## You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3

Question

You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 150 kPa , and the depth of the water will be 13.6 m . The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 kPa .
A) Find the net downward force on the tank’s flat bottom, of area 2.15 m^2 , exerted by the water and air inside the tank and the air outside the tank.

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3 years 2021-09-04T16:02:04+00:00 1 Answers 109 views -1

630.93 kN of force.

Explanation:

Pressure inside the tank is 150 kPa

The acceleration due to gravity on Mars g is 3.71 m/s^2.

The depth of water h is 13.6 m.

Pressure due to air outside tank is 93 kPa

The density of water p is 1000 kg/m^3

Pressure of the water on the tank bottom will be equal to pgh

Pressure of water = pgh

= 1000 x 3.71 x 13.6 = 50456 Pa

= 50.456 kPa.

Total pressure at the bottom of the tank will be pressure within tank and pressure due to water and pressure outside tank.

Pt = (150 + 50.456 + 93) = 293.456 kPa

Force at the bottom of the tank will be pressure times area of tank bottom.

F = Pt x A

F = 293.456 x 2.15 m^2 = 630.93 kN