Applications of USLE
Applications of USLE
◼ Used for estimating quantity of soil erosion annually
◼ Used for designing soil conservation measures and
structures
◼ Popularly used for estimation of the values of C and P
◼ Knowing R,K,L,S the values of C & P are optimized to
keep soil loss (A) within a permissible limit
◼ Used to assess the impact of soil conservation
measures implemented
◼ Compare soil losses before and after implementation
Drawbacks of USLE
◼ It was developed based on American conditions only
◼ May not be valid in other regions
◼ It is not applicable to large watersheds
◼ Non-availability of measured data limits application
◼ It can not be used for estimating soil loss from individual
rainfall events
◼ It considers only rainfall erosivity, but gully erosion &
wind erosion is not considered
◼ In case of high intensity rainfall it underestimates soil loss
Drawbacks of USLE
◼ Being an generalized empirical equation it suffers
from predictive errors
◼ It does not compute sediment deposition during
transportation
◼ The equation has conceptual limitations in
considering interdependency between the variables
(eg. some effects are counted twice; terracing effect is
counted in both L and P)
◼ In computation of rainfall erosivity drop-size
distribution is not considered
◼ Runoff volume and velocity can affect soil erosion
significantly; but it is not taken into account
Modified USLE (MUSLE)
◼ Rainfall Erosivity Factor replaced by a Ruoff Factor
◼ Sediment yield can be calculated for individual
rainfall events
◼ Applicable to large watersheds up to 70 km2 area
◼ The MUSLE is given as:
Y = 11.8 (Q qp)0.56 K L S C P
◼ Where, Y = sediment yield from an individual rainfall
Q = total runoff volume (m3)
qp = peak or maximum rate of runoff (m3/s)
◼ Other terms are same as USLE
Numerical Problems on USLE
◼ Problem 1:
◼ Determine the value of soil erodibility factor for a soil sample
which was analyzed for the following parameters:
(Silt + fine sand) = 65%, Sand = 10%, Gravel = 0%,
Organic matter = 3%. Soil structure is fine granular
and soil permeability is moderate
◼ Solution:
◼ Given: Soil structure code (b) = 2 (for fine granular)
Permeability class (c) = 3 (for moderate)
% clay = [100 – (65+10+3)] = 22%
M = (% silt + % fine sand)(100 - % clay) = 65 × (100 – 22) = 5070
K = 2.8 × 10-7 (5070)1.14 (12 – 3)
= 0.042 (ton/ha)/(MJ.mm/ha.h) (Answer)
◼ Problem 2:
◼ Determine topographic factor if length of slope is 400
m with a 9% land slope. Assume m = 0.5 for 9% slope.
◼ Solution:
◼ Given: Lp= 400 m, s = 9% and m = 0.5
◼ We know, topographic factor (LS) is given as:
LS = (Lpm/100) (1.36 + 0.97s + 0.1385s2)
= (4000.5/100) (1.36 + 0.97 × 9 + 0.1385 × 81)
= 4.25 (Answer)
◼ Problem 3:
◼ In a 20 ha catchment soil erosion is to be evaluated. The
information available are: rainfall erosivity 1000 (ton-
m/ha)(mm/h) per year, soil erodibility 0.25 ton/ha per unit
rainfall erosivity and topographic factor is 0.1. The value of C
for the crops cultivated is 0.5. Two conservation practices are
adopted: Contour farming in 12 ha (P = 0.6) and Strip
cropping in 8 ha (P = 0.3). Estimate annual soil loss from the
catchment. What will be soil loss if no conservation methods
were adopted?
◼ Solution:
◼ Given: R = 1000 (ton-m/ha)(mm/h) per year
K = 0.25 ton/ha per unit R
LS = 0.1
C = 0.5
◼ Average value of P has to be determined
◼ We calculate weighted average of P as:
◼ P = [(12 × 0.6) + (8 × 0.3)] / 20 = 0.48
◼ Annual soil loss (A) = 1000 × 0.25 × 0.1 × 0.5 × 0.48
= 6 ton/ha (Answer)
◼ In case no conservation practice adopted: P = 1
◼ Annual soil loss (A) = 1000 × 0.25 × 0.1 × 0.5 × 1
= 12.5 ton/ha (Answer)
◼ Problem 4:
◼ The soil loss from a field with a given set of
conditions is 5 ton/ha/year for a 60 m length of slope
and 10% slope. What will be the soil loss from a field
with 240 m slope length under identical conditions.
◼ Solution:
◼ Given: Land slope = tanθ = (s/100) = (10/100)
θ = tan-1(10/100) = 5.7106 o
m = (sinθ)/[sinθ + 0.269 (sinθ)0.8 + 0.05] = 0.5183
◼ For the first plot: Lp = 60 m and soil loss (A1) = 5
ton/ha/year
◼ Soil loss = A1 = RK (Lp/22.13)m SCP
or, 5 = RKSCP (60/22.13)0.5183 …(1)
◼ For the second plot all other conditions are same
except Lp = 240 m
◼ Soil loss = A2 = RKSCP (240/22.13)0.5183 …(2)
◼ Dividing equation (1) by equation (2) we get:
(5/A2) = (60/240)0.5183
or, A2 = 10.2574 ton/ha/year (Answer)
◼ Problem 5:
◼ In a 20 ha watershed soil loss is required to be kept at
25 ton/ha/year. Suggest a conservation practice factor
for the area if rainfall erosivity is 100 (t-m/ha)(mm/h)
per year, soil erodibility factor 6.3 t/ha/unit R,
topographic factor 0.3 and crop cover factor 0.5
◼ Solution:
◼ Given: R = 100 (t-m/ha)(mm/h) per year, A = 25 t/ha
per year, K = 6.3 t/ha/R, LS = 0.3, C = 0.5
From USLE we can solve for (P):
P = [25 / (100 × 6.3 × 0.3 × 0.5)] = 0.26
Note: P = 0.26 can be achieved by adopting contour strip
cropping in the entire watershed
