Energy Conversation States

TMA03 Covering Block 3 Question 1 (a)An impactor mass of 45 kg is used to represent the weight of child reasonably regarded to be involved in an accident with glass or plastics. (b)The BS standards gives the manufacturers a clear set of standards that their products need to achieve to be safe and fit for the purpose that they have been designed for. It also gives the purchaser the expectation that the item has reached the standards set down by the BSI and will be a safe for the expected life of the item. (c) (i)The energy on impact is calculated by using the following equations:Potential Energy (PE) = mass x gravity x height This gives the potential energy at the height when the ball is held at the start of the test. This can be used as a check to for the Kinetic energy (KE) equation as the law of energy conversation states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant. So the PE when the ball is held at height wil l be the same as the KE just before the impact with the glass. To calculate the KE use the equation KE = ? mv2 (ii) Using the KE equation from question (i)KE = ? mv2 |u = initial velocity | |v = final velocity | |a = acceleration | |s = distance | Insert the constants of Mass = 45kg To calculate the v2 using the equation v2 = u2 + 2 x a x s For test 1 KE = ? mv2 For the v2 v2 = 02 + 2 x 9. 8 x 305 Gives 5978 Insert in to KE = ? mv2 to give KE = ? x 45 x 5978 To give 134505 = 135J to 3 sf To check use PE = mgh 45 x 9. 8 x 305 = 134505 Round up to 3 sf to give 135 J So PE =KE 135J is as given in BS 6206:1981. For test 2KE = ? mv2 For the v2 v2 = 02 + 2 x 9. 8 x 457 Gives 8957. 2 Insert in to KE = ? mv2 to give KE = ? x 45 x 8957. 2 To give 201537 = 202J to 3 sf To check use PE = mgh 45 x 9. 8 x 457 = 201537 Round up to 3 sf to give 202 J So PE =KE 202J is as given in BS 6206:1981. For Test 3 KE = ? mv2 For the v2 v2 = 02 + 2 x 9. 8 x 1219 Gives 23892. 4 Insert in to KE = ? mv2 to give KE = ? x 45 x 23892. 4 To give 537579 = 538J to 3 sf To check use PE = mgh 45 x 9. 8 x 1219 = 537579 Round up to 3 sf to give 538 J So PE =KE 538J is as given in BS 6206:1981 iii) The velocity that the impactor strikes the glass when it is dropped from a height of 1219 mm is calculated as above using: v2 = u2 + 2 x a x s v2 = 02 + 2 x 9. 8 x 1219 = 23892. 4 v = (23892. 4= 154. 6 m s-1 Question 2 (a)There are 3 main features of an invention to make it patentable. It has to have something new about it; this could be an improvement on an existing item. Also it must have a purpose (useful) and be able to be manufactured but this is not as important with today’s technologies as a software program can be patented. b) (i) The advantages of using a hollow shape for lintels are that a solid lintel is both heavy and cumbersome when it is in transit and when being manoeuvred into position at the build site. The added weight of the lintel will also require the supporting wall to be at a required strength to support the lintel and the load above it. They can be considered that they can be over designed for the job that they are intended for. It is possible to remove material from a beam without compromising its strength, as the material removed will be from areas that the stress is negligibly small is not a large volume.The stiffness of the beam will depend upon the properties of the material used in its construction and the component geometry of the design of the beam. (ii) In the Dorman Long patent the suggested material of construction plate/sheet steel. The use of plate/sheet steel is favoured due to its stiffness compared to its weight and that it can be easily folded or rolled in the construction of the lintel. (c) (i) Refer back to claim 1 of the Catnic patent as discussed in Block 3 Part 2.List the ‘essential integers’ of the Catnic lintel, and identify which component part is absent from the Dorman Long patent. a first horizontal plate or part adapted to support a course or plurality of superimposed units forming part of the inner skin and a second horizontal plate or part substantially parallel to the first and spaced there from in a downward vertical direction and adapted to span the cavity in the cavity wall and be supported at least at each end thereof upon courses forming parts of the outer and inner skins respectively of the cavity wall adjacent an a perture, and first rigid inclined support member extending downwardly and forwardly from or near the front edge adjacent the cavity of the first horizontal plate or part and forming with the second plate or part at an intermediate position which lies between the front and rear edge of the second plate or part and adapted to extend across the cavity, and a second rigid support member extending vertically from or from near the rear edge of the first horizontal plate or part to join with the second plate or part adjacent its rear edge. ii) The supporting member between th e two Suggest what effect the extra part is likely to have on the performance of the Catnic lintel compared to the Dorman Long lintel. (6 + 2 = 8 marks) Question 3 a) Although in a perfect environment there would be no risk to any persons or property and to remove any risk would mean stopping the processes that give rise to risk. However that is not the case so there will always risk involved in everything that we carry out.I have widely based this answer on the nuclear power industry where the risks involved are both acceptable and less acceptable. |Risk more acceptable | |No alternatives available |This could be classed as the use of a radio active fuel in a nuclear power | | |station.The risk of its use would be classed as acceptable as there is not| | |a suitable alternative as a fuel. | |Risk known with certainty |The use of a radio active fuel and its risks are known with certainty and | | |should be factored in to the design and management of the facility. | |Risk less accep table | |Effect delayed |Prolonged exposure to a radio active element will have a delayed effect | | dependant upon the length and magnitude of the exposure. So if these | | |exposures are not monitored and controlled the risk would be less | | |acceptable. | |Consequences irreversible |Again as the damage done from high, prolonged exposure to a radio active | | |element to the human body can be irreversible.Also a spillage or | | |accidental release to the environment could lead to the area being of no use| | |to the local population for a considerable time. | b) The principle of ALARP where improvements to the systems or process to reduce the risks are shown to be greater than the costs in the production compared to the benefits gained. The extra costs may be balanced against the risk reduction, for example, reducing the risk of exposure to the environment and humanity from hazardous chemicals or ionising radiation.The ALARP assessment in figure 3 shows that the acceptable risk fo r electrical control systems has been broken in to 3 defined harm categories. The inverted triangles show that as the frequency of those injured rises then the risk becomes less tolerable. The area shaded â€Å"broadly acceptable† shows that the processes carried out do not pose a risk to those involved, risk is negligible and it will be necessary to maintain checks and safety futures to keep the injuries at this level.In the area marked â€Å"tolerable if ALARP† shows the area that the risk of injury to those involved in the process ahs increased, if the risk was under taken then there could be a benefit. This can be broken into 2 further subsections, if the frequency is low the it can be tolerable if the cost of the risk reduction exceeds the improvement, as the frequency increases then it would only be tolerable if the risk reduction is impractical or the cost disproportionate to the gains in improvement.The area shaded â€Å"intolerable† is the area where th e frequency is at it highest and therefore the risk cannot be justified unless in exceptionable circumstances, for example working on a live high voltage electrical system. Question 4 Table 4. 4 in Block 3 Part 4 shows the various stages in energy conversion for fuel used to power a computer. It shows a hypothetical balance sheet for energy conversion from chemical energy (in a fuel) to light energy (light emitted by a computer display). a) The law of energy conversation states that energy may neither be created nor destroyed but transformed into different forms of energy i. . to heat, light, or noise energy. Therefore the sum of all the energies in the system is a constant. Explain briefly the principle of the conservation of energy, and how it applies at each stage in the energy conversion process. (2 marks) (b) Calculate the percentage of the total available energy that is converted to: (a) heat (b) noise (c) electrical energy. In each case, show your working fully. (4 + 4 + 4 = 12 marks) †¢ break into three parts†¦ energy required for ice to reach a temp of 0 q1 = mcT q1 = 11. 75 g(2. 09 J/g ·Ã‚ °C)(-5. 00 °C-0 °C) q1 = -122. 8J second part nergy required for change of states from solid to liquid q2 = n*? Hfus n = 11. 75g / 18. 02g/mol n = 0. 65mol q2 = 0. 65mol * 6. 02 kJ/mol q2 = 4. 0kJ careful units!!!! q2 = 4000J third part.. energy required for liquid water from 0 to 0. 500 °C†¦ q3 = mcT q3 = 11. 75 g(4. 21 J/g ·Ã‚ °C)(0. 500 – 0) q3 = 24. 7J ENERGY IN TOTAL†¦ E = q1 + q2 + q3 E = -122. 8J + 4000J + 24. 7J E = 3902J The final 10 marks for the assignment are awarded for presentation – see the guidance in the introduction to this booklet. These will be scored on the PT3 form as Question 5. Assignment Booklet