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Applying first law of thermo to a closed spring piston

  1. Feb 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A cylinder having a piston restrained by a linear spring (of spring constant 15 kN/m) contains 0.5 kg of saturated vapor water at 120ºC. Heat is transferred to the water causing piston to rise. If the piston’s cross sectional area is 0.05 m2 and the pressure varies linearly with volume until a final pressure of 500 kPa is reached,

    1-Find the final temperature in the cylinder.

    2-Calculate the amount of heat transfer.

    3-Calculate the change in the internal energy of water.

    4-Calculate the work is done against the spring.

    5-Sketch the process on a T-v phase diagram.

    2. Relevant equations
    ΔU=Q+W
    W=∫ PdV
    F=kx , linear spring displacement equation
    v=V/m
    u=U/m


    3. The attempt at a solution

    So I'd like some fresh opinions to just verify that my approaches are sensible and I correctly understand everything here. Firstly - I'm imagining this system initially as a "saturated vapor of 100% steam quality".

    Finding the final temperature

    Using the given temperature I look in the steam tables for saturated water and find a specific volume of vg=.891 m3/kg. I multiply this by the mass
    .5 kg to get an initial volume of .4459 m3. I then use the final pressure
    and area of the piston face (to find force) with the spring equation to find that the spring is compressed by 1.666m. This times the piston area gives a newly added volume of .529 m3. I add this to the original volume to find new total volume. I find the new specific volume, and
    look in the steam tables to verify that it is greater than saturation specific
    volume of gas at the new final temperature, confirming the system ends
    up as a superheated gas. I look in the appropriate steam table to locate
    this new specific volume and final pressure, finding that the temperature is approximately 875 C after some extrapolation.

    Finding the change in Internal Energy ΔU
    Going back to the steam table for the initial state, I find that the specific internal energy ug (since it's 100% gas), is 2529.24 KJ/kg. Using the
    mass, I find an initial internal energy U1 of 1264.6 KJ. Using the data
    for the final state found earlier in the superheated region, I find a new specific internal energy of 3805.7 KJ/kg. Using the constant mass,
    I find a final internal energy U2 of 1902.8 KJ. The change in internal
    energy ΔU is therefore 638.18 KJ.


    Finding the amount of Work done
    I make a line graph of Pressure vs Volume, plotting the initial and final
    states. I found the initial pressure by looking at the saturated steam
    tables, finding it to be 198.5 KPa. Using the listed equation for finding
    Work, I understand that the Work is equal to the area under the curve of
    P vs V. However, because I don't have two endpoints and the function
    is linear, instead of determining the function relating P and V and integrating it I just segment the area under the curve into a triangle and a
    rectangle. I find the area of each and sum them, coming out to a Work of
    20.7 KJ.
    It's positive because the system has expanded and done work
    against its surroundings. (left out all the numbers for sake of brevity/clarity)

    Finding the heat transfer Q
    To find Q, I rearrange the first law of thermodynamics to find that
    Q=ΔU-W. Then simply use the values of ΔU and W from earlier to find
    a Q of 617.475 KJ.

    To sum up - I expected energy change, work, and heat transfer to be positive as the system underwent a large increase in pressure while increasing in volume. These numbers make sense to me, at least as far as their signs. If you see any glaring issues with my approach to the problem please let me know. Any and all advice is appreciated, thank you.
     
  2. jcsd
  3. Feb 18, 2016 #2
    I got a work of 29.1973 KJ

    Area Under Curve=0.5(P1+P2)(V2-V1)=(0.5)(0.529-0.4454)(500-198.5)=29.1973 KJ
     
  4. Feb 18, 2016 #3
    I think you're right. I think since I set the baseline of my pressure axis at 100 KPa, I used 98.5 as the height of the rectangle instead of 198.5. Thanks for catching that mistake!
     
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