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An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to the local atmospheric pressure) in kPa inside of the diving chamber required to keep water from entering the chamber. (Assume SG=1.03)

User RTB
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2 Answers

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Final answer:

To keep water from entering an open diving chamber at 60 meters depth with sea water of specific gravity 1.03, the air pressure inside the chamber must be equal to the water pressure at that depth, which is 603.84 kPa.

Step-by-step explanation:

To calculate the air pressure (gage pressure) inside the diving chamber that prevents water from entering, we will use the pressure-depth relationship in fluids. The gage pressure is the pressure relative to atmospheric pressure. Given that the chamber is at a depth of 60 meters underwater and the specific gravity (SG) of the sea water is 1.03, we know that the sea water is denser than pure water (whose SG is 1). To prevent water from entering the chamber, the air pressure inside the chamber must be equal to the water pressure at that depth. The pressure due to a column of liquid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity (9.8 m/s²), and h is the height (or depth) of the liquid column.

Since the water's specific gravity is 1.03, its density (ρ) can be calculated as ρ = SG * ρwater, where ρwater is the density of pure water (1000 kg/m³). So, ρ = 1.03 * 1000 kg/m³ = 1030 kg/m³. The depth (h) is 60 meters.

Now, we can calculate the pressure:

P = ρgh = (1030 kg/m³)(9.8 m/s²)(60 m) = 603,840 Pa or N/m².

To convert this to kilopascals (kPa), we divide by 1,000:

P = 603.84 kPa

This is the gage pressure inside the chamber required to prevent water from entering.

User Connor Hollis
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4 votes

Answer:

Gauge Pressure required = 606.258 kPa

Step-by-step explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals


P(z)=P_(0)+\rho _wgh

Now the gauge pressure is given by


P(z)-P_(0)=\rho _wgh

Applying values we get


P(z)-P_(0)=\rho _wgh\\\\P_(gauge)=1.03* 1000* 9.81* 60Pa\\\\P_(gauge)=606258Pascals\\\\P_(gauge)=606.258kPa

User AudioDroid
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