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To find out what happens to the efficiency of a motor as I change the mass it lifts.

Extracts from this document...

Introduction

Coursework – Energy

Aim: To find out what happens to the efficiency of a motor as I change the mass it lifts.

METHOD

When devices transfer energy, only part of it is USEFULLY TRANSFERRED to where it is wanted and in the form that it is wanted. The rest is transferred in some non-useful way and therefore it is ‘wasted’. The ‘wasted’ energy and the ‘useful’ energy are both eventually transferred to the surroundings. The greater the proportion of energy supplied to a device, THAT IS USEFULLY TRANSFERRED, the more efficient we say the device is.

        A motor is a device that transfers electrical energy into rotational kinetic energy, which can be used to lift a load. We are going to try and find out how the efficiency of a motor differs as we change the mass that it is required to lift. To do this we will let a small electric motor lift a small load 0.5m off the ground and work out it’s efficiency, increasing the weight of the load it has to lift by 0.1N each time we run it. Below is a diagram showing how the circuit for this experiment will be set up:

 As you can see, the motor has to be connected to the ammeter, voltmeter and the power supply. The ammeter is placed in series and the voltmeter is placed in parallel. The motor should be clamped tightly onto a stand over one metre off the ground. A piece of string capable of reaching to the floor should be attached to the spindle of the motor, whilst the other end should be attached to the mass hook.

When the experiment is run, a stopwatch should begin timing as the power supply is switched on.

...read more.

Middle

Energy has to be transferred from one form to another, e.g. a hairdryer turns electrical energy from a mains supply into movement energy (the fan), heat energy (to heat the air as it passes through) and sound energy (waste energy). You cannot create it or destroy it. Energy efficiency is how much of the energy you put into an appliance or machine is transferred into the useful energy that you are trying to get out. All machines in the real world have an efficiency that is less than 1 (or 100%). In the case of the motor above, part of the electrical energy put in is transferred into the useful movement energy, however, the machine also transfers it’s energy into two other waste forms: it creates a little heat and a little sound, caused by the force of friction on it’s moving parts, as in all machines. The greater the proportion of energy supplied to a device, that is usefully transferred, the more efficient the device.        

        To calculate the efficiency of any device we need to use to the following formula:

EFFICIENCY = USEFUL ENERGY TRANSFERRED BY DEVICE

                       TOTAL ENERGY SUPPLIED TO DEVICE

… So in the case of an appliance that coverts 200 joules of electrical energy per second into 150 joules/sec of waste heat energy, 20 joules/sec of useful light energy and 30 joules/sec useful sound energy… EFFICIENCY = 50 ÷ 200 [× 100] = 25%image04.png

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We are trying to find out how the efficiency of a motor changes as we increase the mass it has to lift, so we need to remember that the efficiency of a motor is determined by how much of the electrical energy put in is transferred into useful energy output. The useful energy output is gravitational potential energy (GPE).

...read more.

Conclusion

there would be no useful energy being transferred, all electrical energy inputted would be transferred into non-useful waste forms. It would have been interesting to find out what this weight is. My graph displays a line of best fit which makes it difficult to predict how it will continue. It is very hard to tell from my data whether the line should slope downwards in a straight line, or decrease slowly at first, then more quickly, etc. Therefore collecting more data would provide me with more informative results.

Carrying out further tests around 0.6N – the apparent optimum weight

It would be interesting to try and find out exactly where my line of best fit should peak, and what the precise optimum weight for the efficiency of this motor is. By collecting further, accurate results I could determine what weight this motor will lift most efficiently, e.g. 0.62N.

By plotting more points around the area where my line of best fit peaks on my graph, I would be able to tell exactly how my line of best fit should be shaped around it’s peak.

Lifting the loads higher and finding out at what point is a motor most efficient

We could try and find out how much a motor’s efficiency differs depending on how much of the way through a lift it is. For example, is a motor more efficient when it has just begun lifting it’s load, or when it is almost at its full height. We could do this by letting a motor lift a load slowly, whilst measuring the electrical energy input just after it starts to lift, and whilst it has been running for longer.

We could also try and determine whether a motor is on the whole more efficient when lifting to a small height or to a large height, or in other words, how does the length of time it has to lift a load for affect a motor’s efficiency?

...read more.

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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