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How to calculate the power consumption during steam compression

  1. Jun 24, 2016 #1
    Being a man who learned physics in college, I have tried to calculate that in the simplest way. As for example, if we need to know the power consumption to compress steam from 50C saturated pressure level to 1 bar i.e. 100C saturated pressure level, the best way is to know the gross enthalpy of both the levels first and then subtract the lesser from the bigger.
    As for example, enthalpy of of water at 50C is 209.0 J/g and the latent heat of vaporisation is 2381.4 J/g. Therefore, gross enthalpy of steam is 2590.4 J/g. While that of water at 100C is 419 J/g and 2256.3 J/g and that means 2675.3 J/g. Therefore, in ideal case the power necessary is (2675.3 - 2590.4) J/g or 84.9 J/g. With a 70% efficient compressor, it will go to around 121 J/g.
    But, recently a steam expert shows me how to calculate the power. It's the same as gas. But, where I have doubt is here. We all know that gas and vapour are basically different. We can only call a gaseous fluid a "gas" when its temperature is above the "critical temperature". Below the critical temperature, it's vapour. Therefore, how the laws of gas compression can be applied to steam (having critical temperature at 375C) at far below its critical temperature. But, as the opinion comes from an expert, I haven't gone into debate with him. Just want to put my doubts before the readers here.
     
  2. jcsd
  3. Jun 24, 2016 #2

    JBA

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    Since your steam is saturated at the starting pressure of 50c, as opposed to superheated steam which can be treated as a gas, I think using the gas laws for your calculation is questionable.
     
  4. Jun 24, 2016 #3
    Let me understand your question. You have a continuous flow compressor, with steam entering the compressor saturated at 50 C, and you want to run the compressor in such a way that you get saturated steam coming out of the other end of the compressor at 100 C and 1 bar. You want to find the shaft work to run the compressor. Is the compressor supposed to be operating adiabatically, or can heat be transferred through the walls?
     
  5. Jun 24, 2016 #4
    Lets assume that the steam compressor is running adiabatically.
     
  6. Jun 24, 2016 #5
    If the compression is adiabatic (and assumed essentially reversible), the steam at 1 bar can't be saturated. It must be superheated. (So you have a vapor in both states). From your steam tables, what is the enthalpy of superheated steam at 1 atm and having the same entropy as saturated steam at 50 C (you may have to interpolate in the table)?

    Now, let's try it a different way (your friend's method). What is the specific volume of saturated steam at 0.5 bar, and what is the specific volume of the superheated steam at the final state? (We can continue after you have provided this data).
     
  7. Jun 25, 2016 #6
    The gross enthalpy of saturated steam I have already told. To know the enthaply of superheated steam, the degree of superheating is necessary.
    Whatsoever, it seems that the saturated steam becomes superheated very quickly when the compression process begins and they behave like gas during the compression process.
     
  8. Jun 25, 2016 #7
    Yes. As I said, you can get the final temperature by requiring that the entropy per unit mass is the same for the compressed steam at 1 bar as for the initial saturated steam at 0.5 bar. You can get all this from the steam tables.
    Yes. So, in a way, you and your friend were both correct. You can get the work either by getting the enthalpy change from the steam tables or by using the ideal gas equation for an adiabatic reversible compression to get the final temperature. Both results should come out pretty close to the same value.
     
  9. Jun 25, 2016 #8
    No. Both results are very different. From the enthalpy change factor, it's just 84.9 J/g while if we use the gas formula, it will become around 500 kW.
     
  10. Jun 25, 2016 #9
    Get outta town!!! Both methods have to give essentially the same results. How can you compare something having units of J/g with something having units of kW? What is the 500 kW in J/gm of gas.

    I'm going to do it both ways, and prove it to you. Be back later.

    Chet
     
  11. Jun 25, 2016 #10
    Enthalpy Method:

    At the compressor inlet, the thermodynamic parameter values are:
    P = 0.5 bar
    T = 81.3 C
    h = 2646 kJ/kg
    s = 7.594 kJ/kgK

    At the compressor outlet, P = 1.0 bar, and the relevant parameter values are:
    @ T = 120 C
    h = 2717 kJ/kg
    s = 7.467 kJ/kgK

    @ T = 160 C
    h = 2796 kJ/kg
    s = 7.660 kJ/kgK

    We know that, for adiabatic reversible compression, outlet entropy must be equal to the inlet entropy (7.594)
    Therefore, we know that the final temperature lies between 120 C and 160 C. Interpolating to get the final temperature yields:
    $$\frac{(T-120)}{(160-120)}=\frac{(7.594-7.467)}{(7.660- 7.467)}$$
    This gives an outlet temperature of 146.3 C. Similarly interpolating to get the outlet enthalpy, we obtain:
    h = 2769 kJ/kg

    Therefore the change in enthalpy from inlet to outlet is ##\Delta h=2769-2646=123##kJ/kg
    Since the compressor is adiabatic, this is also equal to the shaft work done by the compressor on the gas.

    Ideal Gas Method:

    In the ideal gas region (low pressures), the heat capacities of water vapor (evaluated from the steam tables over the temperature range of interest) are:
    ##C_p=1.90##kJ/kgC

    ##C_v=1.45##kJ/kgC

    So the ratio of the heat capacities is:
    $$\gamma = \frac{1.90}{1.45}=1.31$$

    For an ideal gas undergoing an adiabatic reversible compression, the absolute temperature is proportional to the pressure raised to the ##(\gamma - 1)/\gamma## power. Therefore, treating the steam as an ideal gas, the final temperature is given by:
    $$\frac{T}{(273+81.3)}=\left(\frac{1}{0.5}\right)^{\frac{0.31}{1.31}}$$
    This gives ##T = 417.5\ K=144.5\ C##

    This give a change in enthalpy of ##\Delta h=C_p\Delta T=1.90(144.5-81.3)=120\ kJ/kg##

    Conclusion:

    For this problem, the results obtained using the Ideal Gas Method match very closely those obtained using the (more accurate) Enthalpy Method.
     
  12. Jun 25, 2016 #11
    At 1 bar, the temperature of saturated steam is 100C, how you have considered the temperature to be 120C and 160C. If the process is isentropic, that means it's 100% efficient and the steam will remain saturated throughout the process.
     
  13. Jun 25, 2016 #12
    As we already said, if you compress the saturated inlet steam adiabatically and reversibly, the steam will not be saturated at the outlet of the compressor. It will be superheated.
    Where did you get this strange idea? Isentropic compression doesn't mean that the steam will remain saturated throughout the process. Just look at the entropy of saturated steam in the steam tables and tell me if the tables show that the entropy of saturated steam at 0.5 bar is the same as the entropy of saturated steam at 1.0 bar.

    Chet
     
  14. Jun 25, 2016 #13
    Can you tell me how the temperature will go to 120C or 160C? Probably this is the most important part of this discussion.
     
  15. Jun 25, 2016 #14
    When you compress a gas which exhibits nearly ideal gas behavior adiabatically and reversibly, its temperature rises. Since Q = 0, you have ##du=C_vdT=-Pdv=-\frac{RT}{v}dv##. If you integrate this equation, you obtain the relationship I gave between T and P in the ideal gas analysis portion of my post.

    Also, as I showed in my post, even if you don't assume nearly ideal gas behavior and employ the (non-ideal) steam tables, for adiabatic reversible (isentropic) compression, the compression causes the vapor to increase in temperature (as a result of the adiabatic work that you do on the vapor) to a temperature between 120 C and 160 C. You can see this from the fact that the outlet entropy of the vapor will equal the inlet entropy at a temperature within this range.

    Note that I solved the problem by two entirely different methods and got virtually the same answer.
     
  16. Jun 25, 2016 #15
    That's the basic question I have asked in this thread. My point is that vapour and gas are totally two different entities. Vapour can be liquefied by compressing it while gas can't be liquefied by just compressing it if its temperature is NOT below the "critical temperature". And that's why I just want to know whether gas formulas can be applicable to vapour/steam too or not.
     
  17. Jun 26, 2016 #16
    In my judgment, saying that "vapour and gas are totally two different entities," is waaaaay beyond reasonable. It's only a matter of terminology to call a non-condensed phase above its critical temperature a gas and a non-condensed phase below its critical temperature as a vapor. And very often the terms gas and vapor are used interchangeably.
    Both a gas and a vapor obey the ideal gas formula in the limit where the pressure is low compared to the critical pressure.

    If you compress water vapor isothermally, it will, of course, eventually condense. However, if you compress water vapor adiabatically, it will not condense because the increase in temperature resulting from doing work on the vapor will be too large, and move you away from the saturation envelope rather than toward the saturation envelope. The figure below illustrates this for our problem.
    PHD.PNG
    This is a thermodynamic diagram for water. The green dot on the figure shows the conditions at the inlet to our compressor at 0.5 bar and 81.3 C. Note that this dot lies on the saturation curve for the vapor. The red dot shows the conditions at the oulet of the compressor at 1 bar and ~ 146 C. This dot lies outside the saturation envelope for water, squarely in the superheated region. Note the lines of constant entropy on the diagram. For this adiabatic reversible compression (isentropic), both the red dot and the green dot lie on the same constant entropy line s = 7.594 kJ/kgK.

    Our analysis using the ideal gas law gives virtually the same results as those using the steam tables or the thermodynamic diagram. This in itself speaks for the validity of using the ideal gas law for the vapor.

    Hope this helps.
     
  18. Jun 26, 2016 #17
    Thank you for this detailed answer. But, still the question is how it can be calculated that the temperature of steam will be at 146C. The graph you have given is the pressure-enthalpy chart of steam and I again want to know how to calculate the final temperature.
     
  19. Jun 26, 2016 #18
    I provided the entire analysis for this in post #10, using both the Enthalpy Method and the Ideal Gas Method. The pressure-enthalpy chart shows constant temperature lines, and you can see that the final condition (red dot) lies about half way between the 100 C line and the 200 C line. The key to getting the final temperature is to treat the compression as being adiabatic and reversible.
     
  20. Jun 26, 2016 #19
    The graph is clear, but I just want to know how you have calculated that the final temperature is 160C.
     
  21. Jun 26, 2016 #20
    The final temperature is not 160 C. It is 146 C. At the exit pressure of 1 Bar, you need to determine the temperature that gives you the same entropy as that at the inlet. So getting the final temperature using the Enthalpy Method involves interpolating in the steam tables:

    At the compressor outlet, P = 1.0 bar, and the relevant parameter values are:
    @ T = 120 C
    h = 2717 kJ/kg
    s = 7.467 kJ/kgK

    @ T = 160 C
    h = 2796 kJ/kg
    s = 7.660 kJ/kgK

    We know that, for adiabatic reversible compression, outlet entropy must be equal to the inlet entropy (7.594)
    Therefore, we know that the final temperature lies between 120 C and 160 C. Interpolating to get the final temperature yields:
    $$\frac{(T-120)}{(160-120)}=\frac{(7.594-7.467)}{(7.660- 7.467)}$$
    This gives an outlet temperature of 146.3 C.
     
    Last edited: Jun 26, 2016
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