Question
Download Solution PDFThe average demand of a plant is 55 MW. Find the maximum energy that can be produced if the plant is running at full load according to the operating schedule. The plant use factor is 60%.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFPlant Capacity Factor (Use Factor)
This is the ratio of the actual energy produced to the maximum possible energy the plant could produce if it ran at full capacity.
\(Plant \space use \space factor={Actual \space Energy \space Produced \over Maximum \space possible \space energy}\)
Actual Energy Produced
This is calculated by multiplying the average demand by the total time (in hours). Since no time is provided, we can calculate the energy for a standard period, like one day (24 hours).
Calculation
Given, Average demand = 55 MW
Plant use factor = 60% = 0.6
Actual Energy Produced (1 day) = 55MW × 24hours = 1320MWh
Maximum Possible Energy (1 day) = \({1320\space MWh\over 0,6}=2200\space MWh\)
Last updated on May 29, 2025
-> SSC JE Electrical 2025 Notification will be released on June 30 for the post of Junior Engineer Electrical/ Electrical & Mechanical.
-> Applicants can fill out the SSC JE application form 2025 for Electrical Engineering from June 30 to July 21.
-> SSC JE EE 2025 paper 1 exam will be conducted from October 27 to 31.
-> Candidates with a degree/diploma in engineering are eligible for this post.
-> The selection process includes Paper I and Paper II online exams, followed by document verification.
-> Prepare for the exam using SSC JE EE Previous Year Papers.