A Calibrated Water Tank

The dynamics of a surge tank, used to suppress damaging over-pressure in fluid lines, is described by a simple ordinary differential equation (ODE) the same as eqn. 1 in the recent paper by Spencer and Brasswell, “On the Misdiagnosis of Surface Temperature Feedbacks from Variations in Earth’s Radiant Energy Balance”, so very generally applicable to many systems.

The dynamics has been coded into an Excel spreadsheet by ControlsWiki, with an example of default variables above. Note:

1. The variability of the height of fluid in the tank (red) is suppressed and lagged, but there is no parameter in the model for lag. Lag is an emergent property of the system.

2. Increasing the periodicity of the flow into the tank increases the mean height of the water in the tank (try it). So the average of the water height is not directly related to the average inflow, but also depends on the period, another surprising emergent property.

This simple example shows how simple dynamics can lead to surprising behavior, and why misdiagnosis occurs by failing to account for dynamics when applying a simple linear regression, say. The first step in a correct analysis of a system is a valid physical model.

You can change the parameters and experiment with other aspects of the model.

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0 thoughts on “A Calibrated Water Tank

  1. Hmmm…  Now I see the meaning of áccumulation’ more often than before, when I used to arm wave and claim that there was much confusion between statics and dynamics in this type of climate work. This is a lovely example from our host.

    Tjhere is also a related example, beautiful to watch, at http://sciencedemonstrations.fas.harvard.edu/icb/icb.do?keyword=k16940&panel=icb.pagecontent278364%3Ar%241%3Fname%3Dindepth.html%2Cicb.pagecontent341734%3Ar%241%3Fname%3Dindepth.html&pageid=icb.page80863&pageContentId=icb.pagecontent256665&view=view.do&viewParam_name=indepth.html

    where you have to start a video in the section named ‘Pendulum Wave”. I can’t open it just now because I’m part way through swirching from 32 bit to 64 bit O/S Windows 7’.

    The pendulum wave example seems to incorporate ‘memory’ of its prior motions, which is a topic I”ve seldom seen discussed in climate time series.

  2. Oh heck, URL has played up. Just search these keywords: simple harmonic motion, periodic motion, quantum revival, aliasing, video

  3. Your plot shows a phase lag of 90 degrees, which is the same as you get for an RC electrical circuit, regardless of values for R and C.

    http://en.wikipedia.org/wiki/RC_circuit#Gain_and_phase_angle
    Go to Frequency-domain considerations

    All of the parameters for this circuit can be stated as their thermal equivalent:

    E volts (ºK)
    I amperes (joules/sec)
    R ohms (ºK/(joules/sec))
    C farads (joules/ºK)

    so that thermal phenomena that do not have significant reactance will all have a 90 degree phase lag.

    This is what I found when I plotted integrated sunspot number against satellite TLT temperatures some time ago. At time I wasn’t aware of the phase lag relationship.

    Sunspot_Model.jpg 950×350 pixels

    I had to lag the model temperatures by 36 months to make the solar influence visually evident.  Of course, signals from ENSO and volcanism are present as well, and can dominate.  Now that I know that the correct lag is 11 * 12  / 4 = 33 months, I take credit for a good guess!

      • Very interesting.  I will study it carefully.  Have you ever derived an expression for the system gain given a sinusoidal perturbation of the radiant flux from the sun, as its affects the earth’s surface temperature?  This would require an estimate of the heat capacity of the earth which I imagine would come mostly from the oceans.  I want to apply an 11 year signal and see what I get.  

        I’ve been working on this, but Laplace transforms and such are in my very distant past.

      • Yes. And also see recent shorter archive http://vixra.org/abs/1108.0020.  The gain is simply 1/(1-a) where a is the correlation of each successive time step Ti to Ti+1.  Its around 10 for the surface temperature record.

        That is not the end of it though, as a gain of 10 also implies a decay time of 10 years, but spectral power plots do not suggest any such limit.  It only looks like 10 because you are looking a a part of the system that is losing heat, accumulated by the ocean.  I think that the right way to look at the dynamics, including the deep ocean, is a limit as a->1, i.e. infinite gain, or pure accumulation, close but not quite.  Its then plausible to recover the last 100 years temperature rise from a solar anomaly of +0.1W/m2 by accumulation of heat in a mixed zone of 100m depth.

      • 10 with intrinsic gain.  That is K/K/m^2.  You should be able to use the Planck conversion between W and K.

    • David,

      Using the solution for the transform for a first order sun/earth/space system and a period of the solar cycle of 10 years (0.628 radians/yr) and 100 meters of water as the heat sink, and a 0.26 deg K amplitude sinusoidal signal at the top-of-atmosphere which is the equivalent of a 1 watt/m^-2 flux variation, I get the following:

      A time constant of 3,450 years.

      An amplitude ratio of the TOA temperature to the “earth” temperature of 0.00046 – the temperature variation over a Schwabe cycle as a direct effect of TSI will never be detected.

      And yet the variation is observed.  There may be other actors.

      • The way I work it out is this. The temperature increase of a body of water is

        T = Joules/(Specific Heat water x Mass)

        The accumulation of 1 Watt per sq meter on a 100m colum of water for one year gives an expected temperature increase of

        T = 32 x 10^6/(4.2 x 10^8)

        = 0.08 C

        Given that about one third attenuation of radiation at TOA at the surface, and a duration of solar cycle of 11 years, the amplitude of the solar cycle will be

        Ta = 0.08 x 11 x 0.3 = 0.26 C

        With the expectation of the temperature increase for the direct forcing (no accumulation) using the Plank relationship of 0.3C/W the expected direct amplitude would be 0.09 C.

        Gain = Accumulated/Direct = 0.26/(0.3×0.3) = 3

        For a longer accumulation of solar anomaly, from a succession of strong
        solar cycles such as we saw late last century, the apparent amplification
        will be more. From the AR correlation of surface temperature you get an
        estimate of gain of 10. But this is only apparent amplification, as the
        system is accumulative, the calculated gain increases as the duration of the
        forcing increases. For long time scales gain (and hence solar sensitivity)
        approaches infinity – a singularity, and ceases to be useful.

        Extend this calculation for 1000 years and you can transition between ice
        ages with no other input. The role of GHGs, water vapor and albedo is to
        maintain the heat state of the system, e.g. solar forcing increases
        temperature increase which causes CO2 concentrations to change. But this
        does not mean that more heat can be trapped by an increase in CO2, because
        the system is being heated by accumulation of solar anomaly.

      • The way I work it out is this. The temperature increase of a body of water is

        T = Joules/(Specific Heat water x Mass)

        The accumulation of 1 Watt per sq meter on a 100m colum of water for one year gives an expected temperature increase of

        T = 32 x 10^6/(4.2 x 10^8)

        = 0.08 C

        Given that about one third attenuation of radiation at TOA at the surface, and a duration of solar cycle of 11 years, the amplitude of the solar cycle will be

        Ta = 0.08 x 11 x 0.3 = 0.26 C

        With the expectation of the temperature increase for the direct forcing (no accumulation) using the Plank relationship of 0.3C/W the expected direct amplitude would be 0.09 C.

        Gain = Accumulated/Direct = 0.26/(0.3×0.3) = 3

        For a longer accumulation of solar anomaly, from a succession of strong
        solar cycles such as we saw late last century, the apparent amplification
        will be more. From the AR correlation of surface temperature you get an
        estimate of gain of 10. But this is only apparent amplification, as the
        system is accumulative, the calculated gain increases as the duration of the
        forcing increases. For long time scales gain (and hence solar sensitivity)
        approaches infinity – a singularity, and ceases to be useful.

        Extend this calculation for 1000 years and you can transition between ice
        ages with no other input. The role of GHGs, water vapor and albedo is to
        maintain the heat state of the system, e.g. solar forcing increases
        temperature increase which causes CO2 concentrations to change. But this
        does not mean that more heat can be trapped by an increase in CO2, because
        the system is being heated by accumulation of solar anomaly.

      • To get the amplitude ratio of a sine wave imposed on the input to a system with capacitance such as the earth with its oceans, one solves the input – output = accumulation differential equation using Laplace Transforms.  After transients die out, the solution is:

        Amplitude Ratio = 1/(1+ω²T²)^½  

        where:

        ω = frequency, rad/yr 

        T = time constant, yr

        I believe you calculated T the same way I did since I saw your value in your referenced paper: 3,500 years (I got 3,450 years).

        If ω = 2

      • pochas – I am not completely following you, though I want to.  How did you calculate the time constant Tau? I used the bandwidth relationship, i.e 40,000 years or so divided by 2pi. 

        This time constant would, I think, relate to the whole ocean volume, and so that is how you get the small solar cycle fluctuation (you don’t see them in ocean heat content for instance).  However the surface temperature fluctuations relate to the mixed layer with depth around 100m so a much smaller capacity.  A Tau of around 10 according to me. 

        But the system is one of variable time constants, from top to bottom, and in the limit the long time constant, almost pure accumulation, dominates. 

      • Well, here is the calculation in detail.  If we take the whole ocean as a capacity, and maybe we should, the time constant will blow up!  The point of this is to bring out how important this time constant is.

        The input – output = accumulation equation is

         

        q/4 – kθ = nCp dθ/dt

         

        where q = input flux, watts/(m^2 sec)

        the factor 4 corrects for the disk to sphere
        surface geometry.

        k = relates thermal flux to temperature (see
        below) watts/(cm^2 sec K)

        θ = ocean temperature, ºK

        n = mass of ocean, grams

        Cp = ocean heat capacity J/(gram ºK)

        t = time, sec or years

         

        Rearrangeing to standard form (terms with θ on
        the left side)

         

        nCp dθ/dt + kθ = q/4

         

        divide by k

         

        nCp/k dθ/dt + θ = q/(4k)

         

        The factor nCp/k has units of time and is the
        time constant T in the

        solution via Laplace Transform of the above.

         

        n = mass of water 100 m deep and 1m^2 surface
        area = 10E11 grams

         

        Cp = Joules to heat 1 gram of water by 1ºK =
        4.187 J/gram

         

        k = thermal flux equivalent to blackbody
        temperature, J/(m^2 sec ºK)

         

        Derivation of k

         

        Stefan Boltzmann equation  q = σθ^4

         

        k = dq/dt

         

        differentiating:  dq/dt = 4σθ^3

         

        Evaluating at θ = blackbody temp of the earth,
        -18 ºC = 256 ºK

         

        k = 4 (5.67E-8) 256^3 = 3.8 J/(sec m^2 ºK)

         

        note: the previous posting had k = 1/3.8 =
        0.26.  This time I’ve used the reciprocal
        and moved it to the bottom of the nCp/k group where it belongs.

         

         

        Calculating Time Constant T

         

        T = 
        nCp/k = 10E11 (4.187) / 3.8  = 1.10E11
        sec

         

        T = 1.10E11 / 31,557,600 sec/yr = 3485.7 yr

         

        Erratum: 
        I neglected to include the geometry factor ‘4’ in the original post, so
        replace “0.26 K per century, without feedbacks.”  With “0.065 K per century, corrected for
        geometry, without feedbacks.”

         

        Everything else remains the same.

         

      • 1cm^3 is one gram, and 100 cm in a meter.  So thats 100*100*100 grams in a cubic meter and another 100 for the depth in meters – E8.

      • Thats an interesting calculation.  Of course the depth is the assumption.  I think I got 160m from the time series, and a Tau of 8 or so.  Very close given the Tau will increase with the depth.  Thanks

      • Dave,
        Please delete my previous post. The results are the same but I have fixed a few things, namely that q sin ωt now appears where appropriate, and corrected units statements at a few places.  ThanksHere is the calculation in detail.  If we take the whole ocean as a capacity, and maybe we should, the time constant will blow up!  The point of this is to bring out how important this time constant is.The input – output = accumulation equation is  q sin ωt /4 – kθ = nCp dθ/dt where q = input flux signal amplitude, watts/(m^2 sec)The factor 4 corrects for the disk to sphere surface geometry.k = relates thermal flux to temperature (see below) J/(sec m^2 ºK)θ = ocean temperature, ºKn = mass of ocean, gramsCp = ocean heat capacity J/(g ºK)t = time, sec or yearsRearranging to standard form (terms with θ on the left side) nCp dθ/dt + kθ = q sin ωt /4divide by knCp/k dθ/dt + θ = q sin ωt /(4k)The factor nCp/k has units of time and is the time constant T in the solution via Laplace Transform of the above.n = mass of water 100 m deep and 1m^2 surface area = 10E11 gramsCp = Joules to heat 1 gram of water by 1ºK = 4.187 J/gramk = thermal flux equivalent to blackbody temperature, J/(m^2 sec ºK)Solution after inverse transform, after transients die outAmplitude Ratio = 1/(1+ω²T²)^½  where ω = frequency, rad/yr Derivation of k Stefan Boltzmann equation  q = σθ^4k = dq/dtdifferentiating:  dq/dt = 4σθ^3Evaluating at θ = blackbody temp of the earth, -18 ºC = 256 ºKk = 4 (5.67E-8) 256^3 = 3.8 J/(sec m^2 ºK)note: the previous posting had k = 1/3.8 = 0.26.  This time I’ve used the reciprocal and moved it to the bottom of the nCp/k group where it belongs.Calculating Time Constant TT =  nCp/k = 10E11 (4.187) / 3.8  = 1.10E11 secT = 1.10E11 / 31,557,600 sec/yr = 3485.7 yrErratum:  I neglected to include the geometry factor ‘4’ in the original post, so replace “0.26 K per century, without feedbacks.”  With “0.065 K per century, corrected for geometry, without feedbacks.”Everything else remains the same. 
                             

         

      • Dave,Please delete my previous post and all the garbage below.
        I have fixed a few things, namely that q sin ωt now appears where appropriate, and corrected units statements at a few places.  Thanks

        Here is the calculation in detail.  If we take the whole ocean as a capacity, and maybe we should, the time constant will blow up!  The point of this is to bring out how important this time constant is.The input – output = accumulation equation is  

        q sin ωt /4 – kθ = nCp dθ/dt 
        where q = input flux signal amplitude, watts/(m^2 sec)
        The factor 4 corrects for the disk to sphere surface geometry.
        k = relates thermal flux to temperature (see below) J/(sec m^2 ºK)θ = ocean temperature, ºK
        n = mass of ocean, grams
        Cp = ocean heat capacity J/(g ºK)
        t = time, sec or years

        Rearranging to standard form (terms with θ on the left side) 
        nCp dθ/dt + kθ = q sin ωt /4

        divide by k
        nCp/k dθ/dt + θ = q sin ωt /(4k)

        The factor nCp/k has units of time and is the time constant T in the solution via Laplace Transform of the above.

        n = mass of water 100 m deep and 1m^2 surface area = 10E11 grams
        Cp = Joules to heat 1 gram of water by 1ºK = 4.187 J/gram
        k = thermal flux equivalent to blackbody temperature, J/(m^2 sec ºK)

        Solution after inverse transform, after transients die out

        Amplitude Ratio = 1/(1+ω²T²)^½  
        where ω = frequency, rad/yr 

        Derivation of k 

        Stefan Boltzmann equation  q = σθ^4

        k = dq/dt

        differentiating:  dq/dt = 4σθ^3

        Evaluating at θ = blackbody temp of the earth, -18 ºC = 256 ºK
        k = 4 (5.67E-8) 256^3 = 3.8 J/(sec m^2 ºK)

        note: the previous posting had k = 1/3.8 = 0.26.  This time I’ve used the reciprocal and moved it to the bottom of the nCp/k group where it belongs.

        Calculating Time Constant T

        T =  nCp/k = 10E11 (4.187) / 3.8  = 1.10E11 sec
        T = 1.10E11 / 31,557,600 sec/yr = 3485.7 yr

        Erratum:  I neglected to include the geometry factor ‘4’ in the original post, so replace “0.26 K per century, without feedbacks.”  With “0.065 K per century, corrected for geometry, without feedbacks.”Everything else remains the same. 

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