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Thermodynamics - Internal Energy to Find Heat Added

  1. Sep 26, 2015 #1
    1. The problem statement, all variables and given/known data
    image.png

    2. Relevant equations
    Assumptions
    : Closed System and Quasi-static.
    p = 90 + 73,386*V -> V = (p-90)/(73,386)
    v = V/m
    Q - W = U2 - U1
    mass balance: m = m1 = m2

    3. The attempt at a solution

    At state 1:
    p1 = 110 kPa
    V1 = (110-90)/(73,386) = 2.725*10-4 m3
    v1 = V1/m1 = 2.725*10-4 / 0.1 = 2.725*10-3 m3/kg

    At state 2:
    p2 = 1000 kPa
    V2 = (1000-90)/(73,386) = 124.002*10-4 m3
    v2 = V2/m2 = 124.002*10-4 / 0.1 = 124.002*10-3 m3/kg

    Now I've got two independent properties of the steam at states 1 and 2, pressure and specific volume. I should be able to use these to find their respective internal energies from the steam tables. However, I cannot find where to get these points.. As I'm thinking that by getting the internal energies I can substitute them into the No Flow Energy Equation and find the heat added as there is no work into or out of the system thus:

    Qin = U2-U1


    Any help in the right direction would be appreciated greatly.
     
  2. jcsd
  3. Sep 26, 2015 #2

    SteamKing

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    Well, do you have steam tables for saturated water?

    If you check these tables for saturated water, you should find a specific volume for 100% vapor and 100% liquid at a given pressure. Where does your v1 fall within these two extremes?

    Do you understand what vapor quality means?

    You should be able to determine the vapor quality of the steam at the indicated pressures, and then use this information to find the internal energy.
     
  4. Sep 26, 2015 #3
    I do have the steam tables for saturated water, and yes I understand what vapour quality means. I have found the specific volume at 100% liquid and vapour, and respective vapour qualities at both p1 and p2.

    At p1:
    v1 = 2.725*10-3 m3/kg
    vf = 0.1046x10-2 m3/kg
    vg = 1.549 m3/kg
    x = 0.0011

    At p2:
    v2 = 124.002x10-3 m3/kg
    vf = 0.1127x10-2 m3/kg
    vg = 0.1944 m3/kg
    x = 0.6357

    Am I understanding what you're saying correctly, by knowing the vapour quality at both p1 and p2, I can then use the internal energy at 100% vapour and 100% liquid with the vapour quality to find the amount of internal energy at both points p1 and p2?

    As such:

    At p1 (110 kPa):
    uf = 429 kJ/kg
    ug = 2510 kJ/kg
    u1 = uf + x*(ug-uf) = 429 + 0.0011(2510-429) = 431.2891 kJ/kg
    U1 = u1*m1 = 431.2891 kJ/kg * 0.1 kg = 43.1289 kJ

    At p2 (1000 kPa):
    uf = 762 kJ/kg
    ug = 2584 kJ/kg
    u2 = uf + x*(ug-uf) = 762 + 0.6357*(2584-762) = 1920.2452 kJ/kg
    U2 = u2*m2 = 1920.2452 kJ/kg * 0.1 kg = 192.0245 kJ

    Therefore using the NFEE:

    Qin = U2 - U1 = 192.0245 - 43.1289 = 148.8956 kJ or 149 kJ (3 SF).
     
  5. Sep 26, 2015 #4

    SteamKing

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    Yes, you understand my point exactly.
     
  6. Sep 26, 2015 #5
    Thank you for your help once again.

    If you don't mind, would be able to help me understand how I could get the specific volume for steam as it passes through critical point?

    The information given is: Sealed system of volume 10-4 m3, containing liq/vap mixture in equilibrium at 100 kPa. It states that "The fraction of liquid and vapour is such that when heated the steam passes through the Critical Point". I need to find this specific volume at CP to find the initial dryness fraction.

    I'm assuming that:

    v(critical point) = vf + x(vg-vf), x = [(v(critical point)-vf)/(vg-vf)].
     
  7. Sep 26, 2015 #6

    SteamKing

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    The critical point of water is where liquid and vapor can coexist, but not as a mixture:

    https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)

    For water, the critical point occurs at the end of the saturation line, where P = 22,064 kPa and T = 373.95° C. If you check your steam tables at this point, you will see that vf = vg
     
  8. Oct 3, 2015 #7
    1. The problem statement, all variables and given/known data
    A cylinder fitted with a piston contains 0.1 kg of H2O at 110 kPa. Heat is added in a quasi-static process, during which pressure p (kPa) is related to volume V (m3) by

    p = 96 + 70,968*V

    until the pressure is 1000 kPa. Determine the heat added in kJ, to three significant figures.

    2. Relevant equations
    Assumptions
    : Closed System and Quasi-static.
    p = 90 + 70,968*V -> V = (p-90)/(70,968)
    v = V/m
    Q - W = U2 - U1
    mass balance: m = m1 = m2

    3. The attempt at a solution

    p1 = 110 kPa
    V1 = (110-96)/(70968) = 1.97*10-4 m3
    v1 = V1/m1 = 1.97*10-4 / 0.1 = 1.97*10-3 m3/kg

    p2 = 1000 kPa
    V2 = (1000-96)/(70968) = 127*10-4 m3
    v2 = V2/m2 = 127*10-4 / 0.1 = 127*10-3 m3/kg

    At p1:
    v1 = 1.97*10-3 m3/kg
    vf = 0.1046x10-2 m3/kg
    vg = 1.549 m3/kg
    x = 0.0006

    At p2:
    v2 = 127x10-3 m3/kg
    vf = 0.1127x10-2 m3/kg
    vg = 0.1944 m3/kg
    x = 0.6512

    At p1 (110 kPa):
    uf = 429 kJ/kg
    ug = 2510 kJ/kg
    u1 = uf + x*(ug-uf) = 429 + 0.0006(2510-429) = 430.2486 kJ
    U1 = u1*m1 = 430.2486kJ/kg * 0.1 kg = 43.0249 kJ

    At p2 (1000 kPa):
    uf = 762 kJ/kg
    ug = 2584 kJ/kg
    u2 = uf + x*(ug-uf) = 762 + 0.6512*(2584-762) = 1948.4864 kJ/kg
    U2 = u2*m2 = 1948.4864kJ/kg * 0.1 kg = 194.8486 kJ

    Qin = U2 - U1 = 194.8486 - 43.0249= 151.8237 kJ or 152 kJ (3 SF).

    I have been following Sirsh's step with my own values given but i didnt manage to get it correctly.
     
  9. Oct 3, 2015 #8
    What about the work? You know, the integral of PdV.

    Chet
     
    Last edited: Oct 3, 2015
  10. Oct 3, 2015 #9
    Chet, i still get the wrong answer. Q= (U2-U1) + W
    whereas W= integral PdV = P(V2-V1)

    Based on Sirsh calculation for W = 110( 124.002*10-4 m3 - 2.725*10-4 m3 ) =1.334 kJ
    Is my calculation for W correct?
     
  11. Oct 3, 2015 #10
    That isn't the integral of PdV.

    Integrate the function given in the question to find the volumes for state 1 and 2.
     
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