Question
Download Solution PDFA diesel engine power plant has one 700 kW and two 500 kW generating units. The fuel consumption is 0.25 kg per kWh. Estimate the fuel oil required for a day if the plant capacity factor = 40%.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFPlant Capacity Factor is the ratio of total actual energy produced to the maximum energy produced in that particular time.
Calculation:
Given:
Power plant rating = One 700 kW and two 500kW
Fuel consumption = 0.25 kg per kWh
Plant capacity factor = 40%
Maximum capacity = (500 + 500 + 700) = 1700 kW
Maximum Energy produced in a day = 1700 kW × 24 h = 40800 kWh
Actual energy produced = Plant Capacity Factor × Maximum energy produced in a day
Actual energy produced = 0.4 × 40800 kWh = 16320 kWh
Fuel oil Required = Fuel consumption × Actual energy produced
Fuel oil Required = 16320 × 0.25 = 4080 kg
Last updated on Jul 1, 2025
-> SSC JE Electrical 2025 Notification is released on June 30 for the post of Junior Engineer Electrical, Civil & Mechanical.
-> There are a total 1340 No of vacancies have been announced. Categtory wise vacancy distribution will be announced later.
-> 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.