Question

Re-work Problem 4.10 if the phase spacing is (a) increased by$20\%$to $4.8$ft. or (b) decreased by to localid="1650392776814" $20\%$. Compare the results with those Of Problem 4.10.

Short answer

(a) When the spacing between the conductor increased by$20\%$, the positive sequence inductance and positive sequence inductive reactance are $1.137\times {10}^{-6}\frac{H}{M}$ and $0.428\frac{\Omega }{km}$respectively.

(b) When the spacing between the conductor decreased by$20\%$, the positive sequence inductance and positive sequence inductive reactance are $1.057\times {10}^{-6}\frac{\text{H}}{\text{m}}$ and $0.398\frac{\Omega }{\text{km}}$respectively.

When spacing is increased between the conductors, the positive sequence inductance and positive sequence inductive reactance are increases and vice-versa.

Step-by-step solution

Write the given data by the question:

The frequency of the system,$f=60\text{\hspace{0.17em}Hz}$

The diameter of the conductor,$d=0.5\text{\hspace{0.17em}in}$

The spacing between the conductors,$D=4\text{\hspace{0.17em}ft}$

The increased spacing,$D_{in}=4.8\text{\hspace{0.17em}ft}$

The decreased spacing, $D_{de}=3.2\text{\hspace{0.17em}ft}$

Determine the formulas to calculate the positive sequence inductance in Hmand positive sequence inductive reactance in Ωkm.

The expression to calculate the radius of the conductor is given by,

$r=\frac{d}{2}$ ……. (1)

The expression to calculate the positive sequence inductance when spacing increased to is given by,

$L=2\times {10}^{-7}ln\left(\frac{D_{in}}{0.7788r}\right)$ …… (2)

The expression to calculate the positive sequence inductance when spacing decreased to is given by,

$L=2\times {10}^{-7}ln\left(\frac{D_{de}}{0.7788r}\right)$ …… (3)

The expression to calculate the positive sequence inductive is given by,

$X=2\pi fL$ …… (4)

The conversion from foot to meter.

$1ft=0.3048m$

The conversion from inches to meter.

$1ft=0.025m$

Calculate the positive sequence inductance and positive sequence inductive reactance when the diameter increased to 4.8 ft.

(a) Calculate the increased radius of the conductor.

Substitute $d=0.5\text{\hspace{0.17em}in}$for$d=0.5\text{\hspace{0.17em}in}$into equation (1).

$\begin{array}{l}r=\frac{0.5}{2}\\ r=0.25\text{\hspace{0.17em}in}\end{array}$

Calculate the positive sequence inductance when spacing is increased.

Substitute $4.8\text{\hspace{0.17em}ft}$for $D_{in}$and$0.25\text{\hspace{0.17em}in}$ for into equation (2).

$\begin{array}{l}L=2\times {10}^{-7}\ln \left(\frac{4.8\times 0.3048}{0.7788\times 0.25\times 0.0254}\right)\\ L=2\times {10}^{-7}\ln \left(295.839\right)\\ L=2\times {10}^{-7}\times 5.689\\ L=1.137\times {10}^{-6}\frac{\text{H}}{\text{m}}\end{array}$

Calculate the positive sequence inductive reactance.

Substitute$c$ for$L$ and$60\text{\hspace{0.17em}Hz}$ for$f$ into equation (4).

$\begin{array}{l}X=2\pi \times 60\times 1.137\times {10}^{-6}\\ X=428.63\times {10}^{-6}\\ X=0.428\frac{\Omega }{\text{km}}\end{array}$

Hence the positive sequence inductance and positive sequence inductive reactance are $1.137\times {10}^{-6}\frac{\text{H}}{\text{m}}$and $0.428\frac{\Omega }{\text{km}}$ respectively.

Calculate the positive sequence inductance and positive sequence inductive reactance when the diameter decreased to 3.2 ft.

(b) Calculate the increased radius of the conductor.

Substitute $d=0.5\text{\hspace{0.17em}in}$for$d=0.5\text{\hspace{0.17em}in}$ into equation (1).

$\begin{array}{l}r=\frac{0.5}{2}\\ r=0.25\text{\hspace{0.17em}in}\end{array}$

Calculate the positive sequence inductance when spacing is increased.

Substitute$3.2\text{\hspace{0.17em}ft}$ for $D_{in}$and$0.25\text{\hspace{0.17em}in}$for$r$ into equation (3).

$\begin{array}{l}L=2\times {10}^{-7}\ln \left(\frac{3.2\times 0.3048}{0.7788\times 0.25\times 0.0254}\right)\\ L=2\times {10}^{-7}\ln \left(197.22\right)\\ L=2\times {10}^{-7}\times 5.284\\ L=1.056\times {10}^{-6}\frac{\text{H}}{\text{m}}\end{array}$

Calculate the positive sequence inductive reactance.

Substitute $1.137\times {10}^{-6}\frac{\text{H}}{\text{m}}$for$L$ and$60\text{\hspace{0.17em}Hz}$ for $f$into equation (4).

$\begin{array}{l}X=2\pi \times 60\times 1.056\times {10}^{-6}\\ X=398.10\times {10}^{-6}\\ X=0.398\frac{\Omega }{\text{km}}\end{array}$

Hence the positive sequence inductance and positive sequence inductive reactance are $1.057\times {10}^{-6}\frac{\text{H}}{\text{m}}$and $0.398\frac{\Omega }{\text{km}}$respectively.

When spacing is increased between the conductors, the positive sequence inductance and positive sequence inductive reactance are increases and vice-versa.