Fluid Mechanics-Frank M White Solution Ch3

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Fluid Mechanics-Frank M White Solution Ch3에 대한 보고서 자료입니다.

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P3.1 Discuss Newton’s second law (the linear momentum relation) in these three forms: ∑F=ma ∑F= d dt(mV) ∑F= d dt V ρ d υ 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, e.g. 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.

P3.2 Consider the angular-momentum relation in the form ∑MO = d dt (r×V) ρ d υ 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. 3.1)? In what manner? Solution: These questions are just to get the students thinking about angular momentum versus linear momentum. One might forget that r is the position vector from the momentcenter O to the elements ρ d υ where momentum is being summed. Perhaps rO is a better notation.

P3.3 For steady laminar flow through a long tube (see Prob. 1.14), 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 = 2 π r dr. From Eq. (3.7),
Q= u dA= C 0 R ∫ (R2 −r2)2 π rdr=
π 2CR4 Ans.∫







2


P3.4 A fire hose has a 12.5-cm inside diameter and is flowing at 2.27 m3/min. The flow exits through a nozzle contraction at a diameter Dn. For steady flow, what should Dn be, in cm, to create an exit velocity of 25 m/s?
Solution: This is a straightforward one-dimensional steady-flow continuity problem. 2.27 m3/min = 0.038 m3/s;

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