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eeet-241/_Homework/hw9.md

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Skyler MacDougall

Homework 9: due 3/25/2020

  1. What is the difference between a drip-proof motor and an explosion-proof motor? Drip-proof motors are protected from dripping liquids up to 15 degrees from the vertical. An explosion-proof motor, by contrast, are fully sealed, and the frames are designed to withstand the enormous pressure.

  2. What is the approximate life expectancy for a motor? It depends on the size, speed, horsepower, what voltage and current its run at, and how long it is run at one time.

  3. Explain what a NEMA Design D motor is unsatisfactory for driving a pump. A pump should be a consistent speed, and shouldn't take long to get to full speed. A NEMA Design D motor achieves neither of these.

  4. Identify the motor components shown in Fig. 3

  5. Show the flow of active power in a 3-phase induction motor when it operates

    1. As a motor. Power is being put into the system in the same direction as rotation.
    2. As a brake. Power is being put in in the opposite direction of rotation. As such, a lot of power is lost as heat.

    image-20200401140723691

  6. Will a 3--phase motor continue to rotate if one of the lines becomes open? Will the motor be able to start on such a line? The motor needs all 3 phases to start. It will be able to run with only two phases, but there will be a lot of excess heat, and the motor will wear out faster.

  7. A 300hp, 2.3kV, 3-phase, 60Hz, squirrel-cage induction motor turns at a full-load speed of 590r/min. Calculate the approximate value of rotor I^2R losses. P_{jr}=({120{f\over p}-n\over 120{f\over p}})(3P)\ P_{jr}=({120{60Hz\over 6\ poles}-590rpm\over 120{60Hz\over 6\ poles}})(3(300hp746W/hp))\ P_{jr}=({1200rpm-590rpm\over 1200rpm})(3(223.8kW))\ P_{jr}=(0.508\overline3)*(3(223.8kW))\ \underline{\overline{|P_{jr}=341.295kW|}} If the line voltage then drops to 1944V, calculate the following:

    1. The new speed, knowing that the load torque remains the same. n=1200(1-s({V_1\over V_2})^2)=1200(1-0.508\overline3({2300V\over 1944V})^2)\ n=346rpm

    2. The new power output. P'=P*{n'\over n}=300hp*{346rpm\over 590rpm}\ P'=176.34hp\approx131.6kW

    3. The new I^2R losses in the rotor. P_{jr}=({120{f\over p}-n\over 120{f\over p}})(3P)\ P_{jr}=({120{60Hz\over 6\ poles}-346rpm\over 120{60Hz\over 6\ poles}})(3(131.6kW))\ P_{jr}=0.711\overline6*3(131.6kW)\ P_{jr}=280.6kW

  8. The bearings in a motor have to be greased regularly, but not too often. The following schedule applies to two motors: Motor A: 75hp, 3550r/min; lubricate every 2200 hours of running time. Motor B: 75hp, 900r/min; lubricate every 10000 hours of running time. Motor A runs continually, 24 hours per day. Motor B drives a compressor and operates about 6 hours per day. How often should the bearings of each motor be greased per year? Motor A should be greased about once every quarter, or 4 times in a year. Motor B should be greased about once every 5 years.

  9. A 40hp 1780r/min, 460V, 3-phase, 60Hz, drip proof Baldor Super E premium energy induction motor has a power factor of 86% and an efficiency of 93.6%. The motor, priced at $2243, runs at full-load 12 hours a day, 5 days a week. Calculate the driving cost of the motor during a 3-year period, knowing that the cost of energy is $0.06/kWh. 12\ hours5\ days/week\ 52\ weeks/year3\ years=9360\ hours\ P_{in}={P_{out}\over eff}={40hp746W/hp\over0.936}=31.9kW\ Cost=price/kWhkWh=$0.069360\ hours31.9kW\ Running\ cost=$17,904.00\ Cost\ w/\ motor=$20,147.00