
MidLevel Electrical Engineer
 London
We are looking for an electrical engineer with around 4 6 years of design experience to join and work with an able and talented group of engineers..
 Recruiter: Max Fordham LLP

Engineering Manager  Naval Architect
 England, Cumbria, BarrowInFurness
 Competitive package
As an Engineering Manager  Naval Architecture you will be managing the Whole Boat Architecture and Concepts team tasked with supporting the delivery of the remaining Astute submarines, and developing new technology for future submarine programmes.
 Recruiter: BAE Systems

Nuclear Inspectors
 Bootle, Cheltenham and London
 Competitive + Benefits
With expertise and influence, you’ll set the standard for nuclear safety.
 Recruiter: Office for Nuclear Regulation
IET 
search :
help :
home


Latest News:


Topic Title: Volt Drop 3 Phase unbalanced Topic Summary: Created On: 09 December 2013 12:50 PM Status: Post and Reply 
Linear : Threading : Single : Branch 

Search Topic

Topic Tools

09 December 2013 12:50 PM


Hi all
If i have a 3 Phase distribution circuit which is not balanced, what is the correct way of calculating volt drop? So if Ia = 100 at 0 degree P.f 1 Ib = 50 at 120 degree P.f 0.8 Ic =[ 40 at 240 degree P.f 0.7 I get a Neutral Current of 66.73 at 16.44 degree P.f 0.95 So should i add the voltage dropped in the neutral to the three individual lines and base it on a percentage of 230 V and not 400V as its now referenced to the neutral or is there another way of approaching it. Many Thanks Chris 



09 December 2013 03:56 PM


Forget about the power factor, it is a confusion.
Use V = IZ The drop in each phase will be the same as the magnitude of the source. So if the source is 230V; then the total drop is 230V. So, you need to know the values of Z in the components in which you are calculating the volt drop. Is it the supply cables? Or the loads? Or both. BTW The currents you give are confusing? I assume the angles are Va = 0 degrees Vb = 120 degrees Vc = 240 degrees? Are the current angles in reference to these? i.e. is Ib at 120 to Vb? If so the power factor is 0.5. Assuming the current and voltage waveforms are sinusodial, PF is the phase angle difference between the two. 



09 December 2013 04:47 PM


Forget about the power factor, it is a confusion. Use V = IZ The drop in each phase will be the same as the magnitude of the source. So if the source is 230V; then the total drop is 230V. So, you need to know the values of Z in the components in which you are calculating the volt drop. Is it the supply cables? Or the loads? Or both. BTW The currents you give are confusing? I assume the angles are Va = 0 degrees Vb = 120 degrees Vc = 240 degrees? Are the current angles in reference to these? i.e. is Ib at 120 to Vb? If so the power factor is 0.5. Assuming the current and voltage waveforms are sinusodial, PF is the phase angle difference between the two. Thanks for the reply Yes i can see i the confusion caused The Currents are Ia = 100 at P.f 1 Ib = 50 at P.f 0.8 Ic =40 at P.f 0.7 Ia the reference phase at 0 Degrees, then 120, 240 I get a Neutral Current of 66.73 at 16.44 degree P.f 0.95 So my question is do i calculate using three phase mV/A/m or single phase and should i base it on % of 230V, so as you say just add the VD of the neutral to the volt drop of the individual Lines. Thanks again 



09 December 2013 05:30 PM


I'm still not sure I understand what the question is?
If your supply voltage, Vsupply = 230V then your volt drop, Vdrop, assuming the neutral is at 0V, is also 230V. What you need to know the is impedance of the various loads, and what share of 230V is dropped across each load. You haven't given the impedance of the loads, so there is no way of answering this question! If you consider Ohm's Law V = IZ If you are calculating the volt drop across the suppy cable then because the loads are different per phase, the current is different per phase (as given), and so the volt drop per phase, over the cable is different. But you need to know the impedance of the cable! If you assume a unit resistance (no reactance) for the cable, then the magnitude of the volt drop in ConductorB is half that of ConductorA; and the volt drop in ConductorC is 40% of conductorA. 



09 December 2013 05:44 PM


I'm still not sure I understand what the question is? If your supply voltage, Vsupply = 230V then your volt drop, Vdrop, assuming the neutral is at 0V, is also 230V. What you need to know the is impedance of the various loads, and what share of 230V is dropped across each load. You haven't given the impedance of the loads, so there is no way of answering this question! If you consider Ohm's Law V = IZ If you are calculating the volt drop across the suppy cable then because the loads are different per phase, the current is different per phase (as given), and so the volt drop per phase, over the cable is different. But you need to know the impedance of the cable! If you assume a unit resistance (no reactance) for the cable, then the magnitude of the volt drop in ConductorB is half that of ConductorA; and the volt drop in ConductorC is 40% of conductorA. The mV/A/m values in BS7671 are either single phase so current in the neutral or 3 phase assuming balanced. The distribution is a 3 phase circuit though has neutral current due to being unbalanced. If it were balanced i could use the 3 phase mV/A/m values and design to a 1 % of 400V . So im looking as to what is the most accurate way of determining volt drop in an unbalanced 3 phase circuit. Thanks again 



09 December 2013 05:53 PM


So im looking as to what is the most accurate way of determining volt drop in an unbalanced 3 phase circuit. Volt drop across what? The cable? Or the load? 



09 December 2013 09:09 PM


So im looking as to what is the most accurate way of determining volt drop in an unbalanced 3 phase circuit. Volt drop across what? The cable? Or the load? Volt drop across the cable. Thanks again 



10 December 2013 10:55 AM


Volt drop is relative to the phase in this case, so surely just calculate mV/A/m for each phase giving you the VD for each phase?




10 December 2013 11:12 AM


Hi
Loads: Ia = 100 Amps PF 1 (2.3 + j0) Ib = 50 Amps PF 0.8 ( 0.56 + j4.57) Ic = 40 Amps PF 0.7 (3.99 + j4.14) Va 0, Vb 120, Vc 240 I calculate a Neutral Current of 66.73Amps at 16.44 degree P.f 0.95 I haven't the mV/A/m for the Distribution cable, what im looking to determine is the correct and most accurate way of determining volt drop in this situation. The mV/A/m values in BS7671 are either single phase so current in the neutral or 3 phase assuming balanced and no Neutral current. The distribution is a 3 phase circuit though has neutral current due to being unbalanced. If it were balanced i could use the 3 phase mV/A/m values and design to a 1 % of 400V . So which mV/A/m value should be used and what approach is best suited to this scenario. Many thanks 



10 December 2013 02:55 PM


Hello Chris.
I think your 50A load has a power factor nearer to 0.121. If you want to use the tables then you should use the single phase mV/A/m values because your loads are unbalanced. It depends what you mean by "volt drop". Because the neutral will be at a potential above "0" and at a phase angle (relative to the A phase as you have calculated it) then it is probably best to calculate the load voltages, for example using J Millman, and then subtracting these individually to determine the "volt drop". This will be different from the "volt drop" that you would obtain by multiplying the conductor impedances by the load currents. 



10 December 2013 05:39 PM


I think your 50A load has a power factor nearer to 0.121. I think that is because the original data is a little vague (just a little!) The reference angle for the yellow phase is 120° so a power factor of 0.8 will give a angle leading or lagging that of 37°. i.e. the yellow phase current is at 157° or 83° as referenced to the red phase voltage. 



10 December 2013 05:55 PM


I haven't the mV/A/m for the Distribution cable, what im looking to determine is the correct and most accurate way of determining volt drop in this situation. Can you not see the contradiction here? You cannot accurately determine, or even guess, the volt drop unless you know the impedance of the cable! Yo have so little data it is impossible to model anything without making assumptions. Let's make some reasonable assumptions, so you can at least get started. 1) The currents quoted are the currents through the phase windings of the distribution transformer. 2) The voltage at the head of the cable is 240V per phase 3) The load is star connected to a star wound TX 4) The neutral is at 0V 5) The load is simply connected in series with the cable  i.e. you have a three phase cable, and at the end of each conductor is an impedance. These three impedances are different giving the different phase currents. 6) The impedance of the cable does not vary with load nor temperature. It is fixed at all times. We have to make one further assumption  which is the main one  and that is the impedance of the cable! Let's assume it is purely resistive and is equal to 1ohm. Start with the Red Phase The total volt drop, from the output of the TX through the cable, through the load to the neutral is 240V. Using Kirchoffs Law and working through the calculation in steps Total Volt Drop = Volt Drop Across Cable + Volt Drop Across Load Vdrop = (Ia * Zcable) + (Ia * Zaload) 240V = (100A * 1 ohm) + (100A * Zaload) 240V = 100 V + (100A * Zaload) Therefore the volt drop across the cable is 100V and the volt drop across the load is 140V. Now the Yellow Phase: The yellow phase current is 40+30J in reference to the yellow phase voltage 240V = (40+30j * 1 ohm) + (40+30j A * Zbload) 240V = 40+30j V + (40+30j A * Zbload) Therefore the volt drop across the cable is 40+30j V (50V) Rearranging for Zbload we get the volt drop across the load as 20030JV (202V) Blue Phase You can work that one out. You should get a cable drop of 40V and a load drop of 214V. cheers W Edited: 10 December 2013 at 06:01 PM by Zuiko 



10 December 2013 06:34 PM


Thankyou for Your replies
Ill go over check my calcs, had numbers everywhere Hpvompac, thanks ill have a nosey at millmans therom and see if I can fathom it out, Thankyou Zuikio, sorry for the confusion, I realise I needed the cable mV/A/m, it was more the method to use more so. Thanks for your in depth reply ill have read and digest it all. Thanks again Chris 



10 December 2013 09:07 PM


I have read the OP as a 4 wire circuit and therefore the voltage across the load must be the difference between the source voltage and the common neutral voltage  this is a vector relationship.
Power factor of the load is R/Z  the B load of is therefore 0.121. The various load voltages are best worked out using Millman or Maxwell. A bit long winded but you could always use superposition. Best regards. 



10 December 2013 09:10 PM


I have read the OP as a 4 wire circuit and therefore the voltage across the load must be the difference between the source voltage and the common neutral voltage  this is a vector relationship.
Power factor of the load is R/Z  the B load of is therefore 0.121. The various load voltages are best worked out using Millman or Maxwell. A bit long winded but you could always use superposition. Best regards. 



10 December 2013 09:16 PM


I have read the OP as a 4 wire circuit and therefore the voltage across the load must be the difference between the source voltage and the common neutral voltage  this is a vector relationship.
Power factor of the load is R/Z  the B load of is therefore 0.121. The various load voltages are best worked out using Millman or Maxwell. A bit long winded but you could always use superposition. Best regards. 



11 December 2013 06:51 AM


yes, the original question was vague, but it was cleared up later...the three references were 0, 120, 240 and the power factors 1, 0.8 and 0.7 respectively.
You don't really need to use superposition unless you have loads across different voltages, it adds needless complications. I think using plain old Ohms Law, V = IZ, as above, is the easiest. It only looks long winded because every stage of the calc was broken down. You can do it on a calculator in 30 seconds. cheers W Edited: 11 December 2013 at 10:27 AM by Zuiko 



11 December 2013 03:16 PM


Hello W.
I think that there may be an error in your method to answering this question. You posted "Start with the Red Phase The total volt drop, from the output of the TX through the cable, through the load to the neutral is 240V. Using Kirchoffs Law and working through the calculation in steps Total Volt Drop = Volt Drop Across Cable + Volt Drop Across Load Vdrop = (Ia * Zcable) + (Ia * Zaload) 240V = (100A * 1 ohm) + (100A * Zaload) 240V = 100 V + (100A * Zaload) Therefore the volt drop across the cable is 100V and the volt drop across the load is 140V" The load Z is 2.3 ohms. This would therefore give a load voltage of 230V as opposed to 140V. Best regards. 



11 December 2013 05:00 PM


Hello W. I think that there may be an error in your method to answering this question.. The numbers used were only for illustration. I assumed a voltage of 240 V at the transformer, which is reasonable,(point 2 above) and a cable impedance of 1 ohm (which is an easy number to multiply!) Whilst the parameters used may not be realistic, there is no error in the method! cheers W 



11 December 2013 05:11 PM


Hello W.
Again, if using your method, the voltage across the yellow phase load would be 230V not 202V. The correct method needs to allow for the magnitude and the vector relationship of the displaced load neutral. Best regards. 



IET
» Wiring and the regulations
»
Volt Drop 3 Phase unbalanced

Topic Tools

FuseTalk Standard Edition v3.2  © 19992016 FuseTalk Inc. All rights reserved.