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Homework Help: Power consumption for maintaining Temp in a control volume

  1. Mar 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Question 1:

    An insulated 8-m3 rigid tank contains air at 600 kPa and 400 K. A valve connected to the tank is now opened and air is allowed to escape until the pressure inside the tank drops to 200 kPa. The air temperature is maintained constant by an electric heater during the process.

    1-Calculate the amount of discharged air in kg.

    2-Calculate the power consumed by the electric heater in watt.

    3-Explain how the enthalpy of discharged air is calculated if its temperature changes during the process.

    Tables for specific energy/enthalpy of air as an ideal gas are given for 100 KPa and 25 C in book.
    2. Relevant equations
    ideal gas law - PV = mR(air)T
    v = V/m
    for a general control volume, dE/dt = Qcv(per sec) - Wcv(per sec) + Σmi(inflow per sec)*(hi + .5(Vi^2) + gZi) - Σme(outflow per sec)*(he + .5*(Ve^2) + gZe)

    power = work per second = Wcv(per sec) [there's no way to insert a dot directly above a letter, to represent work per second)

    first law of thermo - ΔU = Q - W

    3. The attempt at a solution
    For part 1 -

    I assume it's an ideal gas because no critical constants for air are given to me and I don't know if it's accepted to go outside the book and given info here, so I can't use a Z factor chart. And I'm not given masses or specific volumes anyway, so this is a must as far as I can tell.

    I rearrange the ideal gas law to get PV/RT = m, and solve for the initial and final mass inside the box.
    Mi = 41.81 Kg
    Mf = 13.94 Kg

    While I'm here, I use the specific internal energy and enthalpy at 400k (u/h doesn't vary with P much) to find initial and final U and H for the contents of the box.

    Taking the differences gives me
    ΔU = 7984.47 KJ
    ΔH = 11184.23 KJ

    M(discharged) = difference between initial and final mass in box = 27.87 Kg

    For part 2 -

    Firstly, this is where I'm stuck, but i'll go ahead and state what I tried.

    Since I know that specific internal energy doesn't really vary with pressure, I assume it to be constant since the temperature is held constant by the generator. So by that logic, and considering the box is insulated (thus Q = 0), I say that the change in internal energy of the box is due to a loss of air mass.

    So I guess this is where I'm stuck perhaps both conceptually and mathematically. For a general control volume I have the giant formula above, but since heat transfer = 0 due to insulation, no air flows IN, and I'm assuming kinetic/potential energy are negligible (no velocities /areas / heights given) , I whittle it down and simplify it to

    dEcv/dt = - Wcv(per sec) - Σ me(outflow per sec) * (he) = -Wcv(per sec) - He(enthalpy outflow per sec)

    Now, I wasn't given any time frame for this to happen during, so the easy way of figuring out flow rates isn't possible here. To try and take at least some step forward I simplify further and say that if not accounting for actual rates and instead just looking at total change, (basically just multiplying everything by whatever unknown total time this process takes)

    ΔEcv = -W - He ?

    And I'm assuming that the change in energy is simply ΔU, essentially? Since I'm not accounting for any kinetic or potential energy? Do I need to keep in mind that the air in the box does boundary work when it goes from high pressure to the outside low pressure?

    I'm pretty sure what I want is the work per second, because that would be equivalent to the energy supplied by the generator in order to keep temperature constant. Is this the correct interpretation?
    And if so, how exactly do I get to it if I have no time frame? I'm not given any velocities or areas so I'm unsure how else to establish a mass flow rate, or any flow "rate" for that matter. Do I just need to assume it all happens in 1 second or something, make some kind of arbitrary assumption or something?

    Basically, how do I need to visualize this, and where are my errors in interpretation or conceptualization? Any and all help is greatly appreciated, thanks.

  2. jcsd
  3. Mar 15, 2016 #2


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    The problem statement is bad - you cannot calculate power as you don't know how long the process takes. Your approach looks fine.
  4. Mar 16, 2016 #3
    Part 1 looks good.

    For part 2 your simple equation to follow is W=mehe+(mfuf-miui), also ui=uf since there is no heat transfer. Then to express your answer in watts you just take your answer which should be in KJ and use the conversion 1KWh=3600KJ. You probably already noticed that you did not need a time to calculate the power.

    For part 3, we know that enthalpy is h=u+Pv
    We also know(because volume is constant) that the specific heat is C(v), which should be on "Tables for specific energy/enthalpy of air as an ideal gas are given for 100 KPa and 25 C in book".
    Also, uf-ui=C(v)(Tf-Ti)
    modifying the first equation we get: hf-hi=(uf-ui)+(Pf-Pi)v, substituting in from equation on previous line we get: hf-hi=C(v)(Tf-Ti)+(Pf-Pi)v
    v is the specific volume of the air, but since we are assuming air is an ideal gas then the specific volume is equal to the reciprocal of the density or 1/p which can be found on the same table where you found the specific heat at constant volume.
    To answer part 3, you simply state that in order to find the enthalpy of discharged air if the the temperature changes during the process, one must use equation
    hf-hi=C(v)(Tf-Ti)+(Pf-Pi)v and solve for hf assuming we know every variable but hf.

    I hope this helps.
  5. Mar 17, 2016 #4

    This is all incredibly helpful, thank you very much. If you don't mind I'd just like to clarify a point or two for part 2.

    So after looking at the KWh to KJ conversion I see now that I just need to confirm which terms represent which "process" happening here, then I can just plug my numbers.

    You rearranged the energy formula to get W=mehe+(mfuf-miui), I want see if putting this into words still makes sense.

    "The work that occurs (power consumed) is equal to the enthalpy lost by the box plus the box's change in total internal energy"

    I guess I'm confused on the mehe part. You're saying "the mass that exited, multiplied by it's exit specific enthalpy", correct? That would just be the discharged mass multiplied by air's specific enthalpy at exit conditions (which I assume are atmospheric)? Or is enthalpy constant enough with pressure that I can just say the enthalpy lost by the box is equal to mihi - mfhf ?

    Thank you again for your help!
  6. Mar 17, 2016 #5
    Yes, I got that equation by dropping the terms for kinetic energy, potential energy and mass coming in from the energy equation.
    However, remember that we have isothermal and adiabatic process for part 2 since temperature is constant and there is no heat transfer. Thus the change in specific internal energy and specific enthalpy is 0.
    So actually the work done by the heater(power consumed) is equal to the enthalpy of air in general at 400K plus the change in mass times the specific internal energy of air in general at 400K.
    Let's see if rewriting the equation helps:
    W(power consumed)=He+(mf-mi)u
    He= enthalpy of air in general at 400K
    u=specific internal energy of air in general at 400K
    You can now see that we actually use the only thing changing is the masses.

    So for the mehe part that you say you are confused about, all I did was to use he instead of hi or hf(because in this case hi=hf=he) to keep my format the same.
    The box does not have any change in internal energy or enthalpy.
  7. Mar 19, 2016 #6

    I see, thank you for clarifying.

    On second look, how did you get that arrangement of the energy equation though? I initially got to ΔU = -W - He, so how did you get to W = He + ΔU ?

    If I go by your equation, using the numbers I have, it would come out to W = -11184.23 KJ + 7984.47 KJ = -3200 KJ (numbers are the change in total internal energy and total enthalpy, or in other words the energy/enthalpy carried out by the exiting mass. up in original post)
    The work is negative because work was done on the system by the generator, or in other words, work was done by the generator in order to keep temperature constant. I'd then take the negative of that to be +3200 KJ, that is the energy consumed by the generator correct? Am I correct in assuming that you just went ahead and flipped the sign of Work in the energy equation to account for this ahead of time? I hope I'm understanding this right. If I had rearranged for W in my equation, I'd have gotten W = - ΔU - He, the opposite of yours.

    Anyway, if using your formula I come out to... 3200 KJ * ( 1 KwH / 3600 KJ) = .888 KwH
    .888 KwH * (1000 wH/ KwH) = 888 wH
    888 wH is just 888 watts for the duration of an hour, so the answer is 888 watts. Sound about right? Sorry for all the questioning, I've heard this class is notorious for ruining GPA's etc etc, so I'm being as cautious as possible.
  8. Mar 19, 2016 #7


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    Where does the 1 hour come from?
    If you don't have a time frame given, you cannot convert the energy to a meaningful, unique power value.
  9. Mar 19, 2016 #8
    etrevino94 said I could convert KJ to KwH, where 1 KwH = 3600 KJ. I looked online for unit converters for more peace of mind and they said the same thing, so then I worked it out myself.

    A Kw is 1 KJ/second, so a KwH is a KJ/second for a continuous hour, or A KJ/second for 3600 seconds. That would be 3600 KJ after an hour right?

    I'm also concerned because this problem has been posted for a week now and no announcement has been made correcting or changing any part of it
  10. Mar 19, 2016 #9


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    You can convert kJ to kWh, sure, but not to Watts.
    1 kW over 1 hour is 3600 kJ, sure. 2 kW over 30 minutes is also 3600 kJ. If you just know the energy value, you cannot determine power.
  11. Mar 19, 2016 #10
    I see what you mean. So you think the problem statement is a genuinely bad one? Do you think it'd be acceptable to say the answer is .888 kWh ? I've tried but I just can't see any way to get farther than that without any sort of time frame or velocity of exiting gas, something like that.
  12. Mar 19, 2016 #11


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    If post 1 has the full problem statement, it is impossible to solve.

    An answer in J is the reasonable thing to do. kWh... as additional value if you like, but J is better.
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