Calculating current capacity of Aluminum

Status
Not open for further replies.
C

cjphelps

Member
Hi Everyone ,

I am doing what I though was a simple calulation and getting a funny result. Maybe someone could help me out with this:

I need to calculate the cross section of solid Aluminum needed to handle 150Amps @ 20V, used by my group for a 75 cm long piece of solid square aluminium. Assume Temperature K = 298

What Ive done so far:

1.V=IR 20 = 150*R R= 20/150 R=0.1333
2.R=pl/A

RESISTIVITY:
p= 2.709X10^-8 OHM*m
p(cm) = 2.709X10^-6OHM*cm

LENGTH:
l=75cm
I substitute R from Ohms Law into equation and solve for A.

A=pl/R A= 1.5698X10^-3 cm*cm

This seems to small.

I compare this to the awg of Aluminum required to handle 150A = Size 0
TO CALCULATE CROSS SECTION OF 1/0 WIRE:
n = −(m−1) where m=1, n = 0, plugging into d = .005"*92^(36-n/39)
D=diameter=0.324"
AREA=(PI*D^2)/4 = 82.88X10^-3"^2 = cross section for 1/0 wire which handles 150Amps.


this is much bigger then my calc.
Why is my calculation so off?

Thank you in advance!
 
You are using the wrong calculation for your application. Your calculation is correctly giving an aluminium link with a length of 75 cm which will show a voltage drop of 20V when carrying 150A at room temperature. a) it is _very_ unlikely that you want to dissipate your 20V in the aluminium link and b) such a link would not stay at room temperature very long, since it would be dissipating 20V * 150A = 3000W

Presumably you have a 20V system, and need to deliver 150A to a load, and you want this load to consume the bulk of the 20V. There are two constraints to the size of the bus bar. 1) The bar needs to be big enough to have a low enough voltage drop for your application and 2) The bar needs to be big enough that its resistive self heating doesn't make it too hot.

-Jon
 
You are calculating R as if you know the voltage drop across the conductor will be 20 volts. This is probably not what you mean by 150Amps @ 20V.

To calculate the ampacity, you have to equate power in (I??R) with heat lost from radiation, conduction, and convection. Your equations do not do this. Equations that do are not the simple calculations that you expected.
 
winnie said:
You are using the wrong calculation for your application. Your calculation is correctly giving an aluminium link with a length of 75 cm which will show a voltage drop of 20V when carrying 150A at room temperature. a) it is _very_ unlikely that you want to dissipate your 20V in the aluminium link and b) such a link would not stay at room temperature very long, since it would be dissipating 20V * 150A = 3000W

Presumably you have a 20V system, and need to deliver 150A to a load, and you want this load to consume the bulk of the 20V. There are two constraints to the size of the bus bar. 1) The bar needs to be big enough to have a low enough voltage drop for your application and 2) The bar needs to be big enough that its resistive self heating doesn't make it too hot.

-Jon
Click to expand...


Thats right, I want to deliver the 20V to the load, which means little-no voltage drop acrossthe aluminum bar. What crude calculation can I use?
I need to calculate the x-sectional area of a 75cm piece of aluminum that would do the trick, but at the same time I cannot oversize it too much.

Any extra calculation info would be super helpful!
 
jghrist said:
You are calculating R as if you know the voltage drop across the conductor will be 20 volts. This is probably not what you mean by 150Amps @ 20V.

To calculate the ampacity, you have to equate power in (I??R) with heat lost from radiation, conduction, and convection. Your equations do not do this. Equations that do are not the simple calculations that you expected.
Click to expand...

Which formula's would help me do this please?

I will assume now that the aluminum bar should be in a circuit as a resistor, and that I need to deliver the 20V across the load, so I must make the resistor as small as possible. I am not quite understanding how to take into account the heat losses. Coul you point me towards a solution?

Thank you !
 
1) You must decide how much voltage drop is to be permitted in the bar. There is no equation for this, it is a design consideration based upon things like the supply voltage, the variability of the supply voltage, the requirements of the load, etc. If the load will function with 19V delivered, then your answer will be quite different than if your load requires no less than 19.97V. Also remember the _complete_ circuit; if you have a bus bar from supply to load, you presumably have some connection from the load back to the supply.

You cannot say 'make the resistor as small as possible' and get a useful answer; you need to clearly describe the constraints: how much voltage drop can you tolerate.

2) The issue with heat losses is the _basis_ of ampacity calculation. Conductors self heat when they carry current, and conductors have a maximum permitted temperature. The conductor temperature will be set by how much heat is produced by current flow and how much heat is carried away to the surroundings. The more current flow, the more heat produced, and the higher the temperature where the conductor will be in equilibrium with the surroundings.

Here is an article on the basic calculation http://www.electrician2.com/articles/ampacity.htm

Here is a table of ampacities for _aluminium_ bus bars. http://www.ems-industrial.com/main/aluminum_bus_bar.asp

-Jon
 
Thanks Jon or the helpful info.

As a rule , when designing a conductor to transport power to a load, is there,
what is considered a decent Voltage drop, because If I specify my voltage drop across the load as 0.001V, I dont know if thats being reasonable in practice
 
The NEC is not a design guide, but it has some useful information. Two methods I would use as guides are:

Size out an Al wire per 310.16. Convert to mm^2.
60C >> #2Al = 33.62 mm^2
75C >> #3Al = 26.67 mm^2
Conversions to mm^2 per NEC table 88

Your application with bare Al busbars would run cooler than the table 310.16 ampacity temperature for insulated wire.

Second method is:
Use NEC 366.23, and size Al at 1.09A/mm^2.
For 75A this is 68.8 mm^2

Why the difference in size? Temperature.

For a 75cm length, I'm pretty sure VD won't be a problem no matter which you pick. You have the formulas to figure it out.

I'm not a thermodymanics/heat transfer kind on a guy so I would probably pick one of the above. In the past, for battery applications, I have used insulated building wire with crimped lugs, sized per T310.16, and then checked the VD to see if I needed to go to a larger a wire size.

Just curious about the application: Science fair, college project, industrial giant designing secret weapons, home builder making a gadget?

carl
 
cjphelps said:
Thanks Jon or the helpful info.

As a rule , when designing a conductor to transport power to a load, is there,
what is considered a decent Voltage drop, because If I specify my voltage drop across the load as 0.001V, I dont know if thats being reasonable in practice
Click to expand...
I don't know what you mean by "voltage drop across the load". I'm going to assume you are interested in voltage regulation at the load - you don't want the voltage at the load to change much. So how close the you need the voltage to be regulated?
This is a design question. Well, ask your equipment what is reasonable. The equipment you are driving has a voltage spec - meet that.

carl
 
cjphelps said:
As a rule , when designing a conductor to transport power to a load, is there,
what is considered a decent Voltage drop
Click to expand...

For most loads that I've dealt with, a few percent is fine.

But it is really application specific.

I suspect that if you select a minimum size (based upon current capacity) bus bar, your voltage drop will be quite small. Do the calculation for a 1/8" by 1" bar, and you'll see what I mean.

-Jon
 
There are no simple equations to use. Look at the table linked by Jon for aluminum bus bar. Notice two things:


  • Fatter bars with the same area have lower ampacity - compare 1/2 x 3 to 1/4 x 6. This is because there is less surface area.

  • The smallest listed bar has twice the ampacity that you require.
I suggest that you use a 1" x 1" bar which will have more ampacity than the 308A listed for 1/4" x 1". The difference in cost for a 75 cm length will be far less than the worth of the time trying to figure out the smallest possible bar.
 
cjphelps said:
Hi Everyone ,

I am doing what I though was a simple calulation and getting a funny result. Maybe someone could help me out with this:

I need to calculate the cross section of solid Aluminum needed to handle 150Amps @ 20V, used by my group for a 75 cm long piece of solid square aluminium. Assume Temperature K = 298

What Ive done so far:



I compare this to the awg of Aluminum required to handle 150A = Size 0
TO CALCULATE CROSS SECTION OF 1/0 WIRE:
n = −(m−1) where m=1, n = 0, plugging into d = .005"*92^(36-n/39)
D=diameter=0.324"


Why is my calculation so off?

Thank you in advance!
Click to expand...

Using what you did above why not just use a .324 x .324 for a 0.105 sq in x 0.75cm aluminum piece? A little larger than the circular wire that handles that current but close enough.
If you wanted to get it closer divide .105/1.27 (ratio of circular to sq mils) and go with .083 sq in?
 
jghrist said:
...I suggest that you use a 1" x 1" bar which will have more ampacity than the 308A listed for 1/4" x 1". ...
Click to expand...
366.23 suggests that 700A/in^2 is okay for AL. 1 x 1 seem way past overkill. 1/4 x 1 is overkill.

carl
 
Status
Not open for further replies.
Top