Air Bubble Velocity in Clean Water - 10 ft/1ft³ in 70-80F

[ENRG]

Hello everyone:

Please give me some guidance on the measurement of air bubble velocity in clean water (not salt water); the water depth would be not more than 10 feet: for simplicity use a bubble capacity of one cubic foot; consider the water temp to be 70 to 80 degrees farrenheit; Assume also that the air bubble temp is the same as the water; I know you'll want to give formulas which is OK but calculate an example where said bubble travels 10 ft. how long does it take? Would the velocity be the same for one cubic meter?:

 

[LightHero]

The velocity of an air bubble in clean water is determined by the equation: V = (2gH)^0.5where V is the velocity, g is the gravitational constant (9.8 m/s2), and H is the height of the bubble (in meters). For a bubble with a volume of one cubic foot, traveling 10 ft (3.048 m) in water with a temperature of 70-80°F (21-26.7°C), we can calculate the velocity as follows:V = (2*9.8*3.048)^0.5 = 5.84 m/sTherefore, it would take the bubble 0.52 seconds to travel 10 ft. The velocity would be the same for one cubic meter, since the equation remains the same.

 

[charng]

Thank you for your question. The velocity of an air bubble in clean water can be calculated using the following formula:

V = (2gΔρd)^1/2

Where:
V = velocity of the bubble (m/s)
g = acceleration due to gravity (9.8 m/s^2)
Δρ = density difference between the air bubble and water (kg/m^3)
d = diameter of the bubble (m)

Assuming a one cubic foot bubble of air at a temperature of 70-80°F, we can calculate the density difference using the ideal gas law:

Δρ = (P1-P2)/(R*T)

Where:
P1 = pressure of the air bubble (atmospheres)
P2 = pressure of the surrounding water (atmospheres)
R = gas constant (0.0821 L*atm/mol*K)
T = temperature (K)

For simplicity, let's assume the bubble is at a depth of 10 feet (3.048 meters) and the surrounding water pressure is 1 atmosphere. Using the ideal gas law, the pressure of the air bubble would be approximately 1.1 atmospheres. Plugging these values into the density difference equation, we get:

Δρ = (1.1-1)/(0.0821*298) = 0.00005 kg/m^3

Now, we can plug this value into the velocity formula:

V = (2*9.8*0.00005*0.3048)^1/2 = 0.0034 m/s

Therefore, for a one cubic foot bubble of air at a temperature of 70-80°F, it would take approximately 2,941 seconds (or 49 minutes) to travel a distance of 10 feet in clean water.

For a one cubic meter bubble at the same temperature and depth, the velocity would be the same as long as the diameter of the bubble remains the same. However, if the diameter of the bubble were to increase, the velocity would also increase.

I hope this helps answer your question. Please let me know if you have any further inquiries.