ABB Group. Leading digital technologies for industry — ABB Group.Branch-Circuit, Feeder, and Service Load Calculations | UpCodes

The subject matter covered in most electrical licensing examinations is: Grounding and bonding, Overcurrent protection, Wiring methods and installation, Boxes and fittings, Services and equipment, Motors, Special occupancies, Load calculations, Lighting, Appliances, Box and raceway fill, Hazardous locations, Trade knowledge and Electrical theory.

For complete practice, try the Master Electrican Practice Exam Kit with questions and fully explained answers. With Tests. Check with your local code official to determine which code applies to your location. View Answers as You Go. According to Section This calculation method applies to a single dwelling unit, whether it is a separate building or located in a multifamily dwelling.

Generally, the optional calculation method provides an easier and less complex calculation method than the standard general method for computing the total demand load for feeders and service-entrance conductors. Incorrect answer. Please choose another answer. Due to the fact all the equipment on a farm will not be used simultaneously for an extended length of time, it is permitted to apply the demand factors shown in Table Table The Class I, Division 1 location below grade extends from the floor of the pit to the service area floor level.

The requirements for the location of conduit seals are to prevent an explosion from traveling through the conduit to another enclosure and to minimize the passage of gases or vapors from hazardous locations to nonhazardous locations. In Class 1, Division 1 locations, if the conduit enters an enclosure that contains arcing or high-temperature equipment, as required by Section As addressed in Section The barriers are required between the switches fed from two different phases of this system because the voltage between the phase conductors would be volts and would exceed the volt limit.

Study Online Instantly. For the purpose of determining box size and fill, Table However, when sizing pull and junction boxes containing conductors of size 4 AWG and larger, the box dimension is to be calculated based on the diameter of the raceways entering the enclosure and the rules specified in Section Column 3 of Table The intent of Section When electrical conduit or tubing enclose conductors of the same size, with the same type of insulation, where the raceway is more than 24 inches in length, Tables C.

Table C. The general requirement for selecting the minimum size equipment grounding conductor is to select directly from Table In this situation, the circuit breaker has a rating of amperes, therefore a minimum size 8 AWG copper equipment grounding conductor is required to be routed with the air-conditioning unit supply conductors. Part II of Article provides the general requirements for overcurrent protection and overcurrent protection devices not over 1,volts.

Section This text parallels the requirement of Section The warning ribbon reduces the risk of an accident, such as electrocution or an arc-flash incident during excavation near underground service conductors that are not encased in concrete, because these circuits are not protected from short circuit and overload.

Article establishes the requirements for interconnecting power production equipment, such as solar photovoltaic PV systems, that operate in parallel with a utility supplied service. As per Section The provisions of Article cover the electrical installations within a recreational vehicle park. As noted in Section The demand factors shown in Table Article address the mandates for electrical equipment and wiring for all voltages in Zones 0, 1, and 2 hazardous classified locations where fire or explosion hazards may exist due to the presence of flammable gases, liquids, or vapors.

Conduit seals are to prevent an explosion from traveling through the conduit and minimize the passage of gases or vapors from hazardous to non-hazardous locations. It is practically impossible to make threaded conduit joints gastight. The conduit seal is intended to prevent the passage of flammable liquids, gases or vapors that can exist in Class I, Division 1 locations.

The provisions included in Article applies to any temporary installation whether it is at a construction site, retail mall parking lot, or a local art and crafts fair in a previously unoccupied open lot or field.

Therefore, the volt receptacles are to have GFCI protection also. This GFCI protection for the temporary installed receptacles is intended to protect personnel from shock hazards that may be encountered during construction or maintenance of equipment. Part III of Article addresses the requirements for grounding of electrical systems. To ensure an adequate grounding electrode system, many times the grounding electrode system may consist of multiple driven electrodes ground rods rather than reliance on a single driven ground rod.

This practice is very common when encountering sandy soil conditions. To solve this problem first find the VA rating. Article introduces general requirements for wiring methods and materials. The purpose of this rule is to prevent the accumulation of debris between the conduit and the mounting surface and help to prevent the conduit from rusting. Prev Next Finish. Quality starts with who wrote the material. Our practice exam writer s.

Ray is a certified instructor of electrical trades. His classes are presented in a simplified, easy-to-understand format for electricians. He is an active member of the International Association of Electrical Inspectors and a retired member of the International Brotherhood of Electrical Workers. Bookmark Page. Class I, Division 1. Class I, Division 2. Class II, Division 1. Class II, Division 2.

None of these. Given: You are to install an underground run of trade size 2 in. The PVC will not cross under any public streets, roads, driveways or alleys. Calculate the minimum demand load, in amperes, on the ungrounded service-entrance conductors for a recreational vehicle park that has 28 RV sites all with ampere outlets at each RV site.

I only. II only.

amp feeder calculation | Mike Holt’s Forum

I pretty sure the formula given is correct. Within the body of the post I have added an example of how to derive the formulae. Please check this and if you find any errors or mistakes let me know. How to use the formula and which formula to find it out..??? You need to be cognisant of what you are supplying and if this will affect the sizing for example if the load is non-linear. Not directly related to the post, but this is a great question.

Would it be possible for you to ask this in our question section. I think you will get a better response and help our other users. I will also put together a proper answer when I get some time.

Can we use in Asia volt 50 Hz? I would say you cannot use the motor at 50Hz, V. The motor would be running faster due to the frequency. In addition the voltage would be increase the torque and current for which the motor is likely not designed. I have the same issue with our Contractor supplying pump motor 50hz, where-in our system here is 60 hz.

To make the situation confusing, the contractor got the confirmation from the manufacturere. ABB that according to their calculation, that 50hz motor would be fine for 60hz system.

Need your excellent advise Mr. Thank you so much! I deal with unbalanced systems on a frequent basis and am trying to figure out how to calculate the amps drawn on each leg. Strictly speaking you should be using kVA which you can obtain by dividing the watts by the power factor.

Also make sure you are really dealing with a delta system and not a star-connected system i. Steven, let me give you a few more details. I work in the entertainment industry and definitely work with a star connected wye system where all legs measured to neutral is V and leg to leg is V. What I need to know is how many amps I’m pulling on each individual leg. We have a lot of loads that are connected line-neutral and figuring out those loads is easy.

Where I have issues is when we have loads connected across two phases. If the system is perfectly balanced I again have no issues figuring out the amps drawn on each leg. But in the example I gave you the loads are quite obviously not balanced. Now the calculation you gave me gives me the total amps on A-B but it doesn’t tell me how many amps I’m pulling on A and how many amps I’m pulling on B.

I know when you are dealing with loads connected across phases figuring out the current isn’t as straight forward as using ohms law. I just don’t know how to go about it. I understand a little more now perhaps not fully. I think you can still convert to three single phase problems, with phase A having halt of the AB load and half of the AC load.

Thanks for the write up. In your example where you break down a 3-phase problem to a single phase problem with the 36kW load, you get 60A.

That 60A is total rms current for a single phase, not the peak current, correct? So is it as simple as multiply by 3 to get the total, 3-phase rms current? Your correct it is the RMS current. You could do the calculation in terms of peak current if you wanted for a sine wave the relationship between the peak and RMS is the square root of two.

You do not need to multiply by 3. The current calculated is the line current – if you place an ammeter in the circuit, this is what you will measure. So it is the second part of your reply that I don’t understand. In a three phase system, the amount of current going through one of the phases is not a fraction of the amount of current going through all three?

If I was to create a 3 phase fault by shorting all three phases together, how much current would go through the fault towards the load? Also, thanks again for posting and all of your help. Try to look at the following post. In the post, the motor is a three phase load, where as the socket is single phase – in each case case you have the line current which is common for example, the current flowing in L1.

If you create a fault the current will increase you have reduced the circuit impedance with the fault. Current won’t flow through the fault to the load or very little will. At the fault location, because the phases are shorted, they will be at nearly the same voltage close to zero depending on the impedance. The load sees this voltage and no current will flow.

Hope this makes sense. For a three phase AC generator. Hi all, i have problem in total current computation from ;three phase and single phase, the system is v 3-phase, line to neutral is The three phase current as you are trying to calculate it does not exist. In the A phase you have You still have your three phase system, with the current in each line being slightly different if all the line currents were the same it would just be a balanced three system.

Because you have an unbalanced system, you will have some current in the neural line. You can calculated this be summing the A, B and C phases, taking into account the phase difference easiest to do this using complex notation. Three phase simply means you have three windings instead of one; giving you the A, B and C phases and if star connected a neutral conductor.

In a single phase system you would only have one winding, giving you a single phase live and neutral. Hope this makes some sense. You give the worse case as I’m not sure of where you are getting your If the This is done all the time fridge or toilet extract fan in your house for example. Just try to balance your single phase loads as much as possible across all three phases. Power factor 0. Kindly show the computation. I like to use kVA as this is just the voltage multiplied by the current.

If you know the kVA and voltage, it is very easy to work out the current. If you wanted to do all your calculations in kW, you could – but you would need to include the power factor as well in any calculation. Thanks Steven, it helps a lot. But if you don’t mind, I would like to ask a few more question about the apparent power and complex power. What is the difference between the two. I have a book here about circuits it tells that the key to solving problems about three phase is to find the current.

I know that it is right but how can i used it to find the value of complex power?? And the value of the impedance and the reactance if required. The following note touches briefly on the topic of complex power: alternating-current-circuits Have a read of the note and if you are still unclear, you can add a comment to post.

I know this probably is easy to understand, but I got myself to a completely confused state. Thank you! In general when connecting in delta you have a higher voltage on each leg, so I would say you consume more power. I try not to think like that as it confuses me as well and there is sometimes more to it than that. My advice would be to look at your actual situation and analysis it to see the effect of changing from star to delta. Also be aware that if draw more power, you equipment may not have been designed for it.

Sir, when a voltage dip occurs current increases, so instead of dimming, a bulb should glow brighter but the reverse thing happens, why? I believe that as you lower the voltage across the lamp, the current would reduce and the brightness think of the lamp as a resistance. Hi Steven, I drive a centrifugal water pump with a three phase motor. I measure the Power kW required by the electrical motor.

I expect it to stay the same. Am I correct? Hi Steven, Thanks for sharing your knowledge, it really helped me understanding the theory. And now I would like to know how does the calculation works when a generator has to be used?

Does Power Factors plays any role? It works the same a for a consumer of power. Just multiply the current by the phase-neutral voltage to get the power per phase and then by 3.

The power factor only comes into play if you want to see the kW just multiply the kVA by the power factor. Hi steven sir, I want to know in motors we use capacitor, we use star connection for starting 3 phase, is it because the current lags the voltage by 90 degree in an coil or is it for a different reason..

Regards Abdulla. Sorry if this is a simple question. What is the difference between three phase power factor and single phase power factor.

The power factor is the ratio of the real work to the apparent power. The same definition holds for both three phase and single phase systems. Generally when people talk about power factor, they are concerned with the overall system. If you are looking at it on a phase by phase you may need to be a little careful.

Thank you. Assuming V is Line Voltage and load is balanced. Many questions sent in to the site are in connection with motor starting and in particular star-delta. For all but the simplest application, there is If two dissimilar metals are touching and an external conducting path exists, corrosion of one the metals can take place. Moisture or other materials The correct sizing of current transformers is required to ensure satisfactory operation of measuring instruments and protection relays.

Several methods IEC ‘Protection against lightning’ requires a risk assessment be carried out to determine the characteristics of any lightning protection system It wasn’t so long ago I was telling someone that I don’t use rules of thumb as most things are easily calculated anyhow. As it turns out I last week Straight forward list of some common motor faults.

If I have missed any other common faults, please take a bit of time to add them in as a comment below It a good demonstration of innovative Collection of links to various places with useful motor information.

Fault calculations are one of the most common types of calculation carried out during the design and analysis of electrical systems. These calculations Anyone specifying or using electric motors should have a basic understanding how the insulation is related to temperature. Three classes of insulation If you have some expert knowledge or experience, why not consider sharing this with our community.

By writing an electrical note, you will be educating our users and at the same time promoting your expertise within the engineering community. If conductor is oversized a lower temperature can be used. See NOTE 4. This value only used for adjusting conductor resistance.

PVC conduit aluminum conduit steel conduit no conduit. Length of cable in feet one-way distance. Load current in amperes. This is recommended by the NEC A common rule of thumb is to check the voltage drop when the one-way circuit length in feet exceeds the system voltage number. Therefore, using this rule one would check the drop for a V system if the circuit length exceeds feet.

For refining the calculation the conductor operating temperature can be estimated as follows: If the operating current equals the ampacity listed in NEC Tables If the operating current is less than the ampacity listed then the temperature will be less.

This results in a slight reduction of the voltage drop and may be useful for marginal calculations. Are YOU ready for the next power outage?

We welcome your feedback or questions. Click here to contact us.

Electrical calculation tools | Schneider Electric Global

 
Apr 26,  · x = V drop / x = % drop; Conclusion: % voltage drop is very acceptable. (See NEC Article , which suggests that a voltage drop of 3% or less on a feeder is acceptable.) To select minimum conductor size: Determine maximum desired i voltage drop, in volts. Divide voltage drop by ii (amperes x circuit feet). Jul 22,  · ABB is a pioneering technology leader that works closely with utility, industry, transportation and infrastructure customers to write the future of . Raza Haider, Chul-Hwan Kim, in Integration of Distributed Energy Resources in Power Systems, Overcurrent protection scheme. The overcurrent protection scheme is used to protect the distribution lines of electric grids integrated with DER. This protection scheme is further classified into two categories, the phase overcurrent protection and the ground overcurrent .

POWER SYSTEM PROTECTION.Branch-Circuit, Feeder and Service Calculations | Electrical Contractor Magazine

The provisions of Article cover the electrical installations within a recreational vehicle park. As noted in Section The demand factors shown in Table Article address the mandates for electrical equipment and wiring for all voltages in Zones 0, 1, and 2 hazardous classified locations where fire or explosion hazards may exist due to the presence of flammable gases, liquids, or vapors.

Conduit seals are to prevent an explosion from traveling through the conduit and minimize the passage of gases or vapors from hazardous to non-hazardous locations. It is practically impossible to make threaded conduit joints gastight. The conduit seal is intended to prevent the passage of flammable liquids, gases or vapors that can exist in Class I, Division 1 locations.

The provisions included in Article applies to any temporary installation whether it is at a construction site, retail mall parking lot, or a local art and crafts fair in a previously unoccupied open lot or field. Therefore, the volt receptacles are to have GFCI protection also. This GFCI protection for the temporary installed receptacles is intended to protect personnel from shock hazards that may be encountered during construction or maintenance of equipment.

Part III of Article addresses the requirements for grounding of electrical systems. To ensure an adequate grounding electrode system, many times the grounding electrode system may consist of multiple driven electrodes ground rods rather than reliance on a single driven ground rod. This practice is very common when encountering sandy soil conditions. To solve this problem first find the VA rating.

Article introduces general requirements for wiring methods and materials. The purpose of this rule is to prevent the accumulation of debris between the conduit and the mounting surface and help to prevent the conduit from rusting. Prev Next Finish. Quality starts with who wrote the material. Our practice exam writer s. Ray is a certified instructor of electrical trades. His classes are presented in a simplified, easy-to-understand format for electricians. To use Table In the NEC, the previous language in The new note under the table makes it clear that the table can only be used if adjustment or correction factors are not required.

Below is a preview of the NEC. ORG for the complete code section. N A Services. Instant access to millions of ebooks, audiobooks, magazines, podcasts and more. The SlideShare family just got bigger. Enjoy access to millions of ebooks, audiobooks, magazines, and more from Scribd. Successfully reported this slideshow. Your SlideShare is downloading. Download Now Download. Next SlideShares.

You are reading a preview. Activate your 30 day free trial to continue reading. Continue for Free. Upcoming SlideShare. Fundamentals of Power System protection by Y. Paithankar and S. Protective devices and their coordination in power systems. Embed Size px. Start on. Show related SlideShares at end. WordPress Shortcode. Share Email. Top clipped slide. Download Now Download Download to read offline. Education Business. Power system protection devices.

Improving the reliability of feeder protection using digital relays. Power system protection topic 1. Power system-protection-presentation-datedintegrated-protection-c Protection schemes and zones. Power System Faults and Protection System. Er rahul sharma feeder protection. More Related Content Slideshows for you. Protection of transmission lines encrypted. Power system Protection Services in India. Fundamentals of Microprocessor Based Relays.

Power System Protection basics.