A “U” shaped tube (with a constant radius) is filled with water and oil as shown

Question

A “U” shaped tube (with a constant radius) is filled with water and oil as shown. The water is a height h1 = 0.37 m above the bottom of the tube on the left side of the tube and a height h2 = 0.11 m above the bottom of the tube on the right side of the tube. The oil is a height h3 = 0.31 m above the water. Around the tube the atmospheric pressure is PA = 101300 Pa. Water has a density of 103 kg/m3.

1)

What is the absolute pressure in the water at the bottom of the tube?  

Pa

2)

What is the absolute pressure in the water right at the oil-water interface?  

Pa

3)

What is the density of the oil?  

kg/m3

4)

Now the oil is replaced with a height h3 = 0.31 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?

Pa

5)

How much higher is the top of the water in the tube compared to the glycerin? (labeled d in the diagram)

m

6)

What is the absolute pressure in the water at the very bottom of the tube?

Pa

Explanation / Answer

P_absolute = P_atm + P_fluid

P_fluid = density * g * h

1) P_absolute = Patm + density of water * g * height of the left side

2) P_absolute on one side = P_absolute on the other side

P_left = P_right

P_atm + density of water * g * height of left side = P_atm + density of water * g * height of the water on right side + density of oil * g * height of oil on right side

Solve for density of oil

P_at the oil water line = P_atm + density of oil * g * height of oil on right side

3) You solved for the density earlier

4) Not listed

5) Pressure on left side = pressure on right side

P_atm + density of water * g * height of water on left side = P_atm + density of water * g * height of water on right side + density of glycerin * g * height of glycerin

d_h20 * (h_left - h_right) = d_glycerin * h_glycerin

(h_left - h_right ) = d_glycerin / d_h20 * h_glycerin

Difference = (h_left - h_right) - h_glycerin

6) Not sure how you book would solve but I would solve it like this
You need to find the height of water on either side

Total height of the water = .50 m from the 1st part

density of water * g * height of water on left side = density of water * g * height of water on right side + denisty of glycerin * g * height of glycerin on right side

d_h20 * h_left = d_h20 * h_right * d_glycerin * h_glycerin

h_left + h_right = .50 m
h_right = .50 m - h_left

d_h20 * h_left = d_h20 * (.50 m - h_left) + d_g * h_g

h_left = .50 m - h_left + d_g / d_h20 * h_g
2*h_left = .50 m + d_g / d_h20 * h_g

h_left = 1/2*(.50 m + d_g / d_h20 * h_g)

P_abs = P_atm + d_h20 * g * h_left

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