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Chapter 2: Problem 6

An unknown immiszible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the urknown liquid is \(1.5 \mathrm{m}\) and the depth of the oil (specific weight \(=8.5 \mathrm{kN} / \mathrm{m}^{3}\), floating on top is \(5.0 \mathrm{m}\). A pressure gage connezted to the bottom of the tank reads 65 kPa. What is the specific gravity of the unknown liquid?

Short Answer

Expert verified
The specific gravity of the unknown liquid is 1.53.

Step by step solution

01

Calculate the pressure contribution of the oil

The pressure at any depth in a fluid is given by the specific weight of the fluid multiplied by the depth. This relationship can be represented as \(P = \gamma h\), where \(P\) is the pressure, \(\gamma\) is the specific weight of the fluid, and \(h\) is the depth.\\ Substituting \(h=5.0 \mathrm{m}\) and \(\gamma=8.5 \mathrm{kN/m^3}\), we get \(P_{oil} = (8.5 \mathrm{kN/m^3})(5.0 \mathrm{m}) = 42.5 \mathrm{kPa}\).
02

Calculate the pressure contribution of the unknown liquid

The total pressure at the bottom of the tank is the sum of the pressures contributed by the oil layer and the unknown liquid layer. Hence, the pressure contribution from the unknown liquid can be calculated by subtracting the oil's pressure from the total pressure. \\Given that the total pressure, \(P_{total}\), is 65 kPa, the pressure contribution of the unknown liquid, \(P_{liquid}\), is \(P_{liquid} = P_{total} - P_{oil} = 65 \mathrm{kPa} - 42.5 \mathrm{kPa} = 22.5 \mathrm{kPa}\).
03

Calculate the specific weight of the unknown liquid

The specific weight of the unknown liquid, \(\gamma_{liquid}\), can be found by dividing its pressure contribution by its depth. Given that the depth, \(h_{liquid}\), is 1.5 m, then \(\gamma_{liquid} = P_{liquid} / h_{liquid} = 22.5 \mathrm{kPa} / 1.5 \mathrm{m} = 15 \mathrm{kN/m^3}\).
04

Compute the specific gravity of the unknown liquid

The specific gravity of the unknown liquid is the ratio of its specific weight to the specific weight of water. The specific weight of water is approximately \(9.81 \mathrm{kN/m^3}\). Therefore, the specific gravity, \(SG_{liquid}\), can be found by \(SG_{liquid} = \gamma_{liquid} / 9.81 \mathrm{kN/m^3} = 15 \mathrm{kN/m^3} / 9.81 \mathrm{kN/m^3} = 1.53\).

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Most popular questions from this chapter

A 5 -gal, cylindrical open container with a bottom area of 120 in. \(^{2}\) is filled with glycerin and rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the container when the elevator has an upward acceleration of \(3 \mathrm{ft} / \mathrm{s}^{2}\). (b) What resultsnt force does the container exert on the floor of the elevator during this acceleration? The weight of the container is negligible. (Note: 1 gal \(=231\) in. \(^{3}\) )

An iceberg (specific gravity 0.917 ) floats in he ocean (specific gravity 1.025 ). What percent of the volume of the iceberg is under water?

A closed, 0.4 -m-diameter cylindrical tank is completely filled with oil $(S G=0.9)$ and rotates about its vertical longitudinal axis with an angular velocity of 40 rad/s. Determine the difference in pressure just under the vessel cover between a point on the circumference and a point on the axis.

A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by \(W\). When in salt-free riverwater. it floats 0.25 ft deeper than when in seawater \(\left(\gamma=64 \mathrm{lb} / \mathrm{ft}^{3}\right) .\) Find the weight \(W\).

A studant needs to measure the air pressure inside a compressed air tenk but does not have ready access to a pressure gage. Using materials already in the lab, she builds a U-tube manometer using two clear 3 -ft-long plastic tubes, flexible hoses, and a tape measure. The only readily avai able liquids are water from a tap and a bottle of corn syrup. She selects the corn syrup because it has a larger density \((S G=1.4) .\) What is the maximum air pressure, in psia, that can be measured?
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