Problem 15 - Fundamentals of Physics

Question

What gaugc pressure must a machine produce in order to suck mud of density \(1800 \mathrm{~kg} / \mathrm{m}^{3}\) up a tube by a height of \(1.5 \mathrm{~m} ?\)

Step-by-Step Solution

Answer

The gauge pressure needed is 26,487 Pa.

Step 1: Understand the problem

We need to find the gauge pressure required to lift mud with a density of \(1800 \text{ kg/m}^3\) up a height of \(1.5\text{ m}\). This involves calculating how much pressure is needed to counteract the weight of the mud column.

Step 2: Use the hydrostatic pressure formula

The pressure required to lift a column of fluid can be calculated using the hydrostatic pressure formula: \[ P = \rho \cdot g \cdot h \] Where \( P \) is the pressure, \( \rho \) is the density of the fluid (\(1800 \text{ kg/m}^3\)), \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)), and \( h \) is the height (\(1.5 \text{ m}\)).

Step 3: Substitute values into the formula

Substitute the known values into the hydrostatic pressure formula:\[ P = 1800 \cdot 9.81 \cdot 1.5 \]

Step 4: Calculate the pressure

Perform the multiplication:\[ P = 1800 \times 9.81 \times 1.5 = 26,487 \text{ Pa} \]So, the gauge pressure required is \(26,487 \text{ Pa}\).

Key Concepts

Gauge Pressure

Gauge pressure is a measure of pressure relative to the local atmospheric pressure. Unlike absolute pressure, which accounts for the pressure exerted by the atmosphere, gauge pressure only considers the pressure generated above this local atmospheric level.
For instance, when a machine needs to lift mud through a tube, it must produce sufficient gauge pressure to overcome the weight of the mud and the atmospheric pressure acting on it.
When calculating the gauge pressure required to lift a fluid, the objective is to find how much additional pressure needs to be applied by the machine. This is essential for any system operating under conditions where the external atmosphere also acts on the substance being moved.

• Gauge pressure is convenient because many instruments measure pressure in this way, simplifying practical applications.
• It's crucial in fields like fluid mechanics and engineering where it aids in designing systems capable of lifting or moving liquids efficiently.
• In our problem, the gauge pressure of 26,487 Pa indicates how much extra pressure the machine must exert to elevate the mud 1.5 meters upward.

Fluid Density

Fluid density is an essential characteristic of a fluid, describing how much mass a fluid has in a given volume. It's expressed as \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. The density is typically expressed in units of kilograms per cubic meter (kg/m³).
This property significantly influences the behavior of the fluid under various pressures and forces, such as in our tubing scenario for mud lifting.
In the given exercise, the mud has a density of 1800 kg/m³.

• High density usually means more mass in the same volume compared to a fluid with lower density.
• Affects the gravitational pull on the fluid, as denser fluids require more pressure to lift the same height.
• Understanding fluid density is crucial in applications where substances must be moved, such as in piping systems and pumps.

In our example, knowing the density helps compute how much pressure is needed to lift the mud, as denser substances naturally exert more gravitational force.

Pressure Calculation

Calculating pressure involves determining the force exerted over a particular area, and when dealing with fluids, one often uses the hydrostatic pressure formula: \[ P = \rho \cdot g \cdot h \] where \( \rho \) is fluid density, \( g \) is acceleration due to gravity, and \( h \) is height or depth.
In this problem, substituting the known values allows us to find the specific gauge pressure required.
When solving the problem, we calculate:

• First, understand each component: \( \rho = 1800 \, \text{kg/m}^3 \), \( g = 9.81 \, \text{m/s}^2 \), \( h = 1.5 \, \text{m} \).
• Substituting values gives \( P = 1800 \times 9.81 \times 1.5 \).
• Calculate the product to get a final gauge pressure of 26,487 Pa.

This calculation is crucial for any such lifting system, providing insight into how much force must be generated by a machine or a pump to accomplish the task efficiently.