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Problem with boiling water and raising piston

  1. Jul 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a cylindrical tank closed by a movable piston with mass ##m=3 kg##. The radius of the cyclinder is ##r=7.5 cm##. In the tank there is a mass ##m'=2 kg## o water at temperature just below ##100°C##. At the base of the cyclindrical tank there is an electrical heater which provides heat to the water with power ##P=100W##. Treat vapor as an ideal gas.

    a) When water starts boiling, what is the velocity of raising of the piston?
    b) After all the water becomes vapor, what is the velocity of raising of the piston?

    Answers : ##\bigg[ v_a=4.19 mm/s \, , \, v_b=12.6 mm/s \bigg]##
    2. Relevant equations


    3. The attempt at a solution
    I have less problems with part b). In part b) it should just be the isobaric expansion of an ideal gas, so I could write
    $$P=\frac{dV}{dt} \frac{c_p}{R} p=\frac{7}{2} \frac{dh}{dt} (3 \cdot 9.81)=\frac{7}{2} v_b (3 \cdot 9.81)\implies v_b=0.97 m/s$$
    Which is not the correct result (I suppose vapor biatomic, but also trying with monoatomic does not give the correct answer)

    I also tried for point a) but I'm quite unsure of the procedure. In that case I think that

    $$P=\lambda \frac{dm}{dt}=\lambda \rho_{vapor} \frac{dV}{dt}=\frac{\lambda p \mathscr{M}}{R T} \frac{dh}{dt} \pi r^2=\frac{\lambda (3 \cdot 9.81) \mathscr{M}}{R T}v_a \implies v_a=0.26 m/s$$

    Where ##\frac{dm}{dt}## is the mass of water per unit time that becomes vapor, ##\mathscr{M}## is the molecular mass of water and ##\lambda## is the enthalpy of vaporization .

    Also in this case the result is not correct.

    Could you highlight the conceptual mistakes in what I have tried for this exercise?

    Thanks a lot in advice
     
  2. jcsd
  3. Jul 14, 2016 #2
    I can see what you did, and it's almost right. Did you look up the heat capacity for water vapor on Google? They give Cp = 1.996 kJ/kgC. What value did you use for the gas pressure? Did you remember to include the atmospheric pressure outside the container in getting the pressure? I got 48.3 mm/s for part b.

    Chet
     
    Last edited: Jul 14, 2016
  4. Jul 14, 2016 #3
    Thanks a lot for the answer!

    For part b)

    The pressure on the vapor equals the (constant) pressure due to weight of the piston ##p=\frac{(9.81 m/s^2 \cdot 3 kg)}{\pi r^2}##

    $$P=\frac{dV}{dt} \frac{c_p}{R} p$$
    Here I used the fact that, in isobaric process the heat can be written as##Q=n c_p dT=\frac{c_p}{R} p dV##, I do not see where the number of moles is missing..

    ##R=8.314 \frac{J}{K mol}## is universal constant of gases and I now corrected the value of ##c_p## using yours, which gives ##\frac{c_p}{R}=240##

    $$P=\frac{c_p}{R} \frac{dh}{dt} (3 kg \cdot 9.81 m/s^2)=\frac{c_p}{R} v_b (3 kg \cdot 9.81 m/s^2)\implies v_b=0.014 m/s=14 mm/s$$

    It's closer to the result give, but still not equal.. Do you think that there are other problems?

    ------------------------------------------------------------------------------------------------------------------------------------------
    For point a)

    $$P=\lambda \frac{dm}{dt}=\lambda \rho_{vapor} \frac{dV}{dt}$$
    Here I just used the fact that the heat given to the water translates into the vaporization of a mass ##m## of water such that ##Q=m \lambda##, where ##\lambda =22.6 \cdot 10^5 J## is the entalpy of vaporization of water. I consider the heat and the mass per unit time (power ##P## and ##\frac{dm}{dt}##) and I have rewritten ##dm## as ##dm=\rho_{vapor} dV##, where ##\rho## is the density.


    $$=\frac{\lambda p \mathscr{M}}{R T} \frac{dh}{dt} \pi r^2=\frac{\lambda (3 \cdot 9.81) \mathscr{M}}{R T}v_a \implies v_a=0.26 m/s$$
    I used equation of ideal gases ##\rho R T=p \mathscr{M}##, where ##R=8.314 \frac{J}{K mol}## is universal constant of gases, ##T=373.15 K## is temperature, ##p## is pressure ##\mathscr{M}=0.018 kg## is the molar mass of water. I also used the fact that ##dV= dh S=dh \pi r^2##. Then, looking at product ##p S## i rewrote it as ##pS=F= (3 \cdot 9.81 )N##. Finally I used ##\frac{dh}{dt}=v_a##.
     
  5. Jul 14, 2016 #4
    Yes. the pressure you used is incorrect. The gas pressure should be 101325 Pa + 6655 Pa. Also, for Cp/R, I got (1.996)(18)/8.314 = 4.32

    ------------------------------------------------------------------------------------------------------------------------------------------
    For part a, it would be better to use the steam tables. Do you know how to use the steam tables?
     
    Last edited: Jul 14, 2016
  6. Jul 14, 2016 #5
    In part (a), for the saturated liquid and saturated vapor at 108000 Pa, let

    mL = mass of saturated liquid

    mV = mass of saturated vapor

    vL = specific volume of saturated liquid

    vV = specific volume of saturated vapor

    uL = specific internal energy of saturated liquid

    uV = specific internal energy of saturated vapor

    The first law tells us that $$\frac{dU}{dt}=P-p\frac{dV}{dt}$$
    where P is the heater power. We know that the total mass of liquid and vapor in the container is equal to 2 kg. Thus:
    $$m_L+m_V=2$$
    In terms of the parameters in the above table, what is the total internal energy U of the cylinder contents, and what is the total volume V of the cylinder contents?

    Do the specific internal energies and specific volumes in the above table change as the liquid evaporates within the cylinder to form vapor?
     
  7. Jul 14, 2016 #6
    Actually, your solution to part (a) seems like it would be correct also if you had used the correct pressure to calculate the density of the vapor.

    I got about 16 mm/sec for part a and 48 mm/sec for part b. So, my answers are each about 4 times the given answers.
     
    Last edited: Jul 15, 2016
  8. Jul 16, 2016 #7

    TSny

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    I get close to the answers provided by assuming the ideal gas law holds for the steam. (Maybe this is not a good assumption since the steam is near 100 oC).

    Use the ideal gas law to express height of steam in terms of mass of steam, temperature, and pressure of steam.

    For (a), the temperature and pressure remain constant. The rate of change of the height of the steam is therefore determined by the rate of change of the mass of the steam, which is determined by the rate of heat input to the system and the latent heat (enthalpy) of vaporization. There will also be a rate of change of the height of the water, but you can check that this will be negligible compared to the rate of change of the height of the steam.

    For (b), the mass of steam and pressure remain constant. The rate of change of the height of the steam is then determined by the rate of change of temperature, which is determined by the rate of heat input to the system and the specific heat at constant pressure.
     
  9. Jul 16, 2016 #8
    This is interesting. Here is the continuation of the analysis of part (a) that I started in post #5.
    $$U=m_Lu_L+m_Vu_V$$
    $$V=m_Lv_L+m_Vv_V$$
    From the first law, this gives (subject to ##dm_L/dt=-dm_V/dt##:
    $$(u_V-u_L)\frac{dm_V}{dt}+p(v_V-v_L)\frac{dm_V}{dt}=P$$or, equivalently,
    $$(h_V-h_L)\frac{dm_V}{dt}=\lambda\frac{dm_V}{dt}=P$$
    where ##\lambda## is the heat of vaporization at the system temperature T=102 C and saturation pressure P=107.98 kPa
    From the volume equation, we have:
    $$\frac{dV}{dt}=(v_V-v_L)\frac{dm_V}{dt}$$
    Therefore, combining these equations, we get:
    $$\frac{dV}{dt}=\frac{(v_V-v_L)}{\lambda}P$$So the velocity of the piston is:
    $$\frac{dh}{dt}=\frac{(v_V-v_L)}{\lambda}\frac{P}{A}$$
    If the specific volume of the liquid is neglected relative to the specific volume of the vapor and employ the ideal gas law to get the latter, we can easily calculate the piston velocity. If TSny got a result close to the book answers, I must have made an arithmetic error in my calculations.
     
  10. Jul 16, 2016 #9

    TSny

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    This is equivalent to what I got.
     
  11. Jul 16, 2016 #10
    Excellent. Two great minds at work!!!!
     
  12. Jul 16, 2016 #11
    I see where I went wrong now. I took the diameter to be 7.5 cm. rather than the radius. That accounts for my factor of 4.
     
  13. Jul 16, 2016 #12

    TSny

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    OK. Good to know we're getting the same numbers.
     
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