ERGY8070. Week 6: Sample Problems

Week 6 ERGY8070 Into to Energy Management Practice Problems 1) What is the power draw in watts of a 3-phase motor rated at 15 A and 575 V with a power factor of 0.85? 2) Can energy savings be directly measured? 3) Historically, what has been cheaper, energy efficiency or renewable energy? 4) A system uses a 10 hp motor running at 78% efficiency when at full load. • • What is the output power in kW? What is the input electrical power in kW? 5) An exhaust fan uses a 3-phase, 7.5 hp motor. At full-load, the motor is 91.7% efficient. What is the motor’s electrical input power in kW? 6) Using the nameplate values, calculate the annual energy usage in kWh of this motor if it operates 8 hours a day, 365 days a year: Model No. Serial No. HP Volts Amps RPM Hz Ph. Ser. F. NEMA Nom. Eff. PF Rating ENCL ABC 123 25 230 59 1760 60 3 1.15 90.2% 88% 40C AMB CONT TEFC 7) A client is looking to implement an ECM that is a retrofit of their facility’s exterior lights, which operate 12/7/365 (12 hours per day, 7 days per week, 365 days per year). They currently have five 150 W HID lights and want to replace each of them with an equivalent 40 W LED fixture. What is the annual energy savings from this ECM? 8) What is the rated voltage of a 32 W fluorescent lamp (i.e. fluorescent tube) with a rated current of 0.25 A and a power factor of 0.95? The lamp is single phase. 9) A client is looking to reduce the electricity usage of one of their facilities. After an energy audit, they found that lighting was one of the main users of energy that they could easily reduce. Before they can implement an ECM, they first have to understand their current Week 6 ERGY8070 lighting energy usage. They counted 150 T8 florescent ceiling tubes, each with an input wattage of 30W, which are running for approximately 18 hours a day (although staff do have the option to turn them off and on at will) year-round. Calculate their baseline energy use. 10) For the scenario in question 8, the client has determined that they can either replace the lights with LEDs - which have an input wattage of 17.55W - or they can install occupancy sensors to reduce the operating hours of the lights by an estimated 6 hours a day. • They only have the budget for one ECM, so which one will result in greater energy savings in kWh? 11) A single-family home uses an 18 kW electric furnace to provide space heating. Assume it runs 2 hours a day for half of the year. If the total home’s annual electricity consumption is 20,000 kWh, what percentage of the total electricity is used for heating? Week 6 ERGY8070 ANSWERS 1) What is the power in watts of a 3-phase motor rated at 15 A and 575 V with a power factor of 0.85? Recall from Slide 20 that: 𝑃𝑜𝑤𝑒𝑟 𝑓𝑜𝑟 3 − 𝑝ℎ 𝑙𝑜𝑎𝑑 = 𝑉𝑜𝑙𝑡𝑠 ∗ 𝐴𝑚𝑝𝑠 ∗ 𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 ∗ √3 𝑃𝑜𝑤𝑒𝑟 = 575 𝑉 ∗ 15 𝐴 ∗ 0.85 ∗ √3 𝑃𝑜𝑤𝑒𝑟 = 12,698 𝑊 Note that √3 is approximately equal to 1.732 2) Can energy savings be directly measured? No, as per slides 27 & 28. 3) Historically, what has been cheaper: energy efficiency or renewable energy? As per slide 13, energy efficiency has been historically cheaper than renewable energy. 4) A system uses a 10 hp motor running at 78% efficiency when at full load. • What is the output power in kW? 𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑀𝑜𝑡𝑜𝑟 ℎ𝑝 ∗ 0.746 𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 10 ℎ𝑝 ∗ 0.746 𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 7.46 𝑘𝑊 • What is the input electrical power in kW? 𝑀𝑜𝑡𝑜𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝜂!"#"$ 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 7.46 𝑘𝑊 0.78 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 9.56 𝑘𝑊 5) An exhaust fan uses a 3-phase, 7.5 hp motor. At full-load, the motor is 91.7% efficient. What is the motor’s electrical input power in kW? 7.5 ℎ𝑝 ∗ 0.746 0.917 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 6.10 𝑘𝑊 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = Week 6 ERGY8070 6) Using the nameplate values, calculate the annual energy usage in kWh of this motor if it operates 8 hours a day, 365 days a year: 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟%&''('")* = 25 ℎ𝑝 ∗ 0.746 / 0.902 𝑀𝑜𝑡𝑜𝑟 𝐼𝑛𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟%&''('")* = 20.68 𝑘𝑊 OR 1𝑘𝑊 = 20.68 𝑘𝑊 1000𝑊 = 20.68 𝑘𝑊 𝑀𝑜𝑡𝑜𝑟 𝑃𝑜𝑤𝑒𝑟%&''('")* = 230𝑉 ∗ 59𝐴 ∗ 0.88 ∗ √3 ∗ 𝑀𝑜𝑡𝑜𝑟 𝑃𝑜𝑤𝑒𝑟%&''('")* 𝐴𝑛𝑛𝑢𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑒 = 𝐸𝑞𝑢𝑖𝑝𝑚𝑒𝑛𝑡 𝑃𝑜𝑤𝑒𝑟 ∗ 𝑅𝑢𝑛𝑡𝑖𝑚𝑒 ℎ𝑟𝑠 𝑑𝑎𝑦𝑠 𝐴𝑛𝑛𝑢𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑒 = 20.68 𝑘𝑊 ∗ 8 ∗ 365 𝑑𝑎𝑦 𝑦𝑒𝑎𝑟 𝐴𝑛𝑛𝑢𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑒 = 60,385.6 𝑘𝑊ℎ 7) A client is looking to implement an ECM that is a retrofit of their facility’s exterior lights, which operate 12/7/365. They currently have five 150 W HID lights and want to replace each of them with an equivalent 40 W LED fixture. What is the l annual energy savings from this ECM? 𝑊 ℎ𝑟𝑠 𝑑𝑎𝑦𝑠 ∗ 12 ∗ 365 𝑙𝑖𝑔ℎ𝑡 𝑑𝑎𝑦 𝑦𝑒𝑎𝑟 𝐵𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝐸𝑛𝑒𝑟𝑔𝑦 = 3,285,000 𝑊ℎ = 3,285 𝑘𝑊ℎ 𝐵𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝐸𝑛𝑒𝑟𝑔𝑦 = 5 𝑙𝑖𝑔ℎ𝑡𝑠 ∗ 150 𝑊 ℎ𝑟𝑠 ∗ 4380 𝑙𝑖𝑔ℎ𝑡 𝑦𝑒𝑎𝑟 𝑃𝑜𝑠𝑡 − 𝐼𝑛𝑠𝑡𝑎𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 = 876,000 𝑊ℎ = 876 𝑘𝑊ℎ 𝑃𝑜𝑠𝑡 − 𝐼𝑛𝑠𝑡𝑎𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 = 5 𝑙𝑖𝑔ℎ𝑡𝑠 ∗ 40 𝐴𝑛𝑛𝑢𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 𝐵𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝐸𝑛𝑒𝑟𝑔𝑦 − 𝑃𝑜𝑠𝑡 − 𝐼𝑛𝑠𝑡𝑎𝑙𝑙𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 𝐴𝑛𝑛𝑢𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 3,285 𝑘𝑊ℎ − 876 𝑘𝑊ℎ 𝐴𝑛𝑛𝑢𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 2,409 𝑘𝑊ℎ 8) What is the voltage of a 32 W lamp with a rated current of 0.25 A and a power factor of 0.95? 𝑃𝑜𝑤𝑒𝑟 (𝑙𝑖𝑔ℎ𝑡𝑖𝑛𝑔 𝑖𝑠 𝑠𝑖𝑛𝑔𝑙𝑒 − 𝑝ℎ𝑎𝑠𝑒) = 𝑉𝑜𝑙𝑡𝑠 ∗ 𝐴𝑚𝑝𝑠 ∗ 𝑃𝐹 32 𝑊 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 0.25 𝐴 ∗ 0.95 Week 6 ERGY8070 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 = 134.74 𝑉 9) A client is looking to reduce the electricity usage of one of their facilities. After an energy audit, they found that their lighting was one of the main users of energy that they could easily reduce. Before they can implement an ECM, they first have to understand their current energy usage. They counted 150 T8 florescent ceiling tubes, each with an input wattage of 30W, which are running for approximately 18 hours a day (although staff do have the option to turn them off and on at will). Calculate their baseline energy use. 𝑊 ℎ𝑟𝑠 𝑑𝑎𝑦𝑠 ∗ 18 ∗ 365 𝑡𝑢𝑏𝑒𝑠 𝑑𝑎𝑦 𝑦𝑟 𝐵𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝐸𝑛𝑒𝑟𝑔𝑦 = 29,565,000 𝑊ℎ = 29,565 𝑘𝑊ℎ 𝐵𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝐸𝑛𝑒𝑟𝑔𝑦 = 150 𝑡𝑢𝑏𝑒𝑠 ∗ 30 10) For the scenario in question 8, the client has determined that they can either replace the lights with LEDs - which have an input wattage of 17.55W - or they can install occupancy sensors to reduce the operating hours of the lights by an estimated 6 hours a day. • They only have the budget for one ECM, so which one will result in greater energy savings in kWh? 𝑊 ℎ𝑟𝑠 𝑑𝑎𝑦𝑠 𝐿𝐸𝐷 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑎𝑔𝑒 = 150 𝑡𝑢𝑏𝑒𝑠 ∗ 17.55 ∗ 18 ∗ 365 𝑡𝑢𝑏𝑒 𝑑𝑎𝑦 𝑦𝑟 𝐿𝐸𝐷 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑎𝑔𝑒 = 17,295,500 = 17,295.5 𝑘𝑊ℎ 𝐿𝐸𝐷 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 29,565 𝑘𝑊ℎ − 17,295.5 𝑘𝑊ℎ 𝐿𝐸𝐷 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 12,269.5 𝑘𝑊ℎ 𝑊 ℎ𝑟𝑠 𝑑𝑎𝑦𝑠 ∗ 12 ∗ 365 𝑡𝑢𝑏𝑒 𝑑𝑎𝑦 𝑦𝑟 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑎𝑔𝑒 = 19,710,000 = 19,710 𝑘𝑊ℎ 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 29,565 𝑘𝑊ℎ − 19,710 𝑘𝑊ℎ 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 𝑆𝑎𝑣𝑖𝑛𝑔𝑠 = 9,855 𝑘𝑊ℎ 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝐸𝑛𝑒𝑟𝑔𝑦 𝑈𝑠𝑎𝑔𝑒 = 150 𝑡𝑢𝑏𝑒𝑠 ∗ 30 Therefore, the LED lighting replacement results in greater energy savings. 11) A single-family home uses an 18 kW electric furnace to provide space heating. Assume it runs 2 hours a day for half of the year. If the total home’s annual electricity consumption is 20,000 kWh, what percentage of the total electricity is used for heating? 𝑑𝑎𝑦𝑠 ℎ𝑟𝑠 365 𝑦𝑟 𝐻𝑒𝑎𝑡𝑖𝑛𝑔 𝐸𝑛𝑒𝑟𝑔𝑦 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 = 18 𝑘𝑊 ∗ 2 ∗ 𝑑𝑎𝑦 2 𝐻𝑒𝑎𝑡𝑖𝑛𝑔 𝐸𝑛𝑒𝑟𝑔𝑦 𝐶𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 = 6,570 𝑘𝑊ℎ 6,570 𝑘𝑊ℎ ∗ 100% 20,000 𝑘𝑊ℎ 𝐻𝑒𝑎𝑡𝑖𝑛𝑔 % 𝑜𝑓 𝑇𝑜𝑡𝑎𝑙 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑖𝑡𝑦 = 32.85 % 𝐻𝑒𝑎𝑡𝑖𝑛𝑔 % 𝑜𝑓 𝑇𝑜𝑡𝑎𝑙 𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑖𝑡𝑦 =