Fluid Statics/depth in a cylinder

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The discussion revolves around calculating the depth of mercury in a cylindrical container filled with water, given a total pressure of 1.3 atmospheres at the bottom. The user is attempting to apply fluid statics equations but is struggling to combine them effectively. Key equations include the total pressure equation and the individual pressures of water and mercury based on their densities and heights. The solution involves recognizing that the height of the water can be expressed as the total height of the cylinder minus the height of the mercury. The user ultimately gains clarity on how to proceed with the calculations.
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Homework Statement


Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder is 0.8 m tall and the pressure at the bottom is 1.3 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36
multiply.gif
104 kg/m3.)

Homework Equations


Ptotal=Patmosphere+Pwater+Pmercury
P=density x g x height
1 atmosphere=1.013e5 Pa
density of water=1 x 103 kg/m3
pressure of atmosphere=1 atm

The Attempt at a Solution


I first subtracted out the pressure of the atmosphere, so
.3 atm=Pwater + Pmercury
.8m=Hwater + Hmercury
Pmercury=1.36 x 104 x 9.8 x Hmercury
Pwater=1 x 103 x 9.8 x Hwater

I am having difficulty putting all 4 of these equations together. I feel like there is substitution needed, but which equations should I use?
 
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Perhaps you should be more systematic, Starting with your equation
Ptotal=Patmosphere+Pwater+Pmercury
substitute
Pwaterwaterg hwater
Pmercurymercuryg hmercury
and solve for hmercury.
 
tennisgirl92 said:

Homework Statement


Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder is 0.8 m tall and the pressure at the bottom is 1.3 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 View attachment 195815 104 kg/m3.)

Homework Equations


Ptotal=Patmosphere+Pwater+Pmercury
P=density x g x height
1 atmosphere=1.013e5 Pa
density of water=1 x 103 kg/m3
pressure of atmosphere=1 atm

The Attempt at a Solution


I first subtracted out the pressure of the atmosphere, so
.3 atm=Pwater + Pmercury
.8m=Hwater + Hmercury
Pmercury=1.36 x 104 x 9.8 x Hmercury
Pwater=1 x 103 x 9.8 x Hwater

I am having difficulty putting all 4 of these equations together. I feel like there is substitution needed, but which equations should I use?
You have 4 linear algebraic equations in 4 unknowns. Are you familiar with Gaussian elimination?
 
kuruman said:
Perhaps you should be more systematic, Starting with your equation
Ptotal=Patmosphere+Pwater+Pmercury
substitute
Pwaterwaterg hwater
Pmercurymercuryg hmercury
and solve for hmercury.

Ok, if I do that
.3atm=(1000 x 9.8 x Hwater) + (1.36 x 104 x 9.8 x Hmercury)
I still need to find the height of water, right? Where would that come from?
 
Chestermiller said:
You have 4 linear algebraic equations in 4 unknowns. Are you familiar with Gaussian elimination?
No, never heard of that. What is Gaussian elimination?
 
tennisgirl92 said:
I still need to find the height of water, right? Where would that come from?
Since the cylinder is filled to the top,
Height of water = height of cylinder - height of mercury
 
kuruman said:
Since the cylinder is filled to the top,
Height of water = height of cylinder - height of mercury

thank you! Got it!
 
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