2.25 ADVANCED FLUID MECHANICS
Fall 2004
FINAL EXAM SOLUTIONS
Each problem is graded out of 20 points
(40 points total)
Please see Prof Sonin with grading queries regarding Question 1
Please see Prof McKinley with grading queries regarding Question 2
SOLUTION: PROBLEM 1
(Ain Sonin)
PART A
From the definitions of the lift and drag coefficients in the problem statement,
R
R
±
1
Lift
=
2
²
Ldr
=
2
²
c
L
2
³
(
´
r
)
2
bdr
=
c
L
³´
2
bR
3
(1)
3
0
0
PART B
Equating lift to the total helicopter weight
Mg
that is kept aloft, we obtain, using Part A, that the
required rotation rate is
3
Mg
±
=
²
c
L
bR
3
(2)
PART C
Apply the steadystate angular momentum theorem,
±
rV
²
V
rn
dA
=
(
T
z
)
CV
,
(3)
CS
to a control volume that envelops the rotor blades and jet engines and “cuts” through the
frictionless bearing and the exit planes of the two jet engines. At the jet exit planes, the

direction gas velocity
relative to the inertial reference frame of the ground
is
V
=
R
±
V
0
, and
V
=
V
0
. The torque due to aerodynamic drag may be deduced from the formula for drag given in
rn
the problem statement:
R
1
(
T
z
)
CV
=
±
2
²
c
D
2
(
r
)
2
brdr
=
±
c
D
²³
2
bR
4
.
(4)
4
0
Inserting this into (3), and using (2) for the angular rotation rate, we cast (3) into the simple form
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Final Examination Solutions
V
Mg
0
=
±
R
+
3
c
D
(5)
4
c
L
²
0
V
0
A
0
where A
0
is the exit area of a jet. Equation (5) reduces to a quadratic equation for V
0
, which can be
solved straightforwardly. Putting the equation in the form (5) is useful since it shows at a glance
that the torque due to aerodynamic drag becomes negligible in the limit
3
c
D
Mg
<<
R
(6)
4
c
L
0
V
0
A
0
PART D
The power required for hovering flight is the torque exerted on the two rotor blades —Eq.
(4)—multiplied by
.
NOTE:
The above solution accounts for the dynamic effects of the jet exit velocity but neglects the effects
of the inflow into the jet engines. This is a reasonable approximation: the hot exhaust gases are
much less dense than the cool inlet air. Mass conservation implies that, with inlet and exit areas of
the same order of magnitude, both the velocity and the momentum flux will be much higher at the
exit than in the cool gas at the inlet.
Problem 2: Lubricated Pipelining (
Gareth McKinley
)
Model the fluid motion as flow in a
²,µ
2
water
cylindrical pipe of radius
R
with a
core
of
thickness
R
1
consisting of viscous liquid
1
R
1
oil with viscosity
µ
1
surrounded by a shell
of water or other low viscosity fluid) of
Q
o
oil
R
thickness
=
R
±
R
1
that is density
matched (so that
1
=
2
=
) with
viscosity
µ
2
<
1
. The interfacial tension between the two liquids is denoted
. The average
velocity of the oil through the pipe is denoted
v
o
=
Q
o
R
1
2
a)
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 Fall '05
 GarethMcKinley
 Fluid Dynamics, vo R1 R1, vo2 R1 R1

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