AIMS zState and apply the law of conservation of energy yFixed amount in closed systems yChange form not create or destroy zUnderstand need to transform energy yExplain any losses zUse systems diagrams to account for energy changes zIdentify energy forms and changes within a system zCalculate energy transfers

The Law of Conservation of Energy zThe conservation of energy is a fundamental concept of physics. yAlong with the conservation of mass and momentum. yDerived from first law of thermodynamics. zWithin a closed system, the amount of energy remains constant and energy is neither created nor destroyed. yEnergy can be converted from one form to another but the total energy within the domain remains fixed.

Energy Transformation zHow energy can be converted to other forms is important to technologists zSome forms are directly interchangeable yDropping a stone yPotential  Kinetic zOthers require several stages yCoal burnt in a power station to produce electrical power yChemical  heat  kinetic  electrical

Systems Approach zSystems diagrams can be used to summarise energy changes zConsider a light bulb (simplified) zProduce system diagrams for an electric motor and an electric generator LIGHT BULBELECTIRCALLIGHT

Energy Transformation Examples State the energy form at points A, B, C and D A B C D A: Potential energy B: Kinetic Energy (linear motion) C: Kinetic Energy (rotary motion) D: Electrical Energy

Energy Transformation Examples 1.State the energy form at points A to H in the diagram opposite. 2.Describe the energy changes that take place within the system A: Potential B: Electrical C: Sound D: Electrical E: Light F: Electrical G: Electrical H: Potential

Energy ‘losses’ during transformation zWe accept that energy cannot be created or destroyed zThis tells us that the energy output of a system equals the energy input zHOWEVER, not all the energy is used to do USEFUL work zWhen a conversion takes place there is always a loss yExamples are sound, friction or heat zGo back to the energy conversion diagrams for the bulb, motor and generator and add any losses to the output side

Energy Losses in a Wind Turbine zA turbine can be used to generate electricity. The generator can be connected to it in two ways. coupled directly to vanescoupled via shafts and gears

Energy ‘losses’ during transformation 1.List the energy conversions that take place during its operation 2.Describe the energy losses in both systems 3.Which do you think is more efficient?

Worked Example zA body of mass 30 kg falls freely from a height of 20 metres. Find its final velocity and kinetic energy at impact. First calculate the initial potential energy. EP = mgh = 30  9.8  20 = 5.88 kJ zThis potential energy is converted or transferred into kinetic energy, which means that the kinetic energy at impact is equal to 5.88 kJ. zTo calculate the final velocity of the body we begin by taking EK = 5.9 kJ. Ek = ½mv² 5.88  10³ = ½  30  v² v² = v = 19.8 ms- 1

Pupil Problems zA 5 kg mass is raised steadily through a height of 2 m. What work is done and what is the body’s potential energy relative to the start? zA body of mass 30 kg is projected vertically upwards with an initial velocity of 20 m/s. What is the initial kinetic energy of the body and to what height will it rise? zA mass of 20 kg is allowed to fall freely from a certain height above a datum. When the body is 16 m above the datum, it possesses a total energy of 3,531 J. What is the starting height of the object?

Efficiency zCalculating efficiency The efficiency of an energy transformation is a measure of how much of the input energy appears as useful output energy. The efficiency of any system can be calculated using the equation: Efficiency,  =Useful energy output Total energy input

Worked Example An electric lift rated at 110 V, 30 A raises a 700 kg load a height of 20 m in two minutes. zBy considering the electrical energy input and the potential energy gained by the mass, determine the percentage efficiency of this energy transformation.

Worked Problem Ee = ItV = 30  120  110 = 396 kJ Potential energy gained is calculated as follows. EP = mgh = 700  9.8  20 = kJ Percentage Efficiency = Useful Energy Output  100% Total Energy Input =  100% = 34.65% 396

Pupil Problems (1) An electric kettle is rated at 240 V, 10 A. When switched on it takes three minutes to raise the temperature of 0.5 kg of water from 20  C to 100  C. Determine: The electrical energy supplied in the three minutes The heat energy required to raise the temperature of the water The efficiency of the kettle.

Pupil Problems Boxes in a factory are transferred from one floor to another using a chute system as shown above. The boxes start from rest at the top of the chute and during the decent there is a 40 per cent loss of energy. The boxes weigh 10 kg each. Calculate the velocity of the boxes at the bottom of the chute.

Energy Audits z Your Teacher will show you how to construct an energy audit

Revision Questions zA load is raised through 6m. The load is 1kg in mass. What is the load’s potential energy at its final height? zWhat work is done in lifting the above load? zIf the lifting system is only 60% efficient. What input energy is required to lift the load? zA kettle boils water in 1 minute. If the kettle uses 100KJ of energy boiling the water, what is the current rating of the kettle? (240V supply to kettle) zA process raises the temperature of water from 6 0 C to 70 0 C in 12.6 seconds. If the initial mass of water is 2kg. What energy is used in the process? zIf the energy of a moving body is 20MJ and it has a mass of 300kg, what is the bodies velocity?

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