Extra Long Power Transmission

[Khonwan]

Crossy, elkangorito, Dr Naam, et al

This is for real – no jest, no testing of skills. My only related-qualification is a Scottish Higher (roughly equivalent to an English A Grade) in physics, gained 31 years ago. However, I installed my own electrics to my house from my transformer – a run of 1.556 km – over a year ago (3-phase). This being Thailand, there is, of course, a longer story involved but I’ll not bore you with the details. Concerned about voltage drop, I put together an equation (based on bits of acquired knowledge – I could find no source of assistance) to calculate voltage loss (on the basis of single-phase) and therefore the required cable size. I determined a maximum load of 17 kW (assuming a balanced 5,666 W per phase). My equation was U=2LrW/V/s, where

U: lost voltage (Volts)

L: single cable length (km)

r: rated aluminum conductor resistance (Ohms per sq.mm per km)

W: power load (Watts)

V: no-load voltage supply (Volts)

S: conductor section (sq.mm)

Where L=1.556 km; r=30 Ohms per sq.mm per km; W=5,666 Watts; V=230 Volts; s=50 sq.mm

U=2*1.612*30*3*3463 / 235 / 50

= 46 volts lost

Therefore, only 184 V would remain (= to 84% of a nominal 220 V).

Alternatively, conductors of 110 sq.mm should be required to provide a resultant voltage of 209 V (or 220 sq.mm to provide 220V).

I used PVC-insulated overhead cables rated at 75deg. C., 700 V.

And now finally to my question: Can you confirm my equation (and rated aluminum conductor resistance) is correct? I’ve tested “known” loads (single-ph appliances) up to 5 kW and the resulting voltage readings seem to bear it out.

[bdenner]

Using my “Olex Cables” hand book I come up with this solution:-

Voltage Drop = 53.66

I=P/E

I=5666/230

24.63Amps

50mm Aluminium drops 1.4mV per Amp. Meter (From the table in the hand book)

Volt drop= 24.63 x 1556 x .0014

Volt Drop = 53.66

You mention a single phase operating voltage of 230 but are using 235 in your equasion?

PM me if you want a scanned copy of the Page concerned.

[Crossy]

OK. Elkangorito has MUCH more recent experience of these calculations than myself but here goes.

50mm2 Al has a resitance of about 0.6ohm/km

110mm2 Al has a resitance of about 0.27ohm/km

For 110mm2 lets assume

L=1.556km V=220 W=5666 Power Factor=0.8

Load current = 5666/(220*.8) = 32.2A

Resistance per leg = 1.556*.27 = 0.42ohm

voltage drop per leg = 32.2*0.42 = 13.5V * 2 legs = 26.5V giving us 193.5V at the load.

EDIT Ties up reasonably well with bdenner's solution for 50mm2 cable

[Khonwan]

Crossy, elkangorito, Dr Naam, et al

This is for real – no jest, no testing of skills. My only related-qualification is a Scottish Higher (roughly equivalent to an English A Grade) in physics, gained 31 years ago. However, I installed my own electrics to my house from my transformer – a run of 1.556 km – over a year ago (3-phase). This being Thailand, there is, of course, a longer story involved but I’ll not bore you with the details. Concerned about voltage drop, I put together an equation (based on bits of acquired knowledge – I could find no source of assistance) to calculate voltage loss (on the basis of single-phase) and therefore the required cable size. I determined a maximum load of 17 kW (assuming a balanced 5,666 W per phase). My equation was U=2LrW/V/s, where

U: lost voltage (Volts)

L: single cable length (km)

r: rated aluminum conductor resistance (Ohms per sq.mm per km)

W: power load (Watts)

V: no-load voltage supply (Volts)

S: conductor section (sq.mm)

Where L=1.556 km; r=30 Ohms per sq.mm per km; W=5,666 Watts; V=230 Volts; s=50 sq.mm

U=2*1.612*30*3*3463 / 235 / 50

= 46 volts lost

Therefore, only 184 V would remain (= to 84% of a nominal 220 V).

Alternatively, conductors of 110 sq.mm should be required to provide a resultant voltage of 209 V (or 220 sq.mm to provide 220V).

I used PVC-insulated overhead cables rated at 75deg. C., 700 V.

And now finally to my question: Can you confirm my equation (and rated aluminum conductor resistance) is correct? I’ve tested “known” loads (single-ph appliances) up to 5 kW and the resulting voltage readings seem to bear it out.

Sorry guys - I forgot to ammend my draft! This should have read:

U=2*1.556*30*5666 / 230 / 50

= 46 volts lost

[Crossy]

Sorry guys - I forgot to ammend my draft! This should have read:

U=2*1.556*30*5666 / 230 / 50

= 46 volts lost

Did you install 50mm2 or 110mm2 in the end?

I did the sums for 110 with a Power Factor of 0.8 and 220V just to throw in another variable

EDIT is your Tx 230v output to account for a long feed?

[chownah]

Using my “Olex Cables” hand book I come up with this solution:-

Voltage Drop = 53.66

I=P/E

I=5666/230

24.63Amps

50mm Aluminium drops 1.4mV per Amp. Meter (From the table in the hand book)

Volt drop= 24.63 x 1556 x .0014

Volt Drop = 53.66

You mention a single phase operating voltage of 230 but are using 235 in your equasion?

PM me if you want a scanned copy of the Page concerned.

I have a question.

Assuming that I=P/E means that the current (I) through the load is equal to the power consumed at the load (P) divided by the voltage across the load (E)....then I think you have not found the answer...you are using 230V as the voltage across the load in your calculations but clearly because of the voltage drop this is not correct...shouldn't you be useing something like 170V in your calculations? Doesn't this type of calculation have to be done by the method of iteration?

Chownah

[Khonwan]

Sorry guys - I forgot to ammend my draft! This should have read:

U=2*1.556*30*5666 / 230 / 50

= 46 volts lost

Did you install 50mm2 or 110mm2 in the end?

I did the sums for 110 with a Power Factor of 0.8 and 220V just to throw in another variable

EDIT is your Tx 230v output to account for a long feed?

Using my equation on the basis of 220V as assumed by you, Crossy, I come to a load voltage of 198V. My calculations appear much closer to yours than bdenner. What is your power factor of 0.8 based on?

When you go down the high voltage / low voltage own-transformer route, PEA contracts to provide a supply at 230V. Due to the distance in my case, they increased the transformer setting to 235-240V. I usually measure 230V at the house at non-load.

I installed 50 sq.mm cables!

I have always "messed around" with electricity (DIY, for my own properties only) but never had any experience (or understanding) of 3-phase (do now though!). It was going to cost me almost one million baht to have PEA route the electricity (high voltage) to my house so I asked them about the consequences of having a transformer connected at the roadside next to the high voltage supply cables, with me then running the low voltage cables back to my house (1.556km). They agreed to this, stating that single-phase would not be possible but that three-phase would be suitable. They (PEA, Lat Yao office, Nakhon Sawan - top manager and several field staff) told me that 50 sq.mm per leg would be fine - I specifically told them I calculated a maximum load of 17kW would be needed. I asked them to explain to me how the voltage would not reduce under my calculated loads. I tried to show them my equation, but, of course, they would not even look at it. They merely stuck to their mantra that I would not suffer voltage loss with this 3-ph supply. I explained that I accepted that higher voltage (around 400V in this case) would not suffer so much but that nearly all of my appliances would be single-ph. They stuck to their mantra. I drove to PEA's head office in Bangkok and asked to see an engineer but was told they had none at that office. They advised me to go to Klong Toey office. I spoke with their engineer, who was completely disinterested once he realised I lived outside his area. He also refused to consider my concerns or my arithmetic (my equation). He also stuck to the mantra that 3-ph, 50 sq.mm per phase, over 1.556km would be no problem for 17kW. Local electricians told me the same. Even my own father, a retired marine electrician, could see no problem with it. I gave in!

I trucked the cable direct from a factory in Sammutprakan and had PEA install the 50kVA 3-ph transformer. In total, it cost me over 300,000 baht (a LOT cheaper than the million PEA would have charged). I should add that all the land between my house and my roadside transformer is mine so I was able to erect my own posts.

Once up and running, I applied known loads and measured the resultant voltage only to find that my concerns were well founded - my equation appeared correct, unfortunately. The idiots from PEA (and they are!!!) couldn't believe it. They checked my connections. They measured my cables. They observed my load measurements. They then decided that perhaps I should have had a single-phase supply after all. They suggested they experiment by trial installing a 30kVA s-ph transformer. I told them that my arithmetic suggested that that would be acceptable...with cables sized at 700 sq.mm rather than the 50 sq.mm cables I had already installed (as per their advice). They actually came out again and replaced my transformer with a borrowed 30kVA s-ph transformer and checked my load results again. Surprise, surprise...they were as my arithmetic predicted.

Then they concluded that the distance was too great! Their only solution: for me to pay them to move the transformer closer to my house!

I have not moved the transformer, or added further cables. I just put up with the inconvenience. I use a 3-ph motor for my pig-pen washing; don't have an electric shower; never intended having air-conditioners.

Lesson: don't trust PEA's so called "electricians". Much better advice is obtainable at this forum!

[Crossy]

Using my equation on the basis of 220V as assumed by you, Crossy, I come to a load voltage of 198V. My calculations appear much closer to yours than bdenner. What is your power factor of 0.8 based on............................

................... never intended having air-conditioners.

Lesson: don't trust PEA's so called "electricians". Much better advice is obtainable at this forum!

Typically Thai

I thought things looked about right in your original calcs. Certainly better than PEA came up with. My only reason for banging in a PF of 0.8 is that domestic loads are rarely PF 1.0 and it only takes a few induction motors (fridge, aircon, pump) to bugger it, actually in your case you might score with some PF correction capacitors at the house end of the feed, you'll need to measure the real PF first though.

I'd also suggest an AVR and/or on-line UPS for your technology (TV, computer etc) so the no-load spikes (which there are bound to be) don't kill anything expensive

[Khonwan]

Using my equation on the basis of 220V as assumed by you, Crossy, I come to a load voltage of 198V. My calculations appear much closer to yours than bdenner. What is your power factor of 0.8 based on............................

................... never intended having air-conditioners.

Lesson: don't trust PEA's so called "electricians". Much better advice is obtainable at this forum!

Typically Thai

I thought things looked about right in your original calcs. Certainly better than PEA came up with. My only reason for banging in a PF of 0.8 is that domestic loads are rarely PF 1.0 and it only takes a few induction motors (fridge, aircon, pump) to bugger it, actually in your case you might score with some PF correction capacitors at the house end of the feed, you'll need to measure the real PF first though.

I'd also suggest an AVR and/or on-line UPS for your technology (TV, computer etc) so the no-load spikes (which there are bound to be) don't kill anything expensive

Thanks for the confirmation. I use a UPS-unit for my PC and did indeed plan to buy some more for TVs and sat decoders. etc

[bdenner]

I guess there are a number of ways of doing this, the bottom line is your cable may be too small to meet the requirements – anyway I’ve been doing a bit more.

Lets try this assuming you require 17Kw at load, maximum, in a 3 phase configuration:

E = Lets use the nominal ph to ph voltage of 380 – 10% voltage drop = 342, considered reasonable.

Pf = 0.8 (Crossy’s figure) many variables, even the capacitive reactance in your long feeders help a very small bit.

Kw = I x E x 1.732 x pf / 1000 therefore

I = Kw x 1000 / E x 1.732 x 0.8

I = 17 x 1000 / 342 x 1.732 x 0.8

I = 17000/473.87

I = 35.87 amps

This is well within the current limits placed on 50mm2 cable in an aerial installation. 130 amps quoted in the book.

Crossy’s figure of 32.2 amps works here if we use 380 Volts. Here I will use them:-

VL = Voltage at Load

50mm2 has 0.6 ohms per Km

VL = Distance x 2 x 0.6 ohms x I /380

1.556 x 2 x 0.6 x 35.87 / 380 = 313.03

313.03/1.732 = 180.73 Volts phase to neutral

180.73 volts using 50mm2 cable

193.5 volts using 110mm2 cable Crossy based his calcs on this.

[Crossy]

I guess there are a number of ways of doing this, the bottom line is your cable may be too small to meet the requirements.

Looks like the case, but the OP is stuck with what's been installed

I am certainly going to look long and hard at what is proposed for my house although the load is not anticipated to be anything like 30+ A per phase.

EDIT. I'd certainly look at using an AVR (Automatic Voltage Regulator) for your house stuff (except the big power aircon, stove etc), light bulbs and electonics like stable power

[Khonwan]

I guess there are a number of ways of doing this, the bottom line is your cable may be too small to meet the requirements – anyway I’ve been doing a bit more.

Lets try this assuming you require 17Kw at load, maximum, in a 3 phase configuration:

E = Lets use the nominal ph to ph voltage of 380 – 10% voltage drop = 342, considered reasonable.

Pf = 0.8 (Crossy’s figure) many variables, even the capacitive reactance in your long feeders help a very small bit.

Kw = I x E x 1.732 x pf / 1000 therefore

I = Kw x 1000 / E x 1.732 x 0.8

I = 17 x 1000 / 342 x 1.732 x 0.8

I = 17000/473.87

I = 35.87 amps

This is well within the current limits placed on 50mm2 cable in an aerial installation. 130 amps quoted in the book.

Crossy’s figure of 32.2 amps works here if we use 380 Volts. Here I will use them:-

VL = Voltage at Load

50mm2 has 0.6 ohms per Km

VL = Distance x 2 x 0.6 ohms x I /380

1.556 x 2 x 0.6 x 35.87 / 380 = 313.03

313.03/1.732 = 180.73 Volts phase to neutral

180.73 volts using 50mm2 cable

193.5 volts using 110mm2 cable Crossy based his calcs on this.

Thanks for this, bdenner. Yes, I realised that Crossy's 194V is based on 110 sq.mm. My own equation returns 198V based on a similar 110 sq.mm.

[Khonwan]

I guess there are a number of ways of doing this, the bottom line is your cable may be too small to meet the requirements.

Looks like the case, but the OP is stuck with what's been installed

I am certainly going to look long and hard at what is proposed for my house although the load is not anticipated to be anything like 30+ A per phase.

EDIT. I'd certainly look at using an AVR (Automatic Voltage Regulator) for your house stuff (except the big power aircon, stove etc), light bulbs and electonics like stable power

How far is your house from the supply, Crossy?

I don't think I've seen an AVR - any idea of cost? Mind you, been running my systems this way for well over a year without problems. In fact, the only problem I had was when an electrical storm blew apart my kWh meter: the PEA "electricians" rigged me up a meter-by-pass supply (it was during a weekend and a new meter could not be procured) but crossed the cables over, blowing up one of my fridges! They tried to wiggle out of responsibility but paid up for it in the end (inc. my transportation costs).

[Crossy]

How far is your house from the supply, Crossy?

I don't think I've seen an AVR - any idea of cost?

Actually only about 50 metres, but the village is on the end of a long feed at 220x3, I've not located the actual transformer as the feed heads off across the rice paddies but I reckon it's a good 1.5 km or so to the main road where the 25kV poles are. I have no idea as to the size of the incoming 220 cable or the tx rating.

Problem is that as the place has grown the supply doesn't seem to have been uprated. Wifeys son has a restaurant by the river in which we have a stake (that's why we're moving there) and the lights dip up and down during the evening load peak.

An AVR is basically a motorised variac with some electronics to sense and adjust the output, price varies with rating, needs more research.

http://www.dianxing-kebo.com/china/Products.asp?CategoryId=7

The nice thing is they will happily step up a low supply for ever (no batteries to discharge as an on-line UPS would do) if the supply is low, it simply draws more current. You still need a UPS incase the supply goes off completely though.

[Khonwan]

How far is your house from the supply, Crossy?

I don't think I've seen an AVR - any idea of cost?

Actually only about 50 metres, but the village is on the end of a long feed at 220x3, I've not located the actual transformer as the feed heads off across the rice paddies but I reckon it's a good 1.5 km or so to the main road where the 25kV poles are. I have no idea as to the size of the incoming 220 cable or the tx rating.

Problem is that as the place has grown the supply doesn't seem to have been uprated. Wifeys son has a restaurant by the river in which we have a stake (that's why we're moving there) and the lights dip up and down during the evening load peak.

An AVR is basically a motorised variac with some electronics to sense and adjust the output, price varies with rating, needs more research.

http://www.dianxing-kebo.com/china/Products.asp?CategoryId=7

The nice thing is they will happily step up a low supply for ever (no batteries to discharge as an on-line UPS would do) if the supply is low, it simply draws more current. You still need a UPS incase the supply goes off completely though.

1.5km is a long way for PEA. I was under the (obviously false) impression that supply lines were not longer than around 700 metres. They would not allow me to connect to their 220V supply lines at that distance.

Thai supplier of AVR: http://www.siliconthai.com/

[chownah]

Khonwan,

If you are still interested in discussing the validitiy of your equation I am willing to do that. If so can you write down the equations you started from and I'll try to verify that the math you used to combine them was correct.

Chownah

[Crossy]

1.5km is a long way for PEA. I was under the (obviously false) impression that supply lines were not longer than around 700 metres. They would not allow me to connect to their 220V supply lines at that distance.

Thai supplier of AVR: http://www.siliconthai.com/

I've not actually followed the lines, so it's not impossible that there's a 25kV Tx sitting behind the trees. That said we do experience some pretty signifficant voltage fluctuations. Once we start construction I'll leave it to PEA to decide what they do but I have no intention of paying for a Tx or 25kV extension.

Cheers for the link

[Somtham]

KW - since you just have the one 3 phase motor at your place couldn't you change the transformer back to single phase and then connect 2 of the 50mm^2 wires for the neutral and connect the other 2 for the line essentially doubling your size from 50 to 100 mm^2 and halving your voltage drop to 23?

rgds

[Khonwan]

Khonwan,

If you are still interested in discussing the validitiy of your equation I am willing to do that. If so can you write down the equations you started from and I'll try to verify that the math you used to combine them was correct.

Chownah

Will you be relying on your own mathematical skills or your research prowess of the Internet? Sorry Chownah, but as I've just written in another thread (in General Forum), "Respect breeds respect". You are too into yourself. Once bitten, twice shy...and all that.

[Khonwan]

KW - since you just have the one 3 phase motor at your place couldn't you change the transformer back to single phase and then connect 2 of the 50mm^2 wires for the neutral and connect the other 2 for the line essentially doubling your size from 50 to 100 mm^2 and halving your voltage drop to 23?

rgds

Hi Somtham (hope your makhua and chillies are recovering) - already considered that. Even made sure PEA did exactly that on their wast-of-time trial of the s-ph 30kVA transformer.

You are basing your suggestion on my use of 5,666W - you forgot that my actual target was 17,000W (which works out at 5,666W per phase.

Based on the fact that, though contracted at 230V, I usually receive a no-load supply of 235V...and based on my equation...I could switch to s-ph using 200sq.mm cables to provide me with 201V at a maximum load of 17kW (it would be 195 load Volts at 230 no-load Volts).

I could solve my (paper) problem by adding three more 50sq.mm cables (at a cost of just over 100,000 baht). I say "three" since the neutral would not need to be increased: the neutral cable could be grounded every 200 metres instead (hel_l of a saving). This would result in 212V at a maximum load of 5,666W per ph (it would be 207 load Volts at 230 no-load Volts).

I suppose I could similarly consider 100sq.mm neutral (grounded every 200 metres) plus 200sq.mm single-phase. That would result in the cheapest alteration in terms of cable (around 70,000 baht) but I doubt if PEA would willing simply swap a 30kVA s-ph transformer for my existing 50kVA 3-ph transformer (remembering that this is my property).

As is, I can still operate 3-ph appliances (at loads right up to 50kVA if necessary) to overcome most problems. I don't think I'll be forced (by needs) to upgrade my current system. It is a wee bit annoying that I can't connect a normal s-ph electric welder, or use a s-ph electric shower more than 3-4kW but 3-ph equivalents are available.

[chownah]

Khonwan,

If you are still interested in discussing the validitiy of your equation I am willing to do that. If so can you write down the equations you started from and I'll try to verify that the math you used to combine them was correct.

Chownah

Will you be relying on your own mathematical skills or your research prowess of the Internet? Sorry Chownah, but as I've just written in another thread (in General Forum), "Respect breeds respect". You are too into yourself. Once bitten, twice shy...and all that.

Yeah, you seem to be too full of yourself too... I don't bite so I'm not sure what "Once bitten" refers to....if you think I've bitten then please send me a link in a PM and I'd be glad to look at it and see what you mean.

Back on topic. I guess what I would do is to first analyze your equations to see if they "fit" together mathematically and dimensionally....if they didn't then it would be up to you if you wanted to pursue finding out why....if I thought I could go to the internet and find a site that analyzed this kind of stuff I'd probably go check it out...if anyone knows of a site that can do this kind of stuff then let me know as I'd be really interested in checking it out.

Its really up to you...if you "don't like me" so you don't want my advise then that's fine with me....no problem there.

Chownah

[lannarebirth]

Hello electrical engineering wizards. I'd like you to check my figures for an installation I'm making if you would be so kind.

I have a 45 amp meter at the entrance of my property connected to a single phase 240V power supply line. I would like to run electric power into my land to a workshop 135 meters from the meter. The workshop will have 3 circuits. One each for lights, plugs, pump. From the wokshop I'd like to continue on to my staff house which is 50 meters more distant, where I will install one circuit each for lights and plugs.

By my calculations, assuming I did it correctly, I need to run 2 - 35mm squared aluminum wires (or AWG 2) to the shop and on to the house.

I figure that's a voltage drop of about 17V to 223V, which should be within acceptable limits?

Thank you very much for any responses confirming or pointing out my error.

[soundman]

Voltage drops are not constant and are determined by the load.

You are all assuming a constant load on the opposing end of the line to the transformer.

No load (assuming infinate resistance on the insulators) = no voltage drop.

Voltage drop is a factor of resistance, voltage & load based on these formula.

P=VxI - Power (Watts) = Volts x Amps

R=V/A - Resistance (Ohms) = Volts / Amps

Lets assume a 1.5km stretch of 50mm2 cable has a resistance of 1 ohm.

Lets assume the load per phase is 5kw.

Lets calculate a voltage drop for one 220V phase.

Using P=VxI we can calculate a required amperage of I=5000/220 - 22.7 Amps. (Well within current carrying capacity of 50mm2 aluminium cable.

We can now determine the resistance of the load using R=V/A - R=220/22.7 - 9.69 Ohms.

Voltage drop is now a calculation of the resistance of the supply line & the resistance of the load.

Voltage = 220-((R1/R2)x220) where R1 = the resistance of the supply line & R2 = the resitance of the load. = 197V or a voltage drop of 23V at 5kw.

Lets rework these formula for a load of 500W.

I=500/220 - 2.3 amps R=220/2.3 97 Ohms 220-((1/97)x220) = 217.7 Volts or a drop of 2.26V at 500W.

Double the size of the supply wire to 100mm2 & the voltage drop will half. These calculations will work with any wire provided the saturation point of conductivity is not exceeded.

(These caculations are assuming a three phase suppy with load balanced over three pahses, where neutral resistances are negligable.)

Cheers,

Soundman.

[chownah]

Lannarebirth,

I use this on line calculator to do these kinds of computations:

http://www.stealth316.com/2-wire-resistance.htm

I entered 270 metres as the length of wire (use total circuit length) and 45 amps as the load. For AWG 2 COPPER cable this gave a voltage drop of 6.4 V and to convert for ALUMINUM you multiply by 1.64....giving about 10 V of drop. I assumed that the extra 50 metres would not draw the entire 45 amps so I modeled it as a circuit off of the first 135 metre run. If you want to analyze for the cable run to go the entire 185 metres, then enter 370 metres and 45 amps in the calculator and you get 8.7V of drop for COPPER and multiplying by 1.64 to adust for ALUMINUM you get about 14 volts of drop.

As to whether this is acceptable...depends on what equipement is being run......also remember that you will probably not be having 240 volts at the source all the time. I think that 5% is considered acceptable voltage drop usually so it looks like you would be just over the line if you were drawing 45 amps at the 185 metre distance and be under the limit at the 135 metre distance.

I'm not an electrician but it seems to me that your plan will work fine....I wouldn't run a 45 amp motor on this line though as it would almost assuredly reduce its life span.

Also, the calculator I reference can be downloaded onto your computer and run without an internet connection.

Chownah

[lannarebirth]

Thanks chownah,

I used a similar calculator, and my figures include an extra 20 meters from actual as a "fudge factor". Thank goodness I can use aluminim as copper is just unbelievably expensive right now. I think it should be fine. Thanks again.

[elkangorito]

I've just figured why I missed this topic. At the time of posting, I was on a 3 month semester break with no internet access.

Anyway, even though it's "shutting the gate after the horse has bolted", I did a calculation (3 times to make sure) on Konwans supplied info. The results are as follows;

Given info;

1] Supply - 380 volt, 3 phase, 50 Hz supply.

2] Cable - Single PVC insulated (75 degree) Aluminium, aerial, not bunched, installed in direct sunlight.

3] Length - 1 500 metres approx.

4] Size - 50 mm squared.

5] Power Factor - negligible.

6] Maximum allowable voltage drop as per AS3000:2000 - 5% of supply voltage (3 phase - 19 volts).

7] Estimated maximum load - 32 Amps.

8] Maximum cable current rating as per installation method (aerials in sun, PVC sheathed, 1 m/s wind) - 185 Amps.

The results;

1] mV/A.m of installed cable - 1.33.

2] Maximum allowable mV/A.m for installed cable under estimated load/installation conditions - 0.229.

3] Calculated maximum voltage drop - 64 volts per phase. Best case scenario - 55 volts per phase.

4] Minimum cable size is estimated at no less than 300mm squared.

These figures are as per AS/NZS3000:2000. Please be aware that these figures assume that at any given point in time, a maximum current of 32 Amps will be flowing through each or any phase (which is unlikely).

A better way to determine the cable size in this scenario would be to do a Maximum Demand calculation first, since it is unlikely that the consumer will actually use 17kW of energy at any instantaneous time. Actually, this is what most of these types of calculations are based upon - Maximum Demand, since it is unrealistic to add up all of the loads that you have installed.

[Khonwan]

Voltage drops are not constant and are determined by the load.

You are all assuming a constant load on the opposing end of the line to the transformer.

No load (assuming infinate resistance on the insulators) = no voltage drop.

Voltage drop is a factor of resistance, voltage & load based on these formula.

P=VxI - Power (Watts) = Volts x Amps

R=V/A - Resistance (Ohms) = Volts / Amps

Lets assume a 1.5km stretch of 50mm2 cable has a resistance of 1 ohm.

Lets assume the load per phase is 5kw.

Lets calculate a voltage drop for one 220V phase.

Using P=VxI we can calculate a required amperage of I=5000/220 - 22.7 Amps. (Well within current carrying capacity of 50mm2 aluminium cable.

We can now determine the resistance of the load using R=V/A - R=220/22.7 - 9.69 Ohms.

Voltage drop is now a calculation of the resistance of the supply line & the resistance of the load.

Voltage = 220-((R1/R2)x220) where R1 = the resistance of the supply line & R2 = the resitance of the load. = 197V or a voltage drop of 23V at 5kw.

Lets rework these formula for a load of 500W.

I=500/220 - 2.3 amps R=220/2.3 97 Ohms 220-((1/97)x220) = 217.7 Volts or a drop of 2.26V at 500W.

Double the size of the supply wire to 100mm2 & the voltage drop will half. These calculations will work with any wire provided the saturation point of conductivity is not exceeded.

(These caculations are assuming a three phase suppy with load balanced over three pahses, where neutral resistances are negligable.)

Cheers,

Soundman.

Sorry, I've just noticed your reply.

Your first assumption is incorrect. Where has it been assumed that there is a constant load? Your opening statements seem obvious to me and have been demonstrated in the equation I provided in my OP. They are also demonstrated by others via their replies. Obviously the voltage drop will depend on the load, which is why I specified an example load.

Your arithmetic agrees 100% with my own arithmetic in my OP. You, however, assume 1 ohm resistance over 1.5km where I assume 1.8 ohms. In fact, at 20 degrees C., the resistance would be 0.641 ohms per km (http://www.bangkokcable.com/catalog/BCC_CATALOG/THWAEN.HTML).

Cheers

Khonwan

[Khonwan]

A general point to all: many contributors appear to believe that electricity in Thailand is always supplied at 220/380V but, in fact, consumers with their own transformers have a contracted supply of at least 230/400V, which can easily be uprated at the transformer by PEA/MEA to 240/415V.

Regards

Khonwan

[Khonwan]

A better way to determine the cable size in this scenario would be to do a Maximum Demand calculation first, since it is unlikely that the consumer will actually use 17kW of energy at any instantaneous time. Actually, this is what most of these types of calculations are based upon - Maximum Demand, since it is unrealistic to add up all of the loads that you have installed.

17kW was, in fact, my maximum demand calculation; it was not simply the sum of all potential loads. The calculation was based on serving the needs of my farm (incorporating workers' houses and potential heavy machinery) rather just my home.

Regards

Khonwan

[elkangorito]

It appears as though you have things all sorted out. Happy farming and....good luck.