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2

# Manometer And A Sealed Cylindrical Cavity

5 replies to this topic
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### #1 Dudesons123

Dudesons123

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Posted 09 August 2017 - 08:16 AM

Hi Scientists and Engineers,

I would like to ask the following question I have about fluid flow. I have tried doing a force balance on it and I have also used Bernoulli's equation but to no avail.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 bar 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 H and base radius R is πR2H/3.

A manometer which is open to the atmosphere contains 50 cm of mercury. If the condensate level h is equal to 10cm, what is the height of the water level in the manometer with respect to the condensate level (i.e. in the diagram)?

Thanks.

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Edited by Dudesons123, 09 August 2017 - 09:16 AM.

### #2 Dudesons123

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Posted 09 August 2017 - 12:40 PM

I'm not entirely sure where to go from here. All I know is that I have reservoir manometer and I am working with a static fluid.

Edited by Dudesons123, 09 August 2017 - 12:49 PM.

### #3 Francisco Angel

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Posted 09 August 2017 - 01:56 PM

Hi:

Didn't a pressure balance at the condensate level do the trick?

In the "manometer" side you will have:

P_atm+P_mercury+P_condensate( Proportional to l )

I the "condensate chamber" side you will have:

P_steam.

In following questions try to include the statement of the problem (in the attached pictures it seems cut off),  sometimes it contain information that is not indicated in the picture.

Also academic problems sort of "guide" you in the steps to solve them by the order of the subquestions a), , c), etc. So include that information also.

Best regards.

### #4 Dudesons123

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Posted 09 August 2017 - 02:31 PM

Hi Francisco,

Here is what I have:

Patm + Pmercury +Pcondensate = Psteam

1 bar + 0.667 bar +1000 kgm^-3*9.81 kgms^-2*l= 2 bar

l = the difference in height between the top water level height inside the smaller tube and the top water level inside the sealed cylindrical cavity.

Re-arranging gives me:

l = (2 bar -1 bar - 0.667 bar)*10^5/(1000kgm^-3*9.81 kgms^-2) = 3.39 m

Which is obviously weird considering that we are given fractions of a meter as our known variables.

Edited by Dudesons123, 09 August 2017 - 02:33 PM.

### #5 Francisco Angel

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Posted 09 August 2017 - 03:23 PM

The procedure seems right, provided the assumptions (like the value of the pressure of the steam over the liquid surface) are right.

### #6 Dudesons123

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Posted 14 August 2017 - 04:43 AM

Hi,

Why did we not take into account the cross sectional areas of the tube and the sealed cylindrical cavity to do a force balance on it?