EML2322L-Electric Motors and Drives

# EML2322L-Electric Motors and Drives - EML 2322L MAE Design...

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EML 2322L -- MAE Design and Manufacturing Laboratory Electric Motors and Drives To calculate the peak power and torque produced by an electric motor, you will need to know the following: Motor supply voltage, V s [V] Peak motor current, I s [amps] Peak motor velocity, N [rpm] The power produced by the motor can be calculated as: P [watts] = V s [volts] * I s [amps] (Eq. 1) Converting power to units of horsepower: P [hp] = P [watts]* 0.001341 [hp/watt] (Eq. 2) Since power is equal to work divided by time , we can use the definition of horsepower to calculate the torque produced by the motor: 1 [hp] = 550 [lb-ft/s] = 33,000 [lb-ft/min] = 750 [W] (Eq. 3) Now let’s select a motor to work with as an example. Going to the course homepage; Parts Description link (or Suppliers/Resources link and finding the motor on the vendor’s site): http://www.mae.ufl.edu/designlab/main.htm We will select one of the DC right angle gear motors we have in the lab. http://www.surpluscenter.com/item.asp?UID=2004061505500522&item=5- 1367&catname=electric

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44 RPM 12/24 VDC 1/70 HP ENTSTORT GEARMOTOR • 44 RPM at 12 VDC no load • Voltage 12/24 DC • Amps 1-1/2 no load • Reversible • Duty continuous • Shaft 10 mm x 1 1/4" with 8 mm threaded end • Mount 3 bolt on 2" B.C. • Size 6 3/4" x 2 5/8" x 4" From the basic equations presented above: P = 12 V * 1.5 A = 18 W = 0.024 hp Comparing this value to the rated power of the motor (1/70 = 0.014 hp ), we see the actual power produced is substantially less than the computed electrical power. This is due to the electrical efficiency of the motor. This efficiency will be denoted by η motor and for the purpose of this course, we will assume the following: η motor 60% Now the power can be more accurately computed as follows: P = V * I * η motor = 12 V * 1.5 A * 0.6 = 10.8 W = 0.014 hp Using the definition of hp (Eq. 3): 0.014 hp = 7.9 ft-lb/s = 95.0 in-lb/s Therefore, this motor should be able to lift a 7.9 lb load at the rate of 1 foot per second. This is equivalent to lifting a 1 lb load at 7.9 feet per second. Note the difference is you trade torque for speed or vice versa. Looking at it in units of in-lb instead of ft-lb, this motor should be able to lift a 95 lb load at the rate of 1 inch per second. This is the same as lifting a 1 lb load at 95 inches per second.
Suppose we wish to calculate the velocity of our robot if we consider using these motors to power our drive wheels. First, we need to select a wheel diameter. Returning to the links presented above, let’s arbitrarily select an 8” diameter wheel to get a baseline. In that case the linear velocity can simply be calculated using the circumference of the wheel: V [in/min] = N [rev/min] * π D [in/rev] (Eq. 4)

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