By Volker Simon, Bernhard Weigand, Hassan Gomaa

This monograph offers the basics of dimensional research and illustrates the strategy by way of a variety of examples for a large spectrum of purposes in engineering. The e-book covers completely the basic definitions and the Buckingham theorem, in addition to the alternative of the method of easy devices. The authors additionally comprise a presentation of version idea and similarity options. the objective viewers essentially contains researchers and practitioners however the publication can also be compatible as a textbook at college level.

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**Extra resources for Dimensional Analysis for Engineers**

**Sample text**

One thus obtains W ¼ f ðU; L; g; q; gÞ. ð2:17Þ Now n ¼ 6 and r ¼ 3 so that the problem is described by d ¼ 3 dimensionless products, and one obtains W ¼f q U 2 L2 UL U 2 ; , m gL ð2:18Þ where m ¼ g = q denotes the kinematic viscosity. For brevity, we introduce the drag coefﬁcient cw , the Reynolds number Re and Froud number Fr according to cw ¼ W , q U 2 L2 Re ¼ UL m and Fr ¼ U2 . gL ð2:19Þ In order to achieve complete similarity, all dimensionless products for the model and for full-scale must be equal, thus Re ¼ Re0 , Fr ¼ Fr0 and cw ¼ c0w .

The rod has the elasticity modulus E, the density q, the length L and diameter d. The natural frequency n of the rod is thus a function of the form n ¼ f ðE; q; L; dÞ: ð3:29Þ Therefore the dimension matrix is obtained as L M T n E q L d 0 0 −1 −1 1 −2 −3 1 0 1 0 0 1 0 0 42 3 Illustrative Examples The dimension matrix yields the two dimensionless products Y 1 L ¼ D and Y 2 rﬃﬃﬃﬃ q ¼ nL E ð3:30Þ and the relation equivalent to Eq. 29) follows to rﬃﬃﬃﬃ q L nL : ¼f E D ð3:31Þ An even clearer result is obtained when considering that in the context of the beam theory, the special cross sectional shape, here the diameter d, only enters through the area moment of inertia J, and that only the product E J is important.

71) takes the form Q_ ¼ f ðD#; U; k; q c; d; kB Þ ð1:80Þ The dimension matrix is L M T H Q_ D# U k qc d kB 2 1 −3 0 0 0 0 1 1 0 −1 0 1 1 −3 −1 −1 1 −2 −1 1 0 0 0 2 1 −2 −1 24 1 Some Fundamentals of Dimensional Analysis Just as in the ½LMT system one obtains three dimensionless quantities with Y 1 Y 2 ¼ Q_ ; k D# d ð1:81Þ ¼ qcU d k ð1:82Þ ¼ q c d3 ; kB ð1:83Þ and Y 3 so that Eq. 80) can now be written in the form of Q_ q c d3 q c U d ¼ f ; k D# d k kB ð1:84Þ The apparent contradiction between Eqs.