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trop_wmo

Determines the level of the thermal tropopause.

Prototype

	function trop_wmo (
		p         : numeric,  
		t         : numeric,  
		punit [1] : integer,  
		opt   [1] : logical   
	)

	return_val  :  numeric

Arguments

p

An array of any dimensionality containing input pressure levels. The pressure values must be monotonically increasing (top-to-bottom). If multi-dimensional, it must be the same size and shape as t. The level dimension must be in the rightmost position.

t

An array of any dimensionality containing the temperatures. Missing data are not allowed. The units must be degrees Kelvin. If multi-dimensional, the level dimension must be in the rightmost position.

punit

Integer which specifies the units of p: 0 means the pressure units are millibars [ hectopascals ]; 1 means the pressure units are pascals.

opt

Option flag. Set to False if no optional arguments are specified. If True and the attribute lapsec is present then this value will be used to test the lapse rate. By default, opt@lapsec=2.0

Return value

The return type will be double if p or t is double, float otherwise. The return size and shape are functions of the input arguments ep and t. See examples below.

Description

trop_wmo computes the pressure at the thermal (static) tropopause for pressures and temperatures following the definition of the height of the tropopause.

Specifically: WMO (1992): International meteorological vocabulary, Genf, 784pp.:

       The first tropopause is defined as the lowest level at which
       the lapse rate decreases to 2 deg K per kilometer or less,
       provided also the average lapse rate between this level and
       all higher levels within 2 kilometers does not exceed 2 deg K.

The user may change the lapse rate used to test via the optional atttribute lapse. The default is 2.0. Some people prefer to use other values [eg: 0.0].

In NCL V6.5.0, a bug was fixed in which the return value didn't have a _FillValue attribute set.

This code is a modified version of codes developed by Dominik Brunner, Peter van Velthoven, Thomas Reichle, Christine Land, B. Steil and R. Hein.

Examples

Example 1: A simple example to illustrate what must be done if the units are not correct and the ordering is not correct.

; PRESSURE (MB)
  p  =(/ 1008.,1000.,950.,900.,850.,800.,750.,700.,650.,600., \
          550.,500.,450.,400.,350.,300.,250.,200., \
          175.,150.,125.,100., 80., 70., 60., 50., \
           40., 30., 25., 20. /)
           
; TEMPERATURE (C)   
  t  =(/  29.3,28.1,23.5,20.9,18.4,15.9,13.1,10.1, 6.7, 3.1,   \
          -0.5,-4.5,-9.0,-14.8,-21.5,-29.7,-40.0,-52.4,   \
         -59.2,-66.5,-74.1,-78.5,-76.0,-71.6,-66.7,-61.3, \
         -56.3,-51.7,-50.7,-47.5 /)

  t  = t+273.15      ; must be degrees Kelvin

  p = p(::-1)        ; reverse order 
  t = t(::-1)        ; as noted, the levels must be top-to-bottom

  ptrop = trop_wmo(p, t, 0, False)     ; ptrop=106.05 mb
The return will be a scalar: ptrop=106.05 mb.

Example 2: Same as Example 1 but use several different critical lapse rates.

  opt     = True
  cLapTst = (/2, 1.5, 1.0 , 0.5, 0 /)
  do n=0,dimsizes(cLapTst)-1
     opt@lapsec = cLapTst(n)
     ptrop      = trop_wmo(p, t, 0, opt)  
     print("n="+n +" cLapTst="+cLapTst(n) +"  ptrop="+sprintf("%7.2f", ptrop) )
  end if
The printed results are:
     n=0 cLapTst=2.0 ptrop= 106.05    ; this is the default
     n=1 cLapTst=1.5 ptrop= 103.87
     n=2 cLapTst=1.0 ptrop= 101.74
     n=3 cLapTst=0.5 ptrop=  99.65
     n=4 cLapTst=0.0 ptrop=  97.60

Example 3: Let p(klev) [hPa] and t(ntime,klev) [K], then

  ptrop = trop_wmo(p, t, 0, False)   ; ptrop(ntime)

Example 4: Let p(klev) [hPa] and t(ntime,nlat,mlon,klev) [K], then

  ptrop = trop_wmo(p, t, 0, False)   ; ptrop(ntime,nlat,mlon)

Example 5: Let p(lev) [Pa] and t(time,lev,lat,lon) [K]: This is similar to Example 4, but t does not have level as the rightmost dimension. Hence, the dimensions must be reordered. Also, p has units of pascals (punit=1).

  ptrop = trop_wmo(p, t(time|:,lat|:,lon|:,lev|:), 1, False) ; ptrop(time,lat,lon) 

Example 6: Let p(time,lev,lat,lon) [Pa] and t(time,lev,lat,lon) [K], then, use 1.0 as the critical lapse rate.

  opt   = True
  opt@lapsec = 1.0
  ptrop = trop_wmo(p(time|:,lat|:,lon|:,lev|:) \
                  ,t(time|:,lat|:,lon|:,lev|:), 1, opt) ; ptrop(time,lat,lon) 

Example 7: Read T and hybrid level data from a model.

  f     = addfile("cam35.h0.0008-07.nc" ,"r")

;************************************************
; calculation pressures at each grid point
;************************************************
  hyam = f->hyam
  hybm = f->hybm
  p0   = f->P0
  psfc = f->PS     ; (time,lat,lon);  units Pa    

  p    = pres_hybrid_ccm (psfc, p0, hyam, hybm)  ; p(ntim,klvl,nlat,mlon)
  p@long_name = "pressure at mid levels"
  p@units     =  psfc@units
                                  ; for illustration convert to hPa (mb)
  p           = p*0.01
  p@units     = "hPa"

  T    = f->T
  copy_VarCoords(T,p)             ; copy T coords to "p"
  printVarSummary(p)

;************************************************
; Calculate the tropopause level [p]
;************************************************
  opt        = True
  opt@lapsec = 1.0         ; default is 2.0

  ptrop = p(:,0,:,:)       ; "trick" to create a variable with coordinates
                           ; The units argument is 0 because p is in hPa (mb)
  ptrop = trop_wmo(p(time|:,lat|:,lon|:,lev|:)   \
                  ,T(time|:,lat|:,lon|:,lev|:), 0, opt)

  ptrop@long_name  = "tropopause pressure"
  ptrop@_FillValue = -999.

  printVarSummary(ptrop)            ; ptrop(time,lat,lon)
  printMinMax(ptrop, True)