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First law of thermodynamics applied to a closed system

  1. Aug 31, 2015 #1
    1. The problem statement, all variables and given/known data
    - A well insulated rigid steel tank contains water (only)
    - A resistance heating element (240V, 3.67Amp) designed to heat the water as well as inlet and outlet valves for - the water.
    - At state-1, the water is in a saturated liquid-vapor mixture state at P(abs)=100kPa, measures 5kg and the portion of it in liquid form is 75%, the rest being vapor
    - To reach state-2, more power is supplied to the heater, but just enough to vaporize all the remaining liquid. When all the liquid is vaporized, the heater is switched off.

    1.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-1.
    2.) Calculate the total internal energy (units: kJ) for the water at state-1.
    3.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-2.

    2. Relevant equations
    yavg=yf+x(yfg)
    ΔU=mΔu
    3. The attempt at a solution

    1.)
    I went to the tables and read of the vf and vg data @ 100kPa. For quality, seeing as 75% is liquid that would make it 0.35. I applied that data to the following equation: yavg=yf+x(yfg) and got:
    vavg = 0.001043+0.25(1.6941-0.001043) = 0.4243 m^3/kg

    2.) I did the same thing for this part I read the data for uf and ufg from the tables @ 100kPa and applied it to this equation: yavg=yf+x(yfg) and got:
    uavg=417+0.25(2088.2) =939.65
    Which I then plugged into ΔU=mΔu, seeing as its the same state I only needed uavg:
    U= 5kg*1147.87 = 4695.25 kJ

    3.) In the statement it says at state 2 all the water is vaporized which would make it a quality of 1 and it would be a saturated vapor..so all I did was read of the value for saturated vapor at 100kPa which is 1.6941m^3/kg

    Wasn't 100% on a few of these, any feed back would be great :)
    Thank you in advance
     
    Last edited: Aug 31, 2015
  2. jcsd
  3. Aug 31, 2015 #2

    SteamKing

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    If 75% is liquid and 35% is vapor, where did the extra 10% come from?
     
  4. Aug 31, 2015 #3
    *sigh* Stupid error... how embarrassing haha. Apart from that is the way I went about it correct?
     
  5. Aug 31, 2015 #4

    SteamKing

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    You seemed to start off OK, but I stopped checking when I got to the quality mistake.
     
  6. Aug 31, 2015 #5
    I have gone back and fixed the errors.
     
  7. Aug 31, 2015 #6
    Part 3 is done incorrectly. The pressure is no longer 100 kPa after the water is heated to the point where all the liquid water has vaporized. But, what actually is still constant?

    Chet
     
  8. Aug 31, 2015 #7

    SteamKing

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    No, you still aren't applying the concept of vapor quality correctly to calculate the properties of a mixture of vapor and liquid.

    If x% is vapor, then (100 - x%) will be liquid, and you must calculate the thermo properties of the mixture accordingly.

    https://www.engineersedge.com/thermodynamics/steam_tables.htm
     
  9. Aug 31, 2015 #8
    Hmmm. It looks correct to me. 0.25x1.6941+0.75x0.001043
    What am I missing?

    Chet
     
  10. Aug 31, 2015 #9

    SteamKing

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    Below is the OP's calculation, Chet. Notice something different w.r.t. your calculation?
     
  11. Aug 31, 2015 #10
    I looked on the website you supplied and read this "v=vf+x(vfg)" isn't that what I've done? The way I look at it is..if 75% is liquid its closer to the Sat Liq point (0 quality) on the Tv diagram.

    Also, me and Chet get the same answer?


    The mass would be constant so then I could use v = V/m ...but I don't have volume so I would then need to rearrange ρ=m/V but I don't have ρ...would it be correct to assume temperature is constant as well? since its only moving along (horizontally) the Tv diagram...because if thats the case I can use the temperature @ 100kPa in the tables to get the density for water for that temperature
     
  12. Aug 31, 2015 #11
    It's not really different. It's just doing the factoring differently. The result is the same.

    Chet
     
  13. Aug 31, 2015 #12

    SteamKing

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    You and the OP are missing the point.

    Take a look at the OP's calculation of internal energy:
    The OP is taking 100% of the internal energy of the liquid phase ...
     
  14. Aug 31, 2015 #13
    I'm really confused now haha... I know x = mVapor/mTotal.....if 25% is vapor of the 5kg of mass = 1.25kg/5kg = 0.25?

    Thank you for your patience guys
     
  15. Aug 31, 2015 #14

    SteamKing

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    Yes, 25% of the mass is vapor, and 75% of the mass is liquid. When you figure the thermo properties for a liquid-vapor mixture, you've got to use the proportions of each property to find the property for that combination of liquid and vapor.

    For example, if you want to calculate the internal energy of this mixture of liquid water and water vapor, you can't use 100% of the internal energy of the liquid and 25% of the internal energy of the vapor, you can only have 75% of the internal energy of the liquid, because only 75% of the sample is liquid ...

    That's why I linked to this article:

    https://www.engineersedge.com/thermodynamics/steam_tables.htm
     
  16. Aug 31, 2015 #15
    Oh. I assumed the 2088 was the difference between the liquid and the vapor.

    Chet
     
  17. Aug 31, 2015 #16
    I understand what you're saying, but if I'm to follow this equation "v=vf+x(vfg)" and just above I worked the quality out to be 0.25..I don't see what else I'm supposed to use for x ?

    2088 is the difference "ufg" (2505.6-417)
     
  18. Aug 31, 2015 #17
    Both the mass of water and the volume of the container are constant. So, if that's the situation, how does the average specific volume in case 1 compare with the specific volume of the pure vapor in case 3?

    Chet
     
  19. Aug 31, 2015 #18
    Ok..so I have 5kg of water and my spec volume for state 1 is 0.4243m^3/kg...to get the volume of state one that would be m*v1 = 2.1215m^3/kg

    considering volume is constant V1=V2...therefore v2=V2/m ---> v2=2.1215/5= 0.4243...which should be the case since no mass has left the system and the volume hasn't changed.

    I'm still confused as to whats wrong with my quality equations...is my quality value wrong? or the numbers I've used from the tables? I've used the equation provided. I don't see what is incorrect. How am I using 100% of the liquid phase for u when I've followed the equation uavg=uf +xufg ? :\ was I supposed to multiply 0.75 to uf? if so..then thats not really following the equation.
     
    Last edited: Aug 31, 2015
  20. Aug 31, 2015 #19
    OK. So the specific volume of the vapor for case 3 is 0.4243 m^3/kg. From your steam tables, what saturation pressure, temperature, and specific internal energy of the vapor does this correspond to? For 5 kg of this vapor, what is its internal energy?

    Chet
     
  21. Sep 2, 2015 #20
    Apologies for the delay, had a mid semester exam to study for,
    (Had to interpolate)
    Sat Pressure: 439.229kPa (0.46242-0.41392)/(0.4243-0.41392)=(400-450)/(x-450)
    Temp: 146.982 degrees Celsius (0.46242-0.41392)/(0.4243-0.41392)=(143.61-147.9)/(x-147.9)
    Internal energy (avg): 2556.24 kJ/kg (0.46242-0.41392)/(0.4243-0.41392)=(2553.1-2557.1)/(x-2557.1)

    U = m* uavg = 12781.2 kJ
     
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