A very large cylindrical tank vented to air contains a layer of oil on top of wa

Question

A very large cylindrical tank vented to air contains a layer of oil on top of water. At the bottom of the container there is a horizontal pipe of very small radius and sealed with a valve at the end. The depth of water is 8.0 m. The thickness of oil layer is unknown. Initially the valve is closed. The gauge pressure at the valve is 95,648 Pa. The density of the water and oil respectively are 1,000 and 880 kg/m^3. The air pressure is 101,300 Pa.

A) What is the unknown thickness of oil in the tank?

B) Find the true pressure at any point just below the water-oil interface:

C) When we open the valve, water comes from the valve with a speed v. Find the speed v, assuming that the water's speed just below the oil-water interface is negligibly small.

Explanation / Answer

air pressure=101300 Pa

using air pressure take a oil of height,

density*g*h=101300

h=11.734 m of oil

A)

then by pascal's law pressure at bottom will be

rho(oil)*g*(11.734+h')+rho(water)*g*8=95648+101300 where h' is required height

880*9.81*(11.734+h')+1000*9.81*8=95648+101300

h'=1.989 m

B)

true pressure just below oil=880*9.81*(11.734+h')

=118467.9 Pa

C)

apply bernoulie's equation

p^2/rho*g + v1^2/2g + z1 = p^2/rho*g + v2^2/2g +z2

here take air pressure on both side, so p term will be neglected, and v1 is 0, z1=8+1.98 , z2=0

then

v2^2/2g=9.98

v=13.99 m/s

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