Calculation problem formula to solve?

[ggunn]

Question:
12awg stranded in Oates copper conductors are run 100 ft; the current through the circuit is 20 amps. What is the power loss of the conductors?


Can someone please work this problem step by step and show me how to solve this and with what formula? Thanks.


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Assuming uncoated copper DC resistance using Table 8
R = (100')(1.98 ohms/1000') = 0.198 ohms
I = 20A
P = (I^2)R = (20A^2)(0.198 ohms) = 79.2W

 

[Dennis Alwon]

Assuming uncoated copper DC resistance using Table 8
R = (100')(1.98 ohms/1000') = 0.198 ohms
I = 20A
P = (I^2)R = (20A^2)(0.198 ohms) = 79.2W

I suspect if the run is 100' would you then need to figure this at 200'

 

[ptonsparky]

I suspect if the run is 100' would you then need to figure this at 200'

One of those small details we are left to wonder about.

 

[ggunn]

I suspect if the run is 100' would you then need to figure this at 200'

My answer is per conductor, of course. We don't know how many there are.

 

[gar]

230201-2039 EST

The resistance of copper wire vs temperature is:

R = K ( 15.4 ( 1.0 + 0.0045 ( T2 - T1 ) ) )

where Ts are in deg C, R is in ohms, and K is a constant based on wire size. 15.4 is based on resistivity of copper.
T1 is temperature of wire with no current flow, and T2 is wire temperature at its load current.

See https://physics.info/electric-resistance/

.

 

[winnie]

gar,

You've neglected the adjustment factors for self heating, work hardening during installation, and cosmic ray annealing. *grin*

-Jon

 

[gar]

230202-1453 EST

If the resistivity of the wire changes, then you just change the value of the 0.0045 term. If the wire material changes its characteristics with temperature, then the 0.0045 becomes a more complex function.

When my brother in law was chief scientist at GE aerospace GE won the contract to build the first high power space satellite relay station for TV. This antenna was a large parabolic reflector that had to have a thermal expansion coefficient of zero from near absolute 0 temperature to several hundred degrees. How did they accomplish this?

The reason for this wide temperature range was that as the satellite stayed in its fixed position relative to a position on earth, then the satellite would go from full exposure to the sun to total shading by the earth.

.

 

[ggunn]

230202-1453 EST

If the resistivity of the wire changes, then you just change the value of the 0.0045 term. If the wire material changes its characteristics with temperature, then the 0.0045 becomes a more complex function.

When my brother in law was chief scientist at GE aerospace GE won the contract to build the first high power space satellite relay station for TV. This antenna was a large parabolic reflector that had to have a thermal expansion coefficient of zero from near absolute 0 temperature to several hundred degrees. How did they accomplish this?

The reason for this wide temperature range was that as the satellite stayed in its fixed position relative to a position on earth, then the satellite would go from full exposure to the sun to total shading by the earth.

.

Shading by the earth of objects in geosynchronous orbit is relatively uncommon. Most of the time the shadow of the earth is above or below the orbital path and when the shadow intersects the orbital path the shaded portion is a small portion of the path. For a given satellite, a few minutes of shading per year around the equinoxes, maybe?

 

[winnie]

Geostationary satellites have up to 70 minutes of shadow per day, during a few weeks surrounding equinox.