HW 8 Begin Date: 3/27/2018 11:59:00 AM Due Date: 4/5/2018 11:59:00 PM End Date:

Question

HW 8 Begin Date: 3/27/2018 11:59:00 AM Due Date: 4/5/2018 11:59:00 PM End Date: 5/12/2018 11:59:00 PM (6%) Problem 12: A certain rigid aluminum container contains a liquid at a g ge prcwe of P,-2.02-105 Pa at sea level where the atmospheric pressure is p.-1.01m 10s Pa The volume ofthe contanaris .s-m" ms The maximum difference between the pressure inside and outside that this particular container can w ied beine bunt er tmploding taPma.-2.54 x o' Pa For this problem, assume that the density of air maintains a constant valus of-120 kg/m and that the density of seawater maintains a constant value op, 1025kg/m ia 25% Part (a) The container is taken from sea level, where the pressure of air is Pa 1 .01 × 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. 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. The depth below the surface of the ocean the container can reach before imploding is the same as the altitude in the atmosphere the container can reach before exploding. OThe depth below the surface of the ocean the container can reach befor exploding is greater than the altitude in the atmosphere the container can reach before imploding OThe depth below the surface of the ocean the container can reach before imploding is greater than the altitude in the atmosphere the container can reach before exploding. OThe depth below the surface of the ocean the container can reach before imploding is less than the altitude in the atmosphere the container can reach before exploding. OThe depth below the surface of the ocean the container can reach before exploding is the same as the altitude in the atmosphere the container can reach before imploding OThe depth below the surface of the ocean the container can reach before exploding is less than the altitude in the atmosphere the container can reach before imploding There is not enough information. Submission History 25% Part (d) What is the maximum depth dmax in meters below the surface of the ocean that the container can be taken before imploding? All content 2018 Expert TA

Explanation / Answer

given

Po = 2.02*10^5 Pa

Pa = 1.01*10^5 Pa

Vo = 3.8*10^-4 m^3

dP max = 2.54*10^5 Pa

density of air, rhoa = 1.2 kg/m^3

rhos = 1025 kg/m^3

a. let the altitude be h

then , pressure at that altitude must be P

where P = Pa - (dPmax - Po)

now, Pa = rho*g*H ( where H is macximum height of atmospherre)

P = rho*g(H - h) = rho*gH - rho*gh = Pa - rho*gh

Pa - rho*gh = Pa - dPmax + Po

h = (dPmax - Po)/rho*g = 4417.261297995242949 m

b. if we include decrease of density due to increase in height

dP = -rho*g*dh

now,

P = -rho*R*T

hence

dP/P = -g*dh/RT

hence

ln(P) = -gh/RT + K

at h = 0, P = Pa

ln(Pa) = K

hence

ln(P/Pa) = -gh/RT

P = Pa*e^(-gh/RT)

hence

Pa - (dPmax - Po) = Pa*e^(-gh/RT)

R = 287.058

hence

h = 6307.184100727 m

c. depth below surface of ocean will be less than the altitude as density of water is higher

let it be d

then

1025*g*d = Po + dPmax

d = 45.34944432 m

also, since the pressure is greatest at the ocean, the box will never implode in air, it will always explode

hence option d.

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