second law efficieny of HE
second law efficieny of HE
(OP)
Hello everyone!
If I need to heat water, for example from 50° to 70° with a shell and tube HE, what is the more efficient primary fluid to be used, in terms of second law efficiency? Hot water or Steam (e.g. 3 bar). I would say hot water but I am not so sure. The system considered is only the HE and not the boiler and pipes etc... I only consider the HE in my analysis. And what about firs law efficiency?
thanks
If I need to heat water, for example from 50° to 70° with a shell and tube HE, what is the more efficient primary fluid to be used, in terms of second law efficiency? Hot water or Steam (e.g. 3 bar). I would say hot water but I am not so sure. The system considered is only the HE and not the boiler and pipes etc... I only consider the HE in my analysis. And what about firs law efficiency?
thanks





RE: second law efficieny of HE
TTFN (ta ta for now)
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RE: second law efficieny of HE
By second-law efficiency, I take it you're referring to the exergy efficiency of the heat exchange. Since exergy is a term describing the available work, it's inversely proportional to entropy. As exergy decreases, entropy increases. The entropy of steam changes less over a given temperature range than liquid water, thus I would expect steam would have a higher exergy efficiency in a straight temperature exchange.
As for first law efficiency, or energy efficiency, we would need to look at the change in enthalpy. For example, with superheated steam at a given pressure, enthalpy will change more over a temperature range than liquid water, which makes sense because liquid water has a higher heat capacity than steam. Thus, the energy efficiency for the temperature exchange would be greater with liquid water.
I really had to rack my brain to remember this stuff (haven't touched exergy in years). Someone please correct me if I've mis-stated anything.
RE: second law efficieny of HE
RE: second law efficieny of HE
Liquids have higher convection and conduction coefficients and higher density so less HX surface area is required. If steam is used and it condenses in the HX, massflow required is much lower since the latent heat of vaporisation is a very large number.
je suis charlie
RE: second law efficieny of HE
Each case has a different Qopt.
Us-w > 2 Uw-w
AMTD for steam-to water: (131+60)/2 = 95.5 °c (assuming a 3-bar saturated steam condensing with no subcooling)
AMTD fro water-to-water: assumed ≦ 95.5 °C
Therefore, Qopt,s-w > Qopt,w-w
Since the thermal efficiency refers to an optimum heat transfer, it follows that for the same duty and in the same exchanger, the water-to-water case has a higher thermal efficiency.
file:///C:/Users/%D7%9E%D7%97%D7%A9%D7%91/Downloads/Heat_Exchanger_Efficiency%20(3).pdf
RE: second law efficieny of HE
The second law efficency basically is like KoachCSR said, the exergy, so we are referring about entropy.
in this Paint file I try to explain better what I mean. I figure A the steam condensates at costant temp and water rise temp. Entropy is proportional to the mean temp difference between the fluids (T_steam - (T_water in+T_Water out)/2). In second case entrpy is still proportional at the difference between the two temp fluids, but in this case the temp of two fluids are near each other, so less entropy is produced. In figure A, water enters the HE cold and is heated by steam at high temp producing high entropy, this is due to the temp of condensing steam which remains costant. this doesn't happen in second figure.
So in my opinion it is better in terms of exergy using hot water to heat other hot water as less entropy is produced. But I am not sure of my statement
RE: second law efficieny of HE
TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
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RE: second law efficieny of HE
To look at the change in entropy for your heat exchanger example, you need to look at the sum of the entropy change of both the liquid water being heated and the steam (or other liquid water) being cooled. The equation would look like:
ΔS = (-Q / TH) + (Q / TL) = (Q / (TH*TL)) * (TH-TL)
where Q is the amount of heat transferred between the two, TH is the upper temperature of the fluid being cooled and TL is the lower temperature of the fluid being heated.
Edited to better provide some clarity in the equation above.
RE: second law efficieny of HE
RE: second law efficieny of HE
ΔS = (-QH/TH) + (QL/TL)
You can check the enthalpy change for each fluid to get the heat transfer of the fluid and then calculate the entropy change of the system. When I dig out my steam tables, I'll throw some random temperatures down and see how it comes out.
RE: second law efficieny of HE
According to the equation you wrote, you can also write ΔS = (-QH/TH) + (QL/TL) that is QH=QL=Q if we suppose a HE without losses towards ambient, so ΔS = (-Q/TH) + (Q/TL) = Q (1/TL - 1/TH) = Q (TH - TL) / (TH*TL). This states, as I said, that entropy is proportional to the mean temperature difference of two fluids (TH - TL) and that is proportional to (TH-TL)^-1. So, taken costant TL (hot water to be heated), the closer TH is to TL (using hot water to heat other hot water) the less ΔS is (if TH was equal to TL, ΔS=0). The more TH grows (using steam) the more ΔS grows.
RE: second law efficieny of HE
Heat Exchanger Efficiency
Article in Journal of Heat Transfer · September 2007
RE: second law efficieny of HE
Second law of thermo......you can't break even
RE: second law efficieny of HE
RE: second law efficieny of HE
RE: second law efficieny of HE