! ! $Header$ ! C C================================================================================ C SUBROUTINE CLOUDS_GNO(klon,ND,R,RS,QSUB,PTCONV,RATQSC,CLDF) IMPLICIT NONE C C-------------------------------------------------------------------------------- C C Inputs: C C ND----------: Number of vertical levels C R--------ND-: Domain-averaged mixing ratio of total water C RS-------ND-: Mean saturation humidity mixing ratio within the gridbox C QSUB-----ND-: Mixing ratio of condensed water within clouds associated C with SUBGRID-SCALE condensation processes (here, it is C predicted by the convection scheme) C Outputs: C C PTCONV-----ND-: Point convectif = TRUE C RATQSC-----ND-: Largeur normalisee de la distribution C CLDF-----ND-: Fraction nuageuse C C-------------------------------------------------------------------------------- INTEGER klon,ND REAL R(klon,ND), RS(klon,ND), QSUB(klon,ND) LOGICAL PTCONV(klon,ND) REAL RATQSC(klon,ND) REAL CLDF(klon,ND) c -- parameters controlling the iteration: c -- nmax : maximum nb of iterations (hopefully never reached) c -- epsilon : accuracy of the numerical resolution c -- vmax : v-value above which we use an asymptotic expression for ERF(v) INTEGER nmax PARAMETER ( nmax = 10) REAL epsilon, vmax0, vmax(klon) PARAMETER ( epsilon = 0.02, vmax0 = 2.0 ) REAL min_mu, min_Q PARAMETER ( min_mu = 1.e-12, min_Q=1.e-12 ) INTEGER i,K, n, m REAL mu(klon), qsat, delta(klon), beta(klon) real zu2,zv2 REAL xx(klon), aux(klon), coeff, block REAL dist, fprime, det REAL pi, u, v, erfcu, erfcv REAL xx1, xx2 real erf,hsqrtlog_2,v2 real sqrtpi,sqrt2,zx1,zx2,exdel c lconv = true si le calcul a converge (entre autre si qsub < min_q) LOGICAL lconv(klon) !cdir arraycomb cldf (1:klon,1:ND)=0.0 ! cym ratqsc(1:klon,1:ND)=0.0 ptconv(1:klon,1:ND)=.false. !cdir end arraycomb pi = ACOS(-1.) sqrtpi=sqrt(pi) sqrt2=sqrt(2.) hsqrtlog_2=0.5*SQRT(log(2.)) DO 500 K = 1, ND do i=1,klon ! vector mu(i) = R(i,K) mu(i) = MAX(mu(i),min_mu) qsat = RS(i,K) qsat = MAX(qsat,min_mu) delta(i) = log(mu(i)/qsat) c enddo ! vector C C *** There is no subgrid-scale condensation; *** C *** the scheme becomes equivalent to an "all-or-nothing" *** C *** large-scale condensation scheme. *** C C C *** Some condensation is produced at the subgrid-scale *** C *** *** C *** PDF = generalized log-normal distribution (GNO) *** C *** (k<0 because a lower bound is considered for the PDF) *** C *** *** C *** -> Determine x (the parameter k of the GNO PDF) such *** C *** that the contribution of subgrid-scale processes to *** C *** the in-cloud water content is equal to QSUB(K) *** C *** (equations (13), (14), (15) + Appendix B of the paper) *** C *** *** C *** Here, an iterative method is used for this purpose *** C *** (other numerical methods might be more efficient) *** C *** *** C *** NB: the "error function" is called ERF *** C *** (ERF in double precision) *** C c On commence par eliminer les cas pour lesquels on n'a pas c suffisamment d'eau nuageuse. c do i=1,klon ! vector IF ( QSUB(i,K) .lt. min_Q ) THEN ptconv(i,k)=.false. ratqsc(i,k)=0. lconv(i) = .true. c Rien on a deja initialise ELSE lconv(i) = .FALSE. vmax(i) = vmax0 beta(i) = QSUB(i,K)/mu(i) + EXP( -MIN(0.0,delta(i)) ) c -- roots of equation v > vmax: det = delta(i) + vmax(i)*vmax(i) if (det.LE.0.0) vmax(i) = vmax0 + 1.0 det = delta(i) + vmax(i)*vmax(i) if (det.LE.0.) then xx(i) = -0.0001 else zx1=-sqrt2*vmax(i) zx2=SQRT(1.0+delta(i)/(vmax(i)*vmax(i))) xx1=zx1*(1.0-zx2) xx2=zx1*(1.0+zx2) xx(i) = 1.01 * xx1 if ( xx1 .GE. 0.0 ) xx(i) = 0.5*xx2 endif if (delta(i).LT.0.) xx(i) = -hsqrtlog_2 ENDIF enddo ! vector c---------------------------------------------------------------------- c Debut des nmax iterations pour trouver la solution. c---------------------------------------------------------------------- DO n = 1, nmax do i=1,klon ! vector if (.not.lconv(i)) then u = delta(i)/(xx(i)*sqrt2) + xx(i)/(2.*sqrt2) v = delta(i)/(xx(i)*sqrt2) - xx(i)/(2.*sqrt2) v2 = v*v IF ( v .GT. vmax(i) ) THEN IF ( ABS(u) .GT. vmax(i) : .AND. delta(i) .LT. 0. ) THEN c -- use asymptotic expression of erf for u and v large: c ( -> analytic solution for xx ) exdel=beta(i)*EXP(delta(i)) aux(i) = 2.0*delta(i)*(1.-exdel) : /(1.+exdel) if (aux(i).lt.0.) then c print*,'AUX(',i,',',k,')<0',aux(i),delta(i),beta(i) aux(i)=0. endif xx(i) = -SQRT(aux(i)) block = EXP(-v*v) / v / sqrtpi dist = 0.0 fprime = 1.0 ELSE c -- erfv -> 1.0, use an asymptotic expression of erfv for v large: erfcu = 1.0-ERF(u) c !!! ATTENTION : rajout d'un seuil pour l'exponentiel aux(i) = sqrtpi*erfcu*EXP(min(v2,100.)) coeff = 1.0 - 0.5/(v2) + 0.75/(v2*v2) block = coeff * EXP(-v2) / v / sqrtpi dist = v * aux(i) / coeff - beta(i) fprime = 2.0 / xx(i) * (v2) : * ( EXP(-delta(i)) - u * aux(i) / coeff ) : / coeff ENDIF ! ABS(u) ELSE c -- general case: erfcu = 1.0-ERF(u) erfcv = 1.0-ERF(v) block = erfcv dist = erfcu / erfcv - beta(i) zu2=u*u zv2=v2 if(zu2.gt.20..or. zv2.gt.20.) then c print*,'ATTENTION !!! xx(',i,') =', xx(i) c print*,'ATTENTION !!! klon,ND,R,RS,QSUB,PTCONV,RATQSC,CLDF', c .klon,ND,R(i,k),RS(i,k),QSUB(i,k),PTCONV(i,k),RATQSC(i,k), c .CLDF(i,k) c print*,'ATTENTION !!! zu2 zv2 =',zu2(i),zv2(i) zu2=20. zv2=20. fprime = 0. else fprime = 2. /sqrtpi /xx(i) /(erfcv*erfcv) : * ( erfcv*v*EXP(-zu2) : - erfcu*u*EXP(-zv2) ) endif ENDIF ! x c -- test numerical convergence: ! if (beta(i).lt.1.e-10) then ! print*,'avant test ',i,k,lconv(i),u(i),v(i),beta(i) ! stop ! endif if (abs(fprime).lt.1.e-11) then ! print*,'avant test fprime<.e-11 ' ! s ,i,k,lconv(i),u(i),v(i),beta(i),fprime(i) ! print*,'klon,ND,R,RS,QSUB', ! s klon,ND,R(i,k),rs(i,k),qsub(i,k) fprime=sign(1.e-11,fprime) endif if ( ABS(dist/beta(i)) .LT. epsilon ) then c print*,'v-u **2',(v(i)-u(i))**2 c print*,'exp v-u **2',exp((v(i)-u(i))**2) ptconv(i,K) = .TRUE. lconv(i)=.true. c borne pour l'exponentielle ratqsc(i,k)=min(2.*(v-u)*(v-u),20.) ratqsc(i,k)=sqrt(exp(ratqsc(i,k))-1.) CLDF(i,K) = 0.5 * block else xx(i) = xx(i) - dist/fprime endif c print*,'apres test ',i,k,lconv(i) endif ! lconv enddo ! vector c---------------------------------------------------------------------- c Fin des nmax iterations pour trouver la solution. ENDDO ! n c---------------------------------------------------------------------- 500 CONTINUE ! K RETURN END