Chapter 3 • Integral Relations - Simon Fraser University

Chapter 3 Integral Relations for a Control Volume Discuss Newton's second law (the linear momentum relation) in these three forms: d d . F = ma F = (mV) F = V d . dt dt system .. Solution: These questions are just to get the students thinking about the basic laws of mechanics. They are valid and equivalent for constant-mass systems, and we can make use of all of them in certain fluids problems, the #1 form for small elements, #2 form for rocket propulsion, but the #3 form is control-volume related and thus the most popular in this Chapter . Consider the angular-momentum relation in the form d . MO = (r V) d . dt system .. What does r mean in this relation? Is this relation valid in both solid and fluid mechanics? Is it related to the linear-momentum equation (Prob. )? In what manner? Solution: These questions are just to get the students thinking about angular momentum versus linear momentum.

3.3 For steady laminar flow through a long tube (see Prob. 1.12), the axial velocity distribution is given by u = C(R2 − r2), where R is the tube outer radius and C is a constant. Integrate u(r) to find the total volume flow Q through the tube. Solution: The area element for this axisymmetric flow is dA = 2π r dr. From Eq. (3.7), 22 0 ()2 . R

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  Flows, Axial

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