Question

A certain ngid aluminum container contains a liquid at a gauge pressure off 0-2.02-10° Pa at sea level where the atmospheric pressure is P-1.01 x 105 Pa. The volume of the container is Vo-4.55 x 104m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is APmax 2.32 x 10% Pa. 1.20 kg /m and that the density of seawater For this problem, assume that the density of air maintains a constant value of pa maintains a constant value ofPs = 1025 kg / m3 25% Part (a) The container is taken from sea level, where the pressure ofair is P,-1.01x 105 Pa, to a higher altitude. What is the maximum height h in meters above the ground that the container can be lifted before bursting? Neglect the changes in temperature and acceleration due to gravity with altitude Grade Summary Deductions 0% Potential 100% sin0 coso tan0 cotan asin acoso Attempts remaining: 2 per attempt) detailed view tan) acotan sin0 cosh0 tanh cotanho ODegrees Radians Hints:-% deduction per hint. Hints remaining Feedback: 0% deduction per feedback. 25% Part (b) If we include the decrease in the density of the air with increasing altitude, what will happen? 25% Part (c) Choose the correct answer from the following options. ? 25% Part (d) What is the maximum depth d x in meters below the surface of the ocean that the container can be taken before imploding?

Answer 1

Answer:

a)    h = 1990 m, b) the height we can reach increases , c)  h = 2,329 m

Explanation:

.a) For this fluid mechanics problem let's use the pressure ratio

        P = P₀ + ρ g y

        P-P₀ = ρ g y

Where rho is the fluid density where it is contained, P₀ is the atmospheric pressure.

It does not indicate that the maximum pressure change before exporting is

             ΔP = 2.32 * (0.1 P₀)

Let's look for this pressure

            ΔP = 2.32   0.1    1.01 10⁵

            ΔP = 0.234 10⁵ Pa.

This is the maximum pressure difference, let's look for the height where we reach this value, the container has an internal pressure of 2.02 10 Pa, which we must subtract from the initial difference

            ΔP = ΔP - 2.02 10⁰ = 0.234 10⁵5 - 2.02

            ΔP = 23398 Pa

            ΔP = ρ g h

            h = ΔP /ρ g

            h = 23398 /1.20 9.8

            h = 1990 m

b) If air density decreases the height from 1,225 at sea level to 0.3629 to 11000m, the height we can reach increases

Density must be functioned with height

c) Now we introduce the contain in the ocean, the density of sea water is

            ρ = 1025 kg / m³

            h = 23398 / 1025 9.8

            h = 2,329 m