Problem 3.19
[Difficulty: 2]
Given:
Data on partitioned tank
Find:
Pressure of trapped air required to bring water and mercury levels equal if right air opening is sealed
Solution:
First we need to determine how far each free surface moves.
In the tank of Problem 3.18, the ratio of cross section areas of the partitions is 0.75/3.75 or 1:5.
Suppose the water surface (and
therefore the mercury on the left) must move down distance
x
to bring the water and mercury levels equal.
Then by mercury volume
conservation, the mercury free surface (on the right) moves up (0.75/3.75)
x
=
x
/5.
These two changes in level must cancel the original
discrepancy in free surface levels, of (1m + 2.9m)  3 m = 0.9 m.
Hence
x
+
x
/5 = 0.9 m, or
x
= 0.75 m.
The mercury level thus moves
up
x
/5 = 0.15 m.
Assuming the air (an ideal gas,
pV
=
RT
) in the right behaves isothermally, the new pressure there will be
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 Fall '07
 Lear
 Fluid Mechanics, mercury levels, Pleft, pright

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