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  1. Subscriber joe shmo
    Strange Egg
    02 May '11 02:31 / 1 edit
    I felt like figuring this out, but the units of measure seem to be flawed at my final result, so If you could/will, point out my flaw(s)

    I'm attemting to find the density as a function of height of a column of air

    Im going to assume ideal gas law

    P =p*R*T eq(1)

    P = pressure
    p= density
    R=universal gas constant
    T= temperature(absolute)

    assuming acceleration due to gravity & Area are constant, and the fluid(air) is in static equilibrium
    A= Area(constant)

    dP = dF/A eq(2)

    dF= dm*g eq(3)
    m= mass of air above section

    m=p*V eq(4)

    where
    p=density
    V=volume
    thus

    dm = V*dp+p*dV eq(4'

    Back substitution until eq(2) yeilds

    A*dP = g(V*dp+p*dV) = g(A*h*dp + p*A dh)

    where the Area (A) divides out &
    h=height

    giving

    dP = g(h*dp+p*dh) eq(5)

    Taking the derivative with respect to density from eq(5) gives

    dP/dp = g( h + p* dh/dp ) eq(6)......(This equation could be wrong, not sure if Im taking the derivative correctly)

    taking the derivative of eq(1) with respect to density (Assuming temp is constant) and substituting for left side of eq(6) gives after some rearrangement

    h + p*dh/dp = RT/g= constant

    here I take the laplace transform of both sides (could be where the mistake lies)
    after solving for F(s) I come to

    F(s) = (RT/g*s- h(0))/(s*(s-1)) which is in basic form

    = (C*s - A)/(s*(s-1))

    which breaks into after some fuddleing

    C/(s-1) - A/(s-1) + A/s

    The inverse Laplace of this yeilds

    h(p) = (RT/g)*(e^p) - h(0)*(e^p) + h(0)

    does anyone else agree with the above equation?
  2. 02 May '11 08:16 / 1 edit
    Your ideal gas law is wrong. The pressure in an ideal gas actually does not depend on the mass of the constituent particles because particles are assumed to have no size and gravity is neglected. Instead, use:

    p = n k T

    p = pressure (Pa)
    n = number density - the amount of particles in a certain volume, not their mass per volume! (m^-3)
    k = Boltzmann constant (J K^-1)
    T = temperature (K)
  3. Subscriber joe shmo
    Strange Egg
    02 May '11 13:12
    Originally posted by KazetNagorra
    Your ideal gas law is wrong. The pressure in an ideal gas actually does not depend on the mass of the constituent particles because particles are assumed to have no size and gravity is neglected. Instead, use:

    p = n k T

    p = pressure (Pa)
    n = number density - the amount of particles in a certain volume, not their mass per volume! (m^-3)
    k = Boltzmann constant (J K^-1)
    T = temperature (K)
    I'm going to respectfully disagree with you. The version of the Ideal Gas Law I used is equally valid as the one you proposed, as long as the Gas constant "R" is specifc to the modeled gas.

    start with

    P*V = n*R*T (1)

    n= m/M (2)

    m= mass of gas
    M= molar mass of specific gas

    substitute (2)--->(1)

    P*V = (m/M)*R*T now rearrange

    P*V = m*(R/M)*T

    P= (m/V)*(R/M)*T

    P= p*(R/M)* T

    Where

    R/M is the SPECIFIC gas constant
  4. 02 May '11 14:23
    Equivalently, you can state that pV = nRT, in which case:

    p = pressure (Pa)
    V = volume (m^-3)
    n = number of particles (mol)
    R = universal gas constant (J mol^-1 K)
    T = temperature (K)

    Note that again mass does not appear. By dividing with volume:

    p = (n/V) R T,

    you obtain the formula I mentioned, since R = N_a k, where N_a is Avogadro's number.
  5. Subscriber joe shmo
    Strange Egg
    02 May '11 14:41
    Originally posted by KazetNagorra
    Equivalently, you can state that pV = nRT, in which case:

    p = pressure (Pa)
    V = volume (m^-3)
    n = number of particles (mol)
    R = universal gas constant (J mol^-1 K)
    T = temperature (K)

    Note that again mass does not appear. By dividing with volume:

    p = (n/V) R T,

    you obtain the formula I mentioned, since R = N_a k, where N_a is Avogadro's number.
    I'm confused, are you still disagreeing with me?
  6. 02 May '11 14:57
    Originally posted by joe shmo
    I'm confused, are you still disagreeing with me?
    I'm still confused about your original calculation, where you appear to be confusing molar and mass density.
  7. Subscriber joe shmo
    Strange Egg
    02 May '11 15:36 / 1 edit
    Originally posted by KazetNagorra
    I'm still confused about your original calculation, where you appear to be confusing molar and mass density.
    In my "original post" I didn't state that "R" was specific, ( I labeled it as "The Universal gas constant), so that is my fault...but as for the form of the ideal gas law I used (P = p*R_air*T_abs) not being valid, I derived it for you above.

    begining with

    (P*V = n*R_universal*T ) and substituting (n= m_air/M_air) into it

    leads to

    P= p*R_air*T

    Look at this

    http://en.wikipedia.org/wiki/Ideal_gas_law#Alternate_forms
  8. 02 May '11 15:49 / 1 edit
    Ah, I get you now. I'll have a look at the rest of the calculation.

    Did you check the Laplace transforms using some kind of software?

    You definitely cannot have e^p as the argument of the exponential function must be dimensionless. It probably should be something like e^(p/p_0), where p_0 is some reference density (e.g. at h = 0).
  9. Subscriber joe shmo
    Strange Egg
    02 May '11 16:02 / 1 edit
    Originally posted by KazetNagorra
    Ah, I get you now. I'll have a look at the rest of the calculation.

    Did you check the Laplace transforms using some kind of software?

    You definitely cannot have e^p as the argument of the exponential function must be dimensionless. It probably should be something like e^(p/p_0), where p_0 is some reference density (e.g. at h = 0).
    no, I did it by hand, and perhaps a little hastely...

    As for using software to check, thats a no as well. I don't have any software that does those operations, nor would I know how to use any software that does.

    but you agree up to, and including the differential equation before the transformation?
  10. 02 May '11 16:36
    Originally posted by joe shmo
    no, I did it by hand, and perhaps a little hastely...

    As for using software to check, thats a no as well. I don't have any software that does those operations, nor would I know how to use any software that does.

    but you agree up to, and including the differential equation before the transformation?
    Up to (6) looks okay at first glance. Maybe you can some other technique to solve that equation, substitution perhaps.
  11. Subscriber joe shmo
    Strange Egg
    02 May '11 17:35
    Originally posted by KazetNagorra
    Up to (6) looks okay at first glance. Maybe you can some other technique to solve that equation, substitution perhaps.
    I re-did the transform, and found I made a few arithmatic mistakes, but the problem of the united exponetial still exists. My question to you is why doens't the question of inital density arise in the mathematics, I see that division by p_0 in the exponent is the most likely solution, but I feel I need some minor justification.

    And as for using another technique, I must say that techniques for solving differential equations are not my strong suit.
  12. 02 May '11 18:47
    Originally posted by joe shmo
    I re-did the transform, and found I made a few arithmatic mistakes, but the problem of the united exponetial still exists. My question to you is why doens't the question of inital density arise in the mathematics, I see that division by p_0 in the exponent is the most likely solution, but I feel I need some minor justification.

    And as for using another ...[text shortened]... echnique, I must say that techniques for solving differential equations are not my strong suit.
    Your best bet is to nondimensionalize your differential equation first, so that you solve an equation for e.g. f(x) with f and x dimensionless.

    Unless I'm making some obvious mistake, your solution does not obey the differential equation so something is wrong in calculating the transform. This equation can probably be tackled using substitution, just insert as a try e.g. h(p) = f(p) e^(p/p_0), with f(p) a polynomial function in p - chances are good a function like Ap + B with A, B, p_0 constants will suffice, then solve for the constants using the known constraints of the problem.
  13. 02 May '11 18:55
    Originally posted by joe shmo
    In my "original post" I didn't state that "R" was specific, ( I labeled it as "The Universal gas constant), so that is my fault...but as for the form of the ideal gas law I used (P = p*R_air*T_abs) not being valid, I derived it for you above.

    begining with

    (P*V = n*R_universal*T ) and substituting (n= m_air/M_air) into it

    leads to

    P= p*R_air*T

    Look at this

    http://en.wikipedia.org/wiki/Ideal_gas_law#Alternate_forms
    To check your final answer, try

    http://en.wikipedia.org/wiki/Density_of_air
  14. Subscriber joe shmo
    Strange Egg
    02 May '11 20:53
    Originally posted by KazetNagorra
    Your best bet is to nondimensionalize your differential equation first, so that you solve an equation for e.g. f(x) with f and x dimensionless.

    Unless I'm making some obvious mistake, your solution does not obey the differential equation so something is wrong in calculating the transform. This equation can probably be tackled using substitution, just ...[text shortened]... onstants will suffice, then solve for the constants using the known constraints of the problem.
    yeah, I must be botching up the transfoms

    My TI-89 gives

    h(p) = RT/g*(1 - p_0/p)
  15. Subscriber joe shmo
    Strange Egg
    02 May '11 22:31
    Originally posted by joe shmo
    yeah, I must be botching up the transfoms

    My TI-89 gives

    h(p) = RT/g*(1 - p_0/p)
    yeah, apparently Laplace transforms are ?only? good for DE's with constant coeficients?

    I guess the way to go about this one is using integration factor method for first order linear ODE's.

    Can anybody show me why the Laplace doesn't supposedly work for non-constant coeficients?