Get An Introduction to Fluid Mechanics and Transport Phenomena PDF

By G. Hauke

ISBN-10: 1402085362

ISBN-13: 9781402085369

ISBN-10: 1402085370

ISBN-13: 9781402085376

This publication offers the rules of fluid mechanics and shipping phenomena in a concise means. it really is compatible as an creation to the topic because it comprises many examples, proposed difficulties and a bankruptcy for self-evaluation.

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Extra resources for An Introduction to Fluid Mechanics and Transport Phenomena

Example text

5. Since n is the outward normal vector to the surface, then outgoing flow rates are positive and inward flow rates, negative. 9 (Mean velocity). 11 (Volumetric flow rate for uniform velocity v parallel to the normal vector of the surface A). In this case, the volumetric flow rate is simply Q = vA The mean velocity is v¯ = Q/A = v r R v(r) z Fig. 13. Fully developed laminar flow in a constant section pipe. 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.

The convective transport is due to the macroscopic fluid velocity. The fluid, with its motion, drags the fluid particles and its properties. Mathematically, the net flux by convection is modeled by the convective term of the substantial derivative, ∂c ∂c ∂c + v2 + v3 v1 ∂x1 ∂x2 ∂x3 This type of transport phenomenon is responsible, for example, for the wind transporting fallen tree leaves. At the same time as the wind blows the fallen tree leaves, it transports all the fluid properties, such as temperature, chemical concentration of the chemical species, energy, momentum and so on.

For the flow field of the above example, determine the trajectory of the fluid particle that passes through the point (x0 , y0 ), at t = 0. Solution. 4 Streamlines, Trajectories and Streaklines 25 which implies C1 = x0 /e C2 = y0 /e Finally, the trajectory is given in parametric form through the combination of 2 x = e(t+1) −1 x0 y y0 = e(t−1) 2 −1 This is a valid curve in two dimensions. Sometimes it is possible to eliminate t and write the same curve in explicit form, that is, as y(x). Getting t from the first equation, t ln xx0 + 1 − 1 = and substituting in the second one, y y0 √ ln =e x x0 +1−2 2 −1 which is the equation of the trajectory in explicit form.

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An Introduction to Fluid Mechanics and Transport Phenomena by G. Hauke

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