1. Outline the energy flow through the following devices and identify the input energy, useful energy, and wasted energy:

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1 Exercise B.2.1 Answers 1. Outline the energy flow through the following devices and identify the input energy, useful energy, and wasted energy: (a) Electric heater Electrical energy input, to heat energy as the useful output. Wasted energy is minimum (light) since the heater is the only device that can have a near 100% efficiency (b) Personal stereo Chemical energy from battery is converted to electrical energy for the input energy. This is converted to sound energy as the useful output. Wasted energy is in the form of heat from the electrical resistance within the device. (c) The archer uses chemical energy to draw the string back (kinetic). This energy is stored by the string as elastic potential until release, when it is converted into the kinetic energy of the string and the arrow. The useful output is the kinetic energy of the arrow with wasted energy in the form of heating the molecules of the bow string and the air molecules (drag) (d) A hair dryer uses electrical energy input and outputs kinetic energy (moving air) and heat. Wasted energy is mainly in the form of sound. (e) A television Electrical energy is converted to two useful outputs, light and sound, with wasted energy in the form of heat from electrical resistance. (f) A toy clockwork car Chemical energy is converted to kinetic in winding the toy (input). This energy is stored as elastic potential by an internal spring and then converted to the useful output (kinetic energy of the car moving). Wasted energy in the form of sound and heat. 2. Construct an energy transfer diagram for the following; (a) A person climbing a steep hill Chemical energy Kinetic energy Gain in gravitational potential energy Heat energy

2 (b) A propeller driven aeroplane Chemical Energy (Fuel) Heat Sound Kinetic Energy 3. Chemical energy from the rider peddling is transferred into kinetic energy of the bike moving forward (useful) and heat (friction between the tyres and the road, and moving parts of the bike). The Kinetic energy from the wheel transfers kinetic energy to the wheel of the dynamo (rotating a coil in a magnetic field) and is converted to electrical energy. Wasted energy occurs as heat via the moving parts of the dynamo and electrical resistance. The electrical energy is then converted to light energy and again, wasted heat due to the electrical resistance of the light bulb. 4. There are many energy transfers between the rider peddling and the bicycle light. At each transfer point wasted energy in the form of heat is transferred to the surroundings either by friction or electrical resistance. Thus only a small percentage of the input energy is converted to useful energy (light from the bulb). Exercise B.2.2 Answers 1. 1 square = 200J 2. Sound energy 3. 1 square = 200J 2400J Electrical Energy 1100J Light Energy 300J Sound Energy 1000J Heat Energy

3 4. Green comments what you did well Red comments areas you need to work on Labeled as 30J but the scale shows width of arrow to be 40J Useful output 30J Useful energy should be a straight through arrow not a vertical arrow Correctly labeled according to the width of the arrow and the scale Input energy 120J Wasted energy 55J Correctly labeled according to the width of the arrow and the scale 1 square = 10J Wasted energy should be an up or down arrow Appropriate scale included 5. For an input energy of 150J, the electric motor produces 78J of kinetic energy and 40J of heat energy. Therefore 32J is lost to sound. 150J Electrical Energy 78J Kinetic Energy 32J Sound Energy 40J Heat Energy KEY: 1 square = 20J

4 6. The output energy (both useful and wasted) must equal the input energy. Since 20% is lost via the generator and use within the power station, and 45% is lost to heat, = 35% is converted to electrical energy. The process has a high amount of waste energy making it inefficient. Exercise B.2.3 Answers 1. Below is an example response one might send covering the main points; Dear Sir. Unfortunately for the world energy crisis, machines that produce more useful energy or work than we put in (often known as over unity machines) are a physical impossibility. This is because the Law of Conservation of Energy states that energy can neither be created nor destroyed. Clearly your proposed machine breaks the law of conservation of energy due to the output energy being greater than the input. Further, in an idealized model the best a machine could do is 100% efficiency, with an output equal to the input. However, due to friction, internal resistance, and other loss channels, no machine can be 100% efficient (unless it s a heater which unfortunately has already been invented). Best regards. 2. m = 3000kg h = 15m g = 10m -2 Ein = J η =? We have useful input and in order to calculate the efficiency we need to calculate the useful output energy, which is equal to the GPE transferred to the container. GGGGGG = mmmmh = 3000kkkk 10mmss 2 15mm = JJ eeeeeeeeeeeeeeeeeeee ηη = uuuuuuuuuuuu eeeeeeeeeeee oooooo eeeeeeeeeeee iiii = JJ = 0.64 = 64% JJ 3. The biggest source of wasted energy in any mechanical process is heat generation due to friction. The efficiency of the crane could therefore be improved by ensuring adequate lubrication of all moving parts. This would be beneficial to the company because making the crane more efficient would reduce fuel costs.

5 4. To calculate the input energy we first need to calculate the useful output, which is equal to the GPE transferred to the climber. Since η = h = 25m m = 70kg g = 10ms -2 GGGGGG = mmmmh = 70kkkk 10mmss 2 25mm = 17500JJ eeeeeeeeeeee iiii = eeeeeeeeeeee oooooo eeeeeeeeeeeeeeeeeeee = 17500JJ 0.26 = 67308JJ When the climber reaches the top of the wall 17500J of the input energy has been converted into the climbers GPE = 49808J of energy has been lost to the surroundings. This wasted energy would be in the form of heat radiated from the climber s body, the climber working against friction slightly heating the climbing wall, and also the climber doing work against the air molecules whilst climbing. Challenge Question 5. η = 20% = 0.2 Eout = 175MJ 1L = 1.05 per L, eeeeeeeeeeee iiii = eeeeeeeeeeee oooooo eeeeeeeeeeeeeeeeeeee = 175MMMM 0.2 = 875MMMM Petrol used: 875MMMM 35MMMM = 25LL = 26.25