You are designing a diving bell to withstand the pressure of seawater at a depth

Question

You are designing a diving bell to withstand the pressure of seawater at a depth of 250m .

Part A

What is the gauge pressure at this depth? (You can ignore the small changes in the density of the water with depth.)

Part B

At the 250m depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 40.0cm in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You may ignore the small variation in pressure over the surface of the window.)

Explanation / Answer

well, the pressure due to depth equation is (rho)*g*h

where:
rho = density of the fluid (seawater, a little more than 1, say 1.003 kg/m^3)
g = acceleration due to gravity (9.8 m/s^2)
h = height of water column. (250m)

gage pressure at surface is zero (h = 0)

gage pressure at depth h is rho*g*h =1.003*9.8*250=2457.35pa

given the diameter, you can find the area of the window

A = pi*d^2/4 =0.1256

then Force is pressure / area=19564.88

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