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Template for Engineering Mini-Case Studies. Colleagues may submit mini-case studies in any format; Latex, Word, handwritten etc. However, please adhere to the structure outlined be- low. An illustration of an Engineering Mini-Case Study is presented below. There are four parts to the template: Part 1 Introduction This will include a paragraph or two which introduces the general background area from which the mini-case study is taken. The mini-case study should be set in context and any essential engineering background information and deﬁnitions should be included here. It may be necessary also to include a brief discussion on any fundamental concepts employed in the mini-case study. Part 2 engprob This will have the title Engineering Problem Posed. This will include a statement of the problem in words to include the purpose of the model. This part should introduce any notation. Any underlying assumptions should be noted. This cannot be empty Part 3 mathsolution This will have the title Mathematical Analysis. Immediately following the title the engineering problem will be expressed in mathematical form. This will be placed in a box with title Engineering Problem Expressed Mathematically. Solve maths problem or explain that solution is easily obtained using the methods being dis- cussed. All techniques used in the solution must appear in the present or earlier workbooks. This cannot be empty Part 4 If appropriate interpret maths solution in engineering terms. Outline any shortcomings in modelling process. Include any further comments as is thought appropriate. This should never be empty 1 EngExample 1 (C. Steele) EXAMPLE FOR WORKBOOK 2 OR 3? Introduction A common need in engineering thermodynamics is to determine the radiation emitted by a body heated to a particular temperature at all wavelengths or a particular wavelength such as the wavelength of yellow light, blue light or red light. This would be important in designing a lamp for example. The total power per unit area radiated at temperature T (in K) may be denoted by E(λ) where λ is the wavelength of the emitted radiation. It is assumed that a perfect absorber and radiator, called a black body, will absorb all radiation falling on it and which emits radiation at various wavelengths λ according to the formula C1 E(λ) = 5 C2 /(λT ) (1) λ [e − 1] where E(λ) measures the energy (in W m−2 ) emitted at wavelength λ (in m) at temperature T (in K). The values of the constants C1 and C2 are 3.742 × 10−16 W m−2 and 1.439 × 10−2 mK respectively. This formula is known as Planck’s Distribution Law. Figure 1 shows the radiation E(λ) as a function of wavelength λ for various values of the temperature T . Note that both scales are plotted logarithmically. In practice, a body at a particular temperature is not a black body and its emissions will be less intense at a particular wavelength than a black body ; the power per unit area radiated by a black body gives the ideal upper limit for the amount of energy emitted at a particular wavelength. 16 T = 10000K 14 12 log E(λ) 10 8 6 4 T = 100K 2 log λ − 6.5 −6 − 5.5 − 5 − 4.5 −4 Figure 1 The emissive power per unit area E(λ) plotted against wavelength (logarith- mically) for a black body at temperatures of T = 100K, T = 400K, T = 700K, T = 1500K, T = 5000K and T = 10000K 2 Engineering Problem Posed Find the power per unit area emitted for a particular value of the wavelength (λ = 6 × 10−7 m) and also obtain the temperature of the black body which emits power per unit area (E(λ) = 1010 W m−2 ) at a speciﬁc wavelength (λ = 4 × 10−7 m) Mathematical Analysis Engineering Problem Expressed Mathematically (a) A black body is at a temperature of 2000K. Given formula (1), determine the value of E(λ) when λ = 6 × 10−7 m? (b) What would be the value of T that corresponds to E(λ) = 1010 W m−2 at a wavelength of λ = 4 × 10−7 m (the wavelength of blue light) ? (a) Here, λ = 6 × 10−7 and T = 2000. Putting these values in the formula gives 5 −2 /6×10−7 /2000 E(λ) = 3.742 × 10−16 / (6 × 10−7 ) / e1.439×10 −1 = 2.98 × 1010 W m−2 (to three signiﬁcant ﬁgures). (b) Equation (1) can be rearranged to give the temperature T as a function of the wavelength λ and the emission E(λ). C1 C1 E(λ) = so eC2 /(λT ) − 1 = λ5 [eC2 /(λT ) − 1] λ 5 E(λ) and adding 1 to both sides gives C1 eC2 /(λT ) = 5 E(λ) + 1. λ On taking (natural) logs C2 C1 = ln 5 +1 λT λ E(λ) which can be re-arranged to give C2 T = (2) C1 λ ln λ5 E(λ) +1 Equation (2) gives a means of ﬁnding the temperature to which a black body must be heated to emit the energy E(λ) at wavelength λ. Here, E(λ) = 1010 and λ = 4 × 10−7 so (2) gives, 1.439 × 10−2 T = = 2380 K 3.742 × 10−16 4 × 10−7 ln +1 (4 × 10−7 ) × 1010 3 Comments Interpretation Since the body is an ideal radiator it will radiate the most possible power per unit area at any given temperature. Consequently any real body would have to be raised to a higher temperature than a black body to obtain the same radiated power per unit area. Mathematical Comments It is not possible to re-arrange equation (1) to give λ as a function of E(λ) and T . This is due to the way that λ appears twice in the equation i.e. once in a power and once in an exponential. To solve (1) for λ requires numerical techniques but it is possible to use a graphical technique to ﬁnd the rough value of λ which satisﬁes (1) for particular values of E(λ) and T . 4

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posted: | 8/19/2011 |

language: | English |

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Statement of the Problem in the Case Study Format document sample

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