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Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.
Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.
Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.
Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.
Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?
Answer:
a
The expression is [tex]F_{No} = A [P_{No} - \frac{P_{sea}}{2}][/tex]
b
[tex]F_{No}= 7771.125 \ N[/tex]
c
[tex]F_p = 2.2*10^{6} N[/tex]
d
From the value obtained we can say the that the force required to open the lid is higher at Denver
Explanation:
The altitude of container in Denver is [tex]d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m[/tex]
The surface area of the container lid is [tex]A = 0.0155m^2[/tex]
The altitude of container in New Orleans is sea-level
The air pressure in Denver is [tex]P_D = 79000 \ Pa[/tex]
The air pressure in new Orleans is [tex]P_{ro} = 100250 \ Pa[/tex]
Generally force is mathematically represented as
[tex]F_{No} = \Delta P A[/tex]
So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would
The pressure at sea level is a constant with a value of
[tex]P_{sea} = 101000 Pa[/tex]
So the [tex]\Delta P[/tex] which is the difference in pressure within and outside the container is
[tex]\Delta P = P_{No} - \frac{P_{sea}}{2}[/tex]
Therefore
[tex]F_{No} = A [P_{No} - \frac{P_{sea}}{2}][/tex]
Now substituting values
[tex]F_{No} = 0.0155 [100250 - \frac{101000}{2}][/tex]
[tex]F_{No}= 7771.125 \ N[/tex]
The force to remove the lid in Denver is
[tex]F_p = \Delta P_d A[/tex]
So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would
The pressure at sea level is a constant with a value of
[tex]P_{sea} = 101000 Pa[/tex]
At sea level the air pressure in Denver is mathematically represented as
[tex]P_D = \rho g h[/tex]
=> [tex]g = \frac{P_D}{\rho h}[/tex]
Let height at sea level is h = 1
The air pressure at height [tex]d_D[/tex]
[tex]P_d__{D}} = \rho gd_D[/tex]
=> [tex]g = \frac{P_d_D}{\rho d_D}[/tex]
Equating the both
[tex]\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}[/tex]
[tex]P_d_D = P_D * d_D[/tex]
Substituting value
[tex]P_d__{D}} = 1828.2 * 79000[/tex]
[tex]P_d__{D}} = 1.445*10^{8} Pa[/tex]
So
[tex]\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}[/tex]
=> [tex]\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}[/tex]
[tex]\Delta P_d = 1.44*10^{8}Pa[/tex]
So
[tex]F_p = \Delta P_d A[/tex]
[tex]= 1.44*10^8 * 0.0155[/tex]
[tex]F_p = 2.2*10^{6} N[/tex]
Following are the solution to the given points:
Solution:
In Denver, the container's altitude is:
[tex]\to d_D = 6000 \ ft = 6000 \times 0.3048 = 1828.8\ m[/tex]
The surface of a container lid:
[tex]\to A = 0.0155\ m^2[/tex]
Its container's elevation in Orleans is now at sea level.
The air density in Denver:
[tex]\to P_D = 79000 \ Pa[/tex]
The air pressure at New Orleans:
[tex]\to P_{ro} = 100250 \ Pa[/tex]
In general, force is mathematically described as
[tex]\to F_{No} = \Delta P A[/tex]
And we're told that the level within the box is half that of that at sea level, and as such, the pressure acting just on the tank would've been Its pressure at water level is steady with a value of
[tex]\to P_{sea} = 101000 \ Pa[/tex]
So that the [tex]\Delta P[/tex], which is the pressure difference between the inside and outside of the vessel, is
[tex]\to \Delta P = P_{No} - \frac{P_{sea}}{2}[/tex]
So
[tex]\to F_{No} = A [P_{No} - \frac{P_{sea}}{2}][/tex]
Numbers are now substituted.
[tex]\to F_{No} = 0.0155 [100250 - \frac{101000}{2}] = 7771.125 \ N[/tex]
In Denver, there is indeed a strong campaign to remove the lid.
[tex]\to F_p = \Delta P_d A[/tex]
Therefore, we're told that its pressure within the container is twice that of at sea level, so the pressure pressing on the containers would've been equal to that. Above sea level, the pressure is indeed continuous with such a value of
[tex]\to P_{sea}= 101000\ Pa\\\\[/tex]
Denver's air pressure at sea level is described mathematically as
[tex]\to P_D = \rho g h\\\\\to g = \frac{P_D}{\rho h}\\\\[/tex]
Assume that h = 1 is the elevation at sea level. [tex]d_D[/tex] is the elevation of the airflow.
[tex]\to P_d__{D}} = \rho gd_D\\\\\to g = \frac{P_d_D}\\\\\to {\rho d_D}\\\\\to \rho \ d_D[/tex]is an abbreviation for Rhodium Dioxide.
Considering both
[tex]\to \frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}\\\\\to P_d_D = P_D \times d_D\\\\[/tex]
Substituting a different value:
[tex]\to P_{d_{D}} = 1828.2 \times79000\\\\\to P_{d_{D}} = 1.445 \times10^{8} Pa\\[/tex]
So
[tex]\to \Delta P_d = P_{d} _D - \frac{P_{sea}}{2}\\\\\to \Delta P_d = 1.445 \times10^{8} - \frac{101000}{2}\\\\\to \Delta P_d = 1.44 \times10^{8}Pa \\\\[/tex]
So
[tex]\to F_p = \Delta P_{d A}= 1.44 \times 10^8 \times 0.0155\\\\\to F_p = 2.2 \times 10^{6} N - \frac{P_{sea}}{2}\\\\\to \Delta P_d = 1.445 \times 10^{8} - \frac{101000}{2}\\\\\to \Delta P_d = 1.44 \times 10^{8}Pa\\\\[/tex]
So
[tex]\to F_p = \Delta P_{d A}= 1.44 \times 10^8 \times 0.0155= 2.2 \times 10^{6} N[/tex]
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