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# Conductive Heat Transfer ME 812

MSU

GPA 3.57

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This 4 page Class Notes was uploaded by Princess Rolfson on Saturday September 19, 2015. The Class Notes belongs to ME 812 at Michigan State University taught by Staff in Fall. Since its upload, it has received 31 views. For similar materials see /class/207535/me-812-michigan-state-university in Mechanical Engineering at Michigan State University.

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Date Created: 09/19/15

Lecture 25 31607 7 Dimensional Analysis Similitude and Modeling Notations The units or dimensions for a physical quantity q are denoted by square brackets ie q For example for velocity V V m 51 71 Dimensional Analysis Dimensional homogeneity principle Any equation between physical quantities that represents a physical principle needs to be dimensionally homogeneous in the sense that the dimensions of the equation must match However an equation with matching dimensions does not necessarily represent a physical principle In a given physical problem first count all the physical quantities involved in the problem q1 q2 qk There are k of them Second identify the required units dimensions necessary to describe all the quantities of the problem these are the basic dimensions of the problem These can be the mass length and time MLT system but the MLT system might not be enough eg a problem involving temperature will require the use ofa dimension for temperature such as kelvins Then find the minimum number ofdimensions to describe all physical quantities of the problem this number is denoted by r and is equal to or less than the number of basic dimensions Choose a set of r primary quantities or repeating variables qk1 qk2 qk whose units are dimensionally independent and form the reference dimensions qk1 qk2 qk Note 1 Force can be expressed in the MLT system using Newton39s second law so newtons N are not independent of kg m and s and should not be added as a basic dimension Note 2 Alternatively once you have listed all the physical quantities and their dimensions involved in the problem you can find the maximum number of independent dimensions by computing the rank of the matrix of the dimensions of the physical quantities expressed in terms of the set of basic dimensions This number is also equal to r Dimensional Analysis consists of observing that given the dimensional homogeneity principle the k 7 r secondary quantities q1 q2 114 are then such that their dimensions ql 12 qk can be expressed as a product of the reference dimensions qk1 qk2 with unknown exponents z an Emma 239 e 1 k A r 269 j1 The analysis then consists in determining the unknown exponents zig by expressing the dimen sions of both the primary and secondary quantities in terms of the basic dimensions 61 Example Consider a problem with k physical quantities and r 3 primary quantities qkq 114 qk and associated reference dimensions qk2 qk1 The dimensions for secondary quantities q z39 E 1 k 7 Tl can be expressed using the dimensions of the primary quantities lQil le72lmil1leillmiJlelmilav i E l17 k l 270 where z are the unknown exponents Assuming the MLT system can be the basic dimensions the dimensions for secondary quantities can be expressed as all Mlbi lllei Zllei i 2396 17k 7 l 271 And similarly for the primary quantities le73jl Mlaj llLlaj ZlTlaj 37 j E 173i 27 Substituting this leads to a system of linear algebraic equations 011 131 021 132 031 133 571 112 131 022 132 032 133 572 273 113 131 023 132 033 133 573 In general this can be expressed as the product of the the matrix of exponents corresponding to the reference dimensions expressed as a function of the basic dimensions A multiplied by the vector of unknown exponents x which is then equal to the vector of knoWn exponents of the dimensions of the secondary quantity under analysis expressed as a function of the basic dimensions b A ac b 274 This system has a unique solution provided the primary quantities are indeed dimensionally independent which is guaranteed ifthe determinant ofA the matrix of exponents corresponding to the reference dimensions expressed as a function of the basic dimensions is not zero 011 121 031 A det 112 122 132 7 0 113 123 033 72 VaschyBuckingham or H Pi theorem Theorem Given the previous notations an equation between k physical quantities fQI7QZ77Qk 0 276 can be written as an equation between k 7 r dimensionless products Hi FH1 H2 Hk 0 277 where Hleir 39lm j j1 62 where r is the minimum number of reference dimensions required to describe the variables 7 can also be interpreted as the maximum number of dimensionally independent quantities used in the equation under consideration Alternatively ifyou wish to emphasize a specific physical quantity eg G1 you may write H1 qH2H3Hk 279 All in all this procedure takes you from an equation relating k variables to an equation between k 7 r dimensionless variables It does not give you actual function F or 11 though unless f is known Implementation 1 Problem setup List all physical quantities involved in the problem list unknown quanti ties independent variables and parameters ql qz 114 and their dimensions qll 12 qk There are k of them Determine the basic dimensions of the problem You have or assume the equation relating them fQ17q27 7Qk 0 280 N Choose the primary quantities and associated reference dimensions In general there is typically a set of quantities that are fixed or hard to vary or that you know will be critical to the problem Pick a set of those that are dimensionally independent There are r of them Make sure they are independent by computing the determinant such as in equation 275 Dimensional analysis Form all the kir dimensionless products H as explained earlier by expressing all the dimensions of the secondary quantities as a product of the dimensions of the primary quantities 0 an unknown power then substituting the basic dimensions thus resulting in a system a linear equations such as equation 273 Solve this linear system to get the unknown exponents w 4 Pi theorem Rearrange equation 280 to get either FH17 H2 1114 07 281 or H1 1H2H3 114 282 Non uniqueness of H products H products are not unique depending on how you pose the problem For example if H is one possible dimensionless product its inverse 1Hi is another possibility However there are certain products that come time and again and have become standards 73 Dimensional analysis example Consider the steady flow average velocity V of an incompressible viscous fluid p p in a pipe of diameter D The pipe is not necessarily smooth so let 6 be the average height of the rough ness We are interested in the pressure drop Ap due to the roughness of the pipe along a length L Given the variables DcLppApV we can use the MLT system as our basic dimension system Thus the dimension matrix is 283 Let us pick our primary quantities DpV The associated reference dimensions are thus m kg m g and m 5 They are indeed independent 1 3 1 A 0 1 0 717 0 284 0 0 71 We now need to calculate the dimensionless products for the 4 secondary variables 6 171 and Ap For Ap Ap kg m 1 5 2 Let us write AP HAP W 285 and solve the corresponding linear system 0 x1 1 2 0 3 1 286 The solution is 1 07 2 17 3 2 Therefore A 10 HA 7 f 288 p pvz Similarly c He 7 289 D lt L HL 5 290 M H i 291 M DpV

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