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Statement of the Problem in the Case Study Format

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Statement of the Problem in the Case Study Format Powered By Docstoc
					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 .




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Description: Statement of the Problem in the Case Study Format document sample