As an energy engineer at your company, you find 12 large exhaust fans constantly running to exhaust general plant air (no localized heavy pollution). They are each powered by 30 kW electric motors with loads of 34 kW each. You find they can be turned off periodically with no adverse effects. You place them on a central timer so that two of them are turned off for 10 minutes each hour. At any time, two of the fans are off, and the other ten fans are running. The fans operate ten h/day, 340 days/year. By using the following costs, answer each question.
* Demand charge: 10 USD/kW/month
* Energy charge: 0.08 USD/kWh
a) What is the annual demand saving in USD?
b) What is the annual energy saving in USD?
c) What is the total saving in USD?
It is given that at any time 2 fans are off. So effectively 2 fans worth of savings are there in total. First lets calculate how many kW are being saved and then we will calculate demand and energy charge savings.
It is given that they run 10 hours per day, 340 days a year and the load is 34kW. So, total kWh saved per year
=2*10*340*34= 231200kWh
While the number of kW (without hours) saved per year is
=2*340*34=23120.
With this data, lets calculate savings.
A. What is the annual demand savings
It is given that 10USD is saved perkW in a month. We had calculated kW saved per year as 23120. So per month kW saved is 23120/12. So, annual demand savings
=10*(23120/12)*12= $231200
B. Energy charge is .08USD per kWh. So, total savings
231200*.08=$18496
C. Total savings=231200+18496= $249696
Get Answers For Free
Most questions answered within 1 hours.