answer fastly otherwise gives dislike 149. What is the role of energy management in reducing operational costs of electronic energy systems?
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Question 1: A 500 V, 2-core feeder 0.8 km long is required to supply a constant load of 100 kW. The cost of the cable including installation charges is Rs (6 a + 1.3) per metre where a is the cross-sectional area of each conductor in cm². Interest and depreciation total 10%. Determine the most economical size. Cost of energy is 2P per unit. Specific resistance of copper is 1.75 x 10⁻⁶ Ω per cm² cross-sectional area and 1 cm long. [B.T.E. A.P. Power Systems II April/May 1995] [Ans 2.022 cm²] Question 2: A 2-core, 11 kV cable is to supply 1 MW at 0.8 pf lag for 3,000 hours in a year. Capital cost of the cable is Rs (20 + 400 a) per metre where a is the x-sectional area of core in cm². Interest and depreciation total 10% and cost per unit of energy is 15 paise. If the length of the cable is 1 km, calculate the most economical x-section of the conductor. The specific resistance of copper is 1.75 x 10⁻⁶ Ω-cm. [A.M.I.E. Sec B. Power Systems Summer 1993] [Ans 0.2255 cm²] Question 3: Find the most economical x-section of a 3-core feeder cable 250 m long supplying a load of 80 kW, 0.8 pf (lagging) at 400 V for 4,000 hrs per year. The cost of the cable including installation is Rs (150 a + 250) per metre where a is the x-section of each conductor in cm². Interest and depreciation total 10%. Cost of energy wasted is Rs 1 per unit. The resistance per km of conductor 1 cm² cross-section is 0.173 Ω. [A.M.I.E. Sec B. Power Systems Winter 1996]
Kratika B.
You are working as an engineer at a company. During an energy audit, you realize that your company has a 40 kW transformer that is connected all the time. While inspecting the transformer, you calculate the core losses from the device to be approximately 5% of its rated capacity. Assume that the electrical costs are €10/hour and $10 USD/KW/month of peak demand, that the average building is used 12 hours/month, and that the average month has 720 hours. Estimate the annual cost savings from installing a switch that would energize the transformer only when the building was being used. The transformer is used for 12 hours/month and it can be turned off for the remaining hours of the month. That means the transformer can be turned off for 708 hours/year. The core losses from the transformer are 2 kW. Installing a switch to turn on the transformer only when the building is used results in: - Annual energy savings of 2 kW * 708 hours/year = 1416 kWh/year. - Annual cost savings of 1416 kWh/year * $10 USD/KW/month = $14160 USD/year. Does it make sense to use this strategy?
Manish J.
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