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At what depth below the free surface of oil having a density of \(600 \mathrm{~kg} \mathrm{~m}^{-3}\) will the pressure be equal to 1 bar?

Short Answer

Expert verified
The depth is approximately 16.99 meters.

Step by step solution

01

Understand the Pressure Formula

Pressure in a fluid at a certain depth is given by the formula: \( P = \rho \cdot g \cdot h + P_0 \), where \( P \) is the pressure at depth, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity (approximately \( 9.81 \mathrm{~m/s}^2 \)), \( h \) is the depth below the surface, and \( P_0 \) is the atmospheric pressure acting on the surface.
02

Set Known Values and Conditions

We know that the desired pressure \( P \) is 1 bar, which is \( 100,000 \) Pa. The density \( \rho \) of the oil is \( 600 \mathrm{~kg/m}^3 \), and the acceleration due to gravity \( g \) is \( 9.81 \mathrm{~m/s}^2 \). Assume that the atmospheric pressure \( P_0 \) is 1 bar (\( 100,000 \) Pa), as this is common if not otherwise specified.Given these, the effective formula simplifies to: \[ 100,000 = 600 \cdot 9.81 \cdot h + 100,000 \]
03

Simplify the Equation

Substituting the values into the equation:\[ 100,000 = 600 \cdot 9.81 \cdot h + 100,000 \]Subtract \( 100,000 \) from both sides to eliminate the atmospheric pressure influence (since it is already accounted for):\[ 0 = 600 \cdot 9.81 \cdot h \]
04

Solve for the Depth \( h \)

Solve for \( h \) by dividing both sides by \( 600 \cdot 9.81 \):\[ h = \frac{0}{600 \cdot 9.81} = 0 \]This indicates the pressure only equals \( 1 \) bar at the surface when atmospheric pressure is equal to \( 1 \) bar. For pressure due to the fluid alone to reach \( 1 \) bar in total, it needs to be considered differently, subtracting \( P_0 \) if it's included.
05

Correct for Depth Calculation

Re-evaluate considering the gauge pressure alone:For gauge pressure to be 1 bar (without considering atmospheric pressure already being present at the surface), set:\[ 600 \cdot 9.81 \cdot h = 1 \times 10^5 \]\[ h = \frac{1 \times 10^5}{600 \cdot 9.81} \]Calculate \( h \).
06

Calculate Final Result

Carry out the calculation:\[ h = \frac{100,000}{600 \cdot 9.81} \approx \frac{100,000}{5886} \approx 16.99 \text{ meters (rounded to two decimal places)} \].

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Key Concepts

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

Pressure in Fluids
Pressure in fluids is a fundamental concept in fluid mechanics. It describes the force exerted by a fluid per unit area. This pressure varies due to a combination of factors like fluid density and the height of the fluid column. In a fluid at rest, the pressure at any given point depends primarily on the depth from the surface. As you move deeper into a fluid, the pressure increases. This increase is a result of the weight of the fluid above compressing the fluid below.The basic formula to calculate pressure in a fluid is: \[ P = \rho \cdot g \cdot h + P_0 \] where:
  • \( P \) is the total pressure at the point of interest.
  • \( \rho \) (rho) is the density of the fluid.
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \) on Earth.
  • \( h \) is the depth from the fluid's surface.
  • \( P_0 \) is the atmospheric pressure at the fluid's surface.
Typically, problems will assume standard atmospheric pressure unless stated otherwise.
Density of Fluids
The density of a fluid is a measure of its mass per unit volume, often expressed in kilograms per cubic meter (\( \text{kg/m}^3 \)). Density plays a crucial role in determining the pressure within a fluid. For example, when dealing with liquids like water or oil, the density is a constant value because liquids are incompressible under normal conditions. This characteristic makes calculations straightforward as the density remains unchanged with depth.A higher density fluid will exert more pressure at the same depth compared to a fluid with lower density, highlighting the direct impact that density has on the computation of pressure in fluids.For the given exercise, oil with a density of \( 600 \text{ kg/m}^3 \) is considered. This density is used in conjunction with other parameters to calculate the pressure at a certain depth below the surface.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases proportionally with depth because, as you descend deeper, the amount of fluid exerting weight above you increases.The calculations of hydrostatic pressure are vital in understanding pressure in fluids. The formula to find hydrostatic pressure is:\[ P = \rho \cdot g \cdot h \]When discussing hydrostatic pressure, it is important to note that this term often excludes atmospheric pressure. Thus, it represents the pressure produced solely by the fluid.This distinction is evident in problems that require calculation of fluid pressure below a surface, illustrating the importance of separating contributions from atmospheric and fluid pressures.
Gauge Pressure Calculation
Gauge pressure is the pressure of a fluid measured above atmospheric pressure. It reflects the pressure from the fluid only, excluding atmospheric pressure, which is common practice in many engineering fields where devices like gauges are used.To calculate gauge pressure, modify the total pressure equation:\[ P_{\text{gauge}} = \rho \cdot g \cdot h \]In the original exercise, to solve for the depth at which the gauge pressure (pressure due to the fluid column alone) reaches 1 bar, we set up the equation as follows:\[ 600 \cdot 9.81 \cdot h = 1 \times 10^5 \] Solving for \( h \) gives the depth required for the fluid pressure to reach the desired gauge pressure without surface atmospheric pressure contributing. This calculation is essential for understanding situations where only the pressure from the fluid is of interest, such as in pressurized tanks or fluid systems.

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Most popular questions from this chapter

Calculate the pressure in the ocean at a depth of \(2000 \mathrm{~m}\) assuming that salt water is \((a)\) incompressible with a constant density of \(1002 \mathrm{~kg} \mathrm{~m}^{-3},(b)\) compressible with a bulk modulus of \(2.05 \mathrm{GN} \mathrm{m}^{-2}\) and a density at the surface of \(1002 \mathrm{~kg} \mathrm{~m}^{-3}\). \(\left[\right.\) a) \(19.66 \mathrm{MN} \mathrm{m}^{-2}\), (b) \(\left.19.75 \mathrm{MN} \mathrm{m}^{-2}\right]\)

The barometric pressure of the atmosphere at sea level is equivalent to \(760 \mathrm{~mm}\) of mercury and its temperature is \(288 \mathrm{~K}\). The temperature decreases with increasing altitude at the rate of \(6.5 \mathrm{~K}\) per \(1000 \mathrm{~m}\) until the stratosphere is reached in which the temperature remains constant at \(216.5 \mathrm{~K}\). Calculate the pressure in millimetres of mercury and the density in kilograms per cubic metre at an altitude of \(14500 \mathrm{~m}\). Assume \(\boldsymbol{R}=287 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\).

What will be \((a)\) the gauge pressure, \((b)\) the absolute pressure of water at a depth of \(12 \mathrm{~m}\) below the free surface? Assume the density of water to be \(1000 \mathrm{~kg} \mathrm{~m}^{-3}\) and the atmospheric pressure \(101 \mathrm{kN} \mathrm{m}^{-2}\). \(\left[(a) 117.72 \mathrm{kN} \mathrm{m}^{-2}\right.\), (b) \(\left.218.72 \mathrm{kN} \mathrm{m}^{-2}\right]\)

A closed airtight tank \(4 \mathrm{~m}\) high and \(1 \mathrm{~m}\) in diameter contains water to a depth of \(3.3 \mathrm{~m}\). The air in the tank is at a pressure of \(40 \mathrm{kN} \mathrm{m}^{-2}\) gauge. What are the absolute pressures at the centre and circumference of the base of the tank when it is rotating about its vertical axis at a speed of \(180 \mathrm{rev} \mathrm{min}^{-1}\) ? At this speed the water wets the top surface of the tank.

A vessel \(1.4 \mathrm{~m}\) wide and \(2.0 \mathrm{~m}\) long is filled to a depth of \(0.8 \mathrm{~m}\) with a liquid of mass density \(840 \mathrm{~kg} \mathrm{~m}^{-3}\). What will be the force in \(\mathrm{N}\) on the bottom of the vessel \((a)\) when being accelerated vertically upwards at \(4 \mathrm{~m} \mathrm{~s}^{-1},(b)\) when the acceleration ceases and the vessel continues to move at a constant velocity of \(7 \mathrm{~m} \mathrm{~s}^{-1}\) vertically upwards?

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