Download ASTR 320: Solutions to Problem Set 2 Problem 1: Streamlines

ASTR 320: Solutions to Problem Set 2
Problem 1: Streamlines
Consider a two-dimensional fluid with velocity field ~u = (ax, by), where a, b
are constants and we adopt Cartesian coordinates. Assume that at time
t = 0 the density everywhere is equal to ρ0 .
a) Draw streamlines for the following two cases: (i) a = b > 0, and (ii)
a = −b > 0.
SOLUTION: See Fig 1.
b) Under what condition(s) is the flow incompressible? Under what condition(s) is it curl-free
SOLUTION: A flow is incompressible if ∇ · ~u = 0. Using that
∇ · ~u =
∂ux ∂uy
+
=a+b
∂x
∂y
we see that the flow is incompressible if a + b = 0.
Similarly, a flow is curl-free if ∇ × ~u = 0. Using that
∇ × ~u =
∂uy ∂ux
−
∂x
∂y
!
~ez = (0 − 0) ~ez = 0
Hence, we see that this flow is always curl-free, independent of the values of
a or b.
c) Consider the location A = (2, 2). How does the density at A change as a
function of time in cases (i) and (ii) above?
SOLUTION: The density is related to the divergence of the velocity field
via the continuity equation
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Figure 1: Solution to Problem 1a.
∂ρ
+ ρ ∇ · ~u + ~u · ∇ρ = 0
∂t
Using that ∇ · ~u = (a + b), and using that initially ∇ρ = 0, we have that
∂ρ
= −(a + b)ρ
∂t
which has as solution
ρ(t) = ρ0 e−(a+b)t
where we have used the boundary condition ρ(t = 0) = ρ0 . Note that ρ(t) is
independent of position, which implies that ∇ρ will remain zero at all times.
Hence, the density at location A is described by the above equation.
Now consider a two-dimensional velocity field ~u = (−ay, bx). Again, assume
that at time t = 0 the density everywhere is equal to ρ0 .
d) Draw streamlines for the following two cases: (i) a = b > 0, and (ii)
a = −b > 0
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Figure 2: Solution to Problem 1d.
SOLUTION: See Fig. 2.
e) Under what condition(s) is the flow incompressible? Under what condition(s) is it curl-free
SOLUTION: A flow is incompressible if ∇ · ~u = 0. Using that
∇ · ~u =
∂ux ∂uy
+
= 0+0 =0
∂x
∂y
Hence, this flow is always incompressible, independent of the values of a or
b.
Similarly, a flow is curl-free if ∇ × ~u = 0. Using that
∇ × ~u =
∂uy ∂ux
−
∂x
∂y
!
~ez = (b − −a) ~ez = (a + b) ~ez
Hence, we see that this flow is curl-free if a + b = 0.
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f) How does the density at A change as function of time at A in cases (i)
and (ii)?
SOLUTION: Since we have shown that the flow is incompressible, we have
that ∂ρ/∂t = 0 (this follows from continuity equation and the fact the the
flow is divergence free). Hence, ρA (t) = ρ0 at all t.
Problem 2: Momentum Equations in Conservation Law Form
The continuity equation in Eulerian vector form is given by
∂ρ
+ ∇ · (ρ~u) = 0
∂t
This equation is said to be in ‘conservation law form’. The Euler equation
describing conservation of momentum in Eulerian vector form is
∇P
∂~u
+ ~u · ∇~u = −
− ∇Φ
∂t
ρ
which is not in conservation law form. Show that, for an inviscid fluid, and in
the absence of gravity, this equation can be put in the following conservation
law form
∂(ρ~u)
+∇·Π=0
∂t
where Π is the momentum flux density tensor.
SOLUTION: The momentum flux density tensor, in index form, is given
by
Πij = ρ ui uj − σij = ρ ui uj + P δij − τij
Since our fluid is inviscid, the viscous stress tensor τij = 0. We first write
the Euler equation in index form (which often makes the calculus somewhat
more transparent):
ρ
∂ui
∂P
∂ui
+ ρ uj
+
=0
∂t
∂xj
∂xi
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where we have multiplied by ρ, and made use of the fact that in the absence
of gravity ∇Φ = 0. Using that
ρ
∂(ρui )
∂ρ
∂(ρui )
∂ρuj
∂ui
=
− ui
=
+ ui
∂t
∂t
∂t
∂t
∂xj
where the second equality follows from the continuity equation, we can
rewrite the Euler equation as
∂(ρuj )
∂ui
∂P
∂(ρui )
+ ui
+ ρ uj
+
=0
∂t
∂xj
∂xj
∂xi
∂ρui ∂(ρui uj ) ∂P
⇔
+
=0
+
∂t
∂xj
∂xi
∂ρui ∂(ρui uj ) ∂P
⇔
+
+
δij = 0
∂t
∂xj
∂xj
∂ρui ∂Πij
⇔
+
=0
∂t
∂xj
which, in vector form, reads
∂(ρ~u)
+∇·Π=0
∂t
Problem 3: The Stress Tensor
Consider a fluid in a 2-dimensional, Cartesian coordinate system (x1 , x2 ),
with stress tensor
σij =
σ11 σ12
σ21 σ22
!
and let n̂ and t̂ be the unit normal and unit tangent vectors of a surface S,
for which n̂ is rotated by angle θ with respect to the x1 axis.
a) Express n̂ and t̂ in terms of θ, i.e., what are n1 , n2 , t1 and t2 in n̂ = (n1 , n2 )
and t̂ = (t1 , t2 )?
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SOLUTION: Simple geometry shows that n̂ = (cos θ, sin θ) and t̂ = (− sin θ, cos θ).
Note that n̂ · t̂ = 0, as required for two vectors that are perpendicular to each
other. Note also, one is free to choose t̂ in the opposite direction, i.e., with
t̂ = (sin θ, − cos θ). In that case, the tangential stress will have the opposite
sign as in the case used here.
b) Show that the normal stress, Σn , can be written as
Σn = σ11 cos2 θ + σ22 sin2 θ + σ12 sin 2θ
and derive a similar expression for the shear stress, Σt .
SOLUTION: The normal stress is the stress along the normal. Hence,
Σn =
=
=
=
~ · n̂ = Σi ni = σij nj ni
Σ
σ11 n1 n1 + σ12 n2 n1 + σ21 n1 n2 + σ22 n2 n2
σ11 cos2 θ + σ12 sin θ cos θ + σ21 sin θ cos θ + σ22 sin2 θ
σ11 cos2 θ + σ22 sin2 θ + σ12 sin 2θ
where, in the last step, we have used that the stress tensor is diagonal, so
that σ12 = σ21 .
Similarly, for the shear stress, we have that
~ · t̂ = Σi ti = σij nj ti
Σ
σ11 n1 t1 + σ12 n2 t1 + σ21 n1 t2 + σ22 n2 t2
−σ sin θ cos θ − σ12 sin2 θ + σ21 cos2 θ + σ22 sin θ cos θ
11
σ22 − σ11
sin 2θ + σ21 cos2 θ − σ12 sin2 θ
=
2
σ22 − σ11
=
sin 2θ + σ21 cos 2θ
2
Σt =
=
=
where in the last step we have used once again that σ12 = σ21 .
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c) Consider the case σ11 = σ12 = σ21 = 0. Under what angle θ is the shear
stress on S maximal? For what angle θ does the shear on S vanish? What
are the normal stresses in both cases?
SOLUTION: Using the expression for the shear stress derived under b), we
see that, when σ11 = σ12 = σ21 = 0 we have that Σt = 21 σ22 sin 2θ. This will
be maximal when sin 2θ = ±1, which occurs for θ ∈ {π/4; 3π/4; 5π/5; 7π/4}.
The corresponding normal stress is Σn = σ22 sin2 θ = σ22 /2.
Similarly, the shear stress vanishes when sin 2θ = 0, which occurs for θ ∈
{0, π/2, π, 3π/2}. The corresponding normal stresses are Σn ∈ {0; σ22 ; 0; σ22 }.
d) Answer the same questions as under c), but now for the case with σ11 =
σ22 = 0 and σ12 = σ21 > 0.
SOLUTION: In this case, Σt = σ12 cos 2θ and Σn = σ12 sin 2θ. The
shear stress is maximal whenever cos 2θ = ±1, which corresponds to θ ∈
{0, π/2, π, 3π/2}. The corresponding normal stresses vanish (i.e., Σn = 0).
The shear stress will vanish whenever cos 2θ = 0, which happens whenever
θ ∈ {π/4; 3π/4; 5π/5; 7π/4}. The corresponding normal stresses are Σn =
{σ12 ; −σ12 ; σ12 ; −σ12 }
e) Under what condition are both the normal and shear stress independent
of θ, and what are their values in that case?
SOLUTION: We seek to have Σt = Σn . Using the derivations for Σn and
Σs under b), it is fairly easy to see that the normal and shear stresses are
identical whenever σij = σ0 δij . In words, we have that the normal stress is
equal to the shear stress whenever the matrix representing the stress tensor
is scalar (i.e., when the matrix is diagonal with identical diagonal elements).
Substituting these values for σij in the expressions for Σn and Σt we see that
the shear stress vanishes everywhere, while the normal stress is equal to σ0 .
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