Homework 5 Solution

Professor Florencia CanelliPHY121.3 Vertiefung zu Physik

FS 2018

March 28, 2018

Exercise 1 Solution. Potential Flow. The flow of a fluid can be described using thepotential flow notation. If we denote the flow as φ the speed in any point of the fluid ~u isdefined as:

~u = ∇φ ux =∂u

∂xuy =

∂u

∂y(1)

The so-called freestream flow is defined as:

φ = Ux (2)

a) What is the speed of the fluid in the x and y directions?A cylindrical body immersed in the fluid (i.e. a submarine) can be described as a two sourcesystem inside the fluid at a distance 2a. the potential flow is modified as follow:

φ = Ux− Q

4π

1√(x+ a)2 + (y)2

+Q

4π

1√(x− a)2 + (y)2

(3)

b) What is the speed of fluid now?c) Assuming the fluid density is ρ, what is the pressure of the fluid P(x,y) as a function ofthe position assuming the submarine is traveling at a depth h?Suggestion: Bernoulli equation is always valid for an inviscid fluid

a) Applying definition 1) to equation 2):

ux =∂u

∂x=∂Ux

∂x= U (4)

uy =∂u

∂y=∂Ux

∂y= 0

(5)

The solutions clearly explain why this is called ”freestream” flow potential flow. It describesa fluid that is moving with uniform speed in x direction.

1

b) We have to differentiate equation 3).

ux =∂u

∂x=∂φ

∂x= U +

Q

4π

x+ a

((x+ a)2 + (y)2)3/2− Q

4π

x− a((x− a)2 + (y)2)3/2

(6)

uy =∂u

∂x=∂φ

∂y=

Q

4π

y

((x+ a)2 + (y)2)3/2− Q

4π

y

((x− a)2 + (y)2)3/2(7)

|~u| =√u2x + u2y (8)

In this case the speed of the fluid is no more constant, but a vector function of the positionu=~u(x,y)c) Bernoulli equation state that:

p+ ρu2

2+ ρgh = const (9)

Far away from the submarine, the flow is undisturbed, so the speed of the fluid is simplyu=U. In that case the pressure is:

ph + ρU2

2+ ρgh = const (10)

On the surface of the fluid, the pressure is:

p0 + ρU2

2= const (11)

So ∆ p can be wrote as:∆p = ph − p0 = −ρgh (12)

The submarine is traveling at a depth h. Near it the speed is not anymore u=U but u(x,y)=|~u|as defined in equations (6) (7) and (8). In that case the pressure can be written as:

∆p = ph − p0 = −ρgh+ ρU2

2− ρu(x, y)2

2(13)

Where u(x,y) is the one defined in equation 8.

Exercise 2 Solution. Viscous fluid in a tube. A viscous incompressible Newtonian fluidis defined by the fact that its shear stress vector ~τ is proportional to the the fluid velocitygradient, or in formula:

~τ = µd~v

dr(14)

A viscous incompressible Newtonian fluid is flowing inside a tube of diameter R and lenghtL. If we denote the pressure drop between the two ends of the tube as ∆p and the shearstress vector as ~τ , what is the speed of the fluid as a function of the distance from the centerof the tube?

2

Suggestion: write the equation of the forces acting on the fluid cylinder

The fluid is flowing inside the tube. At the extremity of the tube the pressure differenceis ∆p, so the force exerted by the difference of pressure can be written as:

|~F | = ∆p× S = ∆pπr2 (15)

This difference of pressure is generated by the shear stress. The shear stress is exerted onthe surface of the fluid cylinder, so:

| ~Fshear| = µ∂v

∂r× S ′ = µ

∂v

∂r(L× 2πr) (16)

Combining equation (15) and (16):

∆pπr2 = µ∂v

∂r(L× 2πr)⇒ ∂v

∂r=

∆pr

2µL(17)

So:

v =∆p

4µLr2 + C1 (18)

Now we have to chose C1 so that v=0 for r=R

v(r = R) = 0⇒ C1 =∆p

4µLR2 (19)

In the end we can write the solution as:

v(r) =∆p

4µLR2(1− (

r

R)2

) (20)

1 Points

exercise 1: 5 points divided as follow1 point for point a (eq. 4 and 5)1 point for point b (eq. 6, 7 and 8)3 point for point c divided as follow:1 point for equation 101 point for equation 111 point for equation 13

exercise 2: 5 points divided as follow1 point for equation 151 point for equation 161 point for equation 171 point for equation 181 point for equation 19/20 ( consideration on C1

3