Voltage Drop Calculations for Three Phase 415V , 50 HZSystem

Pl contact an engineering consultancy firm and pay their requisite fees for the ans.

This is not the required answer. The person raising the query is very well aware of this. Can we help from our side from our experience. we do not need a consultant

Look at article 310 of the National Electric Code book. Also at the apendix section of it you will find various examples of how to do such calculation. All what you have to do is change the numbers. Good Luck.

I assume your refering to a DG set so.

This is the calculation i use for rating my circuit breakers:

KVA x 1000/ (√3 x Voltage) = Full Load Current

If anyone can see a mistake please correct.

I take no responsibility for how you use this.

Please refer to the current edition of the Wiring Regulations maintained and published by the Institution of Electrical Engineers.

PLEASE WHERE I GET current edition of the Wiring Regulations maintained and published by the Institution of Electrical Engineers.

contact book supply bureau

www.standardsindia.com

they are sales agency of national and international standards and regulations.

i want a voltage drop formula and then calculate the sq.mm of the voltage drop cable

Hi TJSHAH,

If I understand your question properly there are 2 parts.

1) How much voltage drop will there be across a 100m line run.

2) How do I size the breaker.

The voltage drop will vary depending upon the internal capacitance of the cable and will vary from cable type to type as well as the AWG of the cable or wire, I would contact the manufacture of the cable and see if they can give you the answer. The general equaztion looks like this:

FORMULA TO FIND VOLTAGE DROP {V D} ;

{exact}
2x K x L x I
CM

Voltage Drop [ VD ] {three phase} = 1.732 [representing three phase] {which is the square root of the three wires going out to the load <hot wires>, making up the 1.732}, That 1.732 is multiplied by the resistance {K} [using exact {K} for copper or for aluminum calculated, by you, using the NEC chapter 9 table 8 {normally uncoated if copper}]. That is multiplied by the length {L} [The total distance of the circuit from the source to the load]. That is multiplied by the load measured in amps {I}. Then, divide the total answer of all of the above, by the Circular Mill {wire size }. This {C M } using (1 for solid or 7 for stranded) is found in the NEC Chapter 9, Table 8 which is the wire size of the feeder, or branch circuit being calculated.

To size the breaker you can either contact the manufacture for recommended breaker size or if you know the nominal current then size the breaker 25% above the full load nominal current rating. Imax *1.25. If it is a motor or inductive load then you will need a breaker delayed trip or a fuse with a slow blow characteristic.

Hi TJSHAH,

If I understand correctly, you want to know

1-how to calculate the current of a three ph load

2-how to select the cable (Aluminum) & how to apply temperature derating @45 deg

3-how to size the circuit breaker

Somebody suggested earlier to go to a consultant which is a good advice; there are safety issues involved in selecting the size of cable & breaker. Unless you are fully conversant with the electrical code, you will have to seek help from somebody who knows this stuff.

I can give you an overall view of what is involved.

1- The equipment you refer to must have a name plate; check the full load current, (IL). Use this current to size up the cable & breaker. To calculate here is the formula:

KW= 3sq.rt X I line X V line X Cos phi / 1000

2- Go to the table in the code or cable manufacturer's catalogue and select the cable for the line current. Since the cable length is only 100 meters I do not think that you have to derate it. To be on the safe side consult the code.

3. This is going to be tricky; the breaker ampacity you will know from the full load current but the trip point you will have to coordinate with the inrush current & main breaker tripping characteristic. People have written books on this subject, if you want to learn more, do some search.

jmsk

Thanks for your answer. Selection of Circuit Breaker - MCCB is still not much clear.

Pl guide.

The breaker has two components; the contactor & the trip element (relay). You want to size the breaker such that the contactor does not heat up when in use. So you want the contactor size as heavy as possible but at same time have a trip relay that would trip at the current you want. Bear in mind that too bigger a breaker size (contactor) would be more expensive. So, for a 100 amp load let us put a safety facor of 25% and choose a breaker of 125 amps. Somebody has already suggested this. Now you have to go to your breaker supplier and see whether they have 125 or 150 amp breaker.

Now for the trip setting (relay), you have to consult with your breaker supplier. They have trip sttings for various breaker type/model etc. you have to choose one that suits your application; they will help you in this regards.

You have to know the inrush current of the load. If it is an induction motor, consider the inrush to be 10 times the full load current (the figure varies between 7-14). So for 100 amp full load, the in-rush current will be 1000 amps. Now you have to look at the characteristic of the trip relay or breaker trip curve and make sure you choose the trip curve under 1000 amps. this way the breaker will not trip every time you start the motor. This is all you need to know about your breaker but a word of caution, too small a trip relay setting will trip your equipment every time you start and too large may not clear the short circuit fault and may trip the main feeder or worst start up an electrical fire. So consult with your breaker supplier or a consultant to select the breaker trip; you need to know your inrush current. Not knowing your load it is hard for me to say.

Incidently, the breaker you are selecting is required to clear the branch circuit fault, like a short circuit, etc. Your load (equipment) already comes with its own protection. The two things you have to consider in selecting your breaker are that it does not trip on an inrush current and it clears the branch circuit fault.

Hope this helps. It has been a long time since I did this stuff.

JMSK

Thanks a lot for your answer. Sir, can you tell me how IEC and NEC are compared with specific reference to this ?

Dear Mr. TJSHAH,

You have to assume the following.

1. Earth Fault Level 50 KA for 1 second.

2. De-Rating Factor 0.8

3. Current Density for Aluminium 0.625 Amp.

4. Voltage Drop permissible - 5 V.

5. Circuit Breaker capacity 1.75 x Full Load Current.

6. Power Factor 0.8

Power = (Sq.Rt 3 x 415 x I x PF) / 1000 KW

Therefore Full Load current = (70 x 1000) / 1.732 x 415 x 0.8 = 70.2852 Amps.

Cable Size = Full Load Current / Current Density = (70.2852) / (0.625) = 112.456 Sq. MM.

Allowing for De-Rating Factor 0.8, the cable Size will be (112.456) / (0.8) = 140.57 Sq.MM

We have to use the nearest higher size to 140.57 Sq.MM SAY 185 Sq. MM OF SINGLE RUN (OR 2 RUNS OF 95 Sq..MM

After deriving this, check for voltage drop as per standards table. applicble for Aluminium Conductor. If Voltage drop is with in 5 Volts the above size is OK or else go for further one step higher size cable..

This is the simplest calculation.

DHAYANANDHAN.S