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Jupyter notebook ENSP-338-Homeworks/ENSP-338-HW-03/HW-03.ipynb

Project: ENSP 338
Views: 78
Kernel: Anaconda (Python 3)

Homework 3

This will homework will cover additional circuit concepts and resistance calculations.

Gauging the Wires

Browse the tabulated data in the Wikipedia articles on American Wire Gauge and Electrical Resistivities of the Elements.

Question 1:

From the resistivity of copper, and from the wire dimensions in the AWG table, confirm the resistance of 12-gauge copper wire, in ohms per kilometer, that is listed in the table. Show your work, which will include unit conversions. Assume room temperature.

Hint: if you find the resistance of a 1000 meter length of wire, you will have found the ohms per kilometer.

Question 1: Solution

Strategy:

  • find the radius of 12 gauge wire

  • use the dimensions (area and 1000m) to find resistance

  • note that the resistance of a 1000m length of wire gives you the ohms per kilometer

Resistance = Resistivity*Length / Cross-Sectional Area, or R=ρLA R = \frac {\rho*L}{A}

Final units should look like this: Ω=Ωmm21000meters\Omega = \frac{ \Omega m}{m^2} * 1000meters

Given from Wiki articles :

Resitivity of Copper: 16.78nΩm16.78 n\Omega m

16.78nΩm16.78 n\Omega m = 1.678e-8$\Omega m$

Electrical resistivities of the Elements: https://en.wikipedia.org/wiki/Electrical_resistivities_of_the_elements_(data_page)

Area of 12-gauge wire: 3.31mm2 3.31mm^2

to convert from 1mm2 1mm^2 to 1m2 1m^2 simply square both sides of the fundamental conversion: 1mm=1m1e3mm 1mm = \frac{1m}{1e3mm}

(1mm)2=(1m1e3mm)2 (1mm)^2 = (\frac{1m}{1e3mm})^2 = 1mm2=1m21e6mm2 1mm^2 = \frac{1m^2}{1e6mm^2}

3.31mm2 3.31mm^2 = 3.31e-6$m^2$

12 Gauge Copper wire:https://en.wikipedia.org/wiki/American_wire_gauge

Resistance=ρCrosssectionalAreaLengthResistance = \frac{\rho}{Crosssectional Area} * LengthΩ=1.678e8Ωm3.31e6m21000m=5.07\Omega = \frac{1.678e-8\Omega m}{3.31e-6 m^2} * 1000m = 5.07

This is the resistance of a 1000m 12gauge copper wire, it can be seen that this is also the resistance/Km

Resistivity_nOhmm = 16.78 Resistivity_Ohmm = Resistivity_nOhmm * (1 / 1e9) Area_sqrmm = 3.31 Area_sqrm = Area_sqrmm * (1/1e6) Length = 1000 Q1 = Resistivity_Ohmm/Area_sqrm * Length Q1 print('{:0.2f}'.format(Q1))
5.07

This answer closely approximates the Resistivity of Copper Wire given at 5.211mΩm 5.211 \frac{m \Omega}{m} .

https://en.wikipedia.org/wiki/American_wire_gauge

note: mΩm=ΩKm \frac{m \Omega}{m} = \frac{\Omega}{Km}

Question 2:

Calculate the resistance of 12 AWG aluminum wire in ohms/km at room temperature. Why is copper preferred over aluminum conductor, even though it is more expensive?

Q2: Solution

Same Strategy as Question1. Only, there's a different resistivity.

Wiki-provided ( ρ\rho ) and Area:

Resistivity of Aluminum : 26.5nΩm 26.5 n \Omega m
26.5nΩm=2.65e8Ωm26.5 n\Omega m = 2.65e-8 \Omega m
Area of 12-gauge wire: 3.31mm2 3.31mm^2
3.31mm2=3.31e6m23.31mm^2 = 3.31e-6m^2Ω=2.65e8Ωm3.31e6m21000m= 8\Omega = \frac{2.65e-8\Omega m}{3.31e-6 m^2} * 1000m = ~8
Resistivity = 2.65e-8 Area = 3.31e-6 Q2 = (Resistivity/Area) *1000 print('{:0.2f}'.format(Q2))
8.01

Aluminum has a Resistance of ~8 Ω\Omega/Km

Aluminum versus Copper

As we all know, electricty is tranmitted over a certain distance before it can be used.

For the purposes of transmitting electrical energy effieciently, resistance is a bad thing. Resistance causes some of the energy to be converted into heat.

The total cost comparison between the initial investment of copper and the long-term energy cost of aluminum, makes copper the more cost-effective conductor.

Question 3:

MaterialCostDensity
Copper$\frac{ParseError: KaTeX parse error: Expected 'EOF', got '}' at position 5: 4.75}̲{Kg}8.96103Kgm3\frac{8.96 \cdot 10^3 Kg}{m^3}
Aluminum$\frac{ParseError: KaTeX parse error: Expected 'EOF', got '}' at position 5: 1.73}̲{Kg} 2.7103Kgm3\frac{2.7 \cdot 10^3 Kg}{m^3}

You can get up to date prices at the London Metal Exchange. But please use the values above.

Using the Costs and Densities given above, Calculate the raw material cost per meter for 12 AWG wire in both copper and aluminum.

How does this compare to the cost of wire you can find online?

hint: Density * Volume = Mass

Question 3 Solution

Unit Analysis:

Mass(Kg)Volume(m3)Volume(m3)=Mass(Kg) \frac{Mass(Kg)}{Volume(m^3)} * Volume(m^3) = Mass(Kg)
CostMass(Kg)Mass(Kg)=Cost\frac{Cost}{Mass(Kg)} *Mass(Kg) = Cost

Volume of 1m of 12 gauge wire: ~cylinder = CrossSectionalAreaLength=Volume CrossSectional Area * Length = Volume

Area from previous problems = 3.31e-3 m2m^2

3.31e-6 m2m^2 * 1m = 3.31e-6$m^3$ = Volume

Here's the calculation for copper
4.75(Dollars)(1Kg)8.96e3(Kg)1(m3)3.31e6(m3)=Cost\frac{4.75(Dollars)}{(1Kg)} * \frac{8.96e3(Kg)}{1(m^3)} * 3.31e-6(m^3) = Cost
Copper_Cost = 4.75 Copper_Density = 8.96e3 Volume_12AWG = 3.31e-6 * 1 Copper_CostPerLength = Copper_Cost * Copper_Density * Volume_12AWG print('{:0.2f}'.format(Copper_CostPerLength))
0.14

This calulation approximates the cost of 12-gauge copper wire to be 14 cents per meter

This product from Home Depot is 2 conductors of 12 gauge copper wire and is 47 USD for 250 ft. We approximate as 0.20 cents per foot or 0.60 cents per meter. It looks like the raw copper cost is about 50% of this cost.

Aluminum_Cost = 1.73 Aluminum_Density = 2.7e3 Volume_12AWG = 3.31e-6 Aluminum_CostPerLength = Aluminum_Cost * Aluminum_Density * Volume_12AWG print('{:0.2f}'.format(Aluminum_CostPerLength))
0.02

The cost of 12-gauge aluminum wire is ~2 cents per meter