Page 324 - Intro to Tensor Calculus

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I 6. Assume λ = µ = 0 so that the fluid is ideal or nonviscous. Use the results given in problem 5 and
make the following additional assumptions:
• The density is constant and so the fluid is incompressible.
• The body forces are zero.
• Steady state flow exists.
• Only two dimensional flow in the x-yplane is considered such that u = u(x, y), v = v(x, y)and
w =0. (a) Employ the above assumptions and simplify the equations in problem 5 and verify the
results
∂u ∂u 1 ∂p
u + v + =0
∂x ∂y % ∂x
∂v ∂v 1 ∂p
u + v + =0
∂x ∂y % ∂y
∂u ∂v
+ =0
∂x ∂y
(b) Make the additional assumption that the flow is irrotational and show that this assumption
produces the results
∂v ∂u 1 2 2 1
− =0 and u + v + p = constant.
∂x ∂y 2 %
(c) Point out the Cauchy-Riemann equations and Bernoulli’s equation in the above set of equations.

I 7. Assume the body forces are derivable from a potential function φ such that b i = −φ ,i . Show that for an
ideal fluid with constant density the equations of fluid motion can be written in either of the forms
∂v r r s 1 rm rm ∂v r s 1
+ v v = − g p ,m − g φ ,m or + v r,s v = − p ,r − φ ,r
,s
∂t % ∂t %
1
2
I 8. The vector identities ∇ ~v = ∇ (∇· ~v) −∇ × (∇× ~v) and (~v ·∇)~v = ∇ (~v · ~v) − ~v × (∇× ~v) are
2
used to express the Navier-Stokes-Duhem equations in alternate forms involving the vorticity Ω = ∇× ~v.
(a) Use Cartesian tensor notation and derive the above identities. (b) Show the second identity can be written
∂v 2 k
mj k
mnp ijk
j m
in generalized coordinates as v v = g v v k,j − g pi v n v k,j . Hint: Show that =2v v k,j .
,j j
∂x
I 9. Use problem 8 and show that the results in problem 7 can be written
∂v r rnp rm ∂ p v 2
− Ω p v n = −g + φ +
∂t ∂x m % 2
2
∂ p v
∂v i j k
or − ijk v Ω = − + φ +
∂t ∂x i % 2
I 10. In terms of physical components, show that in generalized orthogonal coordinates, for i 6= j, the rate

1 h i ∂ v(i) h j ∂ v(j)
of deformation tensor D ij can be written D(ij)= + , no summations
2 h j ∂x j h i h i ∂x i h j
3
1 ∂v(i) v(i) ∂h i X 1 ∂h i
and for i = j there results D(ii)= − 2 + v(k) , no summations. (Hint: See
h i ∂x i h ∂x i ∂x k
i k=1 h i h k
Problem 17 Exercise 2.1.)