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Efficiency of an Electric Motor Objectives In this lab you will measure the efficiency with which a very simple electric motor converts electrical energy into mechanical work and you will attempt to identify the primary source of energy loss. Equipment ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ Open, 2-pole electric motor and pulley Power supply and cables Voltmeter (or multimeter) Ammeter (or a second multimeter) Weights and weight hanger Stopwatch Meter stick Introduction The electrical energy in a wire is stored in the electric field between the free electrons in the wire. An electric motor converts this energy to mechanical work, by using it to spin a shaft that can be attached to a large number of mechanical devices (such as the drive axle of a car). The electric motor is a very efficient device. Commercial electric motors can run at efficiencies of about 80-95 percent when properly loaded. In other words, up to 95% of the electrical energy is converted to useful mechanical work. In contrast, a 4-stroke gasoline engine like the one used in most cars has an efficiency around 30%. In other words, less than a third of the chemical energy in the fuel is converted to useful work. In this lab, we will be measuring the efficiency of a very simple 2-pole electric motor (see figure below) that is built for demonstration purposes rather than efficiency. As a result, you can expect far lower efficiencies than seen in typical commercial electric motors. Figure 1. Demonstration Electric Motor In Figure 1, you can see the red and blue permanent magnet, the “stator,” surrounding the copper windings of the two electromagnets that rotate with the central shaft and together constitute the “rotor.” The small disk on the front side of the shaft is called the “commutator” and it has two semi-circular contacts that make contact with and turn under the “brushes” which in this case are simply two springy copper levers that rise up from the power plugs mounted in the plastic housing. When the contacts (or “poles”) of the commutator turn by 180°, the current in the rotor windings is reversed. This reverses the polarity of the rotor electromagnets which causes the rotor to want to turn by180° so that the north magnetic pole of the electromagnet can be lined up with the south magnetic pole of the stator and vice versa. Since a two-pole motor like this one can start with the poles correctly lined up, it sometimes requires a little push to get it going. The efficiency η of any motor is defined by the relationship η= Ein P = in Eout Pout (1) where Ein is the energy consumed by the motor and Eout is the mechanical energy (the work) done by the motor. Since power is energy per unit time, the efficiency can also be expressed in terms of the ratio of power consumed Pin to mechanical power supplied Pout . In the case of an electric motor the power consumed is simply the applied current times the applied voltage, both of which we can (and will) measure. Pin = VI (2) If the motor runs for a time t then the total energy consumed is Ein = VIt In this lab, we will use the motor to raise some weights via a thread and pulley system. The work done to raise a mass m by a height h in against the acceleration of gravity g is simply Eout = mgh (3) And so, the efficiency of our motor can expressed as η= Ein VIt = E out mgh (4) By measuring each of the quantities on the right hand side of Eqn . 4 above, we can find the efficiency of our electric motor. Note that the energy lost, Eloss , during the operation of the motor is just Eloss = Ein − Eout = 1 −η η mgh We will investigate the efficiency and the energy loss as a function of motor speed and loading. This will help us locate the primary cause of energy loss. Procedure Warning: At no time should you apply more than 8 Volts to the electric motor. If the motor won’t start spinning, DO NOT increase the voltage above 8 Volts. Rather, give it a little push with a finger. If that doesn’t work, either you have too much weight on the pulley or something else is wrong. Ask for assistance if you are unsure! 1. Place the motor and pulley assembly so that the pulley extends just beyond the edge of the table. Place the line/thread over the pulley and hang about 200 g (incl. mass hanger) from the ring. 2. With the power supply DISCONNECTED from the motor, set the power supply voltage all the way down to 0 Volts and set the current limit to about ¾ of maximum. Make sure the power supply voltage should read 0! Turn the power supply off and then connect it to the motor terminals. 3. Connect a voltmeter across the motor terminals. 4. Connect an Ammeter in series with the motor (i.e. between the motor and the power supply). Make sure you use the 10 Amp. socket on the multimeter as you will likely blow the multimeter fuse if you connect to the lower current input. 5. Turn the power supply on but do not raise the voltage level above 0V yet. Pull the line to lower the weight to the floor. Then raise the power supply voltage to about 5-6V. Give the rotor a little push with a finger and see if the motor can raise the weights. If not, increase the voltage, but do not exceed 8 Volts! If you cannot get the motor to raise the weights, ask the instructor for assistance. After raising the weights, turn the power supply voltage back down and lower the weights to the floor. 6. Practice raising and lowering the weights until you are adept at the process and can raise the weights in a controlled way with a range of speeds and can read the average voltage from the voltmeter and the average current from the ammeter while raising the weight. Each time you raise the weight, raise it the same distance, from the floor to the height of the table. 7. Choose a weight that the motor can only just raise when the power supply is set to about 4 V. In this step, after you find the right weight, the suspended mass will not be changed. Raise the mass while recording the voltage and current and measure how long it takes to raise the weight. Raise the voltage and perform this step again, once again recording the voltage, current and the time taken to raise the mass. Repeat at 0.1 V intervals until the voltage is about 8 V. Record all your data in a table and don’t forget to record the mass. 8. For each trial in the last step calculate the average speed of the weight as it was raised. Then make a plot of the speed of the suspended mass versus the current. 9. Calculate the efficiency of the motor for each trial in step 7. Assuming the maximum load that the motor can raise at 4 V is approximately the mass you used and that the maximum weight that the motor can raise is proportional to the voltage, calculate the fraction of maximum load at which the motor is operating during each trial in step 7. Plot the efficiency versus fraction of maximum load. 10. Calculate the energy loss E loss for each trial in step 7. Questions 1. Commercial electric motors tend to be most efficient at about 75% of maximum load. In step 9 above, you looked at the efficiency versus load. Can you see a trend? Does the efficiency of this motor peak near 75% of maximum load? 2. Note that the friction between two contacting surfaces is equal to the coefficient of friction between the surfaces times the force with which those surface are pressed together (the normal force). Therefore, the force of friction between rotating parts should be independent of the speed of the rotating parts. In other words, it should be constant. So, the energy lost to friction should be simply the work done against this constant frictional force. Thus, the energy lost to friction should only depend on how many turns the motor shaft makes. Since the number of turns required to raise the weight the fixed distance between the floor and the table top is fixed, the energy lost to friction should be the same each time. Given this model, does the energy loss you calculated in step 10 support the fact that primary source of energy loss is friction? The electrical energy dissipated in a wire with resistance R is given by the expression E loss = I 2 Rt (5) By plotting E loss versus I 2 t determine whether the primary energy loss is due to resistance in the windings. Using an ohmmeter, measure the resistance of the windings directly, does it agree with the slope of the plot?