Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, Binder Ready Version
Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, Binder Ready Version
8th Edition
ISBN: 9781119080701
Author: Philip M. Gerhart, Andrew L. Gerhart, John I. Hochstein
Publisher: WILEY
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Question
Chapter 4.1, Problem 2P
To determine

The Lagrengian description of the velocity of a fluid particle flowing along the surface.

Expert Solution & Answer
Check Mark

Answer to Problem 2P

The Lagrengian description of the velocity of a fluid particle flowing along the surface is

V=V0ΔVeaxΔVeta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

    V=V0+ΔV(1eax)

Here, the constant initial velocity is V0, the constant change in velocity is ΔV, the constant displacement is x and the constant acceleration is a.

Write the expression of velocity.

    V=dxdt (I)

Here, the time is t.

Substitute V0+ΔV(1eax) for V in Equation (I).

    V0+ΔV(1eax)=dxdtdt=dxV0+ΔV(1eax)                                                                                   (II)

Integrate Equation (II) on both sides with respect to time (t) and displacement (x).

    dt=dxV0+ΔV(1eax)t+c=dxV0ΔV(eax1)t+c=dxΔV(eax1)V0                                                                          (III)

Consider

    u=ax                                                                                                          (IV)

Differentiate Equation (IV) with respect to displacement.

    ddx(u)=ddx(ax)dudx=a×1dua=dx

Substitute dua for dx and u for ax in Equation (III).

    t+c=(dua)ΔV(eu1)V0=1aduΔVeuΔVV0                                                                            (V)

Consider,

    1ΔVeuΔVV0=v                                                                                         (VI)

Differentiate Equation (VI) with respect to u on both sides.

    dvdu=ddu(1ΔVeuΔVV0)dvdu=(1ΔVeuΔVV0)2(ΔVeu)dvdu=(1ΔVeuΔVV0)2(ΔVeu)du=(ΔVeuΔVV0)2dvΔVeu

Substitute (ΔVeuΔVV0)2dvΔVeu for du in Equation (V).

    t+c=1a(ΔVeuΔVV0)2dv(ΔVeuΔVV0)ΔVeut+c=1a(ΔVeuΔVV0)ΔVeudvt+c=1adv(V0+ΔV)ΔVeuΔVV0+1                                                         (VII)

Substitute v for 1ΔVeuΔVV0 in Equation (VII).

    t+c=1a1(V0+ΔV)v+1dv                                                                        (VIII)

Consider.

    (V0+ΔV)v+1=w                                                                                           (IX)

Differentiate Equation (IX) on both sides with respect to v on both sides.

    ddv{(V0+ΔV)v+1}=ddv(w)dwdv=ddv(V0+ΔV)v+ddv(1)dwdv=(V0+ΔV)dw(V0+ΔV)=dv

Substitute dw(V0+ΔV) for dv and w for (V0+ΔV)v+1 in Equation (VIII).

    t+c=1adww(V0+ΔV)t+c=1a(V0+ΔV)dwwt+c=1a(V0+ΔV)ln(w)                                                                          (X)

Substitute (V0+ΔV)v+1 for w in Equation (X).

    t+c=1a(V0+ΔV)ln{(V0+ΔV)v+1}                                                (XI)

Substitute 1ΔVeuΔVV0 for v in Equation (XI).

    t+c=1a(V0+ΔV)ln{(V0+ΔV)ΔVeuΔVV0+1}                                      (XII)

Substitute ax for u in Equation (XII).

    t+c=1a(V0+ΔV)ln{(V0+ΔV)ΔVeaxΔVV0+1}                                         (XIII)

Substitute 0 for t and 0 for x for initial condition in Equation (XII).

    0+c=1a(V0+ΔV)ln{(V0+ΔV)ΔVea×0ΔVV0+1}c=1a(V0+ΔV)ln{(V0+ΔV)V0+1}c=1a(V0+ΔV)ln{1ΔVV0+1}c=1a(V0+ΔV)ln(ΔVV0)

Substitute 1a(V0+ΔV)ln(ΔVV0) for c in Equation (XIII).

    t1a(V0+ΔV)ln(ΔVV0)=1a(V0+ΔV)ln{(V0+ΔV)ΔVeaxΔVV0+1}t=1a(V0+ΔV)ln{(V0+ΔV)ΔVeaxΔVV0+1}+1a(V0+ΔV)ln(ΔVV0)ta(V0+ΔV)=ln{(V0+ΔV)ΔVeaxΔVV0+1}ln(ΔVV0)ta(V0+ΔV)=ln[(V0+ΔV)ΔVeaxΔVV0+1ΔVV0]

    eta(V0+ΔV)=((V0+ΔV)ΔVeaxΔVV0+1)(ΔVV0)eta(V0+ΔV)=[ΔVeaxV0+ΔV(1eax)](V0ΔV)                                                  (XIV)

Substitute V for V0+ΔV(1eax) in Equation (XIV).

    eta(V0+ΔV)=[ΔVeaxV](V0ΔV)V=V0ΔVeaxΔVeta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVeaxΔVeta(V0+ΔV).

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12:34

Chapter 4 Solutions

Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, Binder Ready Version

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