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2

# Floating Cone

6 replies to this topic
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### #1 Dudesons123

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Posted 11 August 2017 - 05:54 AM

Hi Scientists/Engineers,

I just want to know if I'm tackling this problem in the right way or if I don't please help me. I was only given the density of mercury at 13600kgm^-3.

Saturated steam at 2 bars enters the top of a sealed cylindrical cavity and condenses, forming a pool of condensate. At the bottom of the cylinder is an exit pipe of diameter 5 cm. This exit pipe is open to the atmosphere, which consists of air at 1 bar. A cone of mass 1 kg blocks this exit, preventing any condensate from draining from the cylinder. The height of the cone that protrudes into the drain is 5 cm.

The volume of a cone of height and base radius is πR2H/3.

A manometer which is open to the atmosphere contains 50 cm of mercury. If the condensate level is equal to 10cm, what if I now ask for the liquid level h where the cone will begin to float, allowing the condensate to exit the cylinder? For this problem, it's ok to ignore any friction that might occur between the cone and the orifice.

I tried equating the weight of the cone which is 9.8N to the weight of the condensate displaced (ρgV). After rearranging for V, I got V= 10^-3 m^-3. So my unknown is height and the radius of the truncated cone in water?

Working:

W = mg = 9.8N = ρgV

=> V = 10^-3 m^3

#### Attached Files

Edited by Dudesons123, 14 August 2017 - 05:50 AM.

### #2 Dudesons123

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Posted 11 August 2017 - 06:12 AM

I also know that this is related to Pascal's law and Archimedes's principle.

### #3 MrShorty

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Posted 11 August 2017 - 09:38 AM

So, it looks like you are at the point where you have V=1E-3 m3=πh(r^2)/3 where you know neither r nor h. I think what you are missing is you need to look more carefully at the geometry of the problem until you come up with a relationship between r and h so you can eliminate one of those variables from the equation. Once you come up with this relationship, then you will be able to solve for r and h.

### #4 Francisco Angel

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Posted 11 August 2017 - 01:32 PM

So, it looks like you are at the point where you have V=1E-3 m3=πh(r^2)/3 where you know neither r nor h. I think what you are missing is you need to look more carefully at the geometry of the problem until you come up with a relationship between r and h so you can eliminate one of those variables from the equation. Once you come up with this relationship, then you will be able to solve for r and h.

Adding to Mr shorty, you can obtain such relation based on the protruded section of the cone, which corresponds to a smaller cone of diameter 5 cm (diameter of the exit tube) and height 5 cm (provided data).

Now you have the submerged volume and the protruded volume, and the aspect ratio of the cone. I think that you need the total volume of the cone, which require you to know the density ( because you know the mass, and volume = mass / density), to completely solve the problem.

Because you don't know to what of these situations the problem corresponds:

1) total volume of cone = protruded volume + submerged volume

2) total volume of cone = protruded volume + submerged volume + volume over condensate

As in your previous thread I advice you to post all the information of the problem, statement and the questions itself. To have a complete picture.

Best regards.

### #5 Dudesons123

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Posted 14 August 2017 - 05:00 AM

I also need to incorporate the fact that the buoyancy force is equal to the weight of the cone before it starts floating. The condensate level, h, can be above the cone or just under the base of the cone.

Edited by Dudesons123, 14 August 2017 - 05:27 AM.

### #6 Dudesons123

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Posted 14 August 2017 - 05:31 AM

Can I use the ratio of the pressures of the steam and that of condensate to calculate the height ratios? Or using the volume, and the radius-height relation?

Edited by Dudesons123, 14 August 2017 - 07:15 AM.

### #7 MrShorty

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Posted Yesterday, 01:36 PM

With two threads working on two different parts of the question, it can be difficult sometimes to keep track of what we are working on here.

I also need to incorporate the fact that the buoyancy force is equal to the weight of the cone before it starts floating. The condensate level, h, can be above the cone or just under the base of the cone.

As it pertains to the "when does the cone float" question, then this seems correct. You need to equate the buoyant force with the weight of the cone. I might argue that h cannot be above the base of the cone, because it would seem that, if the condensate level rose to completely submerge the cone, then the cone will never float. The cone can only float if the condensate level is below the overall height of the cone.

Can I use the ratio of the pressures of the steam and that of condensate to calculate the height ratios? Or using the volume, and the radius-height relation?

I am not certain I know what ratio of pressures you are looking for here. As I think through the problem in the thread (solve for l in the manometer), I don't seem to encounter a "P(steam)/P(condensate)". That does not necessarily mean that there is no algorithm that will use that ratio to solve this problem, but that the solution I thought through did not encounter that ratio.