Evaporating water at 40 Celcius

[rik2011]

Hello

The questions is how much kW is required to lower air pressure of chamber to boil water at around 40 celsius.
Is there is some equation that i could use to calculate different temperatures?

 

[JaredJames]

OK, on first read through it looks like you've just thrown some words together and tried to form a question. You start with asking for a power level to lower the pressure and then you are asking about different temperatures.

Let's clear this up, step by step.

Firstly, do you want to boil water by lowering the pressure?

 

[rik2011]

Sorry for being not clear.

Firstly, do you want to boil water by lowering the pressure?

Yes .

 

[JaredJames]

Ok, so do you have the facilities to lower the pressure that much? You are talking about pretty much creating a vaccuum.

You need a vacuum pump for this, all you need to do is have a look for a chart showing required pressure and then get a pump to achieve this.

Can I ask what you are using this for?

 

[rik2011]

Ok, so do you have the facilities to lower the pressure that much? You are talking about pretty much creating a vaccuum.

You need a vacuum pump for this, all you need to do is have a look for a chart showing required pressure and then get a pump to achieve this.

Can I ask what you are using this for?

@one: I'm trying find out if i have enough energy.
@two: To separate v.small solids from water or reduce amount of water in mixture

The http://www.engineeringtoolbox.com/boiling-point-water-d_926.html" shows that i need 1 PSI or 0.07 bar to boil at 38C.
How much kW (MW?) i would need to create such low pressure?
Lets say i got flat chamber and size: 40cm x 1m x 5m witch is half filled with water(1m^3 of water, and 1m^3 of air).

Thanks

 

[JaredJames]

Well I recommend that now you know what you are working with, you take a look at some vacuum pumps.

You should be looking at one that can reduce the pressure to the required level. That's it really.

The power requirement will be whatever the pump requires to complete the job.

There's not really a set formula because all pumps will vary, depending on pump type.

This site: http://www.coleparmer.co.uk/products/VacuumPumps/vacuum_pumps.asp, gives a selection of pumps capable of near vacuum levels.

 

[RonL]

Well I recommend that now you know what you are working with, you take a look at some vacuum pumps.

You should be looking at one that can reduce the pressure to the required level. That's it really.

The power requirement will be whatever the pump requires to complete the job.

There's not really a set formula because all pumps will vary, depending on pump type.

This site: http://www.coleparmer.co.uk/products/VacuumPumps/vacuum_pumps.asp, gives a selection of pumps capable of near vacuum levels.

https://training.womack-educational...ore_Code=WE&Product_Code=011&Category_Code=TB


The small books that Womack offers, are great for quick reference to facts based on rule of thumb calculations. Energy to pull a vacuum and time of pump down based on tank volume are listed in charts giving a close value of what is needed.

The OP was clear in my mind and his as well in regard to the two questions, KW's is likely the units that will come into play, unless it is a very small system.
This post of your's should have been the first post you made.

Sorry to be a little critical, J, maybe I just got out of bed wrong today.

Ron

 

[JaredJames]

https://training.womack-educational...ore_Code=WE&Product_Code=011&Category_Code=TB

The small books that Womack offers, are great for quick reference to facts based on rule of thumb calculations. Energy to pull a vacuum and time of pump down based on tank volume are listed in charts giving a close value of what is needed.

I'm making things as simple as possible. You can get all the energy values you want, but at the end of the day you buy a pump suitable for the job and it is what the pump uses that matters (EDIT: I understand the energy to draw a vacuum is the same regardless, but in a real life application there are other factors to be considered). That's why I asked what the purpose was, if it was a school project / question then I wouldn't recommend a pump.

The OP was clear in my mind and his as well in regard to the two questions,

I was with the OP until the question of "two different temperatures" was brought up.

KW's is likely the units that will come into play, unless it is a very small system.

I've never argued that.

This post of your's should have been the first post you made.

Again, I was attempting to clarify the situation before replying. Would you suggest I simply post blind without understanding what the OP wants? Whether or not you see something as clear doesn't mean I do.

 

[Curl]

Hello

The questions is how much kW is required to lower air pressure of chamber to boil water at around 40 celsius.
Is there is some equation that i could use to calculate different temperatures?

the question is perfectly clear.

You must keep your chamber at the vapor pressure of water. To do this, you must pump gas from low pressure to high pressure, and the work is just equal to the flow work. At steady state its simply V*deltaP.

At first, the water will boil at higher pressures (since some of the pressure is due to air) but after a while you have pumped out most of the air so you only have water and water vapor inside.

So your answer will be in Joules/grams of water vaporized

So your answer will be in kW/kg of water vaporized.

And you can't get an answer in kW because it depends on the rate of vaporization. It depends on the surface area of liquid you have (do you have the water on a plate or in a cup?). Also it depends on the heat supply rate (whatever is keeping the water at 40 degrees.

 

[JaredJames]

the question is perfectly clear.

And yet you don't answer the second part of the post. A point I have mentioned to Ron earlier.

"Is there is some equation that i could use to calculate different temperatures?"

I perfectly understood the first part, but it is when the second part was introduced it threw me. Hence the "threw some words together" issue and me asking to clarify things.

I took this to mean 'the different boiling temperatures with associated pressures', but couldn't be sure. As before, I requested clarification before blind posting. (And seeing as no one else had posted I figured why not.)

 

[Archosaur]

I think the question makes perfect sense. It's absolutely possible to boil water by lowering pressure. I don't know where your confusion is coming from, jarednjames.

Hahaha, just kidding.

 

[K^2]

7.3 kPa is not vacuum. Pretty low, though. That's the pressure you need to boil water at 40°C.

Energy will depend on what you do with that vapor. I'm guessing you are trying to extract water from something, and just dumping the vapor into atmosphere. In that case, computations are easy. You are taking something from a 7.3kPa container, and dumping it into a 101.3 kPa "container" (on average). That's pressure differential of 96kPa, and that's 96 kJ per cubic meter of vapor. The density of water vapor at 7.3 kPa and 40°C is 51.1g/m³. So you'll need about 1,900 J per gram of water extracted. For comparison, to boil it away at 1 atm and 100°C, you only need 4.2 J per gram.

So if you are looking for a more energy efficient way of extracting water, this isn't it. But if what you are extracting it from cannot be subjected to more than 40°C, like many proteins, this is your best option.

 

[Borek]

Problem is, this question is so vague everyone can read whatever he wants. My first idea was that - as this is asked in general physics - it is a just a homework style problem that requires W=PΔV, perhaps combined with Clausius–Clapeyron.

And I have no idea why OP wants to lower pressure instead of just a boiling at 100 deg C. In both cases most energy goes into evaporation.

 

[rik2011]

The reason behind it was to make alternative and efficient way to vaporise water.
But it seems that its too energy intensive.
Thank you all for replies.

 

[sophiecentaur]

@K^2 your figure or 4.2 (J/gm ) refers to the specific heat capacity, and the units are not right, of course - not the latent heat of vaporisation.
The latent heat of vaporisation of water doesn't change much over a whole range of pressures and is about 2200kJ/kg; vastly more. (Forgive the use of SI units but we are in the 21st Century! ) That is the main figure of interest when you are trying to evaporate water. If you try to evaporate just by reducing the pressure, the temperature will drop and drop until an equilibrium temperature is reached where the rate of thermal energy getting into the container balances the rate of evaporation times the latent heat. To get more evaporation, you need to supply heat. No free lunch.
Of course, if you want to cause rapid evaporation at a low temperature, using reduced pressure is the answer.