2013 Electrical Load Estimation Course Ali Hassan Certified Energy Manager – AEE ‐ USA Copyrights Reserved for www.Electrical‐Knowhow.com About Author Hi, I'm Ali Hassan el‐Ashmawy, I began my career from 1999 as a site electrical engineer then as area manager from 2001 then as electrical designer from 2003 then as senior electrical designer from 2006 and up to date. In my past experience, I designed and construct about 100 projects in different countries like Egypt, Kuwait, Indonesia, KSA, Gabon and Iraq. My designs were approved by many international authorities like USA corps of engineers and USA ministry of exterior – OBO Office. I'm certified energy manger CEM from AEE – USA since 2006 and I hope to become a well‐known designer in the field of electrical design. To contact me please email to [email protected] Page 2 of 41 • Perform the calculations steps of each method for electrical load estimation.com Course Description: This course is intended to prepare the target persons with the ability to recognize. The target Persons: Design engineers. new graduate engineers. • Understand the procedures and logic of each method for electrical load estimation. and perform preliminary electrical load calculation/estimation for any building type by many calculation methods. Skills Development: On completion of this course the target person will be able to: • Recognize different calculation method for electrical load estimation. understand.Copyrights Reserved for www.Electrical‐Knowhow. Page 3 of 41 . under graduate engineering students. 3.1 3.5 3.2 5.4.Electrical‐Knowhow.com Table of Contents S/N 1 2 3 3.1 5.3 5.5.4.6 Item Introduction Importance of Electrical Load Estimation (Preliminary Load Calculations) Definition of Important terms in Load Estimation Connected load Demand load Demand Interval Maximum demand Demand factor (in IEC.6 3.7.3.4 5.1 5.Copyrights Reserved for www.5 5.3 5.2 5. Factor of simultaneity ks) Diversity factor Difference between demand and diversity factor Load factor Methods of Electrical load estimation Preliminary Electrical Load estimate Difference between preliminary and final load estimate Preliminary load calculations sub‐methods Space‐by‐Space Method (functional area method) Usage conditions of Space‐by‐Space Method Area Measurement in space by space method Method of estimation by using Space‐by‐Space Method First Case Second case The Building Method Comparison between space‐by‐space and building methods Usage conditions of Building Method Area Measurement in Building Method Method of estimation by using Building Method First Case Second case Area method Usage conditions of Area Method Method of estimation by using Area Method First: basic method Second: Optional Method (Load centers method) General notes for all methods of electrical load estimations Page No.7 3.1 5.3.1 3.4 5.5.2 5.4.1 5.8 4 5 5.2 5.4 3. 5 5 5 5 5 6 6 6 7 10 11 18 19 19 19 20 21 21 21 21 21 25 29 29 29 29 29 30 31 35 35 36 36 38 41 Page 4 of 41 .4.3 3. Factor of maximum utilization ku) Coincidence factor (in IEC.2 3.3 5. in the draft design (early design) stage.Electrical‐Knowhow. usually expressed in watts. • Apply to Power Company for supply. these terms are: 3. • Review the utility’s rate structure. the main switchboard and the interconnections to the subsidiary or secondary switchboards. The key single‐line diagram should show the sources of power e. • Calculate initial budget for the electrical works. generators. 2‐ Importance of Electrical Load Estimation (Preliminary Load Calculations) Electrical Load Estimation is very important in the draft design (early design) stage because it help to: • Plan the connection to upstream network and MV circuit configurations. • Develop Demand factor relationship between connected loads and the actual demand imposed on the system.Copyrights Reserved for www.1 Connected load It is the Sum of all the loads connected to the electrical system. utility intakes. 3‐ Definition of Important terms in Load Estimation: There are many important terms which must be understood before performing the load estimation. 3.g. • Plan the transformers substation(s) (if any) and the main switchgear room.com 1‐ Introduction At the beginning of the project. • Make roughly a key single‐line diagram and a set of subsidiary single‐line diagrams. • Review The available voltage system types/classes and levels.2 Demand load It is the electric load at the receiving terminals averaged over a specified demand Page 5 of 41 . the electrical design professional should do the following: • Make Analysis of load characteristics. 5 Demand factor (in IEC.. 30 min.Electrical‐Knowhow.. kilo‐vars. the demand factor at the service where the maximum diversity is experienced is usually 60 to 75 percent of the connected load. with particular attention to electric motors. kilo‐watts. 3. even approaching 100 percent. Page 6 of 41 . Demand may be expressed in amperes.com interval of time.Copyrights Reserved for www.. usually 15 min. kilo‐amperes. For most building types. or 1 hour. For utility billing purposes the period of time is generally one month. 3. or 1 hour based upon the particular utility’s demand interval. The demand factor is the ratio of the maximum demand on a system to the total connected load of the system. 15 minutes.3 Demand Interval It is the period over which the load is averaged. 30 minutes or one hour.. 3. Specific portions of the system may have much higher demand factors. 30 min. Demand factor = Maximum demand load / Total load connected Notes: • This factor must be applied to each individual load. or kilo‐volt‐amperes. usually 15 min. Factor of maximum utilization ku) In normal operating conditions the power consumption of a load is sometimes less than that indicated as its nominal power rating. which are very rarely operated at full load.4 Maximum demand It is the greatest of all demands that have occurred during a specified period of time such as 5 minutes. • Demand factors for buildings typically range between 50 and 80 percent of the connected load. 6 Coincidence factor (in IEC. The third level service size. The apparent‐power supply.Electrical‐Knowhow.e.1): 5 storeys apartment building with 25 consumers. being supplied from a distribution or sub‐distribution board).g. Example#1 (see Fig. each having 6 kVA of installed load. i. Coincidence factor = Maximum system demand / Sum of individual maximum demands Notes: • The factor ks is applied to each group of loads (e. Factor of simultaneity ks) It is a matter of common experience that the simultaneous operation of all installed loads of a given installation never occurs in practice. 4. to the sum of the individual maximum demands of the subdivisions. The total installed load for the building will be the sum of the installed loads in the (5) storeys which will be as follows: Ground floor: Page 7 of 41 . The main service size. The coincidence factor is the ratio of the maximum demand of a system.Copyrights Reserved for www. there is always some degree of diversity and this fact is taken into account for estimating purposes by the use of a simultaneity factor (ks). Solution: 1‐ Calculation of The total installed load.com 3. 2. From Fig. Calculate the following: 1. or part under consideration. 3.1. The total installed load. the installed loads in this storey = 6 x 6 = 36 KVA Second Floor: There are (5) consumers. the total installed load for the building = 24 + 36 + 30 + 24 + 36 = 150 kVA Fig (1) Page 8 of 41 .com There are (4) consumers. the installed loads in this storey = 6 x 6 = 36 KVA So.Copyrights Reserved for www. the installed loads in this storey = 4consumers x 6 KVA installed load per consumer = 24 KVA First Floor: There are (6) consumers. the installed loads in this storey = 4 x 6 = 24 KVA Forth Floor: There are (6) consumers. the installed loads in this storey = 5 x 6 = 30 KVA Third Floor: There are (4) consumers.Electrical‐Knowhow. Electrical‐Knowhow. the Factor of simultaneity ks = 0. the cross‐sectional area of the conductors can evidently be progressively reduced from the lower floors towards the upper floors.63 Page 9 of 41 . These changes of conductor size are conventionally spaced by at least 3‐floor intervals.46 = 69 kVA 3‐ Calculation of The main service size The current entering the rising main at ground level (main service size) = (150 x 0. for number of downstream consumers = 10.com Table#1: Factor of simultaneity (ks) for Apartments Block From Table#1 in above. For vertical rising mains fed at ground level. 2‐ Calculation of apparent power From Table#1. it is possible to determine the magnitude of currents in different sections of the common main feeder supplying all floors.Copyrights Reserved for www.46 x 1000) / (400 x √3) = 100 A 4‐ Calculation of The third level service size The current entering the third floor (the third level service size) = sum of currents delivered to third and fourth floors The number of consumers in the third and fourth floors = 4 + 6 =10 consumers From Table#1. since the number of downstream consumers = 25. the apparent‐power supply required for the building = 150 KVA x 0.46 So. the Factor of simultaneity ks = 0. Calculate the Diversity Factor for this feeder? Solution: The diversity factor can be determined as follows: Sum of total demands = 100 + 95 + 85 + 75 + 65 = 420 A Diversity factor = Sum of total demands ÷ Maximum demand on feeder = 420 A ÷ 250 A = 1. • On Tuesday.68 Page 10 of 41 .g. user one reaches a maximum demand of 100 amps. four reaches 75 amps. being supplied from a distribution or sub‐distribution board). Example#2: Consider that a feeder supplies five users with the following load conditions: • On Monday. five reaches 65 amps. the current entering the third floor (the third level service size) = (36 + 24) x 0. • On Friday. two reaches 95 amps.Copyrights Reserved for www. • On Wednesday. three reaches 85 amps.63 x 1000 / (400 x √3) = 55 A 3.com So.Electrical‐Knowhow. • On Thursday. Diversity factor = Sum of individual maximum demands / Maximum system demand Notes: • The Diversity Factor is applied to each group of loads (e. • The feeder’s maximum demand is 250 amps.7 Diversity factor the diversity factor is the reciprocal of the coincidence factor. B & C. 250 and 200 kilovolt‐amperes (kVA) with demand factors of 95. workshops A. 80 and 75 percent respectively.5 = 866 kVA.1 Difference between demand and diversity factor: Most of the electrical engineers confuse between the demand and diversity factors. being supplied from a distribution or sub‐distribution board). 90. which are very rarely operated at full load. using the diversity factor of 1.g.7.300 kVA However. 350. each workshop will include the following loads: Page 11 of 41 .5. don't forget that: • The Demand factor must be applied to each individual load. to solve this confusion.300 kVA ÷ 1. • The Diversity Factor is applied to each group of loads (e. then the size of the main feeder = 1.com Example#3: Calculate the size of a main feeder from substation switchgear that is supplying five feeders with connected loads of 400.Electrical‐Knowhow.300 kVA ÷ 1.Copyrights Reserved for www. 300. 3. Example #4: An industrial building consists of (3) nos.00 = 1. Use a diversity factor of 1.5. the size of the main feeder = 1. 85.300 kVA 3‐ If the feeder were sized at unity diversity. Solution: 1‐ Calculate demand for each feeder: Feeder#1 demand = 400 kVA × 95% = 380 kVA Feeder#2 demand = 350 kVA × 90% = 315 kVA Feeder#3 demand = 300 kVA × 85% = 255 kVA Feeder#4 demand = 250 kVA × 80% = 200 kVA Feeder#5 demand = 200 kVA × 75% = 150 kVA 2‐ Sum all of the individual demands = 380 + 315 + 255 + 200 + 150 = 1. with particular attention to electric motors. • Draw a key single line diagram for this building? • Determine both the demand (utilization) factor and simultaneity factor with the help of tables # 2 & 3 in below? • Calculate the demand load for each level in the key single line diagram? Table#2: Factor of simultaneity for distribution boards (IEC 60439) Page 12 of 41 .6 KVA total. Workshop B: • • • One nos. • 2 nos. Workshop C: • 2 nos.Copyrights Reserved for www. fluorescent lamps on one circuit with 3 KVA total. fluorescent lamps on one circuit with 1 KVA total. pedestal drill with 2 KVA each. Compressor with 15 KVA. 3 nos. 10 nos. 30 nos.Electrical‐Knowhow. 2 nos. • 5 nos. Oven with 15 KVA each.5 KVA each. fluorescent lamps on one circuit with 2 KVA total. sockets outlets 10/16 A on one circuit with 18 KVA total. lathe with 5 KVA each.com Workshop A: • • • • 4 nos. ventilation fans with 2. 5 nos. sockets outlets 10/16 A on one circuit with 18 KVA total. sockets outlets 10/16 A on one circuit with 10. • 20 nos. fig.com table#3: Factor of simultaneity according to circuit function solution: Follow the solution steps in below and in fig.2 Page 13 of 41 .Electrical‐Knowhow.Copyrights Reserved for www.2. IEC gives Ku estimation values for these loads as follows: • • • For motor Ku = 0.1 No.8 4 No.Copyrights Reserved for www.6 1.4 pedestal drill No.1 5 0.1 No.2 5 nos.6 18 3 12 10.KVA Load Type Load No. Step# 4: group same type of loads on one distribution panel/box and this will be the first Level of distribution (LEVEL 1). sockets outlets 10/16 A 30 nos.8 4 1.2 Workshop C: Utilization Factor Max. Apparent Power (KVA) Oven No.com Step#1: List all the loads in each workshop and write the apparent power of each load in KVA beside it. Step#2: write the utilization factor for each load.5 1 2.8 0.5 15 15 18 2 1 2. fluorescent lamps ventilation fan No.2 5 nos. sockets outlets 10/16 A 20 nos. fluorescent lamps Step#3: calculate the Max.5 1 15 1 15 1 18 1 2 No.1 5 2 2 18 3 15 0. Demand apparent power in KVA for each load = apparent power X Ku for each load.8 4 No.8 4 No. sockets outlets 10/16 A 10 nos.6 1 1 1 2.8 1 1 0.8 For socket outlets Ku = 1 (depend on the type of appliances being supplied from the sockets concerned) For light circuits Ku= 1 The Table of Calculation for Steps# 1&2 will be as follows: Workshop Name Workshop A: Workshop B: Apparent Power Demand Max.5 2. fluorescent lamps Compressor 3 nos.Electrical‐Knowhow.2 5 0.6 1 10.8 0. Page 14 of 41 . lathe No.3 5 0. sockets outlets 10/16 A 10 nos.4 pedestal drill No.5 1 15 1 35 1 15 1 18 0.1 5 0. we will have (3) main distribution panel/box for the (3) workshops and this will be the second level of distribution (LEVEL 2).3 1 1 1 1 1 2. Demand apparent power in KVA X simultaneity factor for each distribution panel/box. sockets outlets 10/16 A 30 nos.28 5 1 2 1 2 Step# 7: group the distribution panel/box in each workshop in one main distribution panel/box.8 1 1 0.5 1 2.6 18 3 12 10.8 0. sockets outlets 10/16 A 20 nos.Electrical‐Knowhow.8 0.8 No. write the simultaneity factor for each distribution panel/box and from table # 3 write the simultaneity factor for each for each separate load. fluorescent lamps ventilation fan No. So.1 5 2 2 18 3 15 0.6 1 1 2.2 5 nos.75 14. lathe No.1 No.3 5 0.2 5 0.5 15 15 18 2 No.2 Worksho p C: Appar Utilizat ent ion Power Factor (KVA) Max.2 1 1 3.4 0.5 2. Demand apparent power in KVA for each distribution panel/box = sum of all branch loads’ Max.K nd factor VA Max.1 No.com Step# 5: in level 1 and from table #2.6 1. Step# 6: calculate the Max. Level‐1 Appar ent Appar Power ent Dema simultan Power nd Dema eity Max. Page 15 of 41 .4 4.6 0. The Table of Calculation will be as follows: Worksho p Name Worksho p A: Worksho p B: Load Type Load No.K VA 4 Oven No. fluorescent lamps Compressor 3 nos.2 5 nos.8 4 No.6 3 12 10.8 4 1.8 4 No.Copyrights Reserved for www. fluorescent lamps 0. sockets outlets 10/16 A 20 nos.6 1 10.8 Page 16 of 41 .9 37.8 4 No.5 15 15 18 1 15 1 18 0. aren t Pow er (KVA ) Utiliz App ation aren t Facto Pow r er Max.6 1 3 1 3 15 0.5 1 35 No. KVA Level‐2 simult App aneity aren t factor simult App aneity aren factor t 0.2 3.1 1 2. Dem and Max.com Step# 8: in level 2 and from table #2. write the simultaneity factor for each main distribution panel/box. KVA 18.6 18 0.3 5 0.2 5 0.5 Oven No. sockets outlets 10/16 A 10 nos.1 1 15 No. fluorescent lamps ventilation fan 1 1 1 1 1 No.9 Pow er Dem and Max.4 4. fluorescent lamps 0.2 1 2.8 4 No.3 No. KVA 14. fluorescent lamps Compressor 2 2 18 0. The Table of Calculation will be as follows: Worksh op Name Works hop A: Works hop B: Works hop C: Load Type Load App No.8 12 1 12 10.Electrical‐Knowhow.8 4 pedestal drill No. Demand apparent power in KVA for each main distribution panel/box = sum of all branch distribution boxes’ Max.8 4 No.6 0.6 0. sockets outlets 10/16 A 30 nos.4 5 0. Demand apparent power in KVA X simultaneity factor for each main distribution panel/box.75 Pow er Dem and Max.1 5 0.6 1. Step# 9: calculate the Max.8 0.9 0.5 2.4 1.9 15.1 No.Copyrights Reserved for www.2 2.28 5 2 1 2 1 2 5 nos.2 5 nos.8 1 3 lathe Level‐1 3 nos. 8 1. 2 No.8 4 2 0. sockets outlets 10/16 A 10 nos. fluorescent lamps ventilation fan Level‐1 Level‐2 App aren t Pow er (KVA ) Utiliz ation Facto r Max.6 3 1 3 1 3 simult aneity factor App aren t Pow er Dem and Max.8 1.4 4. 6 2.8 4 5 0. Level‐3 15.3 1 1 1 1 simult aneity factor App aren t Pow er Dem and Max.6 2 0.6 18 1 18 0. KVA 0. 2 5 nos.2 3.9 37. 3 No.5 1 35 0. fluorescent lamps Compressor lathe 3 nos. sockets outlets 10/16 A 30 nos. App aren t Pow er Dem and Max. 9 0.8 12 1 12 10. write the main general distribution board MGDB.9 No.Copyrights Reserved for www. KVA 0.8 4 5 0.8 4 5 0. 4 pedestal drill No. 6 0. KVA simult aneity factor App aren t Pow er Dem and Max.9 65 0. The Table of Calculation will be as follows: Workshop Name Works hop A: Works hop B: Works Load Type Load No. Page 17 of 41 . Step# 11: in level 3 and from table #2. Demand apparent power in KVA X simultaneity factor for main general distribution board MGDB. Demand apparent power in KVA for main general distribution board MGDB = sum of the (3) workshop main distribution boxes’ Max.5 1 2. 1 No.com Step# 10: group the (3) main distribution panel/box in one main general distribution board MGDB and this will be the third level of distribution (LEVEL 3).9 18.75 14. 6 1 1 10. No. Step# 12: calculate the Max. 1 No.Electrical‐Knowhow. KVA 5 0. 4 15 0. fluorescent lamps 8 2.28 5 2 1 2 1 2 3.5 15 1 15 15 1 15 18 1 18 0.Copyrights Reserved for www. Load factor = Average load / Maximum load Free download You can download tables for different factors listed above by clicking the following links: • IEEE Demand Factor Values • Unified Facilities Criteria ‐UFC‐ Demand Factor Values • NEC Demand Factor Values • Demand Factor Values From Other Regulations • Diversity Factor Values • Unified Facilities Criteria ‐UFC‐Load Factor Values • IEC Factor of Simultaneity Values Page 18 of 41 . sockets outlets 10/16 A 20 nos. 2 5 nos. usually 1 year.8 Load factor The load factor is the ratio of the average load over a designated period of time.Electrical‐Knowhow. 1 No.com 1 hop C: No.5 1 2. 2 Oven No. to the maximum load occurring in that period. 3. which are: A‐ Preliminary load calculation This method is subdivided into (3) sub‐methods as follows: 1. Area method. because the preliminary load estimation methods are used in the early design phase while the other two methods are applied in the final stages of design.Electrical‐Knowhow.1 Difference between preliminary and final load estimate before going through the calculation steps for Preliminary Electrical Loads. C‐ Final load calculations. or/and HP (VA/ft2) volt‐amperes per square (horse power) foot Page 19 of 41 . 2. NEC load calculations and Final load calculations will be explained later in course " EE‐3: Basic Electrical design course – Level II ”. 5‐ Preliminary Electrical Load estimate 5. B‐ NEC load calculations. and the two other methods.Copyrights Reserved for www. The following table shows these differences as follows: S/N 1 Preliminary load calculations Final load calculations Units of Loads will be in (W/ft2) Units of Loads will be in KW (kilo‐watt). watts per square foot or/and or/and KVA (kilo‐volt‐ampere). I will explain the preliminary load estimation methods. Space by space (functional area method). we need to highlight the main differences between the load estimation or calculation by the preliminary and final methods. Building method.com 4‐ Methods of Electrical load estimation There are (5) methods for Electrical Load Estimation. Note: In this course. Area method. Demand and load factors values are Real values that will document and reflect the number. Hp will be converted to kVA. and the diversity of operation of equipment served by the common source. Note: A particular design may use one Preliminary load estimate method or a combination from two or even the three methods. and varying). intermittent.Copyrights Reserved for www. the type. the duty rating (continuous. cost effective calculations insuring that items of equipment and materials are adequate to serve existing.2 Preliminary load calculations sub‐methods: As I indicated before. 5 The connected load will be estimated based on area or population 6 Easy and Fast calculations Units can’t used interchangeably.com 2 units are used interchangeably because unity power factor is assumed 3 Unity power factor is assumed 4 Demand and load factors values will be selected from tables based on the designer estimation and they will be Used to calculate the transformer and service size. periodic. 3. and kVA may be multiplied by the estimated power factor to obtain kW if required Different values of power factors according to load types. Building method. short time. and future load demands 5. Actual demand load will be calculated based on summation of individual building connected loads modified by suitable demand and diversity factors economical. 2.Electrical‐Knowhow. this method is subdivided into (3) sub‐methods as follows: 1. Page 20 of 41 . So. new. Space by space (functional area method). and the wattage or volt‐ampere rating of equipment supplied by a common source of power. Copyrights Reserved for www. workshops and etc. conference halls. 5. these load densities in addition to spaces area will be used to estimate the preliminary electrical load of this building as described in below. shops. parking areas.3. corridors and lobbies. Page 21 of 41 .2 Area Measurement in space by space method The square footage is measured from the outside surface of exterior walls to the centerline of walls between interior partitions of the spaces. 5. And the sum of the Gross Interior Area equals the total Gross Area of the building. Second case: availability of individual load density (i. office. The Load density in (W/ft2) or/and (VA/ft2) is prescribed for these different spaces.e.com 5.Electrical‐Knowhow. The Space‐by‐Space Method may be used for any building or portion of a building.3 Method of estimation by using Space‐by‐Space Method in this method.3.3.3 Space‐by‐Space Method (functional area method) In the Space‐by‐Space Method.1 Usage conditions of Space‐by‐Space Method • • The Space‐by‐Space Method is used only for individual spaces in the building. we have two cases as follows: • • First case: availability of grouped load density (i. 5. general power and power loads) in (W/ft2) or/and (VA/ft2) for each space. one value covering all lighting. general power and power loads) in (W/ft2) or/and (VA/ft2) for each space.e. individual values for lighting. First case: Method of estimation by using Space‐by‐Space Method will be as follows: 1‐ Divide the building into different space based on its function (for example. the building will be divided into different space based on its function like offices. above the images of our members. After finishing your registration. .To download your complete copy of this course. asking for your password and I will send it with email reply.com. you must be a member and to be a member you must register by click on the phrase “Join this site" in left bottom side of any page. you will need to enter your password to open the file. After downloading.electrical‐ knowhow.html You can download the Course by just click on its name. send email to [email protected]/2012/12/electrical‐pdf‐courses. To get your password. please Visit the following link: http://www.