! Examination of a singular fuel reacting with two
! different lumped species of air. Each air has the same
! composition and therefore the same products.
!
! Propane + 5*AIR1 = Products 
! Propane + 5*AIR2 = Products
!
! Composition is stoichiometric, so all fuel and both
! AIR should be consumed and all the reamins is products.
! ------------------------------------------------

&HEAD CHID='reactionrate_lumped_two_air', TITLE='Lumped Species Reactions' /
&MESH IJK=4,1,4, XB=0.0,0.1,0.0,0.025,0.0,0.1 /
&TIME T_END=1.0 DT=0.01 /
&MISC INITIAL_UNMIXED_FRACTION = 0.0, SUPPRESSION=.FALSE. /

! define primitive species and component species that define air and products
&SPEC ID = 'PROPANE' /
&SPEC ID = 'OXYGEN',            LUMPED_COMPONENT_ONLY = .TRUE. /
&SPEC ID = 'NITROGEN',          LUMPED_COMPONENT_ONLY = .TRUE. /
&SPEC ID = 'WATER VAPOR',       LUMPED_COMPONENT_ONLY = .TRUE. /
&SPEC ID = 'CARBON DIOXIDE',    LUMPED_COMPONENT_ONLY = .TRUE. /

!create air species -- each which can reaction with fuel
&SPEC ID = 'AIR1', SPEC_ID = 'OXYGEN','NITROGEN', VOLUME_FRACTION =1,3.76, BACKGROUND=.TRUE. /
&SPEC ID = 'AIR2', SPEC_ID = 'OXYGEN','NITROGEN', VOLUME_FRACTION =1,3.76 /

!create products lumped species
&SPEC ID = 'PRODUCTS', SPEC_ID = 'CARBON DIOXIDE', 'WATER VAPOR', 'NITROGEN',VOLUME_FRACTION = 3,4,18.8 /

!initial fuel and non-background air lumped species
&INIT MASS_FRACTION(1) = 0.46981, SPEC_ID(1)='AIR2',MASS_FRACTION(2)=0.06038, SPEC_ID(2)='PROPANE' /

&REAC ID = 'R1',
    FUEL = 'PROPANE',
    SPEC_ID_NU = 'PROPANE','AIR1','PRODUCTS'
    NU=-1,-5,1 /

&REAC ID = 'R2',
    FUEL = 'PROPANE',
    SPEC_ID_NU = 'PROPANE','AIR2','PRODUCTS'
    NU=-1,-5,1 /

&DEVC XYZ=0.05,0.0125,0.05, QUANTITY='MASS FRACTION', SPEC_ID='PROPANE',  ID='C3H8' /
&DEVC XYZ=0.05,0.0125,0.05, QUANTITY='MASS FRACTION', SPEC_ID='AIR1',  ID='AIR1' /
&DEVC XYZ=0.05,0.0125,0.05, QUANTITY='MASS FRACTION', SPEC_ID='AIR2',  ID='AIR2' /
&DEVC XYZ=0.05,0.0125,0.05, QUANTITY='MASS FRACTION', SPEC_ID='PRODUCTS',  ID='P1' /

&TAIL /