Stratospheric Chemistry HS 2016
Solution to Homework Problem Set 9
For questions:
[email protected] (CHN O15.2)
Problem 1: PSC mixture types
a)
Problem 2: Arctic and Antarctic PSCs
a)
Pitts et al. (2011)
From 20 Questions (WMO)
Problem 2: Arctic and Antarctic PSCs
a)
Arctic
Antarctic
STS
34.3 %
21.1 %
Arctic temperature mostly above T frost
→ STS relatively more important than ice
Ice
3.2 %
12 %
Antarctic temperature mostly below Tfrost
→ Ice clouds relatively more important than STS
Mountain
wave ice
0.21 %
0.18 %
More mountain ridges in Arctic than Antarctic
Problem 3: NAT in Arctic PSCs
a)
T - T NAT
Mix1
Mix2
Mix2-enh
Ice
STS
Pitts et al. (2011)
Problem 3: NAT in Arctic PSCs
a) Yes, because NAT PSCs were observed even at times when ice PSCs had
not yet started to form (December 2009)
“No ice clouds were observed by CALIPSO during this early phase, suggesting
that these early season NAT clouds were formed through a non‐ice nucleation
mechanism” Pitts et al. (2011)
Problem 4: Efficiency of denitrification and chemical processing
a) Mixing ratios of gas species
χHNO3 = 10 ppbv
HNO3:H2O ~ 1:5 in STS → χH2O = 50 ppbv (amount that can condense on STS)
Number density of STS
nSTS = 10 cm-3
Mass density and molar masses
ρSTS = 1.5 g.cm-3
MH2O = 18 g.mol-1, MHNO3 = 63 g.mol-1
Air density at 20 km
nair = 2.1018 cm-3 (Brasseur and Solomon)
• Assume all HNO3 is taken up
• Assume mass of H2SO4 can be neglected
Problem 4: Efficiency of denitrification and chemical processing
a)
"Rule of thumb" for
sedimentation velocity
Vsed = 1 m/h ∙ (radius in μm)2
Problem 4: Efficiency of denitrification and chemical processing
b) Mixing ratios of gas species
χHNO3 = 10 ppbv
HNO3:H2O ~ 1:3 in NAT → χH2O = 30 ppbv (amount that can condense on NAT)
Number density of NAT
nNAT = 10-4 cm-3
Mass density and molar masses
ρNAT = 1.5 g.cm-3
MH2O = 18 g.mol-1, MHNO3 = 63 g.mol-1
Air density at 20 km
nair = 2.1018 cm-3 (Brasseur and Solomon)
• Assume all HNO3 is taken up
Problem 4: Efficiency of denitrification and chemical processing
b)
Problem 4: Efficiency of denitrification and chemical processing
c) Reactivity of NAT and STS with respect to chlorine activation is determined by
their surface area density (A) available for reactions
Surface area density = number density ∙ surface area (of one particle)
Assuming spherical particles, surface area = 4π r2 ~ r2
For NAT compared to STS:
nNAT = 10-4 cm-3
rNAT = 18 μm
nSTS = 10 cm-3
rSTS = 0.43 μm
→ STS droplets are about 56 times more efficient than NAT in processing chlorine