In: Physics
A “U” shaped tube (with a constant radius) is filled with water
and oil as shown. The water is a height h1 = 0.36 m
above the bottom of the tube on the left side of the tube and a
height h2 = 0.1 m above the bottom of the tube on the
right side of the tube. The oil is a height h3 = 0.33 m
above the water. Around the tube the atmospheric pressure is
PA = 101300 Pa. Water has a density of 103
kg/m3.
A)What is the absolute pressure in the water at the bottom of the
tube?
B)What is the absolute pressure in the water right at the oil-water
interface?
C)What is the density of the oil?
D)Now the oil is replaced with a height h3 = 0.33 m of
glycerin which has a density of 1261 kg/m3. Assume the
glycerin does not mix with the water and the total volume of water
is the same as before.
Now what is the absolute pressure in the water at the interface
between the water and glycerin?
E)How much higher is the top of the water in the tube compared to
the glycerin? (labeled d in the diagram)
F)What is the absolute pressure in the water at the very bottom of
the tube?
a) pressure at the bottom is found as
P = Patm + gh
P = 101300 + 1000 * 9.8 * 0.36
P = 104828 Pa
-------------------------
b)
pressure at the interface
P = 101300 + 1000 * 9.8 * ( 0.36 - 0.1)
P = 103848 Pa
-----------------------------
c)
pressure in left column = pressure in right column
water g h1 = oil g h3 + water g h2
1000 * 9.8 * 0.36 = oil * 9.8 * 0.33 + 1000 * 9.8 * 0.1
oil = 787.87 Kg/m3
-----------------------------------
d)
now , oil is replaced with glycerin
so,
P ( interface) = 101300 + 1261 * 9.8 * 0.33
P ( interface) = 105378 Pa
--------------------------------------
(e)
p ( left) = p ( right)
101300 + 1000 * 9.8 * h = 105378
h = 0.416 m
so,
top of water = 0.416 - 0.33 = 0.08613 m
-------------------------------
(f)
height of water in right tube = 0.1 - 0.08613 = 0.01387 m
so,
P = 101300 + 1261 * 9.8 * 0.33 + 1000 * 9.8 * 0.01387
P = 105514 Pa