King Fahd University of Petroleum & Minerals MECHANICAL ENGINEERING DEPARTMENT ME 429 – Energy Efficiency and Auditing HOMEWORK # 6 SOLUTION (Instructor: Dr. Esmail M. A. Mokheimer) 1) A five ton air conditioner has an average electric load of 6 kW. What is its average COP and what is its SEER? Solution: Using Equation 6-2 gives: SEER = (5 tons) (12,000 Btu/hr/ton)/(6 kW)(1000 W/kW) = 60,000 Btu/6,000 Wh = 10 Btu/Wh 2) A 100-ton chiller has a COP of 3.5. What is its electrical load? Solution: Using Equation 6-2 and rearranging it gives: electrical load = cooling capacity/EER = (100 tons)(12,000 Btu/hr/ton)/(3.5)(3412 Btu/kWh) = 100.5 kW 3) I) An industrial facility presently has a 5-hp ventilating fan that draws warm air from a production area. The motor recently failed and they think they can replace it with a smaller motor. They have determined that they can reduce the amount of ventilation air by one third. What size motor is needed now? II) What is the electrical load reduction for the smaller fan motor, above, if the 5-hp motor had an efficiency of 84% and the new 1.5-hp motor has an efficiency of 85.2%. Solution: Use Equation 6-14, and note that the ratio of the new to old cfm rate is 2/3. Thus, the new hp needed is: I) New hp = (2/3)3 × 5 hp = 0.3 × 5 hp = 1.5 hp II) Old load = (5 hp) × (0.746 kW/hp)/(0.84) = 4.44 kW New load = (1.5 hp) × (0.746 kW/hp)/(0.852) = 1.31 kW Electric load reduction = 3.13 kW 4) You have measured the ventilation in a large truck bay and have found that you are using 12,000 cfm. An analysis shows that only 8000 cfm are required. Measurements at the fans give the total electrical consumption of the ventilation system as 16.0 kW at the current cfm rates. You are currently ventilating this area 16 hour each day, 250 days each year, including the times of peak electrical usage. Your monthly electric rates are $.045/kWh and $12.00/kW per month of demand. Assuming that both motors efficiency is 90% and both fans are operating at full load, what is the amount of annual savings that you can expect by the proposed reduction in ventilation rates? Solution: Use Equation 6-14, and note that the ratio of the new to old cfm rate is 2/3. Thus, the new power needed is: New Power = (2/3)3 × 16 kW = 4.740741 kW New load = 4.740741 kW/0.9 = 5.26749 kW Old Power = 16 kW Old load = 16 kW/0.9 = 17.77778 kW Electric load reduction = Old load - New load = 17.77778 – 5.26749 = 12.51029 kW Cost saving due to load reduction = Load Reduction × Demand cost Cost saving due to load reduction = 12.51 kW × $ 12/(kW. Month) × 12 Month/year Cost saving due to load reduction = $ 1,801.481 /year Cost saving due to Energy use reduction = Load Reduction (kW) × Time (hr/year) × Electric Tariff ($/kWh) π ππ£πππ ππ’π π‘π πΈπππππ¦ π’π π ππππ’ππ‘πππ = 12.51029ππ × 16 βπ 250 πππ¦ $ 0.045 × × πππ¦ π¦πππ ππβ ππππππ π ππ ππ π¬πππππ πππ πππ ππππππ = $ ππππ. πππ/ππππ πππ‘ππ π ππ£ππππ = $ 1,801.481 /year + $ 22516.852/π¦πππ π»ππππ πππππππ = $ π, πππ. ππ/ππππ
