Hwk Chapter2

Water Wave Mechanics, Spring 2012 Carlos Serrano Moreno

2.1 Consider the following transition section:

(a) The flow from left to right is constant at Q =12π m3/s. What is the total acceleration of a

water particle in the x direction at x=5 m? Assume that water is incompressible and that the

X component of velocity is uniform across each cross section.

Since the discharge and the area are known the velocity can be easily determined:

u ( x )=QA

=12 pp⋅(3−0 .15⋅x )2

Acceleration is defined as the derivative of velocity over time then:

dudt

=¶u¶t

+ ¶u¶ x

¶ x¶t

=¶u¶ t

+u ¶u¶ x

Since the flow is constant:

¶u¶t

=0 and the variation of velocity along the horizontal axis is due to the

truncated conical section:

dudt

( x )=12(3−0 .15⋅x )2

⋅24⋅0.15(3−0 .15⋅x )3

=12⋅24⋅0 .15(3−0.15⋅x )5

dudt

(5 )=12⋅24⋅0 .15(3−0.15⋅5 )5

=43 .257 .67

=0 .749»0 .75m / s2

(b) The flow of water from right to left in given by:

Q ( t )=p⋅t2

Calculate the total acceleration at x=5m for t=2s. Make the same assumptions as before.

Since the flow is going from right to left, the velocity can be now defined as:

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u ( x , t )=QA

= −p⋅t2

p⋅(3−0.15⋅x )2

following the same procedure as before, but taking into account that this time it is not possible to

neglect the partial derivative of velocity over time, the acceleration will be:

dudt

( x , t )= −2⋅t(3−0 .15⋅x )2

− t2

(3−0 .15⋅x )2⋅2⋅t2⋅(−0 .15 )(3−0.15⋅x )2

= −2⋅t(3−0 .15⋅x )2

+ 2 t4⋅0 .15(3−0 .15⋅x )5

dudt

(5,2 )= −2⋅2(3−0 .15⋅5 )2

+ 2⋅24⋅0 .15(3−0 .15⋅5 )5

=−0 .707m/ s2

2.4 If the water (assumed inviscid) in the U-tive is displaced from its equilibrium position, it

will oscillate about this position with its natural period. Assume that the displacement of the

surface is: h (t )=A⋅cos( 2 pT ⋅t)

where the amplitude A = 10 cm and the natural period T is 8s. What will be the pressure at a

distance 20cm below the instantaneous water surface for = +10, 0, and -10cm?

Assume that g=980cm/s2 and =1g/cm3.

As the displacement of the water surface is know the expressions for velocity and acceleration can

be easily obtained:

w (t )=- A⋅2 pT

⋅sin(2 pT ⋅t)dwdt

( t )=- A⋅(2 pT )2

⋅cos (2 pT⋅t)

The fluid is known to be inviscid, then all the shear stress tensions can be neglected and Euler´s

Equation can be applied:

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dwdt

(t )=¶w¶t

(t )+w¶w¶ z

(t )=-1ρ¶ p¶ z

−g

⇒¶ p¶ z

=- ρ⋅g+ ρ⋅A⋅(2 pT )2

⋅cos(2 pT ⋅t )=- ρ⋅g+ρ⋅(2 pT )2

⋅h (t )

∫z0

h ¶ p¶ z

dz=ρ⋅∫z0

h−g+(2 pT )

2

¿h (t )dz=ρ⋅(−g+(2 pT )2

¿h (t ))⋅(h−z0 )

p (h )−p ( z0 )=ρ⋅(−g+(2 pT )2

¿h (t ))⋅(h−z0 )

0−p ( z0 )=ρ⋅(−g+(2 pT )2

¿h (t ))⋅(h−z0 )

Since =1g/cm3, g=980cm/s2, A = 10cm, T is 8s and (h−z0) =20cm and assuming that there is no

pressure in the free surface, p (h )=0 .

Finally, the pressure at the desired points can be easily obtained:

h p (h−20 cm )

10 cm 19476 .62g

cm⋅s2⇒1947 .662 N

m2

0 cm 19600g

cm⋅s2⇒1960 N

m2

-10 cm 19723 .37g

cm⋅s2⇒1972 .337 N

m2

2.6 Derive the following equation for an inviscid fluid and a non divergent steady flow:

−1ρ¶ p¶ z

−g=(¶uw¶ x )+(¶ vw¶ y )+(¶w2¶ z )Again, if the fluid is inviscid, shear stress terms can be neglected. Taking a look to Euler´s

equation:

DwDt=-1ρ¶ p¶ z

−g⇒ DwDz

=¶w¶t

+u ¶w¶ x

+v ¶w¶ y

+w ¶w¶ z=-1ρ¶ p¶ z

−g

As the fluid is non divergent:

¶u¶ x

+ ¶ v¶ y

+¶w¶ z

=0. Playing with the equation above:

¶w¶t

+u ¶w¶ x

+v ¶w¶ y

+w ¶w¶ z

+w(¶ u¶ x + ¶ v¶ y

+ ¶w¶ z )=- 1ρ ¶ p

¶ z−g

Assuming that the fluid is steady and reorganizing the terms and applying the chain rule of

derivatives:

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(u¶w¶ x +w¶u¶ x )+(v¶w¶ y

+w¶ v¶ x )+(w¶w

¶ z+w¶w

¶ z )=-1ρ ¶ p¶ z

−g

(¶ uw¶ x )+(¶ vw¶ y )+(¶w2¶ z )=-1ρ ¶ p¶ z

−g

2.9 For a particular fluid flow, the velocity components u,v and 2 in the x, y and z directivons

respectively are:

u=x+8⋅y+6⋅t⋅z+t4

v=8⋅x−7⋅y+6⋅z

w=12⋅x+6⋅y+12⋅z⋅cos (2⋅p⋅tT )+ t2a) Are there any times for which the flow is non divergent? If so, when?

If the flow is non divergent:

¶u¶ x

+ ¶ v¶ y

+¶w¶ z

=0

¶u¶ x

=1 ,¶ v¶ y

=- 7 ,¶ w¶ z

=12⋅cos (2⋅p⋅tT )⇒1−7+12⋅cos (2⋅p⋅tT )=0⇒ cos(2⋅p⋅tT )=12 ⇒2⋅p⋅t

T=p3

⇒ t=T6

However, since the solution comes from a cosinus: t = T/6 + k·T, (k = 1,2,...) is also valid.

b) Are there any times for which the flow is irrotational? If so, when?

If the flow is irrorational:

|

i j k¶¶ x

¶¶ y

¶¶ z

u v w

|=0=(¶w¶ y −¶ v¶ z

,¶u¶ z

−¶w¶ x

,¶ v¶ x

−¶ u¶ y )= (6−6,6⋅t−12 ,8−8 )

Then t=2s.

c) Develop the expression for the pressure gradient in the vertical (z) direction as a function

of space and time.

DwDz

=¶w¶ t

+u¶w¶ x

+v¶w¶ y

+w¶w¶ z

=-1ρ¶ p¶ z

−g

¶ p¶ z

=- ρ⋅g− ρ⋅(¶w¶ t +u¶w¶ x

+v ¶w¶ y

+w¶w¶ z )

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since

¶w¶t=-24⋅pT

⋅z⋅sin ( 2⋅p⋅tT )+2⋅t , ¶ w¶ x =12 , ¶ w¶ y

=6 , ¶ w¶ z

=12⋅cos( 2⋅p⋅tT )¶ p¶ z=- ρ⋅g− ρ⋅(−24⋅pT ⋅z⋅sin( 2⋅p⋅tT )+2⋅t+12⋅u+6⋅v+12⋅cos( 2⋅p⋅tT )⋅w)

substituting the velocity values:

¶ p¶ z=- ρ⋅g− ρ⋅¿(−24⋅pT ⋅z⋅sin( 2⋅p⋅tT )+2⋅t+12⋅(x+8⋅y+6⋅t⋅z+t4 )+6⋅(8⋅x−7⋅y+6⋅z )+ ¿)¿

¿

¿¿

2.12 The pressures pA ( t ) and pB ( t ) act on the massless pistons containingg the inviscid,

incompressible fluid in the horizontal tube shown below. Develope an expression for the

velocity of the fluid as a function of time = 1 gm/cm3.

Note: pA (t )=C A sin (s⋅t )and pB (t )=CBsin ( s⋅t+α )

where: s=0 .5 rad / s ,α= p

2,CA=CB=10dyn /cm3

Once again, since the fluid is inviscid, shear stress terms can be neglected. Euler´s equation

along the x-axis:

DuDt=-1ρ¶ p¶ x

⇒ DuDz

=¶ u¶ t

+u ¶u¶ x

+v ¶ v¶ y

+w ¶ v¶ z=-1ρ¶ p¶ x

Pressure just depends on time. All convective acceleration terms can be neglected. Then:

¶u¶t=-1ρ¶ p¶ x

¶ p¶ x

=−pB−p A

100=

−10 sin (0 .5⋅t+ p2 )+10 sin (0 .5⋅t )

100

¶u¶t=-11

−10sin(0.5⋅t+ p2 )−10sin (0 .5⋅t )

100= 110 (sin (0 .5⋅t+ p

2 )+sin (0 .5⋅t ))

5


u (t )= 110⋅0 .5 (−cos(0.5⋅t+ p

2 )−cos (0 .5⋅t ))+C

If t = 0 u (0 )=1

5 (−cos ( p2 )−cos (0 .5⋅0 ))+C=0⇒C=15

u (t )=15 (−cos(0 .5⋅t+ p

2 )−cos (0 .5⋅t ))+ 15 cm/s

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Hwk Chapter2

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