380/1.73 = 219 V Line-Neutral (Phase Voltage) output voltage must not be more then 4.5V even when zener diode burnout. Line & Phase Current and Line & Phase Voltage in Delta (Δ) Connection. http://myelectrical.com/notes/entryid/172/three-phase-power-simplified After looking at the above post, please let me know if it makes more sense or if you still need a better explanation. I measure the Power (kW) required by the electrical motor. To measure the current, I will have the resistor in series with one of the phases and then measure the voltage across the resistor. Just try to balance your single phase loads as much as possible across all three phases. Info. Thanks Steven! 746/3 = 249 watts per phase Perhaps the idea of measuring a 3 phase power current sounds intimidating to some of you. By the way, I think you probably wanted to say 230 V (not 210 V) if VLL is 400 V. How to get power factor three phase motor specification as below. Divide kW by the power factor to get the kVA. tanθ=Q/P, Hi all, i have problem in total current computation from ;three phase and single phase, the system is 380v 3-phase, line to neutral is 220. i computed the current from line to neutral AN=80.6, BN= 65.8, CN=73.2a three phase current is 491.5a. The current flowing through motor depends on various factors including … BC=651.85A, AC=651.84A, supply voltage 220 line to line. Introduction to Three Phase Electric Power, Cables for MV Power Distribution - Earthed versus Unearthed Systems, Understanding LV Circuit Breaker Fault Ratings. The power factor only comes into play if you want to see the kW (just multiply the kVA by the power factor). Current won't flow through the fault to the load (or very little will). Later, using Ohms law, I will figure out the current draw. Tap to unmute. Because you have an unbalanced system, you will have some current in the neural line. But in the example I gave you the loads are quite obviously not balanced. I expect it to stay the same. Hence, three wattmeters are required to measure power in three-phase, four-wire system, whereas, only two wattmeters are required to measure the power in 3 phase, 3 wire system. 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). I have now changed it to show kVA. Using these rules it is possible to work out any three phase problem without the need to remember and/or resort to formulas. In this configuration, the total power, watts is accurately measured by the algebraic sum of the two wattmeter values.   I= S 1ph V LN = P 3×pf 3 V LL, Simplifying (and with 3 = √3 x √3):      Also be aware that if draw more power, you equipment may not have been designed for it. Fig (2). Having said that, you can easily use the variables that you know the value of to find any unknown values. Often where single phase loads are involved, residential and commercial premises for example, the system can be unbalanced with each phase have a different current and delivering or consuming a differing amount of power. Now that we've calculated a single phase power current, we can move on to doing the same for 3 phase power currents. John, line voltage is dependant on country (and varies within countries). When the load is star connected load, the diagram is shown in below-For star connected load clearly the reading of wattmeter one is product of phase current and voltage difference (V 2-V 3).Similarly the reading of wattmeter two is the product of phase current and the voltage difference (V 2-V 3).Thus the total power of the circuit is sum of the reading of both the wattmeters. UNBALANCED LOAD : Fig. ... (3 phases plus neutral), this is known as a Wye (Y) configuration. Sorry if this is a simple question. To find the kVA you can use sqrt(kW^2 + kWr^2), where kWr is the reactive power. thanks :), Hi, I believe your equation I = kW / ( √3 x pf x VLL ), in kA Should VLL not be VLN here? While you're certainly welcome to take a peek at Wikipedia's technical entry, we think that you'll find our simple approach to the this task a bit more...suitable for beginners. System voltages and currents can be found by drawing out the circuit in full detail and using Kirchhoff's laws and other network theorems. 249/0.72 = 345 VA per phase If you interested in an introduction you can view our post: Network Theory – Introduction and Review. Is it possible to measure the leakage current of a 3-phase/3-wire or 3-phase/4-wire device? Here is the three phase energy meter modeled in Autodesk Circuits so … Consequently, we'll be taking this time to briefly touch on watts, apparent power, and power factors. But enough of that talk. We can redraw the circuit with polarity marks to better see how the current going into the CT primary polarity mark comes out the secondary polarity mark to match the polarity of the reference. The apparent power is also tripled in a 3 phase system. kVA per phase = kW per phase / power factor To supply the power, 1 amp flows through 6 wires and there are thus 6 units of loss. I understand a little more now (perhaps not fully). If you have some expert knowledge or experience, why not consider sharing this with our community. Similarly given the power in each phase you could easily find the phase currents. Try to look at the following post. Phase Voltage = VLL/√3 Once you know the kVA, you can calculate the power factor form pf=kW/kVA. Using the aforementioned formula, let’s confirm this rating: Follow answered Nov 12 '15 at 20:10. I am using a 1 mOhms shunt resistor. Use a From block to read the currents. Total three phase power = 18.4 + 16.1 + 18.86 = 53.36 kVA. The above method relies on remembering a few simple principals and manipulating the problem to give the answer. I have electric motor from USA, 3 phase 230 Volt 60 Hz. I think you will get a better response and help our other users. There are two reasons for our presumption here, the first of which is that calculating a single phase power current is much simpler than calculating a multiphase or three Hope this makes some sense. Features of Three Phase AC Voltage Measurement using Arduino: Can measure AC voltage of any amplitude. If playback doesn't begin shortly, try restarting your device. tian, There is always more than one way to do things. The same definition holds for both three phase and single phase systems. I try not to think like that as it confuses me as well and there is sometimes more to it than that. In addition the voltage would be increase the torque and current for which the motor is likely not designed. Want to inquire more Here we are listing common materials along with... Cable insulation is used to provide electrical separation between conductors of  a cable. I've always been bad a remembering formulae, so I prefer to remember concepts/methods. I deal with unbalanced systems on a frequent basis and am trying to figure out how to calculate the amps drawn on each leg. If you do a site search for "International Voltages" we have a Wiki page which lists some of these (may be slightly out of date). Dear Steven - I am a student and I've been struggling to understand something: if we have a balanced STAR connection with a certain apparent power consumption (VA), will this power change if we switch the connection to DELTA? They are: The current coil is connected in series with the line carrying current. To put things simply, the crucial difference between the two systems is the voltage; three phase systems have line to line voltage (VLL) and phase voltage (VLN). Within the body of the post I have added an example of how to derive the formulae. You can't divide the current by three (power kW or kVA - yes). Motor rated power is normally the output power, so to be fully accurate you need to consider the efficiency. Even if you can't calculate the exact answer when dealing with an unbalanced system, though, you've at least figured out (numerically, of course) just what makes 3 phase power that popular kid on the playground that many businesses want on their side. Thanks. It is measured in Joules per second, i.e., watts. So is it as simple as multiply by 3 to get the total, 3-phase rms current? More traditionally formulas may be used to give the same result. I line = 0.90 * 1.73 = 1.57 A c) I am looking to find the current going through each phase. Hi Steven, Thanks for sharing your knowledge, it really helped me understanding the theory. units and get the 3 phase power current you need to run more efficiently. In such cases, you'll need to rely on much more complicated math to get an accurate answer. In general when connecting in delta you have a higher voltage on each leg, so I would say you consume more power. Add up the three phases = 48720 VA (or 48.72 kVA). At this point, we must introduce a new formula that will allow us to calculate the current. Where pf is the power factor, I is the current, V is the voltage and P is the power. If you know the kVA and voltage, it is very easy to work out the current. Kilowatts (kW) can also be used to measure this power; one kilowatt is the equivalent of 1000 watts. Not directly related to the post, but this is a great question. Phase A - 1/2 56,000 + 1/2 17,00 = 36,500 W - divide by 120 V gives 304 A in the line conductor - similar for the other phases, giving B=329 A and C=167 A I have assumed the power factor =1 and W = VA. The following note touches briefly on the topic of complex power: Use a label. You might even want it on your side one day. Apparent power (VA or volt-ampere) is calculated by finding the product of the voltage and the current; apparent power can also be measured in kilovolt-amperes (kVA). Thanks for the write up. 2. Power is defined as the work performed in a specific time or simply a rate of doing work. For Example :- Each phase of 3 phase machine takes a load of 16 amps, So how do I calculate how many KVA generator is required ? A kVA is equal to 1000 volt-amperes. Before we get into the math, however, you must understand exactly how a 3 phase system differs from a single phase system. In your example where you break down a 3-phase problem to a single phase problem with the 36kW load, you get 60A. Does Power Factors plays any role? To perform an accurate measurement to three-phase power installations the power of all three individual phases must be measured.  Share. I like to use kVA as this is just the voltage multiplied by the current. As an example, consider a balanced three phase  load consuming 36 kW at a power factor of 0.86 and line to line voltage of 400 V (VLL) : the line to neutral (phase) voltage VLN = 400/√3 = 230 V During the historical development of cables, numerous types... How often installations are inspected is up to the owner of the installation, provided such durations do not exceed any regulatory maximums in force. Hi Steven, how to get the total KVA of the system if Line AB =642.24amp. Knowing that efficiency of power consuming equipment is the output power divided by the input power, again this can easily accounted.for   Reactive power is not discussed in the article and more details can be found in other notes (just use the site search). The power taken by a circuit (single or three  phase) is measured in watts W (or  kW). Hello everybody, I would appreciate if you can help me with this problem I have: What is the current of an 647kW, 577V three-phase load with cosFI=0.91? As a general guide I would just add up the VA, use a 70% load factor. 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? What is the difference between three phase power factor and single phase power factor. Current = W / voltage = 13900/230 = 60 A. Because of this division, then, your answer only reflects the output of a single phase. Single- and three-phase power are both terms describing alternating current (AC) electricity. phase power current. You should then follow this formula up by dividing the kVA by the voltage (your VLN in the case of a 3 phase system) in order to calculate the current. With A-B, B-C and A-C you are looking a loads in a delta connected system. Planning to make the new site live in about a week. I know this probably is easy to understand, but I got myself to a completely confused state.   P 1ph = P 3, Single phase apparent power S1ph (VA):      This is typical of power transmission systems, electrical motors and similar types of equipment. Am I correct? This configuration of voltage sources is characterized by a common connection point joining one side of each source. That 60A is total rms current for a single phase, not the peak current, correct? This unit is used to measure the power a circuit takes in. Balanced three phase system with total power P (W), power factor pf and line to line voltage VLL, Convert to single phase problem:      phase 1 apparent power = 80 x 230 = 18,400 VA = 18.4 kVA      3 Phase Induction Motor current calculator is used to calculate the rated and full load amperes of the motor. As a rule I remember the method (not formulae) and rework it every time I do the calculation. If you create a fault the current will increase (you have reduced the circuit impedance with the fault). It’s rating is 100 KVA. For example, take a 400 V (VLL) three phase system with the following loads: phase 1 = 80 A, phase 2 = 70 A, phase 3 = 82 A, the line to neutral (phase) voltage VLN = 400/√3 = 230 V I hope to get some answers and if any reference on web would be great. Following on from my post on railway electrification voltages, I thought an introduction to traction substations would be a good idea. If  I have a system having Amperes on each phase as 80A, 70A, 82A and VLL is 400V and VLN is 210 V. Kindly share how to calculate its power consumption in KVA. As a practical matter when literature specifies "line" voltage does this correspond to the country standard; ie 480 VAC 3 phase industrial power in the US? The phase voltage is 400/ √(3) = 230 V. To convert kW to kVA you need the power factor (you could just use kW, but you would have some error). Select yes (default) if you want to measure the three-phase currents that flow through the block. You should notice that this is the exact same formula that was used above for single phase systems. Kindly show the computation. Luckily for us, there is a simple one: Using this formula, we simply divide the kVA we've calculated by the voltage (which should be given) in order to calculate the current. to get the total current. The power factor, for instance, can be represented as: The kilo-volt amperes, on the other hand, can be represented as: Although our ultimate goal is to teach you how to calculate a 3 phase power current, we (and most other people) presume that teaching you how to calculate a single phase power current will lay some important groundwork for what you have at your business now and The simple formula to calculate the rating of three phase Transformers is: KVA = (√3. Similarly a transformer (with three windings, each identical) supplying a given kVA will have each winding supplying a third of the total power. The second, more important reason has everything to do with the fact that you can actually use the logic and formula for calculating single phase power currents when calculating multiphase power currents. You give the worse case as 80.6A. phase 2 apparent power = 70 x 230 = 16,100 VA = 16.1 kVA      If the system is perfectly balanced I again have no issues figuring out the amps drawn on each leg. Thank you. dilan, the kVA would be ( 642.24 + 651.85 + 651.84) * 220 = 428102 VA = 428.1 kVA the kW would be 428.1 x 0.86 = 368.2 kW For the breaker, I would likely be looking at 800 A (posibly something like a MasterPact NT or NW, depending on other parameters) and setting the overload protection in the range of 710 A (this is plus 10%). And now I would like to know how does the calculation works when a generator has to be used? I would like to measure the current draw in a toy 3 phase AC motor. Robert, I make the line current as 33.6 A. Kindly I want to know in 3 phase distribution ( example a house), Say I have 29 kVA ... equals to 36,250watts  (PF=0.8). three phase power is 36 kW,  single phase power = 36/3 = 12 kW V Line-Line = 380 V Thank you so much! That being said, let's hop right into things. Up Next. While it is possible to have 3 phase system with only 3 conductors, that requires the load on each phase to be virtually identical, which isn't really possible in a residential system. Autodesk Circuits Simulation. How to use the formula and which formula to find it out..??? Take a three phase motor (with three windings, each identical) consuming a given kW. If the voltages become unbalanced or there are other considerations (i.e. In order to find the output of the 3 phase system you started with, you need only multiply the current you calculated by 3. To me the easiest way to solve three phase problems is to convert them to a single phase problem. If remembering is formulae more easy, just go with that. In order to make this conversion, however, you need to understand that, for our purposes here, a 3 phase system is essentially producing 3 times as many kilowatts as a single phase system; this difference in power produced actually makes it pretty easy to see why some people upgrade to 3 phase power. And who knows? Let's get down to business. Consider three single-phase systems each supplying 100W to a load (Figure 3). LED indication for individual phase voltage. Thanks for pointing out the mistake. To find the power given current, multiply by the voltage and then the power factor to convert to W.  For a three phase system multiply by three to get the total power.