MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 10.2
This problem is from “Advanced Fluid Mechanics Problems” by 2.25 Problem Set Solution — Problem
• (a) Show that if (1) and (2) are two arbitrary points in a steady, inviscid, incompressible flow in a
uniform gravitational field,
P2 +
2
v2
+ ρgy2
2
=
P1 +
v12
+ ρgy1
2
2
+ρ
1
(v × ω) · dl
(10.2-1)
Here, y is measured up against the gravitational field, ω = ∇ × v is the vorticity vector and the last
term represents a line integral along any path between (1) and (2) through the flow.
• (b) Show that if the flow in (a) is a parallel, horizontal flow, that is,
v = u(y)i,
(10.2-2)
as shown in the sketch, it follows from the equation in (a) that the pressure distribution is the hydro­
static one,
P2 + ρgy2 = P1 + ρgy1
(10.2-3)
• (c) Obtain the conclusion of (b) by using an argument based on Euler’s equation in streamline form,
rather than starting with the equation in part (a)
2.25 Advanced Fluid Mechanics
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Vorticity Theorems
2.25 Problem Set Solution — Problem 10.2
Solution:
(a) From Cauchy momentum equation, we can derive the following equations for a steady, inviscid, and
incompressible flow in a uniform gravitational field.
Dv
∂v
=ρ
+ ρv · (∇v) = −∇P + ρ(−∇gz)
Dt
∂t
(10.2-4)
✓
◆
∂v
1
1
2
− v × w = − ∇P − ∇gz
+∇
|v|
∂t
2
ρ
(10.2-5)
✓
◆
1
1
∇P + ∇
| v |2 + ∇gz = v × w
ρ
2
(10.2-6)
ρ
Z
2
1
✓
✓
◆
◆
◆
Z 2✓
1
1
2
∇P + ∇
|v|
+ ∇gz · dl =
v × w · dl
ρ
2
1
(10.2-7)
Therefore, the final form is the same as
✓
v2
P2 + 2 + ρgy2
2
◆
=
✓
v2
P1 + 1 + ρgy1
2
◆
+ρ
Z
2
1
(v × ω) · dl
(10.2-8)
(b)
i
j
k
∂
∂x
∂
∂y
∂
∂z
ux (y)
0
0
ω =∇×v =
i
v × w = ux
0
Therefore we can cancel ∇
✓
1
2
| v |2
◆
=−
∂ux
k
∂y
✓
◆
j
k
1
∂ux
2
0
0
= ux
j=∇
|v|
∂y
2
x
0 − ∂u
∂y
(10.2-9)
(10.2-10)
in the LHS of eq.(10.2-6) and it results in,
P2 + ρgy2 = P1 + ρgy1
(10.2-11)
(c)
Euler’s equation in the normal direction (en ) is −ρ
2.25 Advanced Fluid Mechanics
∂P
Vs2
=−
+ ρgn
R
∂n
2
(10.2-12)
c 2012, MIT
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Vorticity Theorems
2.25 Problem Set Solution — Problem 10.2
Here, R → ∞ because of parallel flow.
Therefore, integration from point 1 to point 2 gives again,
P2 + ρgy2 = P1 + ρgy1
(10.2-13)
D
Problem Solution by KP and BK, Fall 2012
2.25 Advanced Fluid Mechanics
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2.25 Advanced Fluid Mechanics
Fall 2013
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