# The Efficiency of an Electric Motor.

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Introduction

## The Efficiency of an Electric Motor

Aim: To investigate how the efficiency of an electric motor varies when it is lifting different weights.

Summary: I have investigated how the efficiency of an electric motor varies when it is lifting different weights. I found that the efficiency of my motor (using an input voltage of 4v) was maximum when it was lifting 500-800g weights. I also found that the efficiency of the motor depended on its internal resistance and friction.

I also used my results to estimate the internal resistance of the motor and I found it was about 0.5Ω

In my extension experiment I used the electric motor as a generator. I dropped a variety of weights from my pulley system and I measured the amount of energy produced using a joule meter. I found that the efficiency of the generator increased as the mass of the weight I dropped increased. The percentage efficiency of the generator also appeared to tend towards 14%.

Finally I discussed whether my experiment was time reversible. I decided that it was to an extent but not completely. I decided this since I only retrieved about 1-2% of the energy I used to lift the weight when I dropped the weight again, using the motor as a generator.

Experiment 1:

See Diagram1 for Apparatus and circuit diagram

Experimental Method:

I will use a 12v electric motor to lift a variety of weights a fixed height and I will measure the voltage and current in the motor coils. I will also measure the time taken to lift the masses.

How will I make my experiment accurate?

- In my experiments I will keep all variables constant except the one I am investigating.

Middle

30.69

0.978

66.07

7.97

10.093216

2.9

15.27668045

10.86841268

904.8

1.2

4.17

1.65

3.03

26.82

1.119

53.10

8.32

9.088716

2.9

17.11506565

11.21553793

804.8

1.2

4.17

1.55

3.23

21.56

1.391

40.10

8.78

8.084216

2.9

20.15933529

11.6839388

704.8

1.3

3.85

1.49

3.36

17.84

1.682

34.56

8.88

7.079716

2.9

20.48761318

11.78347289

604.8

1.35

3.7

1.4

3.57

15.66

1.916

29.59

9.19

6.075216

2.9

20.52618135

12.09084109

504.8

1.4

3.57

1.3

3.85

13.72

2.187

24.97

9.6

5.070716

2.9

20.30690738

12.50417134

404.8

1.45

3.45

1.2

4.17

12.57

2.387

21.87

10

4.066216

2.9

18.59113562

12.90157737

304.8

1.55

3.23

1.13

4.42

11

2.727

19.27

10.4

3.061716

2.9

15.89139698

13.27785794

204.8

1.6

3.13

1.05

4.76

10.44

2.874

17.54

10.8

2.057216

2.9

11.72924649

13.66046798

Explaining the graph:

In my experiment energy must be conserved, and it follows that the electrical energy I put in to the motor must equal the sum of the energy I get out.

If I first consider the system without friction: the equation below must be satisfied:

Electrical Energy In=(Energy lost in the internal resistance of the motor coils) +(Work done lifting the weight)

IVt=I2Rt+(Work done lifting the weight) t=time, R=resistance of motor coils

The power used lifting the weight is equal to εI, ε is the back ε.m.f resulting from the fact that the motor coils are moving through flux in a magnetic field.

IVt=I2Rt+εIt [1]

If I now consider friction which I am assuming is constant:

IVt=I2Rt+εIt+(Work done against friction)

IVt=I2Rt+ε It+ FD

Here F= frictional force D=distance mass has travelled

Before I can do any further analysis I must first calculate the internal resistance of the motor coils. In my analysis I am going to assume that the resistance is constant as this is a good enough approximation for calculations. In fact the resistance of the coils varies slightly with the speed and temperature of the motor.

### Estimating the resistance of the motor coils

You can only measure the internal resistance of an electric motor when it in motion as a result of the effect of the commutator and other factors.

From my research I found out that for a motor

ω=Angular velocity of motor shaft, K=motor constant, T=torque provided by motor, R=resistance of motor coils, V=voltage.

If I plot ω against T the value of ω at y intercept is equal to and the torque at the x intercept is equal to . Using my recorded values and the intercept values I will be able to calculate the internal resistance of the motor.

1) Finding an expression for ω:

- Average Circumference of cotton reel used to wind up string= 2π*(average radius of cotton reel)

Average radius≈0.02m

⇒Average Circumference=0.04π

- Total Number of rotations of the motor shaft=Length of string wound up/Average Circumference

=6.15/0.04π

- Number of rotations per second=Total No. of rotations/Time taken to lift weight

=

- ω=No. of rotations of motor shaft per second*2π

ω=307.5/t

2) Finding an expression for the torque of the motor: (ignoring friction)

Torque = Moment acting on the shaft

T=Force*Average Radius of Cotton reel

Average Radius=0.02, Force=1/6 of the weight because of the pulley system

⇒T=*0.02

⇒T=

Calculating values for T and ω from my results:

Mass (g) | Time taken to lift mass I.025m | ω (Radians per second) | T (Nm) |

1304.8 | 75.72 | 4.061014263 | 0.042623467 |

1204.8 | 47.84 | 6.427675585 | 0.0393568 |

1104.8 | 40.97 | 7.505491823 | 0.036090133 |

1004.8 | 30.69 | 10.01955034 | 0.032823467 |

904.8 | 26.82 | 11.46532438 | 0.0295568 |

804.8 | 21.56 | 14.26252319 | 0.026290133 |

704.8 | 17.84 | 17.23654709 | 0.023023467 |

604.8 | 15.66 | 19.63601533 | 0.0197568 |

504.8 | 13.72 | 22.41253644 | 0.016490133 |

404.8 | 12.57 | 24.46300716 | 0.013223467 |

304.8 | 11 | 27.95454545 | 0.0099568 |

204.8 | 10.44 | 29.45402299 | 0.006690133 |

Estimation of y intercept=35

X intercept=0.0465

Calculating the internal resistance of the motor coils:

(Derived from the equations quoted above) V≈1.3V, T=0.0465, ω=35

⇒R ≈1.0Ω

I made many assumptions when I was calculating R. For example, I assumed that V and R are constant. I also assumed that ω was constant through out my experiment, when it actually decreased as the radius of the motor shaft increased. As a consequence of this the value of the internal resistance is only an estimation.

##### Can my model predict the amount of energy the motor needs to lift a weight a height of 1.025m?

Predicted Power in using my model= I2Rt+ε It+ FD

Estimation of frictional force≈7N(the force required to just move my pulley system)

R≈1ohm

Using my results to test my model

Mass (g) | Actual amount of energy used (J) | %E | Joules lost in motor coils | %E | Work Done lifting weight | %E | Work Done against friction | %E | Predicted amount of Energy used |

Estimated 1g per 100g weight | "=Vit | "=I*I*R*t | "=M*9.8*1.025 | "=7*1.025 | |||||

Estimated error=1% | |||||||||

1304.8 | 155.9832 | 7.751 | 302.88 | 55.396 | 13.106716 | 3 | 7.175 | 25 | 323.161716 |

1204.8 | 98.93312 | 7.832 | 169.085696 | 55.946 | 12.102216 | 3 | 7.175 | 25 | 188.362912 |

1104.8 | 84.8079 | 7.858 | 132.7428 | 56.288 | 11.097716 | 3 | 7.175 | 25 | 151.015516 |

1004.8 | 66.069432 | 7.968 | 103.904064 | 56.412 | 10.093216 | 3 | 7.175 | 25 | 121.17228 |

904.8 | 53.1036 | 8.316 | 73.01745 | 57.179 | 9.088716 | 3 | 7.175 | 25 | 89.281166 |

804.8 | 40.1016 | 8.784 | 51.7979 | 57.843 | 8.084216 | 3 | 7.175 | 25 | 67.057116 |

704.8 | 34.55608 | 8.883 | 39.606584 | 58.393 | 7.079716 | 3 | 7.175 | 25 | 53.8613 |

604.8 | 29.5974 | 9.191 | 30.6936 | 59.059 | 6.075216 | 3 | 7.175 | 25 | 43.943816 |

504.8 | 24.9704 | 9.604 | 23.1868 | 59.879 | 5.070716 | 3 | 7.175 | 25 | 35.432516 |

404.8 | 21.8718 | 10 | 18.1008 | 60.72 | 4.066216 | 3 | 7.175 | 25 | 29.342016 |

304.8 | 19.2665 | 10.38 | 14.0459 | 61.577 | 3.061716 | 3 | 7.175 | 25 | 24.282616 |

204.8 | 17.5392 | 10.76 | 11.5101 | 62.397 | 2.057216 | 3 | 7.175 | 25 | 20.742316 |

Conclusion

###### Is a the motor time reversible

If my motor was time reversible it should behave in the same way irrespective of the direction of time. For example if you use electrical energy lift a weight with a motor if the system is time reversible you should be able to get the electrical energy back by dropping the weight.

A motor is obviously time reversible to an extent since it can be used both as a motor and a generator. However my results show that for my experiment you are only able to retrieve a small fraction of the energy you used lifting when using the motor as a generator (about 2.5 joules out of 150 or 2%).

This inefficiency can be partially explained by considering the parts of the system that are not time reversible. This includes the friction in the system and the energy lost in the internal resistance of the motor. Here energy is lost as heat and sound that cannot be retrieved.

Conclusion:

In general my experiments went well and I was able to use my result to make some useful conclusions. I was very pleased with the accuracy of my results.

If I had more time I would have taken more experimental reading so that I could get a more complete picture of what was happening. I would also have spent more time calibrating the joule meter since it’s inaccuracy had a large effect on my results.

Bibliography:

Sources used:

- Nuffield Advanced Science Physics student guide 2 unit H to L, Published by Longman, ISBN=0-582-35416-1
- Web page: “Motors” URL=www.srl.gatech.edu/education/ME3110/design-reports/RSVP/DR4/Motors.http
- Encyclopaedia Britannica CD-ROM
- Encarta 99 CD-ROM by Microsoft

This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.

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