Question

For the following problems assume the cost of electrical demand is $14 /kW-mo, the cost of electrical energy is $0.02/kWh and the power factor charge is $0.60/kVAr-month 3) The measured current, voltage and power factor of a premium-efficient, three- phase, 125-hp air compressor motor are shown in the table below. Assume motor is 95.3% efficient at all loads. Loaded 127 476 92 Unloaded 63 476 .81 Current (Amps) Voltage (Volts) Power Factor (kw/kvA) A) Calculate the power draw of the motor in kW when loaded and unloaded. B) Calculate the power output of the motor in hp when loaded and unloaded. C) Calculate the fraction loaded of the motor when loaded and unloaded. D) If the service factor is 1.05, is the motor in danger of premature failure? E) Calculate the annual cost of electrical demand and electrical energy use if the motor cycles between running loaded for 40 seconds then unloaded for 60 seconds, and operates 5,000 hours per year.

Answer 1

Given data: - cost of electrical demand = $ 14/kw month Cost of electrical energy = $ 0.02/kwh 15 mp motor, 75% load lmp = 746 watt motor ratting = 15 times 1746 = 11.2 kw (11.19 kw) = 125.328 since motor is 75% load motor kw = 0.75 times 11.02 = 8.4 kw when motor runs 2000 hour Cost of electrical demand of whole year = 14 times 11.2 times 12 month = $ 1886.6 Cos of electrical energy = p times T times hours = 8.4 times 0.2 times 2000 =$ 336 therefore total cost = 1881.6 + 336 \r\n Hence cost = $ 2217.6 when motor runs for 5000 hour Since demand is same electrical demand Cost is = $ 1881.6 Cost of electrical energy = 8.4 times 0.02 times 5000 = 840 therefore total cost = 1881.6 + 840 = $ 2721.6 when motor runs for 8000 hour Electrical demand is same cost of electrical demand =