91Ó°ÊÓ

A mass (or load) acting downward on a piston confines a fluid of density \(\rho\) in a closed container, as shown in Fig. \(13-2\). The combined weight of the piston and load on the right is \(200 \mathrm{~N}\), and the cross-sectional area of the piston is \(A=8.0 \mathrm{~cm}^{2}\). Find the total pressure at point- \(B\) if the fluid is mercury and \(h=25 \mathrm{~cm}\left(\rho_{\mathrm{Hg}}=13\right.\) \(600 \mathrm{~kg} / \mathrm{m}^{3}\) ). What would an ordinary pressure gauge read at \(B\) ? Recall what Pascal's principle tells us about the pressure applied to the fluid by the piston and atmosphere: This added pressure is applied at all points within the fluid. Therefore, the total pressure at \(B\) is composed of three parts: Pressure of the atmosphere \(=1.0 \times 10^{5} \mathrm{~Pa}\) Pressure due to the piston and load \(=\frac{F_{w}}{A}=\frac{200 \mathrm{~N}}{8.0 \times 10^{-4} \mathrm{~m}^{2}}=2.5 \times 10^{5} \mathrm{P}\) Pressure due to the height \(h\) of fluid \(=h p g=0.33 \times 10^{5} \mathrm{~Pa}\) In this case, the pressure of the fluid itself is relatively small. We have Total pressure at \(B=3.8 \times 10^{5} \mathrm{~Pa}\) The gauge pressure does not in clude atmospheric pressure. Therefore, Gauge pressure at \(B=2.8 \times 10^{5} \mathrm{~Pa}\)

Step by step solution

01

Calculate pressure due to piston and load

To find the pressure due to the piston and load, use the formula: \[ P = \frac{F}{A} \]where \( F = 200 \, \mathrm{N} \) and \( A = 8.0 \, \mathrm{cm}^2 \). Convert the area from \( \mathrm{cm}^2 \) to \( \mathrm{m}^2 \): \( 8.0 \, \mathrm{cm}^2 = 8.0 \times 10^{-4} \, \mathrm{m}^2 \). Therefore, the pressure is:\[ P = \frac{200}{8.0 \times 10^{-4}} = 2.5 \times 10^5 \, \mathrm{Pa} \]

02

Calculate pressure due to the fluid's height

The pressure due to the height of the mercury column can be found using the formula:\[ P = h \rho g \]where \( h = 0.25 \, \mathrm{m} \), \( \rho = 13600 \, \mathrm{kg/m}^3 \), and \( g = 9.8 \, \mathrm{m/s}^2 \). Calculating the pressure:\[ P = 0.25 \times 13600 \times 9.8 = 33320 \, \mathrm{Pa} \approx 0.33 \times 10^5 \, \mathrm{Pa} \]

03

Add atmospheric pressure

Include the atmospheric pressure as given:\[ P_{\text{atm}} = 1.0 \times 10^5 \, \mathrm{Pa} \]

04

Calculate total pressure at point B

Add all the pressures calculated: the atmospheric pressure, the pressure due to the piston and load, and the pressure due to the fluid's height:\[ P_{\text{total}} = P_{\text{atm}} + P_{\text{piston}} + P_{\text{fluid}} \]\[ P_{\text{total}} = 1.0 \times 10^5 + 2.5 \times 10^5 + 0.33 \times 10^5 = 3.83 \times 10^5 \, \mathrm{Pa} \approx 3.8 \times 10^5 \, \mathrm{Pa} \]

05

Determine gauge pressure at point B

Gauge pressure does not include atmospheric pressure. Subtract atmospheric pressure from the total pressure:\[ P_{\text{gauge}} = P_{\text{total}} - P_{\text{atm}} = 3.8 \times 10^5 - 1.0 \times 10^5 = 2.8 \times 10^5 \, \mathrm{Pa} \]

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Key Concepts

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

Pressure Calculation

To understand pressure calculation in fluid mechanics, we start with the basic formula for pressure, given by \[ P = \frac{F}{A} \]where:

• \( P \) is pressure.
• \( F \) is the force applied perpendicular to the surface.
• \( A \) is the area over which the force is distributed.

In the context of fluids, pressure changes with depth because of the weight of the liquid above. For example, if you have a piston applying a force of 200 N on a fluid with a cross-sectional area of 0.0008 square meters, the pressure at this point due to the piston alone can be calculated as:\[ P = \frac{200 \, \text{N}}{8.0 \times 10^{-4} \, \text{m}^2} = 2.5 \times 10^5 \, \text{Pa} \]This explains how even a moderate force can create significant pressure, especially if the area is small.

Pascal's Principle

Pascal's principle is an essential concept in fluid mechanics. It states that any pressure applied to a confined fluid is transmitted equally in all directions throughout the fluid. This principle is what underpins the calculation of pressure in a closed container, like when a piston exerts force on a liquid.
For example, when the piston compresses the fluid, the pressure increase is distributed evenly at all points in the fluid. So, if the atmospheric pressure is 100,000 Pa and the pressure due to the piston is 250,000 Pa, this new pressure level is uniform throughout the fluid.
Pascal's principle allows us to predict the pressure at any location within the fluid, which is very important for devices like hydraulic systems that rely on fluid pressure to operate efficiently.

Gauge Pressure

Gauge pressure is the pressure relative to atmospheric pressure. It is calculated by subtracting atmospheric pressure from the total pressure in the system.
In situations where the absolute pressure includes atmospheric pressure, gauge pressure only considers what is additionally applied.
For instance, if the total pressure at point B in a fluid is 380,000 Pa and the atmospheric pressure is 100,000 Pa, you find the gauge pressure using the formula:\[ P_{\text{gauge}} = P_{\text{total}} - P_{\text{atm}} \]\[ P_{\text{gauge}} = 3.8 \times 10^5 \, \text{Pa} - 1.0 \times 10^5 \, \text{Pa} = 2.8 \times 10^5 \, \text{Pa} \]This is useful in practical applications where only the pressure above atmospheric pressure is needed, such as in tire or blood pressure measurements.

Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of air in the Earth's atmosphere. It typically has a standard value of approximately 101,325 Pa, commonly rounded to 100,000 Pa for easier calculations.
Understanding atmospheric pressure is crucial as it is the baseline from which we measure other pressures. In fluid mechanics problems, it often acts as the reference point for calculating the total or absolute pressure within a system.
When you calculate the total pressure at a point in a fluid, like point B in the exercise, you include atmospheric pressure along with pressures resulting from the fluid's weight and any applied forces. Thus, in our example:\[ P_{\text{total}} = P_{\text{atm}} + P_{\text{other}} \]where \( P_{\text{other}} \) includes the combined influence of piston, weight, and fluid height, showcasing how various elements come together to form the complete picture of pressure dynamics.