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ES 123
Problem Set 6 Solution
Lihua Jin
Mar 26, 2010
1. According to the mass conservation, we have
00
VV
ρ
=
(1)
where
and
V
are the density and volume at any time. Since the crosssection area
keeps a constant, we can cancel area from Eq. (1), i.e.
L
L
=
(2)
where we know
0
L
Lv
t
=
−
(3)
so the dependence of the density on time is
0
L
L
vt
=
−
(4)
You can also use the differential equation of mass conservation to solve this problem.
Actually, Eq. (1) is the ‘integral’ form, which is similar to the integral momentum
conservation law we talked about in class. Eq. (1) can be derived from the differential
equation of mass conservation.
Concept question:
Assume the air in the cylinder is an ideal gas, which satisfies
p
R
T
M
=
(5)
so the force is proportional to the pressure and can be calculated as
()
0
LR
A
M
T
Fp
A
L
vt
==
−
(6)
which can be sketched as shown in Fig. 1. The force increases with the time.
Fig. 1
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View Full Document 2. Concept question:
When the block is accelerated, the faster the block slips, the larger the velocity gradient
in the fluid is, the larger the frictional shear force the fluid can give the block, and the
smaller the acceleration is. So finally the acceleration becomes zero, and the block
reaches the steady terminal velocity.
Call the direction parallel to the slipping the x axis, and the outer normal direction of the
fluid interface y axis. Assume the velocity profile is
( )
( ),0,0
Uv
y
=
and the pressure
p
is independent on x. So the conservation law of momentum in the slipping direction can
be simplified as
2
2
sin
0
v
g
y
ρθ
μ
∂
+
=
∂
(7)
Solving Eq. (7), we get
2
sin
2
g
vy
C
y
D
ρ
θ
=−
+
+
(8)
where C and D are constants. We then need two boundary conditions. First is the nonslip
boundary condition at the bottom of the fluid.
0
0
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This note was uploaded on 11/15/2011 for the course CS 50 taught by Professor Malan during the Spring '08 term at Harvard.
 Spring '08
 malan

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