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Conduction, Convection Or Radiation?


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#1 mixednuts

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Posted 05 May 2013 - 11:02 PM

Hi guys,

 

the situation is like this. I have a tower in which hot exhaust gas is flowing upwards while coming in contact with water droplets being sprayed downwards through a nozzle. May i know what is the main form of heat transfer here? Conductiom, convection or radiation? My guess would be convection. Thank you in advance

 

Regards,

Mixednuts



#2 Dacs

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Posted 05 May 2013 - 11:14 PM

I suppose you have water droplets to quench the hot exhaust gas going up.

 

In that case, then it's simultaneous mass and heat transfer, where phase change cooling drives the heat transfer.

 

Of course, you have to heat up the water droplets first before it boils off, so in this case, it's convection at the exhaust gas side and primarily conduction at the droplet side.

 

Although if the droplets have a line of sight from where the hot exhaust gas is coming from, then you may consider radiation as well, though I feel that it won't matter much unless we're talking about furnace-like temperatures (>600°C)

 

My 2 cents.


Edited by Dacs, 05 May 2013 - 11:17 PM.


#3 mixednuts

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Posted 05 May 2013 - 11:23 PM

If it is a case of conduction and convection, what equations will be applicable to determine the exhaust gas outlet Temperature and the outlet water Temperature?



#4 Dacs

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Posted 05 May 2013 - 11:32 PM

You have to realize that you have simultaneous heat and mass transfer in the process that you're trying to model. I feel that heat effects from those two are smaller (if not much smaller) than latent heat contribution by vaporizing those droplets.

 

Unfortunately, I have little experience designing such system so I can only provide you with much information. To give you a nudge in the right direction though, I think this won't be too different from humidification where you're trying to saturate air with water, only that we're talking about such temperature extreme and with components in your exhaust gas other than air present.


Edited by Dacs, 05 May 2013 - 11:34 PM.


#5 mixednuts

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Posted 06 May 2013 - 12:17 AM

Thanks Dacs for the info.

 

I am asking this question as I feel that in order to determine the amount of water evaporated, I should do an enthalpy balance around the system. And to get the enthalpy, I need to know the Temperatures of the streams in and out of the control volume.

 

Hopefully, someone else can enlighten me on this.



#6 Dacs

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Posted 06 May 2013 - 01:08 AM

I'll give it a shot...

 

I presume that for a counter current flow of hot exhaust gas (flow goes up against gravity) with water droplets (flow goes down with gravity), you'd form two zones: Subcooled zone (Zone A, that starts from the water nozzle spray location) and Evaporative Zone (Zone B) where Zone A is on top of Zone B (assuming that the water droplet terminal velocity is greater than your hot exhaust gas superficial vapor velocity, otherwise you'd have water entrainment all the way up to the stack).

In Zone A, you'd primarily have heat transfer from the (then cooled from Zone B) hot exhaust gas to the subcooled water droplets, bringing its temperature until its boiling point.


When the water reaches its BP then you've entered Zone B where the primary mode of heat transfer is by latent heat (vaporization of water to steam) and this extends until the point where all the water droplets have been evaporated.

 

I can just imagine the complexity of the simultaneous mass and heat transfer that occurs inside both zones, but I think it won't stop you on calculating the minimum amount of water flow to quench the hot exhaust gas from its initial temperature (Ti) until its desired final temperature (Tf). This can be easily done by heat and mass balance.

 

The harder part is the size of the zones (in essence the required volume) where both fluids come into contact. It's also possible that you'd have entrainment if you have water droplet size low enough where gravity settling can't happen and on the other extreme end, your water droplets goes all the way down to the tower without completely evaporating (which I don't think a part of your design intent). In this case, the stack cross section area and the atomizing nozzle design come into play.

 

My 2 cents :)


Edited by Dacs, 06 May 2013 - 01:11 AM.


#7 thorium90

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Posted 06 May 2013 - 01:40 AM

Coulson & Richardson-Chemical Engineering Vol. 1-Fluid Flow, Heat and Mass Transfer, The Chapter on Humidification and water cooling should provide and explain the equations you need.

 

You can also try reading Mass Transfer Operations by Robert Treybal, Chapter 7 on humidification operations.


Edited by thorium90, 06 May 2013 - 01:43 AM.


#8 mixednuts

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Posted 06 May 2013 - 03:10 AM

Dacs,

 

I am still quite unsure about your reply.

 

this is not a cooling tower to be exact. I do not have a desired initial and final temperature requirements. What I do have is the mass flow and temperature of the water stream in, exhaust gas stream in. I am required to find out the mass of water evaporated. So from my understanding, there will be 3 streams out:

1. exhaust gas stream out (T is unknown)

2. water vapour stream out (T is unknown)

3. The rest of the water not evaporated out (T is unknown)

 

I feel that an enthalpy balance will be the easiest to determine the mass of water evaporated. I guess i can assume the temperature of the exhaust gas out and the water vapour out to be the same. To keep things simple, lets assume no change in composition for the water and exhaust streams, meaning only counter current contact and exaporation taking place.



#9 mixednuts

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Posted 06 May 2013 - 03:11 AM

Thorium90,

 

do you know where i can read up on the two texts you have recommended?



#10 Dacs

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Posted 06 May 2013 - 03:31 AM

Let me go through with your points one by one :)

 

this is not a cooling tower to be exact. I do not have a desired initial and final temperature requirements. What I do have is the mass flow and temperature of the water stream in, exhaust gas stream in. I am required to find out the mass of water evaporated. So from my understanding, there will be 3 streams out:

Unless I grossly misunderstood your design intent, my understanding is you have a hot exhaust gas in contact with water (as a mist) and in that regard, you're basically quenching the hot gas.

 

We cannot ascertain what happens to the process at face value. I can think of two cases however:

1. All the water going in the chamber (we'll call it as such for convenience) is getting evaporated by the hot gas. In this case, you only have two streams coming in (walter and hot gas) and one stream going out (cooled gas)

2. Water flow is too much for the exhaust gas to handle (maybe by poor interphase contact or it's just the water flowrate is too much) and in this case, you'd have an additional stream going out (the excess water)

 

I feel that an enthalpy balance will be the easiest to determine the mass of water evaporated. I guess i can assume the temperature of the exhaust gas out and the water vapour out to be the same. To keep things simple, lets assume no change in composition for the water and exhaust streams, meaning only counter current contact and exaporation taking place.

This is the problem.

 

You cannot just assume that they'll be in equilibrium. How long they're in contact as well as how good the contact is will determine the amount of mass and heat transfer between the two streams. And that cannot be determined by heat/mass balance alone. 

 

What you can do however is this:

1. Since you have a fixed amount of water going in, calculate the heat required to ( a ) raise its temperature to boiling point and ( b ) boil of water completely to steam

2. With that amount of heat, calculate the change in temperature of the hot gas from its initial temperature.

 

If the exhaust gas temperature is lower than the water boiling point, then you can straight out conclude that  you won't be able to boil off all the water coming into contact.

 

If the temperature is higher however, then it's possible that all the water will be converted (and be entrained to the exhaust gas) to steam. Take note that I said "possible" because the possibility of this happening lies on how good the interphase condition is.

 

It's similar with heat exchangers. Although thermodynamics allow the transfer of a certain amount of heat from the hot to the cold fluid, the actual amount of heat transferred will be controlled (among other factors) by the heat transfer surface area.

 

Hope I was able to explain myself more clearly :)



#11 mixednuts

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Posted 06 May 2013 - 03:59 AM

Yes, I have calculated the amount of heat required to boil off the water per kg, say Q1, (by using the specific heat capacity and the latent heat of vaporization). That is why I am asking whether the heat transfer is mainly by conduction or convection so that I can use an empirical correlation to determine the heat transfer rate, q. And then determine if water will evaporate by looking at the 2 values.

 

For eg, we know heat transfer by convection follows the equation q = h * A * dT. the convective heat transfer coefficient, h, can be found by the empirical correlation and using the Nusselt number. Area, A, will be the surface area of a sphere, assuming a water droplet diameter. and dT will be the difference in T between the exhaust gas and water. So, I will take Q1 to compare with q(in units of J/s though) and see if the heat transfer by convection is enough to evaporate the water. 

 

Does this sound logical or no?



#12 Dacs

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Posted 06 May 2013 - 06:11 AM

You cannot use heat transfer concepts alone when dealing with evaporation. It's simultaneous mass and heat transfer. Essentially, you're also transferring mass from the water droplet to the hot gas and also transferring heat from the hot gas to the water droplets simultaneously.

 

What you described above might only apply when your water droplets undergo sensible heat change. My ChE concepts might be rusty, but I do remember that there's a concept of HTU and NTU when dealing with simultaneous heat and mass transfer. I suppose you can start on that, especially in how they come up with the basis.

 

:)



#13 thorium90

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Posted 06 May 2013 - 08:30 AM

Guys, what you are discussing is a common problem of humidification that can be found in many textbooks that can be found in libraries or bookstores. There is no need to re-analyse the mechanics and redo the deriviation of the equations involved unless you are gunning for a PhD thesis. It has already been researched and reported on by various researchers in the past. As an engineer, it would be more practical to focus on how to apply the theory and equations to real situations.

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#14 Dacs

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Posted 06 May 2013 - 05:47 PM

thorium90: Not that I don't agree with you (I said humidification in my post way above after all), but I think it'll be more beneficial for him to understand the concept behind the equations so that he'll fully understand the mechanics of what's happening.

 

He has all the time and energy to do that (I believe he's still a student) and it'll benefit him in the long run.



#15 mixednuts

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Posted 06 May 2013 - 07:21 PM

Ok, if the water being sprayed is not pure water, but lets say 40% sodium sulphate solution, Is this still a common problem of humidification?

 

I have read through the coulson excerpt but I am still unclear as to how am I supposed to determine the mass of water evaporated.

 

Guys, thanks for the patience btw.



#16 Dacs

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Posted 06 May 2013 - 11:18 PM

I cannot speak in behalf of others, but I haven't encountered any humidification process that involved using a solution. It's always water (such as cooling towers).

 

As far as the equations are concerned, I'm a bit rusty with it so I won't even pretend to remember those. But maybe others might be in better position to help you :)






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