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Enthalpy of Subcooled Water

  1. Jul 22, 2015 #1
    Hi!

    I was looking at some problems in Geankoplis' book on Transport Processes and Unit Ops and I came across this problem involving water at 85°C, atmospheric pressure. The problem required the enthalpy of the water and so the book used the enthalpy of saturated water at 85°C from a steam table. Clearly, the water is subcooled because the temperature is well below 100°C, thus the enthalpy should intuitively not be based on saturated conditions. Did the book use the enthalpy at saturated conditions as a close approximation of enthalpy at subcooled conditions, or is the enthalpy of subcooled water really just equals the enthalpy of saturated water at the same temperature? If the case is the former, what should be, strictly speaking, the correct way to determine the enthalpy of subcooled water?

    Thanks a lot!
     
  2. jcsd
  3. Jul 24, 2015 #2

    BvU

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    Hi there,

    [edit] withdrawn o:) , see post # 4.
    In an open system, it is not so that water is subcooled if the pressure above it is higher than the saturation pressure. The vapour pressure at 85 degrees is 0.58 Bar, meaning that there is equilibrium if the partial pressure of water vapour above it is 0.58 Bar. So in that case the book is doing the right thing.

    If there is only water in a closed system, you generally look up the compressed liquid water enthalpy. In the absence of a table for that, see at the bottom here for a correction term (basically ##v\; \Delta p##). In the exercise you describe that would amount to a correction of 44 J/kg on a saturated h of 356 kJ/kg. In that case the book is doing practically the right thing by ignoring the correction.

    Now comes the bummer: for your last question the formal answer ## \left ( \displaystyle h = h_s(p) + \int_{T_s}^T c_p \; dT \right )## only moves the difficulties to ##c_p(T)##.
     
    Last edited: Jul 24, 2015
  4. Jul 24, 2015 #3
    The correction to the enthalpy for the effect of the pressure difference is vΔP (per unit mass). You can calculate that and see that it is negligible. Look up in the steam tables the enthalpy of liquid water at 85 C and 1 atm, and compare it to the enthalpy of saturated liquid water at 85C. If there are enough significant figures in the steam table, you can compare the result of your calculation with the difference in the steam tables. It should be the same.

    Chet
     
  5. Jul 24, 2015 #4

    BvU

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    Beg to differ (minimally) with Chet. Again, the correction is an integral ##\int dh = \int T\ ds + v(p) \ dp##.
    [edit] sorry, I should have indicated integration limits (and split the integral): + vdp instead of -vdp​

    It's practically the same, but not exactly (strictly speaking)

    I don't have the numbers for 85 degrees atmospheric, but if I make the p step "a little bigger": from saturation at 80 degrees to 5 Mpa

    s(80, 5) = 1.0720 , s(80, psat) = 1.0753 T = 353 K Tds gives +1.16 kJ/kg
    v(80,5) = 0.0010268, v(80, psat) = 0.001029 vdp gives -4.95 kJ/kg
    so delta h 3.79 kJ/kg

    h(80,5) = 338.85 and h(80, psat) = 334.91, a delta h of -3.94 kJ/kg

    (using tables A-4 and A-7 Appendix A in Yunus Cengel, thermodynamics).
    Of course this isn't just a little change in p, so the correction is over 1%.
    And the Tds term is one quarter of the vdp term - so not completely negligible

    In your exercise the vdp correction is only 0.1% and good enough: the correction to the correction can be ignored.

    ----

    And now comes the realization that I have to withdraw my first paragraph in post #2: the water is at 1 atmosphere, so the situation is the same as in the closed system. My bad !
     
    Last edited: Jul 25, 2015
  6. Jul 24, 2015 #5
    The general expression for dH (which takes into account both TdS and Vdp) is given by:$$dH=C_pdT+\left[-T\left(\frac{\partial V}{\partial T}\right)_p+V\right]dp$$
    If the liquid is considered incompressible and the contribution of the thermal expansion coefficient term is neglected, the correction is VΔp.

    Chet
     
    Last edited: Jul 24, 2015
  7. Jul 24, 2015 #6
    Hi again guys. I redid the calculation that BvU carried out in post #4, and got the same answer, 4.95 kJ/kg. So I tried including the coefficient of thermal expansion term, and it only gave a result that is 8% lower, at 4.58 kJ/kg. So, I am at a loss for explaining the difference between the 3.94 kJ/kg and the 4.58 kJ/kg. Presumably, they employed the same formula to develop the steam tables. I even checked another steam tables, and it gave 3.75 kJ/kg. So, I'm puzzled. Any ideas? Maybe the coefficient of thermal expansion is higher at the high temperature than the value I used.

    Edit: Yes, I checked, and the cte at 80C is much higher than the value at 20C, and exactly enough to explain the difference between the 4.95 and the 3.94. Ain't science great!!!

    Chet
     
    Last edited: Jul 24, 2015
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