A large, open at the top, upright cylindrical tank contains fuel oil with a density of 0.890 ✕ 103 kg/m3.
(a) If the air pressure is 101.3 kPa, determine the absolute pressure (in Pa) in the fluid at a depth of 28.0 m.

(b) Determine the force (in N) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 3.20 cm.

Respuesta :

Answer:

a) Pabs = 345.7 [kPa]

b) F = 278.08 [N]

Explanation:

Manometric pressure in liquids is defined as the product of density by gravitational acceleration by the height of the liquid.

[tex]P_{g} =Ro*g*h[/tex]

where:

Pg = manometric pressure [Pa]

Ro = density = 0.890*10³[kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = liquid height = 28 [m]

Now the absolute pressure is defined as the sum of the atmospheric pressure plus the manometric pressure

a)

[tex]P_{abs}=P_{man}+P_{gau}\\P_{abs}= 101300 + (890*9.81*28)\\P_{abs} = 345765.2 [Pa] = 345.7 [kPa][/tex]

b) To determine the force at that point

The pressure is now defined as the relationship of a force over the area.

[tex]P=F/A\\F = P*A\\[/tex]

But the area of a circle can be calculated as follows:

[tex]A=\frac{\pi }{4} *(0.032)^{2}\\A = 0.0008042[m^{2}] \\F = 345765.2*0.0008042\\F = 278.08 [N][/tex]

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