11 replies to this topic

[AllyK]

I have come up with a design to emit steam from a boiling vessel which will in turn be emitted to the atmosphere (i know it's a simple kettle...).

 

My question is can i use vapor pressures (from Antoine Eqn) when trying to calculate the velocity of the water particles in the density, velocity and dynamic pressure equation (derived from Bernoulli)? Will my dynamic pressure be equal to vapor pressure when the molecules are naturally rising upwards?

 

Sorry i know this is an easy question for some but i'm a postgraduate student and i am a little foggy on my Chemical engineering

 

I'd appreciate the feedback,

 

Alex

[shan]

No.  Vapor pressure and dynamic pressure are totally different concept.

[latexman]

Vapor pressure = static pressure in vapor space of your vessel.

[shan]

The statement "Vapor pressure = static pressure in vapor space of your vessel" is not always true.  For example, when 0 degC water is stored in an atmospheric tank, 760 mmHg is the static pressure in the vapor space of the tank, but water vapor pressure at 0 degC is 4.6 mmHg.

[thorium90]

Recap on Bernoulli's, you should be able to recognise the terms for dynamic pressure (with velocity) and for static (with height).

 

http://en.wikipedia....e_flow_equation

 

Shan, if the vessel is open to the atmosphere like the kettle, then that 4.6mmHg is part of the 760mmHg, unless it was enclosed. So in a way, both you and latexman are right...

 

Think of it another way, if the water was at 90C, the static pressure in the vessel open to atmosphere would still be 760mmHg. Boiling water doesnt cause the pressure of the atmosphere to increase. However if the vessel was enclosed, the pressure would rise.

Edited by thorium90, 25 April 2013 - 04:01 PM.

[latexman]

The statement "Vapor pressure = static pressure in vapor space of your vessel" is not always true.  For example, when 0 degC water is stored in an atmospheric tank, 760 mmHg is the static pressure in the vapor space of the tank, but water vapor pressure at 0 degC is 4.6 mmHg.

 

But that is not the case in this post.  The OP said it was boiling and emitting steam.  Therefore, any atmospheric air has been pushed out .

[Dacs]

In that case, the kettle has to contain only water, which would mean:

1. The temperature is supposed to be the boiling point of water, and

2. The pressure must be dictated by its vapor pressure

 

Since your kettle is open to atmosphere, then it follows that the atmospheric pressure serves as a backpressure at your kettle outlet, which then forces the kettle to operate at the temperature corresponding to your backpressure (plus built-up back pressure, if any).

 

If I understood your design intent, you came up with a design to emit steam (which I suppose is flow controlled by your design) from a boiling kettle to the atmosphere (via a stack/duct I presume).

 

I can't comment further without giving us out some specific details (such as if my assumption is right all along).

 

 

Just in case I got it right, what Antoine can do for you is to determine the vapor pressure of water at a given temperature (or vice versa). If you're after the hydraulics of your system, then you have to:

1. Assume a kettle pressure (which should be above atmospheric obviously)

2.  Use that pressure to calculate the corresponding mass flowrate through your piping/duct/stack.

3. If this doesn't match with your desired flow, then change the duct/piping/stack size and do #1-2 again.

 

When you get your kettle pressure right, you can then back calculate for the water temperature.

Edited by Dacs, 25 April 2013 - 06:40 PM.

[shan]

Anyway, vapor pressure do nothing with  dynamic pressure (1/2*ρ*V^2) in Bernoulli Equation.

[latexman]

Anyway, vapor pressure do nothing with  dynamic pressure (1/2*ρ*V^2) in Bernoulli Equation.

 

Again, that is not the case in this post.  Granted, there was no mention of elevations, but it is the vapor pressure (minus the exit pressure or P2) in this post that is providing the driving force to create the velocity head.  Right?

[shan]

Anyway, vapor pressure do nothing with  dynamic pressure (1/2*ρ*V^2) in Bernoulli Equation.

 

Again, that is not the case in this post.  Granted, there was no mention of elevations, but it is the vapor pressure (minus the exit pressure or P2) in this post that is providing the driving force to create the velocity head.  Right?

In my opinion, steam velocity in a kettle is dependent variable of steam flow and kettle cross area.  Water vapor pressure is a constant for a specified water temperature. 

[AllyK]

Thank you very much for the help guys, and thank you for not blasting me for asking that question  !

 

I was designing a method to take water from the ocean, heat it and emit it as steam, however i can not use shell and tube heating- it must be electrical. To maintain a high velocity i must have a relatively high flowrate of steam exiting because the area of which the steam is passing through must remain fixed. This has been problematic as i am now having half a gallon of water per second to heat and the duty is around 2 MW. Is there any way at all to make an energy efficient kettle? I was thinking i could store a continuous level of heated water which will not fluctuate with entering and exiting fluids, would this help thermodynamic properties? 

 

I would really appreciate it if someone could help me here and i'd be more than happy to credit them in my thesis for the design!

[thorium90]

What you are describing sounds like seawater desalination which is a pretty common topic and many methods can be easily googled online.