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An unknown immiscible liquid seeps into the bottom of an openoil
tank. Some measurements indicate that the depth of theunkown liquid
is 1.5m and the depth of the oil ( specific weight =8.5
kN/m3) floating on top is 5m. A pressure gageconnected
to the bottom of the tank reads 65 kPa. What is thespecific gravity
of the unkown liquid?

Respuesta :

Answer:

1.53

Explanation:

Given:

depth of the unknown liquid  = 1.5m

the depth of the oil floating on top = 5m

specific weight of oil, γ = 8.5 kN/m³

Total pressure at the bottom = 65 kPa = 65 kN/m²

let the specific weight of the unknown liquid be " γ' "

Now,

The total pressure

= Pressure due to the unknown liquid + Pressure due to floating liquid

or

65 = γ' × 1.5 + γ × 5

or

65 = γ' × 1.5 + 8.5 × 5

or

22.5 = γ' × 1.5

or

γ' = 15 kN/m³

Also,

Specific gravity = [tex]\frac{\textup{Specific weight of unknown liquid}}{\textup{Specific weight of water}}[/tex]

specific weight of water = 9.81 kN/m³

or

Specific gravity = [tex]\frac{15}{\9.81}[/tex]  = 1.53

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