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2.25 QUIZ 2, 13 NOVEMBER 2003
ANSWER TO PROBLEM 1
Ain Sonin
NOTE: All extensive quantities are
per unit width
, consistent with a 2D problem.
The answer given here for part (b) is much more complete than was required for the quiz.
Momentum theorem applied to the fixed CV
1
:
d
±
²
v v dV
+
±
v v dA
=
(
F
)
CV
1
(1)
dt
CV
1
x
rn
CS
1
x
rn
x
First, to obtain the outflow velocity
v
(
x
,
t
)
between the plates, we apply mass
x
conservation to CV
1
:
d
±
dV
+
±
v
rn
dA
=
0
dt
CV
2
CS
2
(2)
During the
closing stroke stroke
,
d
±
=
²
³
dt
(3)
and eq. (2) takes the form
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d
±
x
±
x
²
²
³
³
±
x
=
0
(4)
´ +
v
x
dt
µ
2
¶
from which, using (3),
±
x
v
(
x
,
t
)
=
(5)
x
2
(
t
)
Substituting next from (5) into (1), we have
d
L
³
x
´
·
³
L
·
2
µ
¶
·
¸
¸
x
¸
wdx
+
·
¸
¸
Lw
=
F
+
(
p
a
(6)
³
p
exit
)
L
dt
0
¹
2
¹
¹
2
¹
F
is the thrust force exerted in the xdirection on the CV at the “cut” through the
supports, which is equal to the thrust force in the –x direction. The second term on the
right accounts for the possibility that the pressure at the exit plane differs from the
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 Fall '05
 GarethMcKinley

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