Problem 21 - College Physics

Question

(a) Calculate the absolute pressure at the bottom of a fresh-water lake at a depth of \(27.5 \mathrm{~m}\). Assume the density of the water is \(1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and the air above is at a pressure of \(101.3 \mathrm{kPa}\). (b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of \(35.0 \mathrm{~cm}\) ?

Step-by-Step Solution

Answer

The absolute pressure at the bottom of the lake is \(370.8 \mathrm{kPa}\) and the force exerted by the water at this depth on the window is \(35.6 \mathrm{kN}\).

Step 1: Deriving absolute pressure at depth

Lets first calculate the absolute pressure at the bottom of the lake. This can be calculated by adding pressure due to the water column and the atmospheric pressure. The pressure due to a fluid column is given by the equation \(P = \rho g h\), where \(P\) is the fluid pressure, \(\rho\) is the fluid density, \(g\) is acceleration due to gravity and \(h\) is the height of the fluid column. Here, \(\rho = 1.00 \times 10^{3} \mathrm{~kg/m^3}\), \(g = 9.8 \mathrm{~m/s^2}\) (approximated value of gravity on Earth), and \(h = 27.5 \mathrm{~m}\). So, \(P_{Fluid} = \rho g h = 1.00 \times 10^{3} \mathrm{~kg/m^3} \times 9.8 \mathrm{~m/s^2} \times 27.5 \mathrm{~m} = 269.5 \mathrm{kPa}\). The atmospheric pressure \(P_{Atmospheric} = 101.3 \mathrm{kPa}\). Hence the absolute pressure \(P_{Absolute}\) at a depth of 27.5 m can be calculated by adding the atmospheric pressure to the fluid pressure. So, \(P_{Absolute} = P_{Fluid} + P_{Atmospheric} = 269.5 \mathrm{kPa} + 101.3 \mathrm{kPa} = 370.8 \mathrm{kPa}\)

Step 2: Calculate the area of the window

Before determining the force exerted by the water on the window, let's first calculate the area of the window. Given the window is in circular shape, we use the area formula for a circle: \(\mathrm{Area} = \pi(\mathrm{Diameter}/2)^2\). Here, the diameter of the window is \(35.0 \mathrm{cm}\), so the radius is \(Diameter / 2 = 35.0 \mathrm{cm} / 2 = 17.5 \mathrm{cm} = 0.175 \mathrm{m}\), (converted from cm to m). So, \(\mathrm{Area} = \pi * (0.175 \mathrm{m})^2 = 0.096 \mathrm{m^2}\).

Step 3: Calculate the force exerted by the water

We can now calculate the force exerted by the water on the window. The force exerted by fluid pressure is given by the equation \(Force = Pressure \times Area\). Here, Pressure is the absolute pressure we calculated in Step 1 and Area is the area of the window we calculated in Step 2. Hence, \(Force = P_{Absolute} \times \mathrm{Area} = 370.8 \mathrm{kPa} \times 0.096 \mathrm{m^2} = 35.6 \mathrm{kN}\). As the pressure is high, Force is given in kilo Newtons (kN).

Key Concepts

Pressure Calculation

Understanding how to calculate pressure in fluid mechanics is crucial. Pressure is the force exerted by a fluid per unit area on an object's surface. In our exercise, we looked for the absolute pressure at the bottom of a lake. To find this, we add the pressure from the water column to the atmospheric pressure above the surface.

The pressure due to the water column can be calculated using the formula \(P = \rho gh\). Here, \(\rho\) stands for the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the height of the fluid column.

• Density of water \(\rho = 1.00 \times 10^3 \ \text{kg/m}^3\)
• Gravity \(g = 9.8 \ \text{m/s}^2\)
• Depth \(h = 27.5 \ m\)

This gives us the water pressure as 269.5 kPa. Adding the atmospheric pressure of 101.3 kPa, the absolute pressure at the lake's bottom becomes 370.8 kPa.

Fluid Statics

Fluid statics, also known as hydrostatics, deals with fluids at rest. The fundamental principle here is that any part of a fluid at rest is in a state of equilibrium. This means that any pressure applied to the fluid is transmitted uniformly in all directions within the fluid.

In this case, there is fluid (water) exerting pressure due to its weight at a certain depth. This pressure increases with depth because of the weight of the overlaying fluid.

• At greater depths, more water exerts force.
• This is why submarines are built to withstand high pressures.

Fluid statics is essential in designing anything that is intended to be submerged, as it emphasizes how pressures increase with depth.

Force on Submerged Surfaces

When designing structures like the window of an underwater vehicle, calculating the force exerted by the fluid on the submerged surface is crucial. To find this force, you multiply the pressure by the area of the surface.

In our exercise, we calculated:

• The area of a circular window: Using \(\text{Area} = \pi (\text{radius})^2\), we found the area to be 0.096 \(\mathrm{m^2}\) for a window with a 35 cm diameter.
• The force: Calculated using \(\text{Force} = \text{Pressure} \times \text{Area}\), resulting in 35.6 kN.

This exemplifies why precise calculations are needed in engineering. It ensures that structures can tolerate the immense pressures exerted by fluids at depth.