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# 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

### RE: second law efficieny of HE

Typically, your heat exchanger is designed around your process parameters, not vice versa.

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

The advantage of steam over water for heating is that steam can transfer a great deal of heat by condensing without losing temperature. Water loses temperature when transferring heat, which makes much less useful.

### RE: second law efficieny of HE

2nd law is irrelevant for a HX. A HX is generally considered to be an adiabatic system (no heat transferred to or from the surroundings) so all heat removed from the hot fluid is transferred to the cold fluid (1st law). In those terms all HX would be 100% efficient regardless of media choice.

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

As I see it, and I may be mistaken, the thermal efficiency of a heat exchanger is generally expressed as the ratio of the actual heat transferred to the optimum heat transfer Qopt = UA (AMTD).
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.

### RE: second law efficieny of HE

(OP)
thanks everyone for replies

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

Since the heated water exits at the same temperature at the end, the energy transferred must be the same for both cases, so work done must be the same in both cases.

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

giuppi, not sure where you're getting that entropy is proportional to the mean temperature difference; I don't have my old thermodynamics textbook handy, but that's not ringing a bell. As such, I don't believe your sketch is accurate. ΔT of your two fluids is not necessarily reflective of the ΔS of the system; a T-S diagram of the system is what you really need to graphically demonstrate the behavior of the system with respect to its overall entropy change.

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

According to that equation, if hot water and steam have the same TH, for equal heat duties Q, and equal TL, one gets equal ΔS?

### RE: second law efficieny of HE

Good catch - I over-simplified too much. The heat transfer from the hot fluid and heat absorbed by the cold fluid won't be equal. It should really be:

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

(OP)
KoachCSR

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

See

Heat Exchanger Efficiency
Article in Journal of Heat Transfer · September 2007

### RE: second law efficieny of HE

First law of thermo .....the best you can do is break even

Second law of thermo......you can't break even

### RE: second law efficieny of HE

What pressure steam and what temperature hot water is available? Are you talking degrees Centigrade or Farenheit?

### RE: second law efficieny of HE

(OP)
3 barg pressure of steam, water available at 80°C. Celsius

### RE: second law efficieny of HE

Choose hot water because you will get better control than using steam. If th3 steam control valve leaks it can cause the water being heated to boil.

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