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Vapour to liquid heat exchange
- Thread starter Imak
- Start date
You need to know the characteristics of each fluid involved - specific heats, specific gravities, viscosities, etc. Then you require the mass flows of the fluids.
If you provide the required data to a heat exchanger manufacturer, they'll be happy to run the calculations for you, and recommend a unit. Most will have a large amount of data for various fluids, but if you're dealing with an unusual process, they may need you to provide more information on it.
If you provide the required data to a heat exchanger manufacturer, they'll be happy to run the calculations for you, and recommend a unit. Most will have a large amount of data for various fluids, but if you're dealing with an unusual process, they may need you to provide more information on it.
- Thread starter
- #3
The basics is:
Heat capacity for cooling liquid
Inlet temp of cooling liquid and max outlet
Flow of vapout to be condensed
Latent heat of vapor to be condensen
Heat capacity if the condensed liquid is to be sub cooled
Desired liquid outlet temp (in case of sub cooling)
Then use a standard heat balance:
dT_1*Cp_1*Q_1=L*Q_2+dT_2*Cp_2*Q_2
1=cooling media
2=Vapour
dT= Temperature change
Cp=Heat capacity
Q=Flow
L =latent heat. Could'nt remember if there is a "standard letter"
Solve the eq. for Q_1 (or any other that you may have as a unknown).
And thats it.
If you want a more detailed solution (someting like a transfer area) do as suggested: Consult a vendor (or search the web for a vendor estimation tool). Text books on the subject (or more general books such as Perry's could meybe also help with engineering rules of thumb.
Best Regards
Morten
Heat capacity for cooling liquid
Inlet temp of cooling liquid and max outlet
Flow of vapout to be condensed
Latent heat of vapor to be condensen
Heat capacity if the condensed liquid is to be sub cooled
Desired liquid outlet temp (in case of sub cooling)
Then use a standard heat balance:
dT_1*Cp_1*Q_1=L*Q_2+dT_2*Cp_2*Q_2
1=cooling media
2=Vapour
dT= Temperature change
Cp=Heat capacity
Q=Flow
L =latent heat. Could'nt remember if there is a "standard letter"
Solve the eq. for Q_1 (or any other that you may have as a unknown).
And thats it.
If you want a more detailed solution (someting like a transfer area) do as suggested: Consult a vendor (or search the web for a vendor estimation tool). Text books on the subject (or more general books such as Perry's could meybe also help with engineering rules of thumb.
Best Regards
Morten
The quickie formula I use for heat exchanger surface area is: A = Q/U X LMTD.
A = surface area (sq ft)
Q = BTU/hr
U = heat transfer coefficient (maybe 200 for a shell & tube)
LMTD = log of the meam temperature differences
What this won't give you is pressure drops across the HX. You need the manufacturer's data for for his particular HXs to do that. Another thing it won't identify, is if you need a short fat HX, or a long skinny one. Typically, if you require a big temperature change, you'll need a longer unit to have the fluids in thermal contact long enough to take you where you need to go.
A = surface area (sq ft)
Q = BTU/hr
U = heat transfer coefficient (maybe 200 for a shell & tube)
LMTD = log of the meam temperature differences
What this won't give you is pressure drops across the HX. You need the manufacturer's data for for his particular HXs to do that. Another thing it won't identify, is if you need a short fat HX, or a long skinny one. Typically, if you require a big temperature change, you'll need a longer unit to have the fluids in thermal contact long enough to take you where you need to go.
- Thread starter
- #6
If you have 2 variables (outlet temp and e.g. duty) then i assume you will have to guess that you do not subcool the vapour and thats its a single component (steam). Then the condensation temp is equal to saturated steam (vapour) temp. I use steam/vapour in the most general term. If its a single component then the saturated vapour temp is equal to the condensate (e.i. no temp change during condensation). This is of couse puly theretical. In practise the liquid will allways be somewhat sub cooled.
I my equation then dT_2 would be zero.
More that 2 unknow: I would say that the #text book problem" could not be solved. Unless you have some info re. the heat exchanger and from that you were able to get the unknown.
What TBP is talking about is heat exchanger sizing. But that was not stricly your question.
Best Regards
Morten
I my equation then dT_2 would be zero.
More that 2 unknow: I would say that the #text book problem" could not be solved. Unless you have some info re. the heat exchanger and from that you were able to get the unknown.
What TBP is talking about is heat exchanger sizing. But that was not stricly your question.
Best Regards
Morten
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