# How Does Bubble Size Change with Depth in Fluid Mechanics?

• cat_eyes
In summary, the depth of the diver can be calculated by using the equation P = Pair + (rho*g*h) and the absolute pressure at this depth can be determined by using the equation P = Pair + (rho*g*h), where P is the pressure at the depth of the diver, Pair is the atmospheric pressure, rho is the density of sea water, g is the acceleration due to gravity, and h is the depth of the diver. Substituting in the given values, the depth of the diver is 97.8 m and the absolute pressure at this depth is 101325 Pa.
cat_eyes
An air bubble originating from a deep-sea diver has a radius of 1.8 mm at the depth of the diver. When the bubble reaches the surface of the water, it has a radius of 2.6 mm. Assume that the temperature of the air in the bubble remains constant. The acceleration of gravity is 9.81 m/s^2.

A) Determine the depth of the diver. Answer in units of m.

B) Determine the absolute pressure at this depth. Answer in units of Pa.

P=rho(density)
Pair=1.29 kg/m^3
Psea water=1025 kg/m^3
The volume of a sphere = (4/3)(Pi)(r^3)

Any help with this would be greatly appreciated.

Last edited:
A) The depth of the diver can be determined by using the equation P = Pair + (rho*g*h), where P is the pressure at the depth of the diver, Pair is the atmospheric pressure, rho is the density of sea water, g is the acceleration due to gravity, and h is the depth of the diver.Rearranging this equation and solving for h gives: h = (P - Pair)/(rho*g).Substituting in the given values, the depth of the diver is:h = (101325 Pa - 1.29 kg/m^3*9.81 m/s^2)/(1025 kg/m^3*9.81 m/s^2) = 97.8 mB) The absolute pressure at this depth can be determined by using the equation P = Pair + (rho*g*h), where P is the pressure at the depth of the diver, Pair is the atmospheric pressure, rho is the density of sea water, g is the acceleration due to gravity, and h is the depth of the diver.Substituting in the given values, the absolute pressure at this depth is:P = 1.29 kg/m^3*9.81 m/s^2 + 1025 kg/m^3*9.81 m/s^2*97.8 m = 101325 Pa

A) To determine the depth of the diver, we can use the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and inversely proportional to its volume. Since the temperature remains constant, we can use the equation P1V1 = P2V2, where P1 and V1 are the pressure and volume at the depth of the diver, and P2 and V2 are the pressure and volume at the surface.

We know that the volume of a sphere is (4/3)(Pi)(r^3), so we can set up the equation as follows:

(Pair)(4/3)(Pi)(1.8^3) = (Pair)(4/3)(Pi)(2.6^3)

Solving for the depth, we get:

Depth = (Pair)(1.8^3)/(Pair)(2.6^3) = 0.52 m

Therefore, the depth of the diver is 0.52 m.

B) To determine the absolute pressure at this depth, we can use the equation P = Pair + Psea water + Pdepth, where Pair is the pressure of the air at the surface, Psea water is the pressure of the sea water at the surface, and Pdepth is the pressure due to the depth of the diver.

We know that Pair = 1 atm = 101325 Pa, and Psea water = (1025 kg/m^3)(9.81 m/s^2)(0.52 m) = 5280 Pa.

To find Pdepth, we can use the equation Pdepth = (Pair)(1.8^3)/(2.6^3) = (101325 Pa)(1.8^3)/(2.6^3) = 44935 Pa.

Therefore, the absolute pressure at this depth is:

P = 101325 Pa + 5280 Pa + 44935 Pa = 151540 Pa

Therefore, the absolute pressure at this depth is 151540 Pa.

## 1. What is fluid mechanics?

Fluid mechanics is the study of how fluids, such as liquids and gases, behave and interact with their surroundings. This includes analyzing the forces and motion of fluids, as well as their properties and characteristics.

## 2. What are some common applications of fluid mechanics?

Fluid mechanics has many practical applications in our daily lives, such as in the design of airplanes, cars, and ships. It is also important in understanding weather patterns, ocean currents, and the movement of blood and other bodily fluids in living organisms.

## 3. What are the main principles of fluid mechanics?

The main principles of fluid mechanics include the conservation of mass, momentum, and energy, as well as the concepts of pressure, viscosity, and turbulence. These principles are used to analyze and solve problems related to fluid behavior.

## 4. How does fluid mechanics relate to other branches of science and engineering?

Fluid mechanics is closely related to other branches of science and engineering, such as thermodynamics, mechanics, and chemistry. It also has applications in fields such as aerospace engineering, chemical engineering, and biomedical engineering.

## 5. What are some common challenges when solving fluid mechanics problems?

Some common challenges when solving fluid mechanics problems include accurately modeling and predicting the behavior of fluids under different conditions, dealing with complex flow patterns and turbulence, and considering the effects of factors such as temperature and pressure on fluid properties.

• Introductory Physics Homework Help
Replies
25
Views
2K
• General Discussion
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
6K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
811
• Introductory Physics Homework Help
Replies
16
Views
2K