divide the full-load current by the impedance rating of the transformer:
Answer: The available short-circuit current for this transformer is approximately 20,000 A. If a line-to-line short were to occur, this transformer has the capability of delivering more than 20,000 A.
The calculations to determine the short-circuit amperages must be illustrated on the electrical plans to verify that all the components of the distribution system are capable of withstanding any damaging short-circuit and/or fault current values that might occur. There are no set industry standards as to how the calculations must appear on a plan, but they should be clear and concise.
Designers can complete short-circuit calculations manually or with the assistance of specialized software. The method most commonly used for smaller or less detailed projects is the point-to-point method (see the following Calculation Example).
First, the designer must determine the available short-circuit amperage at the terminals of the utility transformer that will serve a facility, and then at the main service. The value of short-circuit amperage available at the main service will be less because of additional impedances based on service entrance conductor type, conductor length, and type of raceway. A conductor’s impedance is chosen from a predetermined value known as the conductor’s C value (see TABLE 6-1). This value is the reciprocal of the conductor’s impedance to 1 ft of cable and varies based on the type of raceway used. With metallic raceways, high levels of amperage are somewhat depleted by counter electromotive forces (also called “back voltages” or CEMFs), which develop by the induced voltages within the metallic raceway. In nonmetallic type raceways, no induced CEMFs are developed, resulting in higher levels of available short-circuit amperages.
Three Single Conductors | ||
---|---|---|
Copper AWG or kcmil | Conduit Type | |
Steel | Nonmagnetic | |
600 Volt | 600 Volt | |
14 | 389 | 389 |
12 | 617 | 617 |
10 | 981 | 981 |
8 | 1557 | 1558 |
6 | 2425 | 2430 |
4 | 3806 | 3825 |
3 | 4760 | 4802 |
2 | 5906 | 6044 |
1 | 7292 | 7493 |
1/0 | 8924 | 9317 |
2/0 | 10755 | 11423 |
3/0 | 12843 | 13923 |
4/0 | 15082 | 16673 |
250 kcmil | 16483 | 18593 |
300 kcmil | 18176 | 20867 |
350 kcmil | 19703 | 22736 |
400 kcmil | 20565 | 24296 |
500 kcmil | 22185 | 26706 |
600 kcmil | 22965 | 28033 |
750 kcmil | 24136 | 28303 |
1000 kcmil | 25278 | 31790 |
Once the value of short-circuit amperage has been determined for the main switchboard, the designer can determine the value of short-circuit amperage available at each component served by the main switchboard, such as panelboards. Again, these values decrease somewhat based on conductor types, lengths, and raceway types.
Additionally, it is important to note that if a short circuit occurs when motors are running, the motors can contribute a substantial amount of counter electromotive force back into the system. Therefore, if the main service supplies motor loads, the designer must calculate the motor contribution component by calculating the inductance to resistance ratio (X/R). Although the derivation of this factor is beyond the scope of this text, it is sufficient for most purposes to multiply the sum of all full-load amperages for the motors by a factor of 4.
As mentioned previously, short-circuit amperages will flow through all components in the path including fusible disconnects, panelboards, and the circuit breakers contained within them. Therefore, all fusible disconnects, panelboards, and circuit breakers are rated by their size in amperage and by the short-circuit amperes that can safely pass through the equipment without causing damage. For example, a standard 20-A single-pole 120-volt circuit breaker used in common panelboards is rated for a maximum allowable current of 20 A with a maximum short-circuit rating of10,000 A. Circuit breakers are also available with short-circuit values of 22,000 A, 42,000 A, or 65,000 A (see FIGURE 6-2). Fuses can have maximum short-circuit amperage values of 200,000 A and more.
Figure 6-2: Circuit breakers are manufactured with various short-circuit ratings (e.g., 10,000 A; 22,000 A; 42,000 A; and 65,000 A)
To determine the available short-circuit current using the point-to-point method for a transformer with the following specifications (as illustrated on the completed sample plan):
750-kVA 3-phase transformer
3.5 percent impedance rating
480-volt secondary voltage
Main service located 100 ft from utility transformer
Main service supplied with two 600 kcmil copper conductors per phase in non-metallic PVC raceway
Determine the full-load secondary amperage for the transformer using Ohm’s law:
Determine the short-circuit amperage available at the transformer:
Determine the f factor, which is an impedance adjustment factor:
Determine the percentage of the initial short-circuit amperage that will be available at the main service:
Determine the short-circuit amperage that will be available at the main service:
Answer: The available short-circuit amperage available at the terminals of the main service will be 21,934.8 A.
If this transformer will serve motor loads, continue with the following steps:
Calculate the motor contribution component. (For this example, assume the main switchboard is serving a motor contribution of 177 A and has an X/R factor of 4.)
Calculate the total available short-circuit amperage:
To determine the available short-circuit current at panelboard P_{1} with the following specifications:
480-volt, 3-phase panelboard
Location 25.9 m (85 ft) from main switchboard
22,642.8 VA available short-circuit amperage at main switchboard
Conductor C value of 28,033
One 600 kcmil copper conductor per phase in non-metallic PVC conduit
Determine the f factor:
Determine the percentage of the initial short-circuit amperage that will be available at the main service:
Determine the short-circuit amperage that will be available at the main service:
Calculate the motor contribution component:
Calculate the total available short-circuit amperage:
Answer: The total available short circuit amperage at panelboard P_{1} is 18,952.7 A.
If short-circuit calculations determine that the short-circuit amperage values available in an electrical system could exceed the maximum short-circuit amperage rating of any electrical equipment or device in the path of a short circuit, the equipment must be rated at or above the calculated available short-circuit value. Alternatively, a greater degree of protection can be provided in the circuit ahead of the equipment to reduce or eliminate these damaging currents.
Circuit breakers do not stop or limit these extreme amperage values; their maximum short-circuit ratings only specify that these devices will safely pass the high amperage levels without damage to the device itself. Fuses are designed to handle short-circuit amperages without damage to the fuse and to block the flow of the short-circuit amperages at the fuse. If a short-circuit calculation determines that a short-circuit amperage value could exceed the maximum short-circuit of a panelboard rating and the circuit breakers contained within the panel-board, a high-quality fuse with the ability to limit the maximum available short-circuit current should be installed in the circuit path ahead of the panelboard. The added current-limiting protection offered by the fuse allows the panelboard to be installed without fear of damage to the electrical equipment.
As electrical demands increase, the electric utility will increase the capacity of the distribution system by providing newer equipment, such as distribution transformers with lower impedance ratings. When distribution systems are updated with transformers with lower impedance ratings, the available short-circuit currents on the system increase because of these greater efficiencies of the new equipment. When utilities upgrade their distribution system, they recommend that customers who will be supplied by the newer equipment have an electrical designer or contractor reevaluate their electrical systems to determine whether they are capable of safely withstanding the higher level of short-circuit current that will be available. If the existing equipment is not capable of safely withstanding the higher short-circuit current levels, appropriate alterations as necessary must be made to ensure that the electrical system meets all requirements.
Tip | The NEC requires that all components in an electrical system must be protected from damaging currents due to short circuits [110.10 and 110.09]. |
Wrap Up: Master Concepts
After the combined load for a facility has been calculated, designers complete a set of load calculations to determine more accurately the actual electrical demands that may be imposed on a facility’s electrical system.
Short-circuit amperages can be very hazardous to electrical equipment and possibly to personnel working on or around electrical equipment.
Designers must complete calculations to evaluate and design components into the electrical system that can reduce or eliminate the hazards associated with short-circuit amperages.
Charged Terms
C value (for conductors) Multipliers that have been developed for conductors that are derived by including both the resistance and the impedance of a conductor (X/R) installed in electrical systems. These multipliers are used in short-circuit calculations and result in calculation of more accurate short-circuit current values.
counter electromotive force (CEMF) An induced voltage that results in a force opposite in direction to the applied voltage; in AC circuits with magnetic properties (such as motors and transformers), this induced voltage can cause the circuit current to lag the applied voltage, resulting in lower power factor values.
demand factor The ratio of power consumed by a system at any one time to the maximum power that would be consumed if the entire load connected to the system were to be operating at the same time.
load calculations A set of calculated values that determine the demand factor for a system and that reflects a more true value of power utilized at any one time compared to calculated values determined during design.
point-to-point method A calculation method to determine the available short-circuit current values at any point in a system.
short-circuit calculations A set of calculated values that determine available short-circuit currents.
transformer impedance rating A voltage drop rating for a transformer given in a percentage (Z) of the full load voltage.
Check Your Knowledge
You are the Designer
Apply the knowledge you have gained from this previous chapter to your own electrical design. In this section you will:
Complete a set of load calculations
Complete all required short-circuit calculations
Determine the size of the main switchboard that will serve your design
About Your Project
To complete your task, you must know the following details about your project:
All panel schedules (previously designed)
Single-line diagram (previously designed)
To develop this part of your design, you need the following resources:
Completed E_{3} (included in the text and on the Student Resource CD-ROM)
Get to Work
At this stage in the design process, you compile several previously designed components to finalize the plan sheet that illustrates the single-line diagram, panel schedules, and load and short-circuit calculations. For each panel schedule, classify the loads by type (e.g., general electrical, specialized electrical, lighting) and complete a set of load calculations as outlined in the chapter. Apply all demand factors as necessary:
For general electrical loads, the first 10,000 VA are calculated at 100 percent, and the remainder of the load is calculated at 50 percent.
For specialized electrical loads, use 100 percent of the load—except for loads with motors, which require 125 percent of the highest-rated motor in the group plus 100 percent for all remaining motors.
For lighting loads, apply an additional 25 percent of the total load for the long continuous load factor.
Compute the final totals and add them to the Main Distribution section of your plan (see E_{3}) to calculate the total amperage. Reference Table 5-1 to determine the minimum size switchboard necessary for your project. Illustrate the size of the main disconnect on your single-line diagram and list the number of poles and main fuse size.
Tip | Lighting loads will be on for three hours or more, therefore they are considered continuous loads and an additional 25 percent must be added to the designed load, as per NEC 215.2(A)(1). |
Next, complete your single-line diagram by listing the method and size of the grounding electrode conductor for the main switchboard (based on NEC Table 250.66). Remember that if the service entrance conductors are in parallel, the grounding electrode conductor is based on the size ofthe total circular mil area for both conductors. In the sample completed plan, the grounding electrode conductor size is referenced as being a size 3/0 conductor, which was obtained from NEC Table 250.66 based on the area of 1200 circular mil, which is the total of the two 600-kcmil conductors installed in parallel.
Tip | General purpose receptacle outlets are not considered to be continuous loads. For these types of loads, demand factors (reductions in designed load) can be taken. |
For the sample project, the following load calculations were applied:
Panelboard P_{1} | Panelboard P_{2} | Panelboard P_{3} | |||
---|---|---|---|---|---|
Lighting load | 28,306 VA | Receptacle load | 9000 VA | Receptacle load | (43,200 VA total) |
Long continuous load (25 percent)^{[*]} | 7077 VA | Motor load (including 125 percent of the highest-rated motor) | 53,995 VA | 100 percent of first 10,000 VA | 10,000 VA |
Motor load (including 125 percent of the highest-rated motor) | 151,500 VA | 50 percent of remaining load | 16,600 VA | ||
Transformer T_{1} | 60,818 VA | Equipment load | 13,080 VA | ||
Total volt-amperes | 247,751 VA | Total volt-amperes | 62,995 VA | Total volt-amperes | 39,680 VA |
Total amperes | 298 A (480-V) | Total amperes | 76 A (480-V) | Total amperes | 47.8 A (480-V) |
Main Distribution | |
---|---|
P_{1} | 247,751 VA |
P_{2} | 62,995 VA |
P_{3} | 39,680 VA |
Motor load (including 125 percent of the highest-rated motor) | 160,475 VA |
Total | 510,901 VA (616 A) |
Next, determine the demand load for the individual loads served from the main switchboard. For example, for panelboard P _{1}:
Determine the total lighting load including the required long continuous load value:
Lighting load from panel schedule
28,306 VA
Long continuous load (25 percent of total lighting load)
+ 7,077 VA
Total lighting load
33,383
Determine the motor load by adding 125 percent of the highest-rated motor in the group and 100 percent of all remaining motors:
125 percent of highest-rated motor
(CNCMill2)
28,233 VA × 1.25 =
35,291 VA
100 percent of all remaining motors
+116,259 VA
Total motor load
151,550 VA
Determine the total volt-ampere load from the transformer:
Add these values together to determine total load:
Use Ohm’s law to determine the total calculated load in amperes:
Answer: The total demand load for P_{1} is 298 A.
Tip | The 25 percent added to the highest motor in the group provides additional capacity for the motor’s inrush current. When the additional percentage is added to the largest motor, this is sufficient for other motors in the group of smaller horsepower. |
The same guidelines were followed for panelboard P_{2} The load on panelboard P_{3} includes both general purpose receptacle loads and specialized equipment loads (e.g., copy machines, vending machines). The total values for each panelboard were added together in addition to the motor loads served directly from the main switchboard to determine the total load for the main switchboard.
The next step is to determine the total connected load for a main switchboard. For this project:
Add the total loads for each panelboard and the motor load served from the main switchboard (including 125 percent of the highest-rated motor):
Calculate the load in amperes using Ohm’s law:
Answer: The total connected load for this switchboard is 616 A.
Once you calculate this total load, the next step is to size the main switchboard using the standard National Electrical Manufacturers Association (NEMA) panelboard sizes presented in Table 5-1. The minimum size switchboard necessary to serve this facility is 800 A.
Tip | Even if the total calculated load was under 600 A, an 800 A service should be designed. Equipment must not be sized at values greater than 80 percent of its rating. |
Now that the load calculations are complete, you must perform the required short-circuit calculations. Carefully complete each calculation as outlined in the text and illustrated on completed plan sheet E_{3} Enter each result in the appropriate area on your single-line diagram. Additionally, you should review the standard short-circuit rating for devices and illustrate the main fuse disconnect and the standard value for the main switchboard. For example, for the completed project it was determined that the available short-circuit amperage exceeded 22,000 short-circuit A; therefore, it was necessary for the equipment to be rated at the next higher available rating of 42,000 short-circuit A.
To determine the short-circuit amperage for distribution panel P_{1} you must start your calculations in step 1 with the maximum available short-circuit amperage at the main switchboard. The available short-circuit amperage at this panelboard (and other panelboards served by the main switchboard) will be derated because of the additional impedance added by conductor sizes, lengths, and raceway used. Follow the same procedure for the other panelboards. Note that on panelboard P_{3}, there is no motor contribution, so no additional amperage was added.
Tip | Standard short-circuit ratings for devices are 10,000; 22,000; 42,000; and 65,000 A. |