What Must the Output of the Powerhouse Be to Power the Factory?

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SUMMARY

The discussion focuses on calculating the necessary output of a DC generator located at a powerhouse to meet the power demands of a factory situated 0.50 miles away. The factory requires 45 kW at a voltage of 110V, while the transmission cables have a resistance of 0.25 Ohms per mile. The total resistance for the two cables is 0.50 Ohms, leading to a voltage drop that must be accounted for in the generator's output. The calculations confirm that the powerhouse must produce at least 50.5V to ensure the factory receives the required voltage after accounting for power loss due to resistance.

PREREQUISITES
  • Understanding of Ohm's Law and power calculations in electrical circuits
  • Familiarity with DC circuit analysis
  • Knowledge of resistance and its impact on voltage drop
  • Basic understanding of electrical power transmission
NEXT STEPS
  • Calculate voltage drop in DC circuits using different resistance values
  • Explore the effects of cable length on power loss in electrical systems
  • Study the design considerations for DC power transmission systems
  • Learn about optimizing generator output for varying load conditions
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Electrical engineers, power system designers, and anyone involved in the design and analysis of DC power transmission systems will benefit from this discussion.

CraZyBryaN
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Hello. Hopefully someone here is able to answer this. I'm not sure if it's easy or hard, but it sure is frustrating... Here is the question...word for word, I would make a sad attempt at a picture, but it would probably be wrong. So...


"A powerhouse near a waterfall has a large DC generator that produces electricity for a factory 0.50 mile away. The energy is transmitted over two cables each with a resistance of 0.25 Ohms/Mile. Given that the factory requires 45kW at a voltage of 110V to run its equipment, what must be the output of the powerhouse"

We haven't talked a lot about resistances yet, so I think I'm probably making mistakes there...but yea. Thanks in advance.
 
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For DC, power P = V * I, so I = P/V.

The total resistance over the line, R, is given by R = [itex]\rho[/itex] L, where [itex]\rho[/itex] is the resistance per unit length and L is the total length of the line.

Now what is the voltage drop and power loss over the line?
 

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