By G. Hauke
This publication offers the rules of fluid mechanics and delivery phenomena in a concise means. it's appropriate as an advent to the topic because it comprises many examples, proposed difficulties and a bankruptcy for self-evaluation.
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Additional info for An Introduction to Fluid Mechanics and Transport Phenomena (Fluid Mechanics and Its Applications)
This has no practical implication since, as shown below, the stress tensor is symmetric. 3 (Normal stress). e. τ11 , τ22 and τ33 . 4 (Shear or tangential stress). e. τ12 , τ13 , τ21 , τ23 , τ31 and τ32 . 3. In Cartesian coordinates, the components of the stress tensor are also denoted by τxx , τyy , τzz , τxy , τxz , and τyz . 3 D fs τ 12 n τ 21 τ 22 τ 23 τ 11 τ 13 P τ 31 C 2 τ 32 τ 33 B 1 Fig. 3. Inﬁnitesimal tetrahedron employed to obtain the stress tensor at the point P. Derivation of the Stress Tensor In order to determine the general expression of the stress at a point P from the stresses on three perpendicular planes, let us select the inﬁnitesimal ﬂuid volume of Fig.
12 (Laminar ﬂow in a circular cross-section pipe). The fully developed laminar axial velocity in a circular cross-section straight pipe, of radius R, obeys r 2 v(r) = V0 1 − R This ﬂow is called Hagen-Poiseuille ﬂow. Determine the volumetric ﬂow rate in the pipe and the mean velocity. Solution. Let us take a section perpendicular to the pipe axis, S. The volumetric ﬂow rate is Q= v · n dS = v(r) dS S S Since the velocity is constant for a given radius, we can take the surface diﬀerential dS = 2πr dr.
What we have done is to obtain all the trajectories of the particles that were injected in the ﬂow ﬁeld before the present time t. 3. Eliminate ξ. 10 (Streakline). In the ﬂow ﬁeld of the previous example, determine the streakline that passes by x0 , y0 . Solution. Integration of the equation of motion yields x = C1 e(t+1) y 2 = C2 e(t−1) 2 26 2 Elementary Fluid Kinematics Now, in order to determine the integration constants C1 , C2 , we search the particles that at time ξ passed by x0 , y0 : 2 x0 = C1 e(ξ+1) y0 = C2 e(ξ−1) C1 = x0 /e(ξ+1) C2 = y0 /e(ξ−1) 2 yielding 2 2 Substituting, x x0 y y0 = e(t+1) 2 −(ξ+1)2 2 −(ξ−1)2 = e(t−1) The parameter ξ represents the diﬀerent particles that make the streakline.