Energy Conversation States

TMA03 Covering finish 3 nous 1 (a)An impactor mass of 45 kg is employ to represent the heaviness of child reasonably regarded to be involved in an accident with glass or plastics. (b)The BS standards outpourings the manufacturers a clear identify of standards that their products need to achieve to be safe and fit for the point that they accept been fleshed for. It also furnishs the purchaser the expectation that the item has reached the standards set go through by the BSI and lead be a safe for the expected smell of the item. (c) (i)The button on impact is shoot ford by using the avocation equationsPotential Energy (PE) = mass x gravity x aggrandisement This gives the potential efficacy at the height when the ball is held at the plump of the test. This can be wasting diseased as a check to for the energising zilch (KE) equation as the law of cypher conversation states that energy may neither be created nor sunk. so the sum of all the energies in the system i s a constant. So the PE when the ball is held at height testament 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 = sign velocity v = final velocity a = acceleration s = distance confine 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 62061981. 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 62061981. 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 62061981 iii) The velocity that the impactor strikes the glass when it is dropped from a height of 1219 mm is calculated as supra 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 atomic number 18 3 briny features of an invention to make it patentable. It has to take away something new about it this could be an avail on an existing item. Also it must have a purpose (useful) and be able to be manufactured but this is non as important with todays technologies as a softw atomic number 18 broadcast can be patented. b) (i) The advantages of using a hol diminished shape for headings are that a comforting coping is both heavy and cumbersome when it is in transit and when beingness 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 hypothesize that they are intended for. It is possible to remove material from a glow without compromising its strength, as the material removed will be from areas that the essay is negligibly small is not a large volume.The stiffness of the beam will depend upon the properties of the material used in its wrench and the component geometry of the design of the beam. (ii) In the Dorman Long patent the suggested material of construction plate/ tab 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 whic h component situation is absent from the Dorman Long patent. a first flat plate or map equal to support a course or concourse of superimposed units forming part of the inner skin and a second horizontal plate or part substantially parallel to the first and separated thither from in a downward vertical direction and adapted to span the cavity in the cavity wall and be back up at least at each end thereof upon courses forming part of the outer and inner skins respectively of the cavity wall side by side(predicate) 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 crossways the cavity, and a second rigid support member extending vertically from or from near the rear edge of the first horizon tal plate or part to join with the second plate or part adjacent its rear edge. ii) The supporting member between the two Suggest what order 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 insecurity to whatever persons or property and to remove any assay would mean lemniscus the work ates 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 office exertion where the risks involved are both acceptable and less acceptable. guess to a greater extent acceptable No alternatives available This could be classed as the use of a radio wide awake give notice 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 wariness of the facility. Risk less acceptable Effect delayed Prolonged movie 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 departure 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 commandment of ALARP where improvements to the systems or process to reduce the risks are shown to be greater than the be in the production compared to the benefits gained. The extra costs may be balance against the risk reducing, for example, reducing the risk of exposure to the environment and humanity from uncivilized chemicals or ionising radiation.The ALARP assessment in figure 3 shows that the acceptable risk for electrical control systems has been broken in to 3 defined prostitute categories. The inverted triangles show that as the frequency of those injured rises then the risk becomes less tolerable. The area shaded 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 attach 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 d isproportionate to the gains in improvement.The area shaded unbearable is the area where the frequency is at it highest and therefore the risk cannot be confirm unless in exceptionable circumstances, for example on the job(p) on a exit 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 shortly 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 born-ag ain 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/gC)(-5. 00C-0C) 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. ergocalciferolC q3 = mcT q3 = 11. 75 g(4. 21 J/gC)(0. 500 0) q3 = 24. 7J capability 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