You are assigned the design of a cylindrical, pressurized water tank for a futur

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 meters per second per second. The pressure at the surface of the water will be 110 kPa , and the depth of the water will be 14.2 m . The pressure of the air in the building outside the tank will be 94.0 kPa .

Find the net downward force on the tank's flat bottom, of area 2.10 m2 , exerted by the water and air inside the tank and the air outside the tank.

Explanation / Answer

here,

accelration due to gravity , g = 3.71 m/s^2

pressure difference , P = 110 kPa - 94 Kpa

P = 16 KPa

depth of water , d = 14.2 m

area of the tank's bottom , a = 2.1 m^2

Pfluid = pgh

Pfluid is pressure in Pa or N/m²
is the density of the fluid in kg/m^3

density of water at 20C = 0.998 g/cm^3 = 998 kg/m^3

g is the acceleration of gravity 3.71 m/s^2

Pfluid = p * g * h = (998)(3.71)(14.2) = 52576 Pa = 52.58 kPa

Add to that 110 kPa to get 162.58 kPa or kN/m^2

and subtract the outside air pressure to get 68.58 kPa or kN/m^2

net force , F = 68.58 kN/m^2* 2.1 m^2 = 144.48 kN

the net downward force on the tank's flat bottom is 144.48 KN

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