calculation for the copper required to deliver 45 amp current

Dear all..

i need a calculation for the copper required to deliver 45 amp current..!!

plz help me out with proper mathematical approch..

i am going to design water load for 10kw systems..

v=230Vac
i 44 amp...

reply with copper required..??

go thru

https://circuitcalculator.com/wordpress/?p=75/
https://www.cable-calculations.co.uk/test/psc.htm
or browse in Google u will get lot of calculators

We used loads like this in kit installed in 1972 for a 50 ohm 15 KW load for a UHF transmitter. It was a tank about 1m diam by about 4 m high with a bit of 50 ohm coaxial cable as its input connector, which attached to a solid copper tube coax inside the tank. The water had Glaubers salt added to it to make it a good 50 ohms. In use the water would slowly boil away and had to be topped up.
To carry 50 A, use a cable with a CSA of 10sqmm. - depends on the temperature of the insulation, if the cable is in a heat conductive medium. . . If you are thinking about the actual load, only the initial bit of it carries 50 A, halfway down it it is only carrying 25 A and at the end - nothing.
Frank

Chuckey...

thanks for the exprience..

would you please tell me water proportion with salt to calculate resistance...

any mathematical explanation..

looking forward to hearing you soon..!!

I can't remember the exact proportions , but the solution has to be super saturated. that is too much G salt is added so a little remains undissolved. the reason is that if you start of with some sort of unsaturated solution, as the water boils away, the saturation will increase and the conductivity of it will fall, so the load impedance will fall. Once the solution is super saturated, as the water boils away, more G salt comes out of solution but the waters conductivity remains constant.
The design of the load must be such that the water can steam freely and escape from the "live" electrode and the container having a large surface area to dissipate the heat. Our tanks used a "reflux" system, which was basically a tall chimney sitting on the tank, so the steam would condense within the chimney and the water droplets would fall back into the tank so minimising the water loss.
How are you intending to get rid of the heat? commercial resistive loads are stuffed with fans which wail like banshies in their attempt to blow enough air on the loads to dissipate the heat. They are compact though. Our set up was BIG but quiet and extremely reliable.
Our "load" was a pair of coax tubes, the inner (live) was say, 50mm diam the outer was 150mm diam and 2m long. the outer one had holes punched in tit to allow the movement of water/steam. We fed the RF into the bottom and the coax pointed upwards in the tank (so the steam can come out and not have to flow through our RF feeder. But in you case the connection will only be a bit of copper cable so that will not get in the way. If you arrange for a raising/lowering system of the tubes independently, then the final resistance of the load can be altered.
Frank