A 3 phase, 60 hz, 138 kv transmission line is 250 miles long and delivers power to a 50 mw load at 132 kv and at 0.92 lagging power factor. The line resistance is 0.175 ohm/mile, the line inductance is 0.302mH/mile and the line capacitance is 0.0245 uf/mile.
Calculate sending end voltage
$Given:\phantom{\rule{0ex}{0ex}}r=0.175ohm/mile=43.75ohm\phantom{\rule{0ex}{0ex}}L=0.302mH/mile\phantom{\rule{0ex}{0ex}}{X}_{L}=jWL=j28.46ohm\phantom{\rule{0ex}{0ex}}C=0.0245\mu f/mile\phantom{\rule{0ex}{0ex}}Y=j2.30\times {10}^{-3}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\varphi =0.92lag\phantom{\rule{0ex}{0ex}}P=50Mw\phantom{\rule{0ex}{0ex}}f=60HZ\phantom{\rule{0ex}{0ex}}{V}_{R}=132Kv\phantom{\rule{0ex}{0ex}}$
$P=\sqrt{3}{V}_{R}{I}_{R}\mathrm{cos}\varphi \phantom{\rule{0ex}{0ex}}{I}_{R}=\frac{50\times {10}^{6}}{\sqrt{3}\times 132\times {10}^{3}\times 0.92}\phantom{\rule{0ex}{0ex}}{I}_{R}=237.7A\phantom{\rule{0ex}{0ex}}{I}_{Y}={V}_{R}\times \frac{Y}{2}\phantom{\rule{0ex}{0ex}}{I}_{Y}=j87.64A\phantom{\rule{0ex}{0ex}}ApplyingKCL\phantom{\rule{0ex}{0ex}}{I}_{z}={I}_{R}+{I}_{Y}\phantom{\rule{0ex}{0ex}}{I}_{z}=218.75\angle -1.45\xb0A$
$NowConsider\mathrm{\pi}\mathrm{model}:\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{\mathrm{s}}={\mathrm{V}}_{\mathrm{R}}+{I}_{z}Z\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{\mathrm{s}}=\frac{132}{\sqrt{3}}\times {10}^{3}+218.75\angle -1.45\left(43.75+\mathrm{j}28.46\right)\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{\mathrm{s}}=76.210\mathrm{K}+9.570\mathrm{K}+\mathrm{j}6.223K-\mathrm{j}0.2419K+0.157\mathrm{K}\phantom{\rule{0ex}{0ex}}{\mathrm{V}}_{\mathrm{s}}=85.937+\mathrm{j}5.981\phantom{\rule{0ex}{0ex}}{\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{86}\mathbf{.}\mathbf{14}\mathbf{\angle}\mathbf{3}\mathbf{.}\mathbf{98}\mathbf{\xb0}\mathbf{KV}\mathbf{\left(}\mathbf{Answer}\mathbf{\right)}$