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Dry adiabatic lapse rate

  1. Aug 29, 2008 #1
    I am aware of the derivation for the dry adiabatic lapse rate using the enthalpy approach: ncpdT+VdP=0, but I can't seem to spot the error in my own derivation. If anyone sees it, I would be especially grateful.

    dU=[tex]\delta[/tex]Q-[tex]\delta[/tex]W

    ncvdT=0 - PdV

    [tex]\frac{dT}{dV}[/tex]=-[tex]\frac{P}{nc_{v}}[/tex]

    [tex]\frac{dT}{dV}[/tex] [tex]\frac{dV}{dz}[/tex]= -[tex]\frac{P}{nc_{v}}[/tex] [tex]\frac{dV}{dz}[/tex]=-[tex]\frac{P}{nc_{v}}[/tex] [tex]\frac{dV}{dP}[/tex] [tex]\frac{dP}{dz}[/tex] by the chain rule

    [tex]\frac{dT}{dz}[/tex] = [tex]\frac{-P}{nc_{v}}[/tex] [tex]\frac{-nRT}{P^{2}}[/tex] [tex]\frac{-MPg}{RT}[/tex] = [tex]\frac{-Mg}{c_{v}}[/tex]

    I obtained [tex]\frac{dV}{dP}[/tex] by differentiating PV = nRT and [tex]\frac{dP}{dz}[/tex] from the hydrostatic equation, substituting
    density = mass/volume = MP/RT, where M is molar mass in g/mol.

    For some reason, I obtain cv in my derivation instead of the correct cp.

    Any thoughts?
     
  2. jcsd
  3. Nov 12, 2008 #2
    You were kind to point out a defect in the Miskolczi model that I gather interferes with his virial theorem claim. I made another criticism of his and most other radiation balance models and hope that you will comment on it. https://www.physicsforums.com/showthread.php?t=261966
    I noted your quandary and an absence of any reply. I offer this analysis to thank you for your Miskolczi comment that encouraged me to go further.

    I spent some time on the web in order to identify past approaches to quantifying temperature change with altitude and noted more and more problems in approach. Your approach was to use a first law expression differentiated by dz as an altitude surrogate. You expected to need cP but made a transcription error of your enthalpy statement into your first law conversion that left you with cV . I want to identify a problem that is universal to all the websites that have attempted this quantitation. The cP solution is cited from a 1980 textbook problem’s solution (reference 8, web citation 1). The ICAO (International Civil Aviation Organization) standard dry atmosphere table (Geigy Scientific Tables,1984, volume 3, pages 27-29) has been modestly revised over the years. Its development cites the perfect gas law, uses a plane parallel model, and gives a fall in temperature with altitude of 15-8.501 = 6.499 oC at 1,000 m, 8.501-2.004 = 6.497 oC at 2,000 m and 2.004- -4.491 = 6.495 oC at 3,000 m. The derivation of the temperature column is not made clear. The Wikipedia treatment gives us 9.8 oC/km, a rather higher number. http://en.wikipedia.org/wiki/Lapse_rate They use a poorly defined γ to get their rate.
    http://pds-atmospheres.nmsu.edu/education_and_outreach/encyclopedia/adiabatic_lapse_rate.htm has posted a NASA approach that gives a value of 9.76 oC/km for Earth and shows that R = cP – cV. Underlying this approach and yours is the concept of specific heat capacity http://en.wikipedia.org/wiki/Specific_heat where cP and cV are in mass units and CP (29.07) and CV( 20.7643) are in molar units (dry air).

    We should start with a reliable statement (. . . . used to maintain spacing, 2 attempted integral signs have a gap)
    . . . . . infinity
    Psurface = ⌠-ρ g* dz with g* the altitude-affected gravitational constant less vector centrifugal force from body rotation
    . . . . . . . ⌡ . . . . . . . . . where Pinfinity is zero and density area expands with body mass center radius squared
    . . . . . surface
    . . . . .. . . . . . . . . . . . . . . . infinity
    which can be generalized to P = ⌠-ρ g* dz with no z directed mass transfer
    . . . . . . . . . . . . . . . . . . . . . . ⌡
    . . . . . . . . . . . . . . . . . . . . altitude
    whose derivative is dP/dz = -ρ g*
    and from the perfect gas law, P = ρ RT M-1, hence dP = ρ R M-1 dT, then
    dT/dz = -g* M R-1 that becomes 34.163 oK/km, a value clearly too high. Using CP instead of R we get 9.77 oK/km and CV 13.68 oK/km. CP is the most attractive but still 50% higher than the ICAO value of 6.50 oK/km. The CP derived fall predicts an upper troposphere temperature well below that observed. We need to be concerned that the real needed thermal change estimate from CP may be substantially higher than the currently accepted number because of the difference between 9.77 and 6.50 oK/km. Obviously 3/2 CP would be a wonderful outcome. Is there a motion reason to make this claim for adiabatic expansion/contraction compared to heat capacity?
     
    Last edited: Nov 13, 2008
  4. Nov 12, 2008 #3

    D H

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    You are assuming a constant volume process on the left hand side (dU=ncvdT) but a constant pressure process on the left hand side (dW=PdV). These two assumptions are at odds with one another.
     
  5. Nov 13, 2008 #4
    jdlawlis wrote
    I noted a second reply and was looking at the NASA dry adiabatic lapse rate citation again in developing a Venus atmosphere analysis. I noticed a feature that deals directly with your problem. The NASA solution started as you did the second time with cV in the first law expression and then substituted cP – cV for R in the gas law equation to allow n cV dT – n cV dT to disappear, leaving only cP in the adiabatic expression. You still have the problem of a 3/2 too large result, but it was not a simple transcription error as I first thought.
     
  6. Nov 18, 2008 #5
    I have spent some time looking into the origin of the ICAO table and sense that the lapse rate there is compromised by specific aviation needs http://www.netweather.tv/forum/index.php?showtopic=49857&pid=1341485&st=0&#entry1341485 . cP is appropriate, not 3/2 cP. The NASA approach takes the derivative directly, while you omitted the T component of the derivative (dT/dP) part in your dV/dP ( n R/P . dT/dP – n R T/ P2). Recovery by substitution is possible, but I found no easy path.
    The most direct approach is
    dQ = dU + δW = n.cV dT+ P dV = 0 , first law
    V dP + PdV = n R dT
    dQ = n cV dT - V dP + n (cP – cV) dT
    dQ = n cP dT - V dP = 0 for adiabatic, V = n <mw> / ρ
    dT/dz = <mw> dP/dz / ρ cP, where dP/dz = - g* ρ
    dT/dz = - g* <mw> /cP = - g* /CP

    I suspect we don’t usually see the predicted low upper tropospheric temperatures because strong clear air downdrafts generate considerable g*-lowering mass transfer associated with high local barometric pressure and horizontal air movement away from the surface site.
     
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