1
OC583 ISOTOPE BIOGEOCHEMISTRY (Spring 2009)
TOPIC 2: HYDROLOGIC CYCLE
References:
1. Chap 3 in Criss’s, Principles of Stable Isotope Distributions, Oxford Press, 1999.
2. Chaps 2 and 3 in Clark and Fritz’s Environmental Isotopes in Hydrogeology, CRC
Press, 1997.
3. Isotope Tracers in Catchment Hydrology, by C. Kendall and L. McDonnell, Elsevier,
1998
3. Classic paper by H. Craig and L. Gordon (1965) “Deuterium and O18 variations in the
ocean and marine atmosphere, In Stable Isotopes in Oceanographic Studies and PaleoTemperatures, ed. E. Tongiorgi, Spoleto 1965 9-130.
4. Classic paper by W. Dansgaard (1964) “Stable isotopes in precipitation”, Tellus 16:
436-468.
I. BACKGROUND
1. Global Water Cycle (Fig. 1)
- magnitude of water fluxes (103 km3/yr)
- water vapor residence time ~10 days in troposphere
- average global precipitation rate about ~1 m/yr
- there is a net transfer of water from the ocean to land surfaces
- there is recycling of water on the continents via evaporation from lakes and
rivers and evapotranspiration from plants
2. Primary source of water input to atmosphere is via evaporation from ocean (Fig. 2)
- most evaporation occurs the tropical ocean (within 30° of equator)
3. There is a net transfer of water from the tropical (E>P) to polar (P>E) latitudes (Fig. 2)
- global wind patterns transfer air masses from tropics to poles
2
- storms and eddies transport water vapor poleward
- as air masses move poleward they cool
4. As air masses cool, by moving poleward or towards continental interiors or by rising, a
temperature is reached where the air is saturated with water vapor (dew point)
-further cooling yields condensation which produces precipitation
5. The empirical relationship between the saturated water vapor pressure (Psat) and
temperature is:
ln Psat (torr) = 21.113 – 5350.5/T(°K), where 1 Torr = 1 mm Hg.
(1)
-thus at 25°C (298 K), then Psat = 23.53 Torr
-if total air pressure is 760 mm Hg (or 760 Torr), then water vapor would make up
~3% of the total pressure (24Torr/760Torr) if the air is fully saturated with water vapor
(100% relative humidity).
II. ISOTOPIC FRACTIONATIONS DURING HYDROLOGIC CYCLE
A. Equilibrium fractionations between different phases of water
1. The gas, liquid and solid phases of water have different 18O/16O and D/H compositions
at equilibrium (Fig. 3)
-liquid water is ~9-12 ‰ enriched in 18O/16O and ~80-110‰ enriched in D/H
compared to the vapor
-ice is ~4 ‰ enriched in 18O/16O and ~20‰ in D/H compared to liquid water
-Notice, in the figure, that they report that α = Rphase1/ Rphase2, but it’s not clear
which is phase 1 or 2.
2. Temperature effects on the equilibrium fractionation (Fig. 4)
- as expected, the equilibrium isotope fractionation effect (ε) for oxygen (and
hydrogen) between water vapor and liquid water decrease with increasing temperature
-the ε for 18O/16O decreases from ~11.5 to 7.5 ‰ between 0 and 50°C
3
-the ε for D/H decreases from 110 to 55 ‰ over 0 to 50°C
-here  = R(liq) / R(gas) and ε (‰) = (-1)*1000
B. Isotope effects in clouds assuming a closed system
1. Assumptions: 1. Isotopic equilibrium is established between water vapor and liquid
water (condensate) in a cloud and then the condensate is removed from the cloud as
precipitation. 2. There are no other sources or sinks of water vapor in the cloud, i.e., the
cloud is assumed to be a closed, not open, system.
2. As condensate is removed from the cloud, the 18O/16O and D/H of the remaining water
vapor in the cloud decreases over time because the condensate is enriched in 18O and D
compared to the vapor.
3. If a cloud doesn’t exchange water with its environment (closed) and undergoes
condensate loss under equilibrium conditions, then the change in isotopic composition of
remaining vapor is described by the Rayleigh Distillation equation for a closed system:
Rt/Ro = f ( - 1)
(2)
where R= isotope ratio initially (o) and at a later time (t), f= fraction of original reservoir
(cloud vapor) remaining at t, = the isotopic equilibrium fractionation effect during
condensation, which equals the isotopic ratio of the product (which is lost) over the
reactant (Rcondensate/Rvapor)
4. Example Calculation: What is the 18O change for the water vapor in a cloud after
50% of the vapor has been lost to condensation when the cloud has reached a temperature
of 15ºC?
a. the relevant equilibrium reaction is:
H2O16(l) + H2O18(g)  H2O16(g) + H2O18(l)
-for which  = (18O/16O)l /(18O/16O)g = 1.010 (at 15°C) (see Fig. 4)
b. in terms of the Rayleigh distillation relationship
4
-since Rt/Ro = f ( - 1)
-thus, ln (Rt/Ro) = (1.010-1)*ln 0.5
-and Rt/Ro = 0.9931
-this means that the 18O/16O of the remaining vapor (Rt), after 50% of the vapor
has been lost to condensate, equals 0.9931* 18O/16O initially (Ro)
c. In terms of  notation
thus  =Rt/Ro = (t/1000+1)*Rstd / (o/1000+1)*Rstd
-the Rstd’s cancel
-multiply by 1000/1000, thus Rt/Ro = (t+1000)/(o+1000)
-thus the Rayleigh relationship is

( t  1000)
= f ( -1)
( o  1000)
d. Since Rt/Ro = 0.9931, then t + 1000 = 0.9931*o + 993.1
-rearranging: t - 0.9931*o = -6.9 ‰
-If o = 0 ‰, then t - o = -6.9 ‰
-However, if o = -10 ‰, then t = -16.83 ‰ and then t - o = -6.83 ‰,
-i.e., this is equivalent to Rt/Ro = 0.98317 / 0.990 = 0.9931
-Thus, the change in 18O (in ‰) approximately, but not exactly (unless Ro
= Rstd, i.e., the initial  = 0 ‰), equals Rt/Ro
-i.e., the accuracy of the t - o change depends on the initial 18O
e. Assumptions in Rayleigh Distillation calculation:
-‘closed’ system, i.e., no exchange of vapor with outside environment
- vapor and condensate always in equilibrium
- is constant over period of observation
-this last assumption isn’t likely since the temperature of the cloud
has to decrease for continuous condensation of water and 
depends on temperature ( α would increase as temperature
decreases)
4. NOTE: Both the water vapor and condensate become more depleted in 18O as the
cloud loses water vapor to condensation (i.e., as f decreases) (Fig. 5)
5
-the offset (in ‰) between water vapor and condensate would remain constant as
the cloud lost water vapor, only if  was constant (i.e.,  independent of temperature)
C. Relationship between 18O and Temperature in Precipitation
1. As cloud temperature decreases there are three effects on the water vapor content and
isotopic composition of water vapor in clouds
a. Loss of water vapor continues because of the decreased equilibrium vapor
pressure of water at colder temperatures (see equation (1))
b. the 18O and D of both vapor (and condensate) decrease because the isotopic
composition of lost condensate is 18O and D enriched relative to vapor
c. the vapor-liquid equilibrium fractionation effect increases with decreasing
temperature (Fig. 4)
2. We can predict the relationship between the 18O (or D) and temperature of
precipitation using the Rayleigh Distillation relationship (as done by Dansgaard in his
1964 paper referenced above)
3. First, we calculate the decrease in water vapor content of the cloud as temperature
decreases using equation (1)
-calculate the fraction of water vapor remaining (f) from some initial temperature
4. Second, we calculate the magnitude of the isotopic equilibrium fractionation factor ()
as a function of temperature
5. Third, we calculate the change in the 18O/16O of water vapor (Rt) as a function of
temperature using the Rayleigh Distillation relationship expressed in equation (2)
-i.e., Rt/Ro = f(-1)
6. Finally, we calculate the 18O/16O of the condensate, where Rcondensate =  * Rvapor
6
7. A rough estimate of the Δ18O/ΔTemp of ~0.60 ‰ per ºC is obtained between +10 ºC
and – 10 ºC using equation (1) to predict the change in water vapor and the Rayleigh
relationship to predict the change in 18O assuming no temperature dependence for .
8. Dansgaard predicted the following isotopic changes over a +20 to -20°C range:
Δ(18O)/ΔT = 0.58 ‰ per °C
Δ(D)/ΔT = 4.8 ‰ per °C
9. The observed relationship between 18O of precipitation and air temperature is fairly
linear over +10 to -50 °C range (Fig. 6)
-observed Δ(18O)/ΔTemp = 0.69 ‰ per °C, i.e. 18O = 0.69* T(°C) - 13.6 ‰
-observed Δ(D)/ΔTemp = 5.6 ‰ per °C, i.e. D = 5.6* T(°C) - 100 ‰
-the observed Δ/ΔT slopes are fairly close to those predicted by the Rayleigh
expression
-this agreement supports the assumptions inherent in the Rayleigh condition
(closed system with equilibrium conditions) as reasonable to describe the isotopic effects
during cloud condensation
10. One can predict the relationship (slope) expected between the D and 18O of water
from the derivation of Dansgaard by dividing the ΔD/ΔT by the Δ18O/ΔT, which yields
a slope = 8.28 (i.e. 4.8/0.58).
11. The observed δD vs 18O slope for global precipitation is ~8.17 (Fig. 7)
-i.e., δD (‰) = 8.17*18O (‰) + 10.56 (r2 = 0.997)
-thus the observe slope is close to the predicted slope
-another indication that Rayleigh conditions are reasonable
7
12. The first empirical look at the δD vs 18O trend measured for precipitation was by
Harmon Craig in the 1960s and was the basis for the term Meteoric Water Line (MWL),
to describe the tight correlation between D and 18O. Craig (1961) defined the MWL as:
δD (‰) = 8*18O (‰) + 10
-thus data from the current global network of precipitation isotope measurements
verified Craig’s original concept, based on the few observations he made of the 18O and
D for water in lakes
-this implies that most meteoric waters (rain, snow, lakes, rivers, groundwaters)
follow a very consistent trend in their D and 18O compositions
13. The good agreement between the predicted and observed slopes for 18O and D
versus Temperature for precipitation indicates that the dominant process affecting global
precipitation is an equilibrium condition between vapor and liquid phases in clouds and
then rapid condensate loss in a closed cloud system
-we’ll see, as discussed below, that there is evidence that evaporation induced
kinetic isotope effects perturbs the MWL from a purely equilibrium situation
14. There are four situations where temperature exerts primary controls on the observed
trends in the 18O (and D) of precipitation
a. 18O (D) versus Latitude (Fig. 8)
b. 18O (D) versus Altitude (Fig. 9)
c. 18O (D) versus distance from coast (Continentality) (Fig. 10)
d. 18O (D) versus Seasonal (Fig. 11)
-What process could explain the 18O increase east of the Rockies in Fig .10?
15. However, attempts to improve the correlation between 18O of precipitation with
factors other than temperature (latitude, altitude, season, etc.) don’t significantly improve
the correlations
-Conclusion: temperature exerts dominant control on the large scale patterns of
18O and D of precipitation
8
D. Effects of evaporation on the 18O and D of precipitation
1. Although the Rayleigh Distillation assumption (closed system with equilibrium
between water vapor and liquid) explains the global patterns of 18O and δD of
precipitation (latitude, altitude, etc.), there are exceptions on local scales
2. Evaporation Effects on Precipitation
-In arid regions with very low precipitation rates (<1cm/month), the observed
D/18O slope for rain is 5-6, rather than 8 for the MWL (Fig. 13)
-one explanation of this deviation from the MWL is due to the isotopic effect of
evaporation on the rain droplet, between the time of condensation in the cloud and arrival
at the ground
-the implication is that there is a non-equilibrium or kinetic isotopic effect during
evaporation which causes deviations from MWL
- this situation occurs mainly during light rains in arid, low humidity, locations.
-this deviation from the MWL would show up in water of local lakes, rivers and
groundwater in these regions
3. Seasonal trends at tropical sites (Fig. 14)
a. At tropical locations where monsoons prevail, the 18O of the precipitation is
lowest during the wettest time (warmest) of the year (opposite to the trend
observed in most continental sites)
b. This inverse correlation between 18O and temperature occurs because during
monsoon conditions, humidity is high and evaporation of rain drops is small
c. In contrast, during non-monsoon conditions, i.e., drier months, at the same site,
evaporation of raindrops enriches the 18O and D of the rain.
E. The Deuterium Excess
1.Globally, there is a very consistent relationship between the δD and 18O of
precipitation that is referred to as the Meteoric Water Line (MWL) (Fig. 7)
9
-the empirical relationship for the MWL is D = 8*18O + 10 (Craig, 1961)
-the slope of the empirical line is close to that expected based on the temperature
dependence of the water vapor pressure, the equilibrium fractionation effects (’s)
between water liquid and vapor for 18O and D, and a closed system Rayleigh
Distillation process
2. Note, however, that the intercept of the MWL doesn't go through D = 0 ‰ (Fig. 7), as
expected if water vapor evaporated from the ocean were in isotopic equilibrium with
surface seawater
-i.e., if water vapor evaporated from the ocean surface under equilibrium
conditions and then immediately condensed (also under equilibrium conditions), this
initial precipitation would have the 18O and D equal to the surface ocean, i.e., ~ 0 ‰
(vs SMOW)
3. That the intercept of the MWL has D = ~ 10 ‰ (Fig. 7) suggests that the source of
water vapor for the global water cycle is not in isotopic equilibrium with the surface
ocean
-this implies that a non-equilibrium (i.e., kinetic) process is affecting the isotopic
composition of water vapor over the ocean
-this process is most likely evaporation
4. To quantify deviations from the MWL, Dansgaard (1964) defined a term called
“Deuterium Excess or d-excess”, which looked for D deviations from Craig’s MWL.
d-excess (‰) = D (‰) - 8*18O (‰)
5. For locations where the precipitation falls on the MWL (Fig. 7), then:
-the d-excess = D – 8*18O
- observed precipitation follows D = 8.17*18O + 10.55
- thus d-excess  10.6 ‰ (when 18O = 0 ‰)
d-excess  8.9 ‰ (when 18O= -10 ‰)
10
6. In certain situations, where evaporation is important, there are significant deviations of
the d-excess from that expected for precipitation following the MWL (Fig. 15)
-precipitation in arid regions can have high d-excess values up to +24 ‰
-precipitation in cold regions can have low d-excess values down to +3.5 ‰
7. An inverse correlation between d-excess and relative humidity is predicted from
calculations of the kinetic fractionation during evaporation of water from the surface
ocean (Fig. 15)
-at a humidity of ~85%, the predicted d-excess about equals the observed 10 ‰
-this is close to the average relative humidity over the ocean (~80%)
8. Thus d-excess values that deviate significantly from +10 ‰, either higher or lower,
imply that evaporation has affected the isotopic composition of precipitation, surface
water (lakes, rivers) or groundwaters
9. Similarly, calculations of the slope of D vs 18O for water affected by evaporation
depend on relative humidity (Fig. 16)
- the D and 18O for lakes in eastern Washington show a significantly lower
slope than for lakes in western Washington (which are close to the MWL)
F. Evaporation effects on the global water cycle
1. If equilibrium conditions were met during the evaporation of water from the ocean then
the first rain produced from that vapor would have 18O and D equal to the 18O and D
of the ocean (i.e., close to SMOW).
-
the implication would be that rain in the tropics, where most the evaporation
occurs, would have 18O and D values around 0 ‰ (vs SMOW)
2. However, the observed trend in the 18O and D of precipitation shows a D value of
+10 ‰ when 18O equals zero (the MWL) (Fig. 7)
11
-this non-zero intercept implies that non-equilibrium processes are occurring
during the evaporation of ocean water into the atmosphere
3. Craig and Horibe (1967) measured the 18O and D of water vapor over the ocean and
found mean values of -13 and –94 ‰, respectively. (Fig. 17)
- vapor 18O and D are is lower than the –9 ‰ and –80 ‰ expected at
equilibrium
4. During a research cruise in Sept 2008, our lab measured the 18O and D of water
vapor in the marine boundary over the N. Pacific and found a range of –14±1 and –108±5
‰, respectively (using a laser-based shipboard instrument).
-again the measured 18O and D are significantly lower than the values expected
at equilibrium
-there are very few measurements of the isotopic composition of water vapor
5. In summary, there are several lines of evidence that indicate that evaporation imparts a
non-equilibrium fractionation on atmospheric water vapor
-since the observed water vapor is depleted in the heavy isotopes (18O, D) relative
to that expected under equilibrium conditions, it is likely that a kinetic isotope effect
occurs during evaporation and makes the evaporating water vapor more depleted than that
expected under equilibrium conditions
-
molecular diffusion of water vapor through an air-sea boundary layer is one
potential source of a KIE
6. The evaporation process can be thought of occurring in three steps: (Fig. 18)
1) water vapor released from the liquid (ocean, lake, raindrop, etc.) to a very thin
layer of air (microns thick?) at the air-water interface is assumed to be in isotopic
equilibrium with the liquid water
2) Adjacent to this microlayer is a “laminar sub-layer” near the air-water boundary
(<cms thick?), in which molecular, rather than turbulent, processes dominate. Water
12
vapor diffuses away from the air-sea interface and isotopically fractionates due to
different molecular diffusion rates for the isotopic species
3) when water vapor reaches the turbulent region (“mixed layer”) further away
from interface, it is mixed into the free atmosphere without fractionation.
-turbulent motions moving parcels of air (or water) do not fractionate isotopic
composition of the parcels
7. The rate of water vapor transport (or flux) through the diffusive boundary layer
depends on the water vapor gradient and the molecular diffusion rate of water vapor in air
-the water vapor gradient depends on the relative humidity of air
-thus the lower the humidity the greater the loss of water vapor and the higher the
evaporation rate
-wind speed also impacts the water vapor loss (evaporation) rate because
turbulence depends on wind speed (actually wind shear) which in turn affects the
thickness of the viscous sublayer.
8. Magnitude of kinetic isotope effect (KIE) during evaporation
a. Merlivat (J. Chem. Phys. 69, 1978) measured the ratio of molecular diffusion
rate (D) of isotopic species of water vapor in nitrogen (close to air) (i.e.,  )
DHDO16 / DH2O16 = 0.9757± .0009
DH2O18 / DH2O16 = 0.9727± 0.0007
b. the closeness of fractionation effect measured for 18O/16O and D/H implies that
it is not simply a mass substitution effect, i.e. 18O substitution increases the mass of the
water molecule 2 (out of 18) and a D substitution increases it by 1 (out of 18)
c. the expected fractionation effect due to isotope substitutions would be:
DHDO16 / DH2O16 =
u H 2O16
u HDO16
= [(18 * 28)/(18  28)] / [(19 * 28)/(19  28)]
= 0.984 or –16 ‰
-whereas u represents the reduced mass
13
-using the same reduced mass equation yields DH2O18 / DH2O16 = 0.969 or
–31 ‰
-
thus the predicted KIE for diffusion of water vapor in air predicts a KIE for
18O of water vapor that is about twice that for D of vapor
-
in contrast, the observed KIEs for D and 18O are about the same magnitude
9. Merlivat (1978) suggests the isotopic substitution of D for H yields a larger than
expected KIE because it results in an asymmetry of the O-H bond length, which in turn
affects the rotational energy state of the isotopically substituted water molecule, which
doesn't occur with the 18O for 16O substitution.
-we’ve mainly discussed isotope fractionation effects due to the differences in
vibrational energy states of isotopologues
10. However, Cappa et al (J. Geophys. Res. 108, 2003) experimentally derived values for
diffusivities of H2O18/H2O16 = 0.969 and for HDO/HHO = 0.984 equal to the expected
values and says Merlivat didn’t account correctly for cooling of ‘skin temperature’ of
water during evaporation.
11.Wind tunnel experiments by Vogt (1976) and Munnich (1978) measured the KIE
during evaporation directly and found similar rates for the two isotopologues:
WH2O18 / WH2O16 = 0.9859±0.0004
WHDO16 / WH2O16 = 0.9878±0.0017
-where W represents the water vapor (gas) transfer rate of each species through
the boundary layer
12. If we now go back to the three-step evaporation model shown schematically in Fig
18, we can estimate the overall isotopic fractionation during evaporation
a. In the monolayer at the exact air-water interface, the atmospheric water vapor
is in isotopic equilibrium with the surface skin of the ocean (or lake, river etc.)
14
b. in the laminar viscous sublayer molecular diffusion fractionates the water
vapor transferring through this layer
c. in the mixed layer, turbulent processes transfer the water vapor away from the
viscous layer without isotopic fractionations
d. In this model the overall fractionation effect is the combination of the
equilibrium fractionation effect and the kinetic fractionation effect, which
would yield 0.976 for 18O/16O (i.e., 0.9902* 0.986) and 0.905 for D/H
(i.e.,0.916*0.988) using the empirical kinetic evaporation ’s measured by
Vogt and Munnich.
e. Cappa et al (2003) experimentally measures overall fractionation effect during
evaporation and finds values of 0.980 and 0.910 for 18O/16O and D/H of water,
respectively, which agrees fairly well with model predictions.
f. Thus the model predicted overall ’s during evaporation –24 ‰ and –95 ‰
for 18O and D of evaporating water vapor, respectively, agrees well with
measured effects of –20 and –90 ‰
13. However, the 3-step model is not realistic yet because it does not take into account the
diffusion of water vapor from the atmosphere back to the surface ocean
-the model considered only the loss of water vapor from the ocean surface,
i.e., the model assumes dry air (humidity = 0 %)
-since the humidity of air over the ocean is typically 75-85%, there is a
significant transfer of water vapor molecules from the air to the ocean surface
-the diffusion of water vapor from air to ocean not only decreases the net
evaporation rate, it also affects the isotopic composition of the net evaporative vapor flux
14. Effect of evaporation’s kinetic fractionation effect on the D vs 18O relationship of
precipitation
- the estimated 18O and D of water vapor leaving the surface ocean would be
15
~ -20 and -90 ‰ for 18O and δD, respectively, using Cappa’s experimental
determination of overall ’s at humidity = 0%) and assuming seawater has 18O and D =
0 ‰ (i.e., SMOW)
- this vapor would plot with a slope of ~4.5 in D vs 18O (-90‰ / -20‰)
- this slope is equal to that found in the bottom plot in Fig. 16 for humidity of 0%
- in Fig 16, the D vs 18O slope increases as you increase the humidity and
approaches the ratio of the equilibrium fractionation effects for D and 18O of ~
8.6 (i.e., -84‰ / -9.8‰) as the humidity approaches 100%
-
this change in slope is a result of the increasing effect of diffusion of water
vapor from air to the water surface as humidity increases
14. Notice that the KIE for 18O (-10 to -14 ‰) during evaporation (is similar in
magnitude to the equilibrium isotope effect (~ -10 ‰) but the KIE for D (-7 to -12 ‰)
during evaporation is much smaller than the equilibrium isotope effect (-84 ‰)
-
this means that the D of the evaporating vapor is higher (enriched) relative to
the 18O of the vapor when compared to the 18O and D expected under
equilibrium conditions
-
thus the KIE during evaporation yields an ‘excess’ in D, relative to 18O, for
the vapor and is the source of the “d-excess” observed in precipitation (Fig.
15)
-
the magnitude of this excess D depends on humidity and varies from 0 ‰ at
humidity = 100% to 30 ‰ at humidity = 60% (Fig. 15),
-
calculating the d-excess of the kinetically fractionated water vapor leaving the
water’s surface (at humidity = 0%) yields a value = +70 ‰ (d-excess = -90 ‰
– 8*(-20 ‰) = 70 ‰, using Cappa’s  values and assuming seawater is 0 ‰ )
15. Since a d–excess = 10 ‰ is observed in global precipitation, this implies that
evaporation occurs (primarily in tropical ocean) with a mean humidity of ~85% (Fig. 15)
III. GLOBAL WATER CYCLE
16
1. On a global scale, the observed D and 18O for most precipitation falls on a meteoric
water line where D (‰) =~ 8.1718O + 10.55 with a range of ~25 ‰ in 18O and ~200
‰ in D between the equator and 75°N and S (Fig. 7).
2. Since the slope of this empirical expression is close to that expected from isotopic
equilibrium, this implies that equilibrium conditions likely prevail during the
precipitation portion of the water cycle
-the expected D vs 18O slope for precipitation at isotopic equilibrium would be
the ratio of the vapor-liquid equilibrium ε’s for D/H and 18O/16O, i.e., -84 ‰ / -9.8 ‰ =
8.6 which is close to the observed slope of 8.2 (see Figs. 3 and 7)
3.However, equilibrium conditions alone cannot explain the entire MWL since there is a
D intercept of ~10.6 ‰ for the MWL at 18O = 0 (Fig. 7).
-this +10.6 ‰ intercept indicates that the D/H of the water vapor in the tropics is
isotopically enriched relative to 18O/16O, i.e., if only equilibrium conditions affected D
and 18O the intercept would be 0 ‰ for both 18O and D
-this relative D enrichment is thought to result from the KIE during evaporation
(as discussed above)
-thus evaporation, and it’s associated KIE, likely controls the isotopic
composition of the water vapor leaving the ocean.
4.The importance of kinetic isotope effects during evaporation is also apparent from a
global atmospheric mass and isotope budget for water vapor.
a. the mean surface temperature on earth is 15 ºC. At this mean temperature, the
18O and D of precipitation is –4 and –22 ‰, based on the 18O vs temperature
correlation (Fig. 6) and D vs 18O relationship (Fig. 7)
b. At steady-state, this implies that the 18O and D of evaporating water must be
–4 and –22 ‰ to maintain an isotope balance for atmospheric water vapor.
17
-An analogous argument is made that the global evaporation rate is equal to
the global precipitation rate to maintain a constant water vapor content at steady-state.
c. In contrast, if isotopic equilibrium conditions existed during evaporation, then the
evaporated vapor would have 18O and D of about –9.8 and –84 ‰, respectively.
- i.e., seawater at 0‰ (for 18O and D vs SMOW) minus equilibrium
epsilons of –9.8 and –84 ‰
d. Thus the steady-state isotopic balance between evaporation and precipitation is
not met under isotopic equilibrium condition during evaporation
5. We can explain the –4 and –22 ‰ values implied for the net evaporative water vapor
flux, by calculating the net rates of H2O16, H2O18 and HDO vapor fluxes.
a. The diffusive flux or transport of water vapor through the diffusive air-sea boundary
layer depends on two criteria, the water vapor gradient and molecular transfer rate W
b. The H2O16 vapor flux (moles/m2/d) = WH2O16 * (H2Osat - H2Oatm),
(4)
-where W = molecular gas transfer velocity (m/d), sat=the saturated water vapor
content of air (moles/m3) and atm=the water vapor content of air (moles/m3)
-in terms of relative humidity (rh), where rh= H2Oatm/H2Osat
-the H2O16 vapor flux (moles/m2/d) = WH2O16 * H2Osat*(1 - rh)
(5)
b. Similarly,
-the H2O18 vapor flux (moles/m2/d) = WH2O18*(H2Osat*Rw*sol - H2Oatm*Ratm),
= WH2O18 * H2Osat*(Rw*sol - rh*Ratm),
(6)
(7)
-where Rw is the isotope ratio of the ocean water, sol is the equilibrium fractionation
between liquid and vapor, and Ratm is 18O/16O of atmospheric water vapor.
6. Calculation of the effect of kinetic and equilibrium fractionation on the 18O of water
vapor flux during evaporation.
a. The ratio of the net H2O16 and H2O18 water vapor fluxes is independent of evaporation
rate [i.e., divide equation (7) by equation (5)] and equal to:
18
-the H2O18 flux / H2O16 flux = w *(1*Rw*sol - rh*Ratm) / (1 - rh)
-where the ratio of the diffusive H2O18 and H2O16 vapor transfer rates
(WH2O18/WH2O16) = αw
-where w = 0.986 (from Vogt, Munnich), sol = 1/1.0098 = 0.9903, the 18O of
atmospheric water vapor = -13 ‰ (based on only few measurements by Craig and
Gordon, Fig. 17) and the 18O of surface water = 1‰ and rh =0.80 (the average
ocean value is 75-85%)
b. Calculate the ratio of the H2O18/H2O16 vapor fluxes
-for a relative humidity (rh) = 0.81
-the H2O18/H2O16 flux ratio = 0.986*(1*1.001*0.9903-0.81*0.987)/(1- 0.81)*Rstd
= 0.9955* Rstd
-if rh = 0.5, the H2O18/H2O16 flux ratio = 0.9816 * Rstd
-if rh =0.9, the H2O18/H2O16 flux ratio = 1.0155 * Rstd
-using  notation, the 18O of the net water vapor flux (18Oflx) is calculated as
follows (for rh =0.81): = -4.5 ‰
- 18Oflx (‰) = (Rsam/Rstd –1)*1000
- 18Oflx (‰) = (0.9955*Rstd/Rstd –1)*1000
- 18Oflx (‰) = (0.9955 – 1)*1000 = -4.5 ‰
-in this case the Rstd represents the 18O/16O of the standard used (SMOW) for the
18O measurements of the surface seawater and atmospheric water vapor (Ratm)
7. What does this calculated 18Oflx value actually represent?
-it represents the 18O/16O (in  terms) of the net water vapor flux from the ocean
-in other words, the net H2O18 flux = net H2O16 flux * (18O/16O)flx, where
(18O/16O)flx = (18Oflx/1000 +1)*Rstd
-Note: one can’t calculate a value for 18Oflx when rh=100%, because there is no
net water flux when rh=100%
19
8. The concept of the isotopic ratio (18O/16O, D/H) of a ‘flux’ is useful because it
quantifies the isotopic signature of the net water vapor flux and thus indicates the isotopic
effect that evaporation has on the water in surface ocean and on the water vapor in the
atmosphere
-based on the above calculations, the net water vapor flux from the ocean is
depleted in 18O (about –4 or -5 ‰) and this water vapor loss will enrich the 18O of the
surface ocean
-thus evaporation is a process that typically enriches the remaining water in 18O
and D (although not the degree predicted by an equilibrium fractionation situation)
9. Importantly, a 18Oflx value of –4.5 ‰ for the net water vapor loss from the ocean
equals the 18O of –4 ‰ observed for the globally average precipitation.
-as a result, there is the isotopic balance needed at steady-state
-Conclusion: the combination of equilibrium and kinetic fractionation effects and
the mean relative humidity over the oceans controls the 18O (and D) of the water vapor
input to the atmosphere via evaporation, which on a global scale, is balanced by the 18O
of precipitation.
-currently, there a very few measurements of 18O and D for water vapor in air
over the ocean which limits our ability to confirm the effects of KIEs during evaporation
as an explanation of d-excess and the isotopic composition of the global net evaporative
water flux
10. In summary, the measured D and 18O of precipitation, and thus the implied D and
18O values for the net water vapor flux from the oceans, indicate that evaporation
kinetically controls the isotopic composition of the water vapor entering the atmosphere
from the surface ocean, whereas vapor-liquid equilibrium processes control the isotopic
composition of the precipitation leaving the atmosphere.
IV. EXAMPLES OF USING ISOTOPE TRACERS IN WATER CYCLE STUDIES
20
A. Paleo-temperature reconstruction using 18O and D measurements of ice cores
in Greenland and Antarctica
1. Basic Approach:
-measure the 18O or D of ice recovered in ice core
-date ice core (either from summer/winter ice layers, ice flow models, comparing
18O record to deep sea sedimentary record, etc.)
-measure today’s 18O or D in precipitation vs temperature relationship near the
ice core site (Greenland and Antarctica)
-estimate paleo-temperature record from measured 18O (D) vs time in ice core
and today’s 18O (D) vs Temperature relationship
- a key assumption is that the 18O (or D) vs temperature relationship for
precipitation today applies back through time. How legit?
-this approach has been used to estimate paleo air temperatures during the last
750K years at polar ice core locations (Greenland and Antarctica)
2. Results from Greenland Ice Core 18O measurements (Fig. 19)
a. Present day Temperature/18O = 1.5 ºC / ‰ for precipitation in Greenland
b. Using today’s Δ18O/ΔT, the 18O ice core record implies a ~ 10°C cooler air temp
during last ice age (~20K yrs ago)
3. However, an independent method to estimate paleo-temperature variations is based on
downcore (borehole) temperature measurements yields temperature changes that are
approximately twice that estimated from 18O-ice changes (Fig. 19)
-The borehole temperature record implies a Temperature/18O = 3.0 ºC / ‰
and temperatures during last ice age of ~ 20ºC colder. (Fig. 20)
-Although the cause for the discrepancy is not fully understood, the ice core
community seems to favor the borehole method which implies there is a problem
with assuming that the 18O vs temperature relationship observed for precipitation
on Greenland today is constant back through time
21
4. Modeling studies of the water vapor transport and its isotopic composition suggest
that the relative importance of large scale circulation vs storm (eddy) transport of
water vapor in the atmosphere impacts the isotopic composition of the vapor
delivered to Greenland (e.g., Hendricks et al., Space and time variations of the
18O and D in precipitation: can paleotemperatures be estimated from ice
cores? Global Biogeochemical Cycles 14: 851-860, 2000)
-in essence, this is a questions of whether the water vapor delivery process
is a closed or open system
5. Ice core 18O record from Antarctica yields temperature changes that are close to
those estimated from borehole temperature measurements (Fig. 21)
-Ice Cores from Antarctica yield temperature record back to 750K yrs
B. Local temporal trends in the isotopic composition of precipitation
1. At most continental sites, the 18O (and D) of the precipitation typically varies with
the annual temperature cycle (as predicted by the MWL) (Fig. 22a)
-thus 18O and D of precipitation is lower in winter than in summer
2. At some sites, however, there are major 18O variations in precipitation which are not
correlated to temperature change (Fig. 22b)
-at these sites changes in the source of the precipitation (air mass) is a likely
explanation
-e.g., the intrusion of polar air masses, with depleted 18O values, may disrupt the
“normal” seasonal 18O cycle associated with local precipitation (air temperature
dominated)
-this implies that the 18O (and D) of local precipitation can be influenced by the
history (source and trajectory) of the air mass
-How could this situation complicate the paleotemperature reconstructions
discussed above?
22
C. 18O as a tracer of water sources in streams and rivers
1. Small rivers and creeks are usually dominated by local precipitation as a water source
a. Thus their 18O (and D) usually show a similar seasonal trend as the local
precipitation, i.e., low 18O in winter and high 18O in summer, which in turn
is controlled by seasonal trends in local air temperature (Fig. 23a)
b. Small streams typically show more temporal variability in isotopic
composition due to events (e.g., snow melt, storms) (see Fig. 23a).
2. Larger rivers, with larger drainage basins can have 18O (and D) temporal trends that
deviate substantially from local precipitation (Fig. 23b)
a. for example, the input of snow melt from higher elevations in the drainage
basin can yield a 18O (and D) minimum in spring and summer
b. in this case, 18O provides a tracer of snow melt versus rain water sources
within a drainage basin
3. For example, in the Rhine River (Fig. 23b), the seasonal cycle in 18O indicate lowest
values in summer when air temps are highest
-implies that snow melt is substantial part of summer water discharge
4. Calculating the fraction of snow melt using a simple mass/isotope water balance
- measured river water 18O = -8.8 ‰ in winter and 18O = -10.5 ‰ in summer
- 18O = -13.5 ‰ for alpine (snow melt) water source (estimated from measured
wintertime temperature?) and -7.7 ‰ for lowland (local precipitation) water
source
Water Mass (H2O16) balance:
Qr = Qa + Ql
-where a is alpine flow, l is lowland flow and r is measured river flow
Water Isotope (H2O18) balance:
Qr*Rr = Qa*Ra + Ql*Rl, where R is 18O/16O.
-One can solve for Ql/Qr and Qa/Qr , if Ra, Rl and Rr are measured:
23
-mass and isotope balances : Qr = Qa + Ql and Qr*Rr = Qa*Ra + Ql*Rl
-substitute: Qa+Ql for Qr
-then: (Qa +Ql)*Rr = Qa*Ra + Ql*Rl
-rearranging: Qa*(Ra - Rr) = Ql*(Rr- Rl)
-finally: Qa/Ql = (Rr - Rl)/(Ra - Rr)
-in summer:
-the river 18O = -10.5 ‰,
-then Qa/Ql = (0.9895 – 0.9923)/(0.9865 - 0.9895) and Qa/Ql = 0.93, i.e., there is
about equal alpine and lowland contribution to river flow (rate of snow melt about
equals lowland precipitation rate)
-since Qa/Qr + Ql/Qr = 1, then Qa/Qr = Ql/Qr = 50%.
(Note: the Rstd’s cancel when solving for Qa/Ql)
-in winter:
-the river 18O = -8.8 ‰,
-then Qa/Ql = (0.9912 – 0.9923)/(0.9865 - 0.9912) and Qa/Ql = 0.23, i.e, the
lowland input is 4x the alpine input (i.e., snow is accumulating, rather than
melting, in the alpine region)
-thus Qa/Qr= 19% and Ql/Qr = 81%, i.e, ~80% is lowland flow
Questions
1. What assumptions are inherent in this simple mass and isotope balance
approach?
2.What measurements primarily affect the accuracy of this approach?
-How would you quantify the uncertainty of the calculations?
3. Would D measurements help?
5. Tracer of Flood Waters in River Systems
a. For river systems that have large drainage basins and multiple sources of
waters (local, highland, mountainous) 18O can be a useful tracer of the source
of flood waters.
24
b. Example: The Mississippi, Missouri and Meramac rivers drain very different
regions (Fig. 24)
-The Missouri R. has significant drainage in the upper central US and
Rocky Mountains, the Mississippi R. drains mainly central US and the
Meramac R. is a much smaller river with drainage restricted to the state of
Missouri (central US)
c. Typically, the isotopic composition of these three rivers is distinct, with the
Missouri having the lowest D and 18O and the Meramac having the highest
D and 18O as expected from the mean annual temperatures of their drainage
basins (see Fig. 24)
d. During the peak of 1995 flood (May 20-23), all three rivers had essentially the
same 18O and D values (Fig. 25)
a. the 18O and D of the small Meramac River changed the least and the
Missouri R. changed the most during the peak flood
b. this observation suggests that peak flood waters were mainly derived
from precipitation in the central US region of the drainage basin
(where the Meramac R. is located)
e. Temporal records of isotopic composition in large river systems, like this one,
should provide validation of models used to predict river flow and response to
storms
5. The key observation to successfully utilizing isotopes as a river water tracer is to
determine whether the potential water sources have different (distinct) isotopic signatures
-if so, great.
-if not, isotopic measurements likely won’t be too useful
6. There are distinct geographic patterns to the d-excess of rivers in the US (Fig. 26)
- d-excess values close to ~10 ‰ (i.e., MWL) are found in the eastern US,
whereas lower values (to –2 ‰) are found in the western US
i. Any ideas on what could cause this pattern?
25
b. potentially, distinctive d-excess could be used as a tracer of specific river
D. Using isotopes to trace groundwater inputs
1. Typically, at some depth in the soil the 18O (and D) of groundwater should
be similar to the mean annual 18O (and D) of local precipitation, assuming
local precipitation is the main groundwater source (Fig. 27b)
2. In general, the seasonal variations in 18O or D of groundwater decrease with
increasing soil depth (Fig. 27a)
a. As you go deeper in soils, mixing has more time to homogenize the
seasonal variations in 18O (D)
b. A similar depth trend exists for soil temperature variations
3. In arid locations, the 18O and D of shallow groundwater can be enriched via
evaporation in dry soil surface layer (Fig. 28a)
a. Remember, evaporation yields an overall fractionation effect of ~ -20
‰ for 18O and –90 ‰ for D (Cappa et al 2003)
b. thus the 18O (D) of the near surface ground waters can be enriched
relative to the local precipitation.
4. There can a situation where the D and 18O of the local precipitation falls on
MWL but the D and 18O of groundwater falls off the MWL (Fig. 28b)
-this implies that evaporation rate within the soils is significant relative to
precipitation inputs and allows KIE during evaporation to affect 18O and
D
5. Potentially, isotopic measurements provide a means to quantify the
contribution of local precipitation versus ground water inputs to streams (or
plants) in this situation where the isotopic composition of these two water
sources differ
26
–If evaporation is important, then coupled D and 18O measurements are
useful because they fall off the MWL
6. Important: The key observation to successfully utilizing isotopes as water
source tracers is whether the sources (e.g., groundwater vs precipitation) have
isotopic signatures that are distinct
E. 18O and D as tracers of water masses in the ocean
1. Generally, the spatial and temporal variations of 18O (and D) in the ocean are very
small compared to trends for precipitation, lakes and rivers because the oceanic water
reservoir size is so large. (Fig. 29)
-for most of the open ocean, the 18O range is only from ~-1 to +1 ‰
2.The 18O and D of surface ocean water is mainly controlled by the rate of evaporation
-in regions of high evaporation, the D and 18O of seawater is high because the
evaporated water vapor is depleted in 18O and D
3.The Atlantic has higher 18O than the Pacific or Indian oceans
-what does this imply about evaporation rates?
4. In regions of very high evaporation rates, like the Mediterranean and Red Seas, the
surface 18O can get as high as +2 ‰ .
5. Generally, the surface ocean 18O (D) correlates linearly with salinity. (Figs. 30, 31)
-not surprising. Why?
6. The 18O of the North Atlantic Deep Water is distinctly higher than Antarctic Bottom
Water (Fig. 31, 32)
27
-why is this expected given the different formation mechanisms for these two
deep water masses?
7. The strong correlation between 18O (D) with salinity dominates the water parcels in
the Deep Sea (Fig. 33)
a. In a sense, this conservative relationship between D (or 18O) and salinity (or
potential temperature) in the Deep Sea has reduced the application of isotopic
measurements as a tracer of water masses in the ocean
c. In other words, you don’t learn much more by measuring the 18O or
D of deep sea water than by measuring its temperature and salinity
(which is much, much easier)
d. The slope of the 18O vs salinity depends on the fractionation effect
during evaporation
e. Since the slope in Fig 31 is about 0.5 (a 1.5 ‰ 18O increase
corresponds to a 3 ‰ salinity increase, this suggests the 18O
fractionation effect during evaporation of about –15 ‰
i. S+ 18O balance: [32(-1 ‰) + 3(18Olost) = 35(0.5 ‰)]
f. This –15 ‰ estimate is close to the –14 to –20 ‰ fractionation effect
estimated from the 3 step model and measured by Cappa et al)
8. In some situations, however, there is some additional information to be gained from
18O (and D) measurements.
-Typically, these situations occur in the coastal ocean where there is significant
input of freshwater from rivers, snow or ice (land)
-especially in polar latitudes where the 18O of the precipitation (and thus land
ice) is low, as seen off Greenland (Fig 31a)
-in these situation, 18O (D) can offer additional information than that derived
from T vs S relationships
-Why isn’t the 18O (and D) likely to be a good tracer of melting sea ice? (Hint:
see Fig. 3)
28
9. Locally, in Puget Sound, I’d guess that the 18O (D) of rivers with large drainage
areas in the Cascades (e.g. Skagit River) would have significantly lower 18O (D) than
rivers with large lowland drainage (e.g. Sammamish R.)
-thus 18O (D) measurements in Puget Sound likely could be used to track
freshwater input from specific rivers that wouldn’t be distinguishable by salinity
-is there local data?
V. REVIEW QUESTIONS
1. What is the isotopic evidence that precipitation is mainly an equilibrium process
and evaporation is mainly a kinetic process in the global hydrologic cycle?
2. Why does the 18O and D of precipitation correlate so well with air temperature?
3. Under what conditions can the 18O and D of precipitation deviate from the
MWL?
4. Explain why kinetic isotope effects during evaporation can cause deviations in the
d-excess value from the ~ +10 ‰ value observed for global precipitation. Why
does the magnitude of the deviation depend on humidity?
5. At steady-state, what is the 18O and D of the global water vapor flux from the
ocean to the atmosphere? What assumption(s) did you make?
6. How is the 18O (and D) correlation with temperature for precipitation used to
reconstruct paleo temperatures from 18O (and D) measurements on ice cores?
What’s a possible complication with this method?
7. What drainage basin characteristics likely make 18O (and D) useful tracers of
riverine water sources?
8. How would a glacial-interglacial change in humidity affect the 18O and D
record in ice cores?
9. Why would the 18O and D of water vapor likely be different during glacial
versus interglacial times?