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A bit of confusion on a simple thermodynamics problem.

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A house with floor space of 2000 ft^2 and average height of 9 ft is heated from an average temperature of 50 degrees F to 70 degrees F. Determine the amount of energy transfered in the house assuming:
    a) The house is air tight.
    b) Some air escapes through the cracks as the heated air in the house expands at constant pressure.

    Given values:
    Floor space = 2000 ft^2
    Height = 9 ft
    Atmospheric pressure = 12.2 psia

    Thermodynamic properties of air:
    Gas constant R = 0.3704 psia ft^3/(lbm R)
    Cv = .171 BTU/lbmR
    Cp = 0.240 BTU/lbmR


    2. Relevant equations

    PV = mRT
    E in - Eout = change in energy of the system


    3. The attempt at a solution

    Part a is relatively simple. Since we are treating the house as an airtight, constant volume, the equation for this would be:

    Qin = mCv*(T2-T1)

    where mass m = PV/RT = 1162.4 lbm, then

    Ein = Qin, Eout = 0:

    Qin - 0 = change of energy in the system

    Qin = 1162*(.171)*(20) = 3974 BTU




    Part B is where I'm a bit unsure of. I feel there is some missing information.

    Since air leaks, we should have a loss of mass in the system, right? So we should have on the energy equation:
    m1*qin - m2*qout = m3*cp*(T2-T1)

    where q denotes heat per unit mass.

    Which would mean energy is lost through mass transfer. Is this correct?

    Using the ideal gas law, m1T1 = m3T3, so the mass remaining in the house after expansion would be:
    m3 = 1118.6 lbm

    Meaning m2 = 1162-1119 = 43 lbm.

    So:
    1162lbm*qin - (43lbm)*qout = (1118.6lbm)*(0.240 BTU/lbmR)*20R
    1162qin - 43qout = 5369 BTU

    But this leaves me with two unknowns and one equation. Should it be assumed that the amount of energy lost through the cracks is negligible?
     
  2. jcsd
  3. May 20, 2013 #2

    TSny

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    For part B, the number of moles of gas remaining in the house will be a function of temperature: n(T).

    The heat required to change the temperature from T to T+dT will be dQ = n(T)CpdT.

    See if you can use the ideal gas law to find an expression for n(T). Then see if you can integrate the expression for dQ.

    [EDIT: Sorry, I'm used to thinking in terms of moles, n, rather than mass, m. Please replace n(T) by m(T) above.]
     
  4. May 20, 2013 #3

    haruspex

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    Seems to me that in part b you need to consider it as a dynamic process and obtain a differential eqn. The air that leaks initially will be cooler (taking away less energy) than the air that leaks later.
    Edit: beaten to the punch by TSny
     
  5. May 21, 2013 #4
    I think involving the changes in mass, changes in temperature and changes in time in one equation will give me a partial differential equation.

    I just found the solution by the author of my textbook and it looks like they neglected air loss. :\ They simply used:

    Qin = m*cp*(T2-T1) = 1162*.240*20 = 5578 BTU

    I don't like where they assume we will assume these things. Many problems in the textbook are like this.

    Now that I see the solution I can see why we can neglect energy loss due to mass loss since, if we were to assume the exiting mass were at the same energy per unit mass as the energy that goes into the house (which would give an upper limit in error):

    q per unit mass in house = 5578 BTU/(1162 lbm) = 4.8 BTU/lbm

    energy leaving house = mlost * q per unit mass = 4.8 * 43 BTU = 206.41 BTU

    so:

    Error = 206.4/5578 = 3.7 percent


    But these things aren't always clear unless you know the solution.

    I strongly dislike looking at a solution manual. :(


    Thanks a ton for taking the time guys. :]
     
  6. May 21, 2013 #5
    For air acting as an ideal gas, the energy content of the air in the room (a definite volume) is independent of the temperature when the pressure must remain fixed. If we subtract the volume of the escaped air we nullify the effect of the increased temperature. So the extra energy provided by the heater is carried away by the air that escapes through the cracks! So paradoxically the energy IN the room remains same! The amount of energy transferred IN the house is therefore zero.
     
  7. May 21, 2013 #6

    haruspex

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    Depends what you think "energy transferred in the house" means. It's a curious expression. To me, all the energy from the heater was transferred in the house, but some later escaped.
     
  8. May 21, 2013 #7
    Oh I assumed "energy transferred in the house" to mean

    (Energy content of air contained in house finally ) - (Energy content of air contained in house initially)

    The problem statement expects us to assume that "energy transferred in the house" is same as the total energy transferred by the heater. With this understanding I agree with the solution given in OP's book. In this case the addition of the extra word "house" leads to ambiguity!
     
    Last edited: May 21, 2013
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