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Author Topic: Cable length, sizes and components  (Read 9942 times)
Stefan (S.T.E.F.)
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« Reply #15 on: January 21, 2009, 06:37:09 PM »

For lay out I would

Have the bank & the inverter very close but in sepperate air spaces to keep the heaviest cables short.
Have the inverter/bank housing as close as possible to the low voltage power producers (IE 48v DC wind turbines with bank at or near the tower).
Have the higher voltage power producers where they will work best (IE solar on the roof running at 140v dc).
Run high voltage (240v ac) to the house from bank/inverter.

That will keep the cable costs down & the volt losses reduced so increasing the power available over power actualy produced.


Justme

That is exactly how I've planned it but I just don't like the idea of having to walk (or fly) 100m through a storm in the rain at 1C to see how my batteries are doing or switch something on/off/over. Keeping in mind the turbine and battery shed location is surrounded by bog land.
Might be ok now, but I am getting older by the minute ;-)

Stefan
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Stefan (S.T.E.F.)
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« Reply #16 on: January 21, 2009, 06:46:03 PM »


If I worked that out correctly I would have a voltage drop of 26V over the 200m cable run costing me 700.  Shocked

Stefan



Like I said vastly under speced. I think for mains 3% is considered as ok. On my dc I have aimed for much much less.

Also quoting 10mm2 as good for bank bus bars is mad. Mine is 120mm2 & only 3m (1.5 red & 1.5 black).

Justme

I worked that out for the 48V, so only 22V would arrive at the batteries...
I have used 50mm in my motor home for battery interconnection, after being told by a salesman that 6mm would do....

That's why I am so keen to get the system commissioned myself, with all of your help of course, so that I don't have to rely on sales people. It would go horribly wrong.

Stefan
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Justme
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« Reply #17 on: January 21, 2009, 06:52:08 PM »

I just don't like the idea of having to walk (or fly) 100m through a storm in the rain at 1C to see how my batteries are doing or switch something on/off/over.
Stefan

You can remote control most stuff. The victron inverter chargers have a port that you can plug a cat 5 network cable (but its not cat 5 coms spec) into to bring the controls inside or use a pc to check on it & activate settings.

Justme
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billi
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« Reply #18 on: January 21, 2009, 08:29:30 PM »

Quote
I worked that out for the 48V, so only 22V would arrive at the batteries...

 Roll Eyes and ?  do you have the turbine already  ? and are you still sure its 48 Volt from the turbine to the battery ?

I got AWG 0  100 meter 53 mm2 then according to here  (http://www.reuk.co.uk/AWG-to-Square-mm-Wire-Size-Converter.htm) from my supplier  with my windturbine (24Volt 1000 watt ) for 300 pounds ,

Quote
If a 110W panel would cost 100 I'd buy the 12 that
I had planned but at 400 it's a bit of an investment with not too much outcome. Would make
more sense in Spain....


We should talk about that again , cause i think PV first.... even on the moon , cause the most reliable source  of free energy i found jet ...... But have to explore water turbines more

Billi
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« Reply #19 on: January 21, 2009, 08:57:54 PM »

Justme, why didn't you say that earlier ?
That sounds like the solution to most of my problems !

Only other problem I can think of right now is dumping.
I can't really do that in the middle of nowhere 100m from the house and the dump loads are supposed to be close to the turbine, right?
I don't really want to heat thin air above bog land.
Any other ideas ?

Stefan
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Justme
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« Reply #20 on: January 21, 2009, 09:13:32 PM »

You could use the virtual switch in the victron to turn on a 240v ac load to draw more amps from the battery instead of a direct 48v dump load. Things like water heater etc. Will mean yet another cable run but could be thin & then use a relay in the house to up the load.

Justme
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2 x Victron Multiplus II 48/5000/70
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« Reply #21 on: January 22, 2009, 08:52:08 AM »

Roll Eyes and ?  do you have the turbine already  ? and are you still sure its 48 Volt from the turbine to the battery ?

No, I don't have it yet and still waiting to hear from the winiwind people about that.

I got AWG 0  100 meter 53 mm2 then according to here  (http://www.reuk.co.uk/AWG-to-Square-mm-Wire-Size-Converter.htm) from my supplier  with my windturbine (24Volt 1000 watt ) for 300 pounds ,
53mm doesn't seem a lot for 24V over 100m, I worked out I need closer to 200mm.... but maybe I am wrong?


We should talk about that again , cause i think PV first.... even on the moon , cause the most reliable source  of free energy i found jet ...... But have to explore water turbines more
Billi

My 150W panel kicks out 4amp in bright sunshine for a couple hours.... on overcast days there was nothing... 12x nothing = nothing. I am not sure if that makes sense to spend 5000 on PV panels that kick out nothing most of the year. Unless I am doing something wrong of course.

I do have 4 13W briefcase panels and they still output 1amp in overcast conditions, but it's a different panel type. They should do large ones of those.... I don't fancy 100 briefcase thingies on my roof.

A water turbine would be great as the water never stands still, but I don't have permission to use the river.... shame.

Stefan
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billi
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« Reply #22 on: January 22, 2009, 09:06:54 AM »

Quote
53mm doesn't seem a lot for 24V over 100m, I worked out I need closer to 200mm.... but maybe I am wrong?

Like i said earlier  the generator in my  turbine  produces upto 300 Volt    not 24 Volt  Shocked  and then rectified  by the controller to 24 DC next to the battery

But i am not the expert if it comes down to all these  electrical details  ( i wounder myself how we manage without the grid since 3 years now  Grin)

About PV  some info here http://www.navitron.org.uk/forum/index.php/topic,3163.0.html

Billi
« Last Edit: January 22, 2009, 09:13:01 AM by billi » Logged

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« Reply #23 on: January 24, 2009, 12:25:24 PM »

Roll Eyes and ?  do you have the turbine already  ? and are you still sure its 48 Volt from the turbine to the battery ?
Malcolm from energistar.com answered me and said the miniwind is 48V at the turbine, but the Merlin has a recifier so could be rund at 300V to the battery bank. Anybody got one of those? Never heard of them before to be honest.

I got AWG 0  100 meter 53 mm2 then according to here  (http://www.reuk.co.uk/AWG-to-Square-mm-Wire-Size-Converter.htm) from my supplier  with my windturbine (24Volt 1000 watt ) for 300 pounds ,

Is that armoured?
I couldn't find anything armoured online above 16mm.


We should talk about that again , cause i think PV first.... even on the moon , cause the most reliable source  of free energy i found jet ...... But have to explore water turbines more
Billi

Which PV panels have you got?


Stefan
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billi
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« Reply #24 on: January 24, 2009, 04:26:16 PM »

Quote
Is that armoured?
I couldn't find anything armoured online above 16mm.

Not armoured , i just ordered it with my 2 windturbines  from Canada ( but turened out that it was a Chinese wind turbine and cable came with it from China so i doubt you find that price here  Roll Eyes )

Same with the Pv  came from china as well .... so nothing special  Grin


Anyhow in an offgrid idea everything helps on a daily charge  and now in winter even i get only 1 -3 kwh a day ( someday ok only 0.5 kwh)

I know PV is costing money , but wind as well ....

In summer  you will (depending on location) see a windturbine standing still for days/week producing next to nothing

In a gridtie idea sure you calculate on a yearly production , but in offgrid situations (due to battery capacity) you think of daily production

Billi
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guydewdney
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« Reply #25 on: January 24, 2009, 06:09:08 PM »

Malcolm from energistar.com answered me and said the miniwind is 48V at the turbine, but the Merlin has a recifier so could be rund at 300V to the battery bank. Anybody got one of those? Never heard of them before to be honest.

I have the 500v (max) versions. I have a few spare, now with SKF bearings instead of the carp chinese things that lasted me four days...
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w0067814
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« Reply #26 on: January 27, 2009, 11:19:15 PM »

Hi Guys,

This is pretty easy to work out.

Code:
P = Power in Watts
I = Current in Amps
V = Voltage in Volts
R = Resistance in Ohms

The current required by the load of a given power can be found using the following formula:
Code:
I = P / V


The voltage drop across the cable can be found using the following formula

Code:
V= IR

The power loss in the cable can be calculated using:

Code:
P = IR

Example:
Work out R using the value from the table below and mulipling by the length of the cable expressed in km. Remember that this if for each cable run, so you shall have to multiply R by two to account for positive and negative cables.

Assuming 25mm cable (as typically used in meter tails) and the 200m cable run with a load of 5000 watts with cable of 230vac...

I = 5000 watts / 230 volts = 21.74 amps
R = 0.8ohms * 0.200km *2runs = 0.32 ohms.
V(loss) = 0.32 ohms *21.74 = 6.96v = 3%
P(loss) = 21.74amps * 21.74amps *0.32ohms = 151.24 watts = 3%

Plug in the values as required for different voltages and powers. I think for the power ratings that have been suggested in ealier posts (4-5kw) you are basically forced to go to 230vac. Remember the losses increase as a square of the current. Thus if you halve the voltage and double the current, you quadruple the losses. Thus as 48v is not reasable for power levels and distances of these magnitudes without having excessive cable size & cost.

Remember that the voltage / power loss across the cable is not available for work at the far end. A resistive load will therefore disapate less wattage than would be the case in no cable losses, and a non-linear load would draw more current (same power) to compensate, thus worsening the cable  losses further.

Below are some approximate resistances in ohms per 1000m for copper cables of varying size.
Code:
AWG  mm2  Ohms per km
16     1.5   13.0
14     2.5    8.54
12     4.0    5.4
10     6.0    3.4
8       10     2.2
6       16     1.5
4       25       .8
2       35       .5
1       50       .4
1/0 55       .31
2/0    70       .25
3/0    95       .2
4/0   120     .16
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billi
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« Reply #27 on: January 27, 2009, 11:28:23 PM »

Quote
Hi Guys,

This is pretty easy to work out.

Code:

P = Power in Watts
I = Current in Amps
V = Voltage in Volts
R = Resistance in Ohms


The current required by the load of a given power can be found using the following formula:
Code:

I = P / V



The voltage drop across the cable can be found using the following formula

Code:

V= IR


The power loss in the cable can be calculated using:

Code:

P = IR


Example:
Work out R using the value from the table below and mulipling by the length of the cable expressed in km. Remember that this if for each cable run, so you shall have to multiply R by two to account for positive and negative cables.

Assuming 25mm cable (as typically used in meter tails) and the 200m cable run with a load of 5000 watts with cable of 230vac...

I = 5000 watts / 230 volts = 21.74 amps
R = 0.8ohms * 0.200km *2runs = 0.32 ohms.
V(loss) = 0.32 ohms *21.74 = 6.96v = 3%
P(loss) = 21.74amps * 21.74amps *0.32ohms = 151.24 watts = 3%

Plug in the values as required for different voltages and powers. I think for the power ratings that have been suggested in ealier posts (4-5kw) you are basically forced to go to 230vac. Remember the losses increase as a square of the current. Thus if you halve the voltage and double the current, you quadruple the losses. Thus as 48v is not reasable for power levels and distances of these magnitudes without having excessive cable size & cost.

Remember that the voltage / power loss across the cable is not available for work at the far end. A resistive load will therefore disapate less wattage than would be the case in no cable losses, and a non-linear load would draw more current (same power) to compensate, thus worsening the cable  losses further.

Below are some approximate resistances in ohms per 1000m for copper cables of varying size.
Code:

AWG  mm2  Ohms per km
16     1.5   13.0
14     2.5    8.54
12     4.0    5.4
10     6.0    3.4
8       10     2.2
6       16     1.5
4       25       .8
2       35       .5
1       50       .4
1/0    55       .31
2/0    70       .25
3/0    95       .2
4/0   120     .16




so to make it short   ,if its so easy  .... what cable then for the  48 volt turbine/generator DC output    for a 100 meter distance to the battery  Roll Eyes


or  does it mean cable costs will  be astronomical



« Last Edit: January 27, 2009, 11:40:07 PM by billi » Logged

1.6 kw and 2.4 kw   PV array  , Outback MX 60 and FM80 charge controller  ,24 volt 1600 AH Battery ,6 Kw Victron inverter charger, 1.1 kw high head hydro turbine as a back up generator , 5 kw woodburner, 36 solar tubes with 360 l water tank, 1.6 kw  windturbine
Bill H
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« Reply #28 on: January 27, 2009, 11:30:37 PM »

Welcome 'W', and nicely nailed   Grin

Have I met you before - just retired from some important overseas job ?   Wink

Best regards

Bill
« Last Edit: January 27, 2009, 11:34:02 PM by Bill H » Logged

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Justme
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« Reply #29 on: January 28, 2009, 09:44:51 AM »

Hi Guys,

This is pretty easy to work out.

edited for brevity


This is easier to work out

A dead easy way to calculate the cross-sectional area required (in mm2) is:

A (mm2) = Length x Current/12

The equation allows for a maximum volts drop of 0.4v and takes in to account the cable's return journey so enter the length (in metres) from the battery to the appliance one way ONLY.

So L (one way) x Amps div by 12 = cable in mm2

To give a higher spec & make the numbers so easy you can do them in your head div by 10 instead.

Justme
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Navitron solar thermal system
30 x 58mm panel 259L TS
1200watts solar 120vdc
FX80 Solar controller
2 x Victron Multiplus II 48/5000/70
Cerbo GX & GX 50 touch
BMV 700
6kva genny
48v 1000ah
Grid Possibly coming soon
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