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Viefleur [7K]
2 years ago
6

A cylindrical tank with a radius of 2-m is filled with oil and water. The water has a density of rho = 1000 kg/m3 while the oil

has a density of rho = 800 kg/m^3 . If the depth of the water is 2.5 m, and the pressure difference between the top of the oil and the bottom of the water is 80 kPa, determine the depth of the oil, in m. Assume that gravity is 9.81 m/s^2 .
Engineering
1 answer:
Keith_Richards [23]2 years ago
8 0

Answer:

Height of oil is 7.06 meters.

Explanation:

The situation is shown in the attached figure

The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by

P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(\rho _{water}h_{water}+\rho_{oil}h_{oil})=P_{bottom}-P_{top}

Applying the given values we get

P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(1000\times 2.5+800\times h_{oil})=80\times 10^{3}\\\\\therefore 800\times h_{oil}=\frac{80\times 10^{3}}{9.81}-2500\\\\\therefore 800\times h_{oil}=5654.94\\\\\therefore h_{oil}=7.06m

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