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Chapter 2: Problem 6

The pressure in the airspace above an oil \((\mathrm{SG}=0.80)\) surface in a tank is \(14 \mathrm{kPa}\). Find the pressure \(1.5 \mathrm{~m}\) below the surface of the oil.

Short Answer

Expert verified
The pressure 1.5 m below the oil surface is 25.772 kPa.

Step by step solution

01

Understanding Given Variables

Identify the known values from the problem statement. We are given the specific gravity (SG) of oil, which is 0.80, and the surface pressure, which is 14 kPa. We need to find the pressure at a depth of 1.5 m below the oil surface.
02

Calculate the Density of Oil

The density of oil is calculated using its specific gravity (SG). Specific gravity is the ratio of the density of the oil to the density of water. Since the density of water is approximately 1000 kg/m鲁, the density of the oil is calculated as follows: \[ \text{Density of oil} = \text{SG} \times \text{density of water} = 0.80 \times 1000 = 800 \text{ kg/m}^3 \]
03

Determine Additional Pressure Due to Oil Column

The pressure at a certain depth due to a liquid column is given by \( P = \rho g h \), where \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity (approximately \(9.81\, \text{m/s}^2\)), and \( h \) is the depth. Here, \(h = 1.5\, \text{m}\), \(\rho = 800\, \text{kg/m}^3\). Calculate the additional pressure: \[ P = 800 \times 9.81 \times 1.5 = 11772 \, \text{Pa} = 11.772 \, \text{kPa} \]
04

Calculate Total Pressure at Depth

The total pressure at 1.5 m below the surface is the sum of the surface pressure and the pressure due to the oil column. Use the formula: \[ \text{Total Pressure} = \text{Surface Pressure} + \text{Pressure due to oil column} = 14 \text{ kPa} + 11.772 \text{ kPa} = 25.772 \text{ kPa} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
In fluid mechanics, calculating pressure is essential to understanding how fluids behave under various conditions. Pressure is the force exerted per unit area of a surface. It can be calculated using various formulas depending on the scenario. In the given exercise, we use the formula to calculate pressure due to a fluid column. Pressure in fluids is additive, meaning it increases with depth due to the weight of the fluid above.
Understanding how different layers of fluid contribute to overall pressure helps solve complex problems. It鈥檚 important to remember that pressure is not just about the force; it also considers the area over which the force is applied.
  • Units: Pressure is measured in Pascals (Pa) or kilopascals (kPa) in metric systems.
  • Key formula: Total pressure at a depth = Surface pressure + Pressure due to fluid column.
Specific Gravity
Specific gravity (SG) is a dimensionless quantity used to compare the density of a fluid to the density of a reference fluid, usually water for liquids. It tells us how much lighter or heavier a fluid is compared to water. This is crucial for understanding the buoyancy and flow characteristics of the fluid.
In this exercise, the specific gravity of the oil is given as 0.80, which means the oil's density is 80% of that of water. Calculating specific gravity is simple and involves dividing the fluid density by the density of water.
  • SG = Density of fluid / Density of water (assumed 1000 kg/m鲁 for water).
  • Helps in determining the density of fluids when working with different liquid types.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases linearly with depth because the weight of the fluid above adds pressure to the fluid below.
To calculate hydrostatic pressure, you can use the formula: \( P = \rho g h \), where \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid column.
This principle is essential in designing structures like dams and submersible vehicles, where materials must withstand water pressure at varying depths.
  • Directly proportional to density (\( \rho \)) and depth (\( h \)).
  • Affects calculations of total pressure in fluid systems.
Density of Fluids
Density is a measure of how much mass a substance has in a given volume. In fluid mechanics, knowing the density of a fluid helps determine how it will interact with forces such as gravity. Density impacts fluid behavior in terms of flow, buoyancy, and pressure.
In the exercise, we calculated the oil's density using its specific gravity: \( 0.80 \times 1000 \text{ kg/m}^3 = 800 \text{ kg/m}^3 \). This knowledge allows us to accurately calculate additional pressures and determine the fluid's impact on its surroundings.
  • Higher density = more mass per unit volume.
  • Plays a vital role in fluid calculations and designs.

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Most popular questions from this chapter

A vertical rectangular gate of height \(h\) and width \(w\) is installed such that the top of the gate is a distance \(d\) below the water surface in a reservoir. Calculate the hydrostatic force on the gate and show that the location of the center of pressure is given by $$ y_{\mathrm{cp}}=d+\frac{L(3 d+2 h)}{6 d+3 h} $$

A liquid is stratified such that the specific gravity of the fluid at the surface is 0.98 and the specific gravity at a depth of \(12 \mathrm{~m}\) is equal to 1.07 . Assuming that the specific gravity varies linearly between the liquid surface and a depth of \(12 \mathrm{~m}\), determine the pressure at a depth of \(12 \mathrm{~m}\). State whether this is a gauge pressure or an absolute pressure.

The weight of a solid object in air is measured as \(40 \mathrm{~N},\) and the weight of this same object in water at \(20^{\circ} \mathrm{C}\) is measured as \(25 \mathrm{~N}\). Estimate the specific weight and volume of the object.

A bubble is released from an ocean vent located \(20 \mathrm{~m}\) below the ocean surface. Atmospheric pressure above the ocean surface is equal to \(101.3 \mathrm{kPa}\), and the ocean temperature down to a depth of \(20 \mathrm{~m}\) is approximately constant at \(20^{\circ} \mathrm{C}\). Estimate the ratio of the density of the air in the bubble at a depth of \(20 \mathrm{~m}\) to the density of the air in the bubble just as it reaches the ocean surface.

A hydrometer with a stem diameter of \(9 \mathrm{~mm}\) is placed in distilled water, and the volume of the hydrometer below the water surface is estimated to be \(20 \mathrm{~cm}^{3}\). If the hydrometer is placed in a liquid with a specific gravity of 1.2 , how far above the liquid surface will the distilled water mark be located?
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