You come across an open container which is filled with two liquids. Since the tw

Question

You come across an open container which is filled with two liquids. Since the two liquids have different density there is a distinct separation between them. Water fills the lower portion of the container to a depth of 0.215 m which has a density of 1.00 × 103 kg/m3. The fluid which is floating on top of the water is 0.328 m deep. If the absolute pressure on the bottom of the container is 1.049 × 105 Pa, what is the density of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s2 and atmospheric pressure is P0 = 1.013 × 105 Pa.

Explanation / Answer

dw = depth of the water = 0.215 m
w = density of the water = 1.00 x 10³ kg/m³
Pw = static fluid pressure of the water = to be determined
du = depth of the unknown fluid = 0.328 m
u = density of the unknown fluid = to be determined
Pu = static fluid pressure of the unknown fluid = to be determined
g = gravity acceleration = 9.81 m/s²
P0 = atmospheric pressure = 1.013 x 10^5 Pa
Pabs = absolute pressure on the bottom of the container = 1.049 x 10^5 Pa
P = Pabs - P0 = 1.049 x 10^5 Pa - 1.013 x 10^5 Pa = 0.036 x 10^5 Pa = 3.6 x 10^3 Pa = 3,600 Pa

Pw = gh = (1.00 x 10³ kg/m³)(9.81 m/s²)(0.215 m)
Pw = 2,109 Pa

P - Pw = 3,600 Pa - 2,109 Pa = 1,491 Pa

1,491 Pa = u·g·h

u = 1,491 Pa / gh = 1,491 Pa / 9.81 m/s²·0.328 m
u = 463 kg/m³ Density of the Unknown Fluid

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