9 replies to this topic

[lonnis]

Hello

 

I need some help calculating the time of cooling down 2000 L of steam (temperature 75 degree Celsius) with cold water at 7 degree Celsius.

 

The situation: We have a tank 5 m^3, pressure at 1 Atm, an steam temperature 75 degree Celsius.. With partial pressure of the steam 38 Kpa we can assume that 38 % of the tank is filled with steam, approx 2000 L steam. The tank is constantly drained, and there will not be any condensate in the tank. By the time we have an inlet flow of cold water. 7 degree Celsius with an inlet flow 6000 L/hr.. The tank is isolated, so we can negligible the heat convection from the tank to the environment. The tank room have free inlet of air at 20 degrees, so there will be no problem with vacuum.

I want to plot a curve of the temperature dependent of the time, and a curve of the inlet flow dependent of the time. 

 

Can you help me and give some tip to calculate this?

 

 

Best regards Sindre

Student from Norway

[ChemEng01]

How can you have steam at 75 degrees C and 1 ATM?

 

Boiling point water is 100 degrees at 1 ATM

[lonnis]

Steam/vapor will occure at 75 deg Celsius, this is simple thermodynamics! I'm talking about the partial pressure of water here, and the vapor quality. At 100 deg Celsius, all of the water will changes it phase and go from liquid to steam (vapour quality of 100%). At 75 deg Celsius you have a vapor quality of 38%, and with 7 deg celius you have aprox 1% vapor.. 

[Ajay S. Satpute]

Ionnis,

 

For water, at 75 oC, saturation pressure value is 38.55 kPa. Meaning if you try compressing it, vapor shall start condensing and temperature shall remain 75 oC until it is condensed 100%.

 

Similarly, at 7 oC, saturation pressure value is 1 kPa. Meaning if you try compressing it, vapor shall start condensing and temperature shall remain 7 oC until it is condensed 100%.

 

As there is no other component considered (air hasn't come yet), you cannot use partial pressure term, as all the pressure is due to water vapor alone. Perhaps you are trying to calculate 38.55/101.325*100 = 38% as partial pressure of water, which is incorrect.

 

Please recheck and re-submit your problem statement. Someone will certainly help.

 

Regards.

 

Ajay S. Satpute

[lonnis]

OK, my situation is that we have a tank of 5000 L, being washed with water at 75 oC. Water is spread by a washing ball and flows in at 6000L/hr. The outlet in the botom of the tank is constantly draining out water. Then after some washing we will flush with water of temperature 7 oC. My main problem is to make sure that there will in no way occur a vacuum in the tank. So we have to ensure that enough air is flowing in by the ventilation canal.  So then I have to make a calculation of the time it takes to cool down, and how fast, and how much air it will take to not give any vacuum or external pressure in the tank.

I

’m sorry if there is something that I’ve don’t figured out the right way. And I am very glad if someone can explain the process, and what is going to happen here. My calculation says that we will need 1.95 m^3 into the tank to not get any external pressure.

 

Sindre

[ChemEng01]

Steam/vapor will occure at 75 deg Celsius, this is simple thermodynamics! I'm talking about the partial pressure of water here, and the vapor quality. At 100 deg Celsius, all of the water will changes it phase and go from liquid to steam (vapour quality of 100%). At 75 deg Celsius you have a vapor quality of 38%, and with 7 deg celius you have aprox 1% vapor.. 

 

You only get steam when water boils!!!  If the pressure of your system was at 38 kPa you would have steam as the water would be boiling at 75 degrees. In your case water will be liquid phase at 75 degrees and 101.325 kPa (1ATM). Water vapour will form in any space within your tank due to evaporation only. (Look at a phase diagram for water).

[lonnis]

When washing tank spaces at approx. 75 deg. C the absolute vapor pressure is equal to 0.40 bar.

 

With temperature of the water equal to approx. 7 gr. C the vapor pressure will certainly be: 0.01 bar .

From the above figures one can make an assessment of the partial pressure caused by steam and what caused the air present in the room during the wash of the tank. It will not be far wrong to say that about aprox 60 % filled the air during the wash.

 

When the tank room is 5 m3 , the vapor volume separately represent 2  m3 . When giving mains water of example 7 gr. C this volume will collapse to approx. 50 L . vapor

 

When the vapor volume collapses , be 1,95 m3 of air flow into the tank to avoid vacuum.

 

 

And to all of you.. At temperatures between 0 -100 deg. C, at a pressure equal to 1 Atm, vapour will occur in a room.

Good example for this is when you look at a warm cup of tea, the water in the tea cup is not at boiling temperature, gas molecules will detach from the fluid. The warmer the cup is, the more gas molecules will detach. 

I really hope some of you can help out here

[ChemEng01]

Maybe a sketch of what exactly it is you are doing and trying to achieve would be useful.

[PingPong]

When the vapor volume collapses , be 1,95 m3 of air flow into the tank to avoid vacuum.

That is only the required air flow due to condensation of the water vapor, but due to the cooling of the 3 m³ (dry) tank air from 75 to 7 ^oC there will be an additional 0.6 m³ of air inflow required.

 

However what really matters is not the total required air quantity but the required air flowrate, which will be largest at the beginning of the cooling process.

How fast the tank cooling will be depends on the contact area between the cold water and the warm moist tank air. But we have no idea how big that area is as it depends on the way the cold water and the moist air are brought into contact with each other.. Your mentioning of a "washing ball" does not make that any clearer to me, or anybody else I suspect.

 

I suggest that when you do this operation you leave a manhole open. That should give sufficient flow area for the incoming air to avoid a vacuum. Having only a small vent valve open may not be sufficient.

[Neelakantan]

Am i missing some thermodynamics gyan here?

 

The tank volume is 5 m3; temp is 75 deg C; how do you say the partial pressure of steam is 38 kPa?  Note  that  you are jumping from vapour pressure of water (not steam!) at 75 deg C which is 289 mmHg =~38.5 kPa, to partial pressure?

 

Granting that the system gas space contains water vapour and air only, you can say the partial pressure of water in the system is 38.5 kPa; then how this "partial pressure" is linked to the volume fraction of the tank??  in ideal situation you can link the partial pressure to the volume fraction of the "vapour" space in the tank and not the total volume.

 

The killer line is that there won't be any condensate! and the tank is filled with 2,000 litres of steam; (ah ha! now the light in the tunnel!) the tank is full of water vapour and air (assuming and granting that the air is water saturated without any water in liquid form!); the total tank is filled with ~ 2 m3 of water vapour and 3 m3 of 'dry' air.

 

Another confusing statement: "the tank is constantly drained"  - and the tank is isolated!  Is the tank in batch process or steady state flow status?..... we continue to unfathom the post!

 

The tank has inlet cold water at 7 deg C and 6 m³/h rate.; the "tank room" (???) have (has) free inlet of air available at 20 deg C;

 

So a tank with 5 m3 capacity is filled with air, water saturated at 75 deg C; it is thermally isolated from the surroundings. cold water at 7 deg C is admitted at 6 m3/hr rate; the water cools the air and keeps running out keeping the tank liquid free; air is admitted at 20 deg C to maintain atmospheric pressure in the tank. The problem is finding the temperature rate (temperature vs time) and the rate of air flow (air flow vs time)

 

phew!

 

regards

neelakantan

 

PS: the rate of cooling will be dependent on how water is admitted, (sprayed?  particle size......), air water contact...