1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Thermodynamics: Piston and Cylinder

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A piston-and-cylinder device contains 5 kg of water initially at 150◦C and 0.20 MPa. The frictionless piston is then pushed slowly in an isothermal process until the volume of water becomes 10% of its initial value. Calculate the heat and work exchanged between the device and the surroundings during this process.

    2. Relevant equations
    Energy balances, W=-P∫dV, etc

    3. The attempt at a solution
    I am stuck at where to start, as I feel like this is a simple problem but I am making it complicated.

    Im not sure what to do since this is a liquid..The work I was doing was for a gas (since I was using the ideal gas law)

    Any help would be greatly appreciated.
  2. jcsd
  3. Apr 5, 2013 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Why do you think it is a liquid?

  4. Apr 5, 2013 #3
    Hm..I actually had a brain fart about that..Thanks!

    I'll treat it as an ideal gas then..

    I am asked to find work and heat..
    ..so I can find the work since W=-RTln(V2-V1)

    ..and the initial U1 is known since it is a function of temperature and pressure, so all that is needed to find the heat is U2..but how would I find this?
  5. Apr 5, 2013 #4

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You can't treat it as an ideal gas either. Water vapour is NOT an ideal gas. During a compression, potential energy is converted to kinetic energy so it is complicated. You will have to use steam tables to determine the change in enthalpy, work out the change in PV and determine the change in internal energy.

    Can you give us the whole question?

  6. Apr 5, 2013 #5
    Thank you so far for your time. That was the entire question given.

    Okay I understand about not being able to treat it as an ideal gas, hence the substitution I made using PV=RT is not valid, correct?

    So then, the work=-∫PdV=P(V2-V1) then?

    Would I be able to make an assumption that this is a reversible process since ΔT=0 ?
    If so, S2-S1=(Q/T) or Q=T(S2-S1)
  7. Apr 5, 2013 #6
    The first step in this problem is to determine the state of the water in the cylinder prior to compression. To do this, you need to find out the equilibrium vapor pressure of water at 150 degrees C. You can find this on the web, look it up in your textbook, or find it in the steam tables, if you have covered steam tables in your course. Next, make a determination of whether the initial pressure of 0.2 MPa is higher or lower than the equilibrium vapor pressure of water at 150 C. Based on this comparison, what is your conclusion as to the state of the water in the cylinder?

    Superheated vapor?
    Saturated vapor and liquid in equilibrium?
    Purely liquid water (unsaturated)?

    After you have done this, you can determine the initial volume of the cylinder (provided you know how to use the steam tables, have an equation of state for water, or approximate the behavior of water as an ideal gas. But the latter may be a little inaccurate at the pressure and temperature involved, as AM has pointed out.

    Once you have made these determinations, get back with us, and we will help you with the next steps in the analysis.

    Last edited: Apr 6, 2013
  8. Apr 5, 2013 #7
    Thanks Chet!

    So, looking at the initial conditions of T=150 degrees C, and P=0.20 MPa I am looking at a superheated vapor..correct?
  9. Apr 6, 2013 #8
    Yes. Correct. Now before starting to write equations, it is helpful to dope out what is happening qualitatively in this problem.

    The superheated vapor starts out at a pressure of 0.2 MPa and is compressed isothermally. As it is compressed, its pressure builds up. What happens when the pressure just reaches the saturation vapor pressure of 0.47 MPa? After this, what is the state of the contents of the cylinder when the contents is compressed further beyond this point? (If we approximate the vapor as an ideal gas, the compression will be about 2.35 X when the pressure reaches 0.47 MPa. So if the total compression is 10X, additional compression will be occurring after the pressure reaches 0.47 MPa)

  10. Apr 6, 2013 #9
    well, once we reach 0.47 MPa we are dealing with a saturated steam now..which means that there must be an equilibrium between the liquid that forms (very little) and the vapor correct? According to my steam table, the specific volume of liquid at 150 degrees C and 0.4758 MPa is 0.001091 m^3/kg, whereas it is 0.3928 m^3/kg for the vapor.
  11. Apr 6, 2013 #10
    Excellent! Excellent! Excellent!

    So you realize now that we are dealing with a two stage process:
    1. Isothermal reversible compression of a superheated vapor up to its saturation point
    2. Isothermal and isobaric (reversible) conversion of saturated vapor to a mixture of saturated liquid and saturated vapor at its equilibrium vapor pressure and temperature.

    You need to consider each of these stages in determining the total heat effects and the total work effects.

    The first thing I recommend your doing is calculating the initial volume within the cylinder prior to compression. If your steam tables are complete, you can look up the initial specific volume of the superheated vapor, and, knowing the mass of water in the cylinder, use this to determine the volume. You can also use the ideal gas law to approximate the initial volume. I recommend that you do both, so you can compare the prediction of the ideal gas law with the (more accurate) prediction from the steam tables. Once you know the initial volume, you will also know the final volume (10X compression). Now, using the specific volume of the saturated vapor and the saturated liquid together with the total mass of water in the cylinder and its final volume, you can precisely calculate the mass of saturated liquid water and the mass of saturated vapor within the cylinder at final steady state. What do you get?

  12. Apr 6, 2013 #11
    Awesome, thanks Chet!

    Okay, I am getting a little confused here, but here is what I have:

    Vi=4.798m3 (using the ideal gas law I got 4.927m3)
    Vf= (0.10)(Vi)=0.4798m3

    Specific Volume of the saturated vapor is 0.3928 m3/kg
    Specific Volume of the saturated liquid is 0.0010191 m3/kg

    From here I am not sure what to do


    I may have figured out the mass of each phase in the final state:

    since the final volume needs to be the sum of the volume of each phase present at the final state,

    (0.4978m3)/5kg =(0.3928m3/kg)(x) + (0.001091m3/kg)(1-x)

    where x=% vapor, 1-x=% liquid

    solving for x=0.241, so 1-x=0.759.

    (5kg)(0.241)=1.205 kg vapor
    (5kg)(0.759)=3.795 kg liquid

    How does this look?
    Last edited: Apr 6, 2013
  13. Apr 6, 2013 #12
    Let x be the number of kilograms of saturated liquid, and 5-x be the number of kilograms of saturated vapor. Since you know the specific volumes of the saturated liquid and the saturated vapor, you can calculate, in terms of x, the volume that each occupies. Together, they must occupy 0.4795 M3. Solve for x.
  14. Apr 6, 2013 #13
    Thanks Chet once again! I'll check to see if this way produces the same answer I got above in the edited post I made..
  15. Apr 6, 2013 #14
    Success! I did it your way and I got the same result!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted