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

Question

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Answer 1

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

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

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Answer 2

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

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

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Answer 3

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

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

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12:34

Answer 4

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

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

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12:34

Answer 5

Chapter 4.1, Problem 2P

To determine

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

Expert Solution & Answer

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

Answer 6

Chapter 4.1, Problem 2P

To determine

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

Expert Solution & Answer

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

Answer 7

Chapter 4.1, Problem 2P

To determine

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

Answer 8

Expert Solution & Answer

Answer to Problem 2P

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

V=V0ΔVe−axΔVe−ta(V0+ΔV).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression of surface velocity of the river.

V=V0+ΔV(1−e−ax)

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(1−e−ax) for V in Equation (I).

V0+ΔV(1−e−ax)=dxdtdt=dxV0+ΔV(1−e−ax) (II)

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

∫dt=∫dxV0+ΔV(1−e−ax)t+c=∫dxV0−ΔV(e−ax−1)t+c=−∫dxΔV(e−ax−1)−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(e−u−1)−V0=−1a∫duΔVe−u−ΔV−V0 (V)

Consider,

1ΔVe−u−ΔV−V0=v (VI)

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

dvdu=ddu(1ΔVe−u−ΔV−V0)dvdu=−(1ΔVe−u−ΔV−V0)2(−ΔVe−u)dvdu=(1ΔVe−u−ΔV−V0)2(ΔVe−u)du=(ΔVe−u−ΔV−V0)2dvΔVe−u

Substitute (ΔVe−u−ΔV−V0)2dvΔVe−u for du in Equation (V).

t+c=−1a∫(ΔVe−u−ΔV−V0)2dv(ΔVe−u−ΔV−V0)ΔVe−ut+c=−1a∫(ΔVe−u−ΔV−V0)ΔVe−udvt+c=−1a∫dv(V0+ΔV)ΔVe−u−ΔV−V0+1 (VII)

Substitute v for 1ΔVe−u−ΔV−V0 in Equation (VII).

t+c=−1a∫1(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=−1a∫dww(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 1ΔVe−u−ΔV−V0 for v in Equation (XI).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−u−ΔV−V0+1} (XII)

Substitute ax for u in Equation (XII).

t+c=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1} (XIII)

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

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

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

t−1a(V0+ΔV)ln(−ΔVV0)=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}t=−1a(V0+ΔV)ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}+1a(V0+ΔV)ln(−ΔVV0)−ta(V0+ΔV)=ln{(V0+ΔV)ΔVe−ax−ΔV−V0+1}−ln(−ΔVV0)−ta(V0+ΔV)=ln[(V0+ΔV)ΔVe−ax−ΔV−V0+1−ΔVV0]

e−ta(V0+ΔV)=−((V0+ΔV)ΔVe−ax−ΔV−V0+1)(ΔVV0)e−ta(V0+ΔV)=[ΔVe−axV0+ΔV(1−e−ax)](V0ΔV) (XIV)

Substitute V for V0+ΔV(1−e−ax) in Equation (XIV).

e−ta(V0+ΔV)=[ΔVe−axV](V0ΔV)V=V0ΔVe−axΔVe−ta(V0+ΔV)

Thus, The Lagrengian description of the velocity of a fluid particle flowing along the surface is V=V0ΔVe−axΔVe−ta(V0+ΔV).

Answer 12

Ch. 4.1 - Prob. 2P Ch. 4.1 - The velocity field of a flow is given by m/s,...Ch. 4.1 - A two-dimensional velocity field is given by u = 1...Ch. 4.1 - Streamlines are given in Cartesian coordinates by...Ch. 4.1 - Prob. 6P Ch. 4.1 - Prob. 8P Ch. 4.1 - Prob. 9P Ch. 4.1 - Prob. 10P Ch. 4.1 - Prob. 11P Ch. 4.1 - Prob. 12P

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

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

Answer to Problem 2P

Explanation of Solution

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