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Given a biomass fuel cell powerplant which has power capacity of 600 kW, and assume it generates at 83% of full power for 40 hours per week, 50 weeks per year. If the installation of the powerplant costs $10 million, and yearly operating costs are $100k, but it saves $400k per year in waste disposal costs, how many years before the powerplant pays back the installation cost? Use the average industrial electricity rate found at https://www.electricitylocal.com/states/california/oxnard/ (

Given a biomass fuel cell powerplant which has power capacity of 600 kW, and assume it generates at 83% of full power for 40 hours per week, 50 weeks per year. If the installation of the powerplant costs $10 million, and yearly operating costs are $100k, but it saves $400k per year in waste disposal costs, how many years before the powerplant pays back the installation cost? Use the average industrial electricity rate found at https://www.electricitylocal.com/states/california/oxnard/ (

Power System Analysis and Design (MindTap Course List)
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter2: Fundamentals
Section: Chapter Questions
Problem 2.31P: Consider two interconnected voltage sources connected by a line of impedance Z=jX, as shown in...
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2.  Given a biomass fuel cell powerplant which has power capacity of 600 kW, and assume it generates at 83% of full power for 40 hours per week, 50 weeks per year.  If the installation of the powerplant costs $10 million, and yearly operating costs are $100k, but it saves $400k per year in waste disposal costs, how many years before the powerplant pays back the installation cost?  Use the average industrial electricity rate found at   https://www.electricitylocal.com/states/california/oxnard/ (

Expert Solution
Step 1

Given

power capacity =600 kW

operation = 50 weeks per year

Installation cost =$10 million =$106

operating cost per year = $100k

saving per year =$400k

Step 2

 

                                            Total output power =0.83×600 kW                                   =498 kWTotal output energy in a week =498 kW× 40 hours                                                    =19.92 ×103 kWhTotal output energy per year =19.92 ×103 ×50         (50 weeks per year)                                                   =996×103 kWh

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