Garden with Insight v1.0 Help: Nitrogen - Organic N Transport by Sediment
A loading function developed by McElroy et al. (1976) and modified by Williams and Hann (1978) for
application to individual runoff events is used to estimate organic N loss. The loading function is
[Equation 164] where YON is the organic N runoff loss in kg/ha, Y is the sediment yield in t/ha, c(ON) is
the concentration of organic N in the top layer in g/t, and ER is the enrichment ratio. The enrichment
ratio is the concentration of organic N in the sediment divided by that in the soil.
Equation 164
YON = 0.001 * Y * C(ON) * ER
Code:
this equation is used to calculate the amount of six materials lost in erosion:
organicNActiveHumus_kgPha, organicNStableHumus_kgPha, organicMatter_tPha,
organicPHumus_kgPha, labileP_kgPha, and mineralPActive_kgPha.
taking the case of ON(a),
propLost = ER * Y / SWT and YON(a) = ON(a) * propLost so
YON(a) = ON(a) * ER * Y / SWT
C(ON) = ON(a) / SWT so
YON(a) = Y * C(ON) * ER
only the 0.001 is lost, possibly because C(ON) is in different units
also an upper bound of 0.9 is placed on propLost (ER * Y / SWT)
Variables:
YON(a) = organicNActiveHumusAdsorbedToSediment_kgPha
ON(a) = organicNActiveHumus_kgPha
propLost = OrganicNOrCOrPProportionLostInSediment_kgPha
C(ON) = organicNConcSurfaceLayer_gPt
Y = totalErosion_tPha
SWT = surfaceLayerSoilWeight_tPha
ER = enrichmentRatioForNPPest
Enrichment ratios are logarithmically related to sediment concentration as described by Menzel (1980).
An individual event enrichment-sediment concentration relationship was developed for EPIC considering
upper and lower bounds. The upper bound of enrichment ratio is the inverse of the sediment delivery
ratio. Exceeding the inverse of the delivery ratio implies that more organic N leaves the watershed than is
dislodged from the soil. The delivery ratio is estimated for each runoff event by using the equation
[Equation 165] where DR is the sediment delivery ratio (sediment yield divided by gross sheet erosion),
q(p) is the peak runoff rate in mm/hr, and r(ep) is the peak rainfall excess rate in mm/hr. Equation 165 is
based on sediment yield estimated by using MUST.
Equation 165
DR = sqrt(q(p) / r(ep))
Code:
same except code has power to 0.56, not 0.5
Variables:
DR = SedimentDeliveryRatio
q(p) = peakRunoffRate_mmPhr
r(ep) = peakRainfallExcessRate_mmPhr
The rainfall excess rate cannot be evaluated directly because the hydrology model predicts only the total
daily runoff volume. An estimate of the rate can be obtained, however, using the equation [Equation 166]
where r(p) is the peak rainfall rate in mm/hr and the runoff-rainfall ratio Q/R is used to account for
infiltration. The Q/R is raised to the 0.1 power to account for intensity effects on infiltration - Q/R is
greater than average near r(p).
Equation 166
r(ep) = r(p) * power(Q / R, 0.1)
Code:
same
Variables:
r(ep) = PeakRainfallExcessRate_mmPhr
r(p) = peakRainfallRate_mmPhr
Q = runoffVolume_mm
R = rainfallForDay_mm
The lower limit of enrichment ratio is 1.0 - sediment particle size distribution is the same as that of the
soil. Thus, 1.0 <= ER <= 1/DR. The logarithmic equation for estimating enrichment is [Equation 167]
where c(s) is the sediment concentration in g/m3 and x(1) and x(2) are parameters set by the upper and
lower limits.
Equation 167
ER = x(1) * power(c(s), x(2))
1.0 <= ER <= 1/DR
Code:
same but 1.0 <= ER <= 5.0
Variables:
ER = EnrichmentRatioForNPPest
c(s) = sedimentConc_gPm3
x(1) = coeffForEnrichmentRatio
x(2) = expForEnrichmentRatio
For the enrichment ratio to approach 1.0, the sediment concentration must be extremely high. Conversely,
for the enrichment ratio to approach 1/DR, the sediment concentration must be low. The simulataneous
solution of equation 167 at the boundaries assuming that sediment concentrations range from 500 to
250,000 g/m3 gives [Equation 168] and [Equation 169].
Equation 168
x(2) = -ln(1/DR) / 2.699
Code:
same
Variables:
x(2) = ExpForEnrichmentRatio
DR = sedimentDeliveryRatio
Equation 169
x(1) = 1 / power(0.1, x(2))
Code:
same
Variables:
x(1) = CoeffForEnrichmentRatio
x(2) = expForEnrichmentRatio
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