Question

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistance per unit length is to be 0.195 Ω/km. The densities of copper and aluminum are 8960 and 2600 kg/m3, respectively. Compute (a) the magnitude J of the current density and (b) the mass per unit length λ for a copper cable and (c) J and (d) λ for an aluminum cable.

Answer 1

(a) Jc = 6.88×10⁵A/m²

(b) λ = mass per unit length = 8960×8.82×10-⁵ = 0.790kg/

(c) Ja = 4.30×10⁵A/m²

(d) λ = mass per unit length = 8960×8.82×10-⁵ = 0.7367kg/m

Step-by-step explanation:

Given:

the current to be carried in the conductors I = 60.7A

Densities of copper and aluminum, 8960kg/m³ and 2600kg/m³ respectively.

R/L = 0.195Ω/km = 0.195×10-³Ω/m

Required to find

(a) J the current density in A/m²

To do this we would need to know what the cross sectional area is. A relation that can help us is that of the resistance of a conductor which is

R = ρL/A

Where R is the resistance of the conductor in Ohms (Ω)

ρ is the resistivity of the (a property) conductor in (Ωm)

L is the length of the conductor

A is the cross sectional area of the conductor

From the formula, the resistance per unit length R/L = ρ/A

So A = ρ ÷ R/L

For copper, resistivity is ρ = 1.72×10-⁸Ωm

So Ac = 1.72×10-⁸/0.195×10‐³ = 8.82×10-⁵ m²

Ac = 8.82×10-⁵ m²

I = 60.7A

Jc = I/Ac = current density

Jc = 60.7/(8.82×10-⁵) = 6.88×10⁵A/m²

(b) λ = mass per unit length = density × Area

Copper: density = 8960kg/m³, Ac = 8.82×10-⁵ m²

λ = mass per unit length = 8960×8.82×10-⁵ = 0.790kg/m

(c) For aluminium, ρ = 2.75×10-⁸Ωm

So Aa = 2.75×10-⁸/0.195×10‐³ = 1.41×10-⁴ m²

Aa = 1.41×10-⁴ m²

Ja = I/Aa= current density

Ja = 60.7/(1.41×10-⁴) = 4.30×10⁵A/m²

(d) Aluminium: density = 2600kg/m³, Ac = 1.41×10-⁴m²

λ = mass per unit length = 2600×1.41×10-⁴ = 0.367kg/m

Answer 2

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Step-by-step explanation:

The expression for electric field of conductor is,

`E = (V)/(L)`

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

`J = (E)/(p)`

Substitute (V/L) for E in the above equation of current density.

`J = (V)/(pL) ------(1)`

Substitute iR for V in equation (1)

`J = (iR)/(pL) ------(2)`

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

`J = ((50) (0.200* 10^-^3) )/(1.69 * 10^-^8 ) \\\\= 5.91 * 10^5A.m^-^2`

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

`p = (RA)/(L)`

`A = (pL)/(R)`

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

`\lambda = d(p)/((R)/(L) ) ------(3)`

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

`\lambda = (8960) ((1.69 * 10^-^8 )/(0.200* 10^-^3) \\\\= 0.757kg.m^-^1`

c) Using the equation (2) current density for aluminum cable is,

`J = (iR)/(pL)`

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

`J = ((50)(0.200*10^-^3) )/(2.89* 10^-^8) \\\\= 3.5 *10^5A/m^2`

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

`\lambda = d(p)/((R)/(L) )`

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

`\lambda = (2700) ((2.82 * 10^-^8) )/((0.200 * 10^-^3) ) \\\\= 0.380kg/m`

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m