To approximate a thermal problem as HEAT EXCHANGER, are the fluids which interacts thermally are to be moving? I have a problem where a tank with a jacket. Tank contains oil at ambient and water flows jacket to heat the oil. Oil has no velocity,but water flows upward. Can this problem be approximated as HEAT EXCHANGER? or any clue to solve this problem?
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Heat exchanger- both fluids should have velocity?
- Thread starter lcatg
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If the oil isn't being forced to flow, then it operates under natural convection, which is similar to what happens when you put a pot of water on the stove.
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- Thread starter
- #4
It would seem to me that that would not be a good approach. Heat exchangers are relying on turbulent flow and good mixing which you don't have here. You have a large tank volume and only a heated jacket. If the tank were being stirred it would probably be a better approximation. Having said all that, I've never tried to do what you're asking so I really can't speak with authority on the subject.
EdStainless
Materials
You could manually calculate the heat transfer coef for the inside, natural convection only, and then apply that to a HX. But it may not work since the HT on the inside will be an order of magnitude less than in a typical HX.
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P.E. Metallurgy, Plymouth Tube
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P.E. Metallurgy, Plymouth Tube
georgeverghese
Chemical
These are the heat transfer coefficients you've got to manually calculate to get the overrall U for heat transfer from the hot water jacket into the oil:
a)Jacket side heat transfer coeff for water - forced convection
b)Oil side heat transfer coeff for oil - natural convection
c)Resistance to heat transfer through tank metal wall -may be ignored in most cases
d)Fouling heat transfer coeffiecents for both water side and oil side - may be ignored if water and oil are non fouling
If the jacket surface facing the tank metal wall is plain and smooth, then this coeff can be calculated from expressions found in heat transfer texts. Else if the surface is dimpled or corrugated etc, you'll have to talk the jacket supplier to get the correlations for this. Correlations for all other coefficents are in heat transfer texts, DQ Kern or McAdams for example.
It is preferable to write out these calcs manually by hand or build them onto a spreadsheet or Mathcad etc.
a)Jacket side heat transfer coeff for water - forced convection
b)Oil side heat transfer coeff for oil - natural convection
c)Resistance to heat transfer through tank metal wall -may be ignored in most cases
d)Fouling heat transfer coeffiecents for both water side and oil side - may be ignored if water and oil are non fouling
If the jacket surface facing the tank metal wall is plain and smooth, then this coeff can be calculated from expressions found in heat transfer texts. Else if the surface is dimpled or corrugated etc, you'll have to talk the jacket supplier to get the correlations for this. Correlations for all other coefficents are in heat transfer texts, DQ Kern or McAdams for example.
It is preferable to write out these calcs manually by hand or build them onto a spreadsheet or Mathcad etc.
georgeverghese
Chemical
Also include water jacket heat loss to ambient, which may be through some heat conservation type insulation.
To get this, the following coeffs should be worked out:
a)Jacket side coeff facing the insulation - forced convection
b)Resistance from jacket metal wall - may be small and could be ignored
c)Resistance from insulation layer
d)External heat loss from the insulation OD - this comprises a natural convection component and a radiation component.
A trial and error procedure will be required to get the U value for the hot end of this heat transfer device and one for the cold end. In most cases, a mean average U can then be calculated. Get some one experienced to help you out with this if you are not familiar with or you dont know how to get these trial and error calcs started.
To get this, the following coeffs should be worked out:
a)Jacket side coeff facing the insulation - forced convection
b)Resistance from jacket metal wall - may be small and could be ignored
c)Resistance from insulation layer
d)External heat loss from the insulation OD - this comprises a natural convection component and a radiation component.
A trial and error procedure will be required to get the U value for the hot end of this heat transfer device and one for the cold end. In most cases, a mean average U can then be calculated. Get some one experienced to help you out with this if you are not familiar with or you dont know how to get these trial and error calcs started.
- Thread starter
- #9
one other thing comes to mind. if the oil in the tank only circulates due to natural convection there might be the possibility for local overheating to the oil that is in prolonged contact with the high temperature surface which in theory could lead to oxidation and forming of lacquer and varnish like products that will adhere to the heated surface - which would rapidly reduce the heat exchange and also possibly could lead to subsequent deterioration of the quality of the oil to be heated. therefor i would suggest to include some way of forced circulation of the oil, even if only a stirrer.
Aside from possible overheating, natural convection only differs in the magnitude of the convection coefficient, and hence, the warm-up time.
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529
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