Question

# A small town with a demand of $800kW$ of electric power at $220v$ is situated $15km$ away from an electric plant generating power at $440v$. The resistance of the two-wire line carrying power is $0.5ohm/km$. The town gets power from the line through a $4000-220volt.$ Step-down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must the plant supply, assuming there is negligible power loss due to leakage? (c) Characterize the step-up transformer at the plant.

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Solution

## Step1: Given dataThe input voltage of the plant is ${V}_{i}=4000volt.$The output voltage is ${V}_{o}=220volt.$The total required electric power is ${P}_{d}=800KW.$Resistance of the wire is $\left(15×0.5\right)ohm.$Step2: Concepts and formulaeThe RMS voltage of an ac circuit is ${I}_{rms}=\frac{Power}{{V}_{rms}}$, where, ${V}_{rms}and{I}_{rms}$are the RMS voltage of the circuit current.Power loss on an ac circuit is $P={I}^{2}R$, where, I is the current through the circuit and R is the resistance of the circuit.From, Ohm's law $V=IR$, where, V is the voltage across the circuit and I is the current through the circuit.Step3: Finding the RMS currentThe RMS current in the coil is ${I}_{rms}=\frac{Power}{{V}_{rms}}$.So, ${I}_{rms}=\frac{Power}{{V}_{rms}}=\frac{800×{10}^{3}}{4000}=200\phantom{\rule{0ex}{0ex}}or{I}_{rms}=200A.$Step4(a): Finding the power lossThe power loss due to the production of heat is ${P}_{L}={I}^{2}R$.So, ${P}_{L}={I}^{2}R={\left(200\right)}^{2}×\left(30×0.5\right)\phantom{\rule{0ex}{0ex}}or{P}_{L}=40000×15\phantom{\rule{0ex}{0ex}}or{P}_{L}=600000\phantom{\rule{0ex}{0ex}}or{P}_{L}=600KW.$Therefore, the power loss in the circuit line is $600KW$.Step4(b): Finding the total power supplyWe know, that the total power supply is the sum of demand power and power loss.So, $P={P}_{d}+{P}_{L}\phantom{\rule{0ex}{0ex}}orP=800+600\phantom{\rule{0ex}{0ex}}orP=1400KW$The total power supply by the plant is $1400KW$.Step4(c): Characterization of the transformer The voltage drop by the circuit is ${V}_{d}=IR$So, ${V}_{d}=IR=\left(200×15\right)=3000\phantom{\rule{0ex}{0ex}}or{V}_{d}=3000volt.$Now, the secondary voltage is ${V}_{s}={V}_{i}+{V}_{d}=4000+3000\phantom{\rule{0ex}{0ex}}or{V}_{s}=7000volt.$According to the question, the power generation takes place at the primary voltage $440volt$.Therefore, the transformer rating is $440-7000volt.$

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