ME 375
HOMEWORK #11 Solution
Fall 2009
PROBLEM 1
(40%)
(a). Given
: Prob. 7.30 in Palm textbook.
Solution
:
In the absence of hydrostatic pressure, the gage pressure at the bottom of the plate is
p
.
Thus, Newton’s law
gives:
mx
pA
kx
=
−
±±
With the plate mass small, this leads to
pA
pA
kx
x
k
=
⇒
=
.
Continuity requires that
1
2
(
)
d
Ax
Ax
q
q
dt
=
=
−
±
,
where
q
1
is the flow through the left resistor, and
q
2
is the flow through the right resistor.
The elemental
equations for the resistors are given by:
1
1
2
2
Rq
p
p
Rq
p
p
=
−
=
−
From the expression for
x
above:
A
x
p
k
=
±
±
Thus,
(
)
(
)
2
1
2
1
2
1
1
A
Ax
p
q
q
p
p
p
p
k
R
R
=
=
−
=
−
−
−
±
±
.
Rewriting, this gives
(
)
2
1
2
2
1
A
p
p
p
p
k
R
R
+
=
+
±
(
)
2
1
2
1
2
2
RA
p
p
p
p
k
+
=
+
±
To find the equivalent fluid capacitance, we need the following relationship:
f
q
Ax
C
p
p
=
=
±
±
±
Substituting
x
±
from above, we get
2
f
A
C
k
=

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