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ESPM 111 Ecosystem EcologyProblem Set III: Soils and Nutrient Cycling

March 1 2013, due March 8

1. We collected 20 g (wet weight) of soil from the field. A 10 g subsample of wet soil was oven-

dried at 105

o

C. The oven-dry weight was 5 g. We extracted the remaining 10 g of wet soil with50 mL of 1 M KCl over 6 h, followed by 50 mL of BaCl over 6 h. We then analyzed thesolutions on the autoanalyzer for K

+concentrations and correct for the dry weight of the soil.

Our analyses gave us 920 mg K+/kg dry soil.

1a. Calculate the cation exchange capacity in cmol+/kg soil.

CEC = 920mgK+/kg * 1mmol/39.1mgK+ * 1cmol/10mmol = 920/391.0 = 2.3529 cmol+/kg

1b. Give two types of soils that might have this level of CEC and explain why

Sandy soils may have this level of CEC because they are bigger particles leading a smalleroverall surface area, and therefor a smaller charge, and these particles lack the ability to hold on

to as many nutrients as compared to fertile clay soils.

1:1 silicate clays may also have this level of CEC because they are highly weathered soils, whichleads to a smaller surface area. As mentioned before, less surface area means less charge and less

ability to retain nutrients.

2a. The effective cation exchange capacity is a measure of the sum of exchangeable cations,generally determined from the subset of dominant cations. Calculate the ECEC in cmol

+/kg soil

given the following concentrations.

Soil 1: Ca2+

= 175 mg/kg, K+

= 95 = mg/kg, Mg2+

= 120 mg/kg, Al3+

= 290 mg/kg

1 mmol+ Ca = 20.039, 175/20.039 = 8.73mmol/kg, 8.73/10 = .873 cmol+/kg Ca2+cmol+/kg K+ = 95/(39.0983*10) = .243 cmol+/kg K+

cmol+/kg Mg2+ = 120/((24.305/2)*10) = .987 cmol+/kg Mg2+cmol+/kg Al3+ = 290/((26.982/3)*10) = 3.22 cmol+/kg Al3+

ECEC = .873 + .243 + .987 + 3.22 = 5.323 cmol+/kg

Soil 2: Ca2+

= 225 mg/kg, K+

= 50 = mg/kg, Mg2+

= 115 mg/kg, Al3+

= 180 mg/kg

cmol+/kg Ca2+ = 225/(20.039*10)= 1.123

cmol+/kg K+ = 50/(39.0983*10) = .128cmol+/kg Mg2+ = 115/((24.305/2)*10)= .946

cmol+/kg Al3+ = 180/((26.982/3)*10)= 2.00

ECEC = 1.123 + .128 + .946 + 2.00 = 4.197 cmol+/kg

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2b. Which soil has higher nutrient availability? Which element in the above examples is not anutrient? What does its concentration tell you about the soil?

Soil 2 has a higher nutrient availability because Al is not a nutrient and therefore should not be

considered when determining the nutrient availability. Soil 1 has a nutrient availability of 5.323-

3.22 = 2.103 where soil 2 has a nutrient availability of 4.197 2.00 = 2.197. Al is not a nutrientand the higher concentration of Al is soil 1 tells you that soil one is more acidic because soil willpreferentially hold on to aluminum and aluminum promotes acidity.

3. Calculate net ammonification, N mineralization, and nitrification from the following values.

Express your answers in g N/g/d. For each rate state if there is net immobilization or netproduction.

To (initial) Tfinal (7 days)

NH4+-N = 1.35 g N/g 3.00 g N/g

NO3--N=0.75g N/g 0.40 g N/g

Net ammonification = (3-1.35)/7 = .2357 ug N/g/d. (net production)

Net nitrification = (.4-.75)/7 = -.05 ug N/g/d (net immobilization)Net mineralization = .2357 + (-.05) = .1857 ug N/g/d (net production)

4. The saturation vapor pressure curve is given as a function of Temperature (Tc) in degrees

Centigrade in the following equation (units, kPa).

17.502( ) 0.611exp( )

240.97

C

s C

C

Te T

T

=

+

a. Compute and plot saturation vapor pressure, es(T), for a range of temperatures between 0and 30 C, at 5 C increments.

Tc es(Tc)

0 1

5 0.839225732

10 0.709235822

15 0.603334102

20 0.516435758

25 0.444645786

30 0.384955849

b. Compute and plot the slope of the saturation vapor pressure curve at 5, 10, 15, 20 and 25C (

( )s

e Ts

T

=

).

0

0.2

0.4

0.6

0.8

1

1.2

0 10 20 30 40

es(Tc)kPa

Tc(C)

Satura0onvaporpressure

Series1

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T Slope

5 -0.032154854

10 -0.025997982

15 -0.021180344

20 -0.017379669

25 -0.01435799530 -0.011937987

c. Compute the relative humidity (RH = ea/es(T) at 5, 15 and 25 C, assuming the vapor

pressure ea is 1200 Pa; make sure your units cancel!

es ea RH(ea/es)

839.225732 1200 1.429889426

603.334102 1200 1.988947742

444.645786 1200 2.698777404

5. Use the information in Problem 4 to assess potential evaporation. It is proportional to

available energy (evaluated here as net radiation, Rnet) timess

s + ,

e.g. the ratio of the slope of the saturation vapor pressure (s) and the sum of s and the

psychrometric constant; ( = 0.0668 kPa C-1

)

In this exercise you will compute potential latent heat exchange (Epot) first:

1.26pot net

sE R

s

=

+

Then you will solve for potential evaporation (Epot) by dividing by the latent heat of evaporation,

.

a. For this temperature range (0, 5, 10, 15, 20, 25, 30 C) compute potential evaporationassuming the daily average net radiation (Rnet) equals 190 W m

-2.

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

0 10 20 30 40

Slope(

kPa/Tc)

T(C)

SlopeofSatura0onvapor

pressure

Series1

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b. Given the latent heat of evaporation ( = 2442 J/g) compute the depth of water evaporated per

day for each temperature class. Other factors to consider include how many seconds per day andthe depth of water per gram, assume 1 gram is 1 cm

3. Watch your units!

From 5a (Wg/m^2J), units = g/m^2s cm^3/m^2/s 3600 cm^3/m^2/d 3600

cm^3/100cm^2/d 36cm/d

T = 5, 36 * (-.090987697) = -3.27556 cm/dT = 10, 36 * (-.06245482) = -2.24837352 cm/d

T = 15, 36 * (-.045515518) = -1.63855836 cm/dT = 20, 36 * (-.034475798) = -1.24112844 cm/d

T = 25, 36 * (-.026840647) = -.96626304 cm/dT = 30, 36 * (-.021332309) = -.76796316 cm/d

c. Compare evaporation rates for an Arctic site with a mean annual temperature near 0 C with

a tropical site near 25 C, assuming the same net radiation balance for this case.

The evaporation rates for the arctic with a temperature near 0 C are negative and lower (-.090987) than those at a tropical site near 25 C (-.026940647). This would make sense because

the negative evaporation rates accounts for a surplus of water and in the arctic there is going to

be little rainfall but even less evaporation because of the cold temperatures. In the tropics, thereare high levels of rainfall and evaporation so the number should be higher and closer to 0.

See spreadsheet

7. The Hubbard Brook Ecosystem study was a large-scale experiment that contributed

significantly to our current thinking about ecosystem structure and functioning. Two leadinginvestigators working on the project published a seminal article on the watershed approach to

nutrient cycling (Bormann, F. H., and G. E. Likens. 1967. Nutrient Cycling. Science 155:424-429) where they proposed the following equation:

(Meteorological Input + Biogeochemical Input)-(Geologic Output + Biogeochemical Output) =

Net System Flux

Below are several ecosystem attributes. Use all or part of the data in this list to answer the

following questions.

Mean annual rainfall: 3500 mm

T

5

10

15

20

25

30

Epot=1.26(s/s+y)*Rnet/wavelength

(Wg/m^2J)

-0.090987687

-0.062464962

-0.045515518

-0.034475798

-0.026840647

-0.021332309

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Mean annual nitrogen in rainfall: 1 kg N/haMean annual dry deposition nitrogen: 0.05 kg N/ha

Mean annual nitrogen in snowfall: 0 kg N/haMean annual nitrogen fixation (plants and free living organisms): 5 kg N/ha

Mean annual nitrogen in litterfall: 50 kg N/ha

Mean annual nitrogen export in stream water: 5 kg N/haMean daily gaseous nitrogen emissions (all forms): 50 ng N/cm2/dMean annual nitrogen loss in sediments (erosional): 1 kg N/ha

1a. Calculate the net ecosystem flux of nitrogen based on the values below (report your answer

in the appropriate units). Based on these data is the ecosystem accumulating or losing nitrogenover a one year period?

Net system flux = ((1 + .05) + (5 + 50) ((5 + 1) + .01825) = 50.03175 kgN/ha/y

50 ngN/cm^2/d- 10^6 cm = 1 ha, 10^12 ng = 1kg, 50ng/cm *365 = .01825 kg/ha/y

Therefore, the ecosystem is accumulating nitrogen.

1b. Assume the standard error of the inputs is approximately 25 % of the mean and the standarderror of the output is approximately 30% of the mean. How does this affect your interpretation

of the result in question 2a?

Inputs = (1+ .05 + 5 + 50) * .25 = 14.0125, +/- 50.03175 = 64.04425/36.01925Outputs = (5 + 1 + .01825) * .3 = 1.805475, +/- 50.03175 = 51.837225/48.226275

Because the inputs are so much larger than the outputs, an error of 25% and 30% does not impact

our overall interpretation of 1a because the ecosystem would still be gaining nitrogen. Even witha negative standard deviation of the inputs and the positive standard deviation of the outputs, the

ecosystem will still gain nitrogen.

1c. The forest managers are considering harvesting the forest for timber. The loggers wouldremove approximately 20 Mg of biomass in the harvest and that biomass would, on average,

have a concentration of 0.15 % nitrogen. How much nitrogen (in kg/ha) would be exported in theharvest? What would the net ecosystem flux of nitrogen be? Would the ecosystem accumulate or

lose nitrogen over the year?

20Mg = 10^3kg = 20000 kg, 20000kg * .0015 = 600 kgN/ha exported in harvest

Adding 600 kgN/ha to outputs = (1 + .05 + 5 + 50) (5 + 1 + .01825 + 600) = -549.96825kgN/ha/y = net flux

The ecosystem would lose a considerable amount of nitrogen over the year

1d. A fertilizer plant opened upwind of the forest and is emitting nitrogen oxides, which are

being deposited on the forest (wet plus dry deposition) at a rate of 10 kg N/ha/yr. Using the datain part (a) what is the net ecosystem flux of nitrogen under this scenario?

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Adding 10 kgN/ha/y to inputs = (1 +.05 + 5 + 50 + 10) (5 + 1 + .01825) = 60.03175 kgN/ha/y= net ecosystem flux