Chilbolton: cold sunny winter day

Posted: February 22, 2013 by tchannon in climate, Clouds, Surfacestation, weather

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This is the first day I have seen were there were no obvious clouds. This reveals the cold sky pyrgeometer values.

However there is a surprise, the pyranometer short wave solar hump is considerable offset in time from the shallow hump of long wave. Eyeballing it looks 2 to 3 hours later. 

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Lidar might give a clue. This shows low altitude return which looks coherent with long wave. A detail if you look closely, very low level higher reflection during the morning, perhaps excess night humidity burning off.

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Air temperature peaks later too from a morning below zero. (from memory it was too dry for a frost, cold previous day)

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This perhaps gives a clue too. There is reduced water vapour 10 to 13hrs a few km up. Could this be solar heating from the top?

Rather than deluging with all the plots, files are provided.

Commentary on what is going on is welcome.

I am putting the day files in an archive, 19th Feb 2013 here chilbolton-2013-02-19.

I thank Chilboton Observatory for their data.

http://www.stfc.ac.uk/Chilbolton/weather/24987.aspx

Post by Tim Channon

Comments
  1. Graeme says:

    This reveals to cold sky pyrgeometer values.

    What does this sentence mean?

  2. tchannon says:

    Word blind. s/to/the/
    Thanks.

  3. RACookPE1978 says:

    OK. So help me out here.

    I’ve worked many nightshifts outside on clear, calm nights – and on overcast, lightly breezy nights. I “know” the calm clear nights are colder because of upward radiation from all exposed surfaces (my jacket and hardhat and gloves and car windows among others!).

    So what are the equations? What are the coefficients and the temperatures?

    Assume daytime air temps are -5 C, 268 K. Nighttime air temps are -15 C, 258 K. Both are completely clear, wind speed is 0.0 Emissivity of a pond of water and the same pond of ice are about the same (0.96 and 0.97). But the pond water will be evaporating about 110 watts/meter sq all the time, right?

    1) So won’t the open water pond be colder than the ice-covered pond if both are exposed overnight to clear skies simply because of the evaporation cooling of the open water, even if all of that heat is going into latent heat of evaporation of the water – which doesn’t change water temperature?

    2) Also, what “temperature” (degree K) and what emissivity are they going to be radiating “to”? The classic Q = e(SB)(Twater^4-Tsky^4) seems to fall apart because the “sky” emissivity cannot be included. Also, this “classic” formula is good only of both substances have an equal “area” both are exposed to – and the water surface “sees” only 1/2 of the sky.

    3) Would not a complete radiation calc have to include the energy radiated to the daytime sky (at a higher temp than the nighttime sky), as well as the assumed daytime absorbed radiation from the sun?

  4. Peter Champness says:

    I don’t know what is going on here, but I will come back and have a look!
    Is a pygeometer different from a pyanometer? Why do they actually measure?

  5. tchannon says:

    Peter,
    Been pushed on time, didn’t give a linkback to what might explain.

    http://tallbloke.wordpress.com/2013/02/11/discussions-on-pyrgeometers-ir-measurement/

  6. tchannon says:

    RACookPE1978,
    I was hoping someone else was going to reply. I’m not a lot of help on those questions.

    Welcome to the settled science, a dire mess.

    The data shown is incomplete, a rule of manipulators and abusers, withhold. We don’t have either complete or accurate information. (nothing to do with Chilbolton, they just happen to have useful data available)

    I hope a little can be deduced from these data when the dust settles.

  7. RACookPE1978 says:

    And that, dear sir, is why I hope I am brave enough (humble enough ??) ask the seemingly “obvious” questions.

    Seems simple, doesn’t it, to ask “How much energy is lost to a clear sky, and to a cloudy sky, of so-and-so air temperature from a flat surface of such-and-such emissivity and blah-de-blah degrees Kelvin?”

  8. tchannon says:

    No need to be humble around here. I don’t know much, why I try and find out… then I know even less.

    In broad terms, not much if there is cloud, yet that is misleading. Cloud can lose some. Other processes can move out heat physically via vapour.

    Hence the claim that CO2 magnifies what clouds do, a workaround, claim the effect is larger.

    We neither have good enough data on cloud nor can model.

  9. How do you know that pyranometer is giving a sensible answer?. Do you know its make up and the nature of the calculation in the instrument? There is no instrument that measures radiation. They all measure a temperature (based on a change of resistance or voltage with temperature) or a relative (or comparative) temperature based on a theoretical emission distribution.
    Prof Claes Johnson has some thoughts here http://claesjohnson.blogspot.com.au/2013/02/big-bluff-of-pyrgeometer-dlr-as.html and more on other posts.
    I have used a large range of temperature measuring instruments and supposed radiation instruments. All have to be calibrated and the zero set. That applies to mercury in glass thermometers which can be several degrees out. Any devices not measuring a solid surface temperature and using assumptions within the Stefan-Boltzman equation is likely to give a wrong answer.

  10. RACookPE1978 says:

    Well, the reason that I am trying to be so precise in actually trying to find real measurements of real inbound radiation under real-world skies is that – without measurements – one is left with “only” theory and hand-waving as theorists try to extrapolate from “pure physics” (ie, computer models of ideal gasses and blackbody radiation from perfect surfaces into perfect absorbers at infinite distances around perfect emitters approximated as perfect flat bodies!) .

    Sure, so pyranometers are not “perfect” receivers of IR radiation. What is better?

    What system actually would measure inbound (AND outbound) radiation more accurately?

    What radiation is actually being absorbed by the sky (from an open body of water or ice) under ALL conditions all 24 hours of a typical day-night/sunny-cloudy conditions, and how would you measure it?

    So far, all of the actual measurements I have found have shown that the so-called “arctic amplification” of ever-warmer Arctic conditions caused by melting arctic sea ice is a myth. The more open water is uncovered by lower arctic ocean sea ice extents in the late summer the greater the heat loss from the water.

  11. RACook, I presume PE means professional engineer so you are likely to understand something about measurements, instruments and and measuring errors.
    You may be interested in Prof Claes Johnson’s new post http://claesjohnson.blogspot.com.au/2013/02/2nd-coming-of-2nd-law.html which seems was prompted by my reference to Lubos Motl’s post about Boltzman -link at the bottom of Claes’s post. or directly here http://motls.blogspot.com.au/2013/02/ludwig-boltzmann-birthday.html
    At last there are coming some proofs that DLR (downwelling longwave radiation) does not exist.if the atmospheric temperature is less than the earth’s surface temperature.
    Next hopefully, more will understand the limits of the Stefan-Boltzman equation and how it should be properly applied.