How to use steam table to solve the following:?

A compressor adiabatically & reversibly compresses a mixture of saturated water and steam from a pressure 1.01 bar and specific volume of 1.5 m^3/kg to 0.313 m^3/kg. USE STEAM TABLES to find the following:

a) fraction of steam at the compressor entrance.

b) the exit entropy in kJ/kgK

c) The exit pressure in bar.

I have the Thermodynamic and Transport Properties of Fluids steam table (by Rogers ad Mayhew). I have no idea how to use it to solve for the above. Please help me. Best answer for the first person to offer help :-) Thanks!

2 Answers

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  • 7 years ago
    Best Answer

    Go to some page entitled "Saturated Water Properties, SI Units" or like that.

    Part(a)

    At exactly 1 bar, the specific volume of the steam portion would be 1.694 m^3/kg.

    At 1.013 bar, the specific volume of the steam portion would be 1.673 m^3/kg.

    Therefore, at 1.01 bar, the specific volume of the steam portion will be around 1.678 m^3/kg.

    The specific volume of the water portion will be very close to 0.001 m^3/kg,

    so if the specific volume of the mixture is 1.50 m^3/kg, you have

    1.678 F + 0.001(1-F) = 1.500

    where "F" is the fraction of steam.

    1.677 F = 1.499 => F = 89.4%

    (b) I think the "adiabatically & reversibly" means that delta-S = 0.

    Therefore, we just have to find the ENTRANCE entropy,

    On the tables I'm looking at, the entropies of water and steam respectively are

    1.307 kJ/(kg K) and 7.358 kJ/(kg K for 1.01 bar.

    For a mixture that's 89.38% steam, the specific entropy would be

    (0.1062)(1.306 kJ/kg K) + (0.8938)(7.358 kJ/kg K) = 6.715 kJ/(kg K)

    This is the entrance entropy, but I believe it's the exit entropy as well.

    (c) So we need to find a pressure where a specific volume of 0.313 m^3/kg

    could correspond to a specific entropy of 6.715 kJ/(kg K).

    At 543.1 kPa, the specific volume of the steam portion is 0.3468 m^3/kg

    and at 617.8 kPa, the specific volume of the steam portion is 0.3071 m^3/kg.

    For the moment, I'm going to fudge the volume of the water portion, which will

    account for no more than 1/300 of the overall volume.

    If the specific volume of steam portion is (let's say) 0.314 m^3/kg,

    then the pressure should be about

    617.8 kPa - (7/38.7)(617.8 kPa - 543.1 kPa) = 604 kPa

    but let's see what steam fraction that gives in entropy-land...

    Specific entropy of saturated steam at 604 kPa would be about 6.758 kJ/(kg K),

    and the specific s of saturated water at 604 kPa would be about 1.934 kJ/(kg K).

    For this to end up as a specific entropy of 6.715 kJ/(kg K),

    we would need

    6.758 F + 1.934 (1 - F) = 6.715

    4.824 F = 4.781 and F = 99%, which makes no sense to me.

    I would be thinking the steam fraction should decrease in the compression.

    So you can't trust my answer to part (c) [604 kPa absolute],

    but I'm pretty comfortable with parts(a) and (b).

  • 6 years ago

    can anyone send me steam table of (by Rogers ad Mayhew)

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