Calculate boiling rate of water

[samsterC]

How would I calculate the rate that water would boil off? I've done a lot of looking into and found an equation but it doesn't seem quite right. What I found states that the KJ/h delivered to the water divided by the latent heat energy gives you the amount of water that will boil off. I tried this but I got a very low amount (.05 kg/h).
I'm boiling 4kg of water in an area that's ambient temperature is 60C under reduced pressure (about 30 torr) in a closed system. The boiling point is 30C and the latent heat energy is about 2438 KJ/kg at this pressure. Is the boiling rate just very slow or should I be using a different equation? Or will I just have to test the system and find out for myself?

 

[jbriggs444]

That should be 2438 kiloJoules per kilogram.

 

[samsterC]

That should be 2438 kiloJoules per kilogram.

Thanks ^-^

 

[mfig]

How are you reaching 60° C if the water boils @ 30° C?

 

[samsterC]

How are you reaching 60° C if the water boils @ 30° C?

The water itself isn't 60C but the thing boiling the water is

 

[mfig]

OK.

To calculate the boiling rate we need to know the heat transfer rate. Have you calculated or estimated this? It sounds like you are reasoning correctly.

 

[samsterC]

The thermal conductivity of water at 30C is about .62 W/MK. Since the whole area around the water is 60C the area would be 452.39 inches or 0.0074 Meters (it's in a cylinder with a radius of 6 inches filled up to the 5 inch mark). the temperature difference is 30C and I'm not sure what to say about thickness since the whole area around the water would be at 60C but from what I know the equation would be 0.62 * .0074 * 30/Thickness correct?

 

[russ_watters]

What do you mean by "whole area around the water"? What, exactly, is at 60C? Is it a metal cylinder that is directly heated? Air outside of the cylinder?

Frankly, though, it is probably easier to measure the heat transfer either with a direct measurement of the input or by weighing the water than it is to try to calculate it from scratch.

 

[samsterC]

The water is in a closed container and that container is being heated from all sides so yeah It could be the air around it and the issue is I can't measure it by the input because this heat isn't being supplied by and electrical force or anything else I can measure like imagine it's in a super hot desert or something, I'm not sure how to measure that input other than the fact that it's 60C :/

 

[russ_watters]

This is all very vague. Are you actually building/testing this? Why don't you have access to your own device/experiment?

Convection is very difficult to calculate accurately and you haven't really even attempted it. The thermal conductivity of the water is not a relevant issue here.

 

[samsterC]

I actually am going to construct and test this which is why I wanted to have some idea what would have before-hand

 

[samsterC]

And I don't see how it's vague, the whole container is going to be 60C as if it were for example submerged in warm water or wrapped up in a heating blanket

 

[russ_watters]

And I don't see how it's vague, the whole container is going to be 60C as if it were for example submerged in warm water or wrapped up in a heating blanket

As if it is wrapped in a heating blanket? Submerged in warm water? In a super hot desert? These are all very different things. Which is it? That's what's not specific (vague) about it. You also haven't been answering all of the questions: Again, why can't you measure the input heat? If it is an electric heating blanket, you can measure the electricity supplied to the heating blanket. Or, can you test the evaporation rate itself by putting it on a scale? And what is the purpose of this device? What are you trying to do/what problem are you trying to solve? Elaboration/specificity would be a big help. We don't know what you are thinking/trying to do -- we can't help you unless you tell us, exactly/specifically/in detail.

Anyway, here's a good link on convective heat transfer: http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html

 

[samsterC]

Ok then specifically I'm putting the container into a compost pile which at it's highest temperature will reach about 60C

 

[russ_watters]

Ok then specifically I'm putting the container into a compost pile which at it's highest temperature will reach about 60C

Interesting. So, the limiting factor will be the heat generation rate of the compost and getting that heat into your water container. If the compost container is insulated, according to this, it generates about 1,000 BTU/hr (300W) per ton of compost:
http://smallfarms.cornell.edu/2012/10/01/compost-power/

 

[AamsterC2]

I can't log into my account for some reason but it's still me. My compost pile would likely be about a metre cubed which would be 700lbs so the energy would be more like 100 watts so how do I calculate the boiling rate from there?

 

[russ_watters]

I can't log into my account for some reason but it's still me. My compost pile would likely be about a metre cubed which would be 700lbs so the energy would be more like 100 watts so how do I calculate the boiling rate from there?

You had the method right in your first post: heat input divided by specific heat of vaporization.

 

[AamsterC2]

Wold the equation to find the heat input be K *A *change in T all divided by D?

 

[russ_watters]

Wold the equation to find the heat input be K *A *change in T all divided by D?

No, 100 watts is the heat input (assuming a well insulated compost pile).

 

[AamsterC2]

Well the container would only be 12" by 10" and the pile would be a 1m^3 so it wouldn't be able to absorb all the heat would it?

 

[AamsterC2]

Hmmm never mind you're right, if the compost pile is insulated all that energy would flow into the water

 

[samsterC]

So that means at this scale at least it would take a day to boil away one gallon of water (100 w/h = 360 Kj/h. 360/2438 = 0.15 Kg/h. 3.79 Kg/0.15 Kg/h = 25 hours)

 

[samsterC]

Although this isn't using the convective heat transfer equation in that link you sent me. I found that equation earlier but was unable to find water's convective heat coefficient is (some sites say between 50-3000 or I suppose 3-100 since the water is boiling http://www.engineersedge.com/heat_transfer/convective_heat_transfer_coefficients__13378.htm , another says 100, this one says 500-10,000 http://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html and another website http://www.tlv.com/global/TI/steam-theory/overall-heat-transfer-coefficient.html gives another range so I have no idea what the coefficient should be if that equation is what I should be using)

 

[russ_watters]

So that means at this scale at least it would take a day to boil away one gallon of water (100 w/h = 360 Kj/h. 360/2438 = 0.15 Kg/h. 3.79 Kg/0.15 Kg/h = 25 hours)

Correct.

To be honest, I was surprised at how much heat compost generates. I would have guessed a tenth of that.

 

[russ_watters]

Although this isn't using the convective heat transfer equation in that link you sent me. I found that equation earlier but was unable to find water's convective heat coefficient is (some sites say between 50-3000 or I suppose 3-100 since the water is boiling http://www.engineersedge.com/heat_transfer/convective_heat_transfer_coefficients__13378.htm , another says 100, this one says 500-10,000 http://www.engineeringtoolbox.com/overall-heat-transfer-coefficient-d_434.html and another website http://www.tlv.com/global/TI/steam-theory/overall-heat-transfer-coefficient.html gives another range so I have no idea what the coefficient should be if that equation is what I should be using)

Further down on the engineering toolbox page, it has a boiling water coefficient of 10-40. But you can see from these broad ranges that it is highly variable and dependent on the specific conditions, such as the size and shape of the vessel.

 

[AamsterC2]

Do you know any specific way to determine its coefficient?

 

[AamsterC2]

While the method of taking the energy produced in an hour and dividing I by the latent heat of vaporization makes sense in some ways it also seems odd because it also means that if the boiling point and pressure were raised that the boiling rate would increase despite the fact that the temperature needs to boil the water are only met instead of exceeded by 30°C. I don't know if I explained that well but that's what I'm hung up on.

 

[russ_watters]

Do you know any specific way to determine its coefficient?

No. I took a course in heat transfer, but didn't do very well; It is very complicated mathematically, since it is fluid dynamics mixed with thermodynamics. What I took away from it is that experimentation is better than trying to figure it out with math.

While the method of taking the energy produced in an hour and dividing I by the latent heat of vaporization makes sense in some ways it also seems odd because it also means that if the boiling point and pressure were raised that the boiling rate would increase despite the fact that the temperature needs to boil the water are only met instead of exceeded by 30°C. I don't know if I explained that well but that's what I'm hung up on.

The enthalpy of vaporization is slightly higher, but remember, you also had to heat the water to raise the temperature. The total energy is - I think - exactly the same. What's missing from that picture is the effective generation of the heat: the compost has a certain optimum operating temperature I'm sure, and if it overheats it probably dies (I'm assuming it is a biological process, not a chemical process...).

 

[Nidum]

Why do you want to do this anyway ?

Even under the most optimistic conditions all that could be produced would be a very small quantity of sub atmospheric pressure steam .

Note also that you would need to have an external energy input to produce the vacuum needed to allow the water to boil at such low temperature .

 

[AamsterC2]

Yeah the maximum temperature is somewhere around 60°C