Question power system 3. Calculate the inductance and capacitance/ph/m for a 3 phase double circuit line whose phase conductors have a radius of 5.3 cm with the horizontal conductor arrangements the spacing between each conductors are 8 m. [Ds -0.995 m. Dm - 14.518 m, L/ph/m= 536x10-7 H) In a single nhasr lineranductors
power system
Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
(2) Given question\begin{aligned}\left.\Rightarrow G_{M P}\right|_{a k a^{\prime}} & =(8 \times 18 \times 32 \times 40)^{1 / 4} \\\left.G_{M P}\right|_{b 4 b^{\prime}} & =(8 \times 8 \times 16 \times 32)^{1 / 4} \\\left.G_{M D}\right|_{C 4 C^{\prime}} & =(8 \times 16 \times 8 \times 16)^{1 / 4} \\G m D_{\text {Total }} & =\left(G M D_{a} \cdot 40 \mathrm{GmD} \cdot G m D_{c} \cdot G m D_{a^{\prime}} \cdot G m D_{b^{\prime}} \cdot G m D_{c^{\prime}}\right)^{1 / 6} \\& =\left((8 \times 16 \times 32 \times 40)^{1 / 2} \cdot(8 \times 8 \times 16 \times 32)^{1 / 2} \cdot(8 \times 16 \times 8 \times 16)^{1 / 2}\right)^{1 / 6} \\G m D_{\text {Todal }} & =14 \cdot 52 \mathrm{mtr} .\end{aligned}\left.\Rightarrow \operatorname{GiR}\right|_{a, a, b, b, 0,0}=\left(\gamma^{\prime} \times 24\right)^{1 / 2}\left.G m R\right|_{b 4 b^{\prime}}\left(\gamma^{\prime} \times 24\right)^{1 / 2},\left.G m R\right|_{C 4 c^{\prime}}=\left(x^{\prime} \times 24\right)^{1 / 2}\begin{aligned}\operatorname{Gim}_{\text {total }} & =\left[\operatorname{Gir}_{2}\left(a<a^{\prime}, b, b^{\prime}, c, c^{\prime}\right)\right]^{1 / 6} \\& =\left(\gamma^{\prime} \times 24\right)^{1 / 2} \\& =\left(0.7788 \times 5.3 \times 10^{-2} \times 24\right)^{1 / 2}\end{aligned}Inductance L=2 \times 10^{-7} \ln \frac{G m P}{G m R} \mathrm{H} / \mathrm{m} / \mathrm{hl}\begin{array}{l}L=2 \times \ln \left(\frac{14.52}{0.995}\right) \times 10^{-7} \\L=5.36 \times 10^{-7} \mathrm{H} / \mathrm{m}\end{array}Caporitance C=\frac{2 \pi \epsilon_{0}}{\ln \frac{G m D}{G M R}} F / m t r.\begin{array}{l}\text { Hele GMR }=(r \times 24)^{1 / 2}=\left(5.9 \times 10^{-2} \times 24\right)^{1 / 2}=1.127 \mathrm{mtr} \text {. } \\C=\frac{2 \pi \times 10^{-9}}{36 \pi \times \ln \left(\frac{14.52}{1.127}\right)} \\C=21.73 \mathrm{pF} / \mathrm{m}) \\\end{array} ...