Samm Shell1

Samm Shelly ABE 224: ABE Principles - Soil and Water Homework # 4 Due Date : Monday, 11/27/2023 1) If the predicted soil loss at Memphis, Tennessee, for a given set of conditions is 12 Mg/ha, what is the expected soil loss at your present location if all factors are the same except the rainfall factor? ( Problem 7.1 of the textbook ) Present location: Urbana, IL 12Mg/ha = soil loss in Memphis, Tennessee USLE (Universal Soil Loss Equation) A = R*K*L*S*C*P A – average annual soil loss [Mg/ha] R – rainfall erosivity index K – soil erodibility factor L – slope length factor S – slope steepness factor C – Cover Management Factor P – Conservation Practice Factor R TN =5000 K TN =K IL L TN= L IL S TN =S IL C TN =C IL P TN =P IL R IL =3000 A TN =12[Mg/ha] = R TN *K*L*S*C*P KLSCP = 12/5000 => KLSCP = 0.0024 => 0.0024*R IL =A IL => 0.0024*3000= 7.2 Mg/ha 2) On a 60-m long slope, what is the relative difference in water erosion between slopes of 2 and 10%, assuming other factors are constant? ( Problem 7.2 of the textbook ) Slope length=60m Slope steepness factor_1=2% Slope steepness factor_2=10% Where RKLCP are constant for both S 1 and S 2 Where: θ = tan − 1 ( s ) θ 1 = tan − 1 ( 0.02 ) = 1.146 θ 2 = tan − 1 ( 0.10 ) = 5.711 Using equation (7.8b) for S 1 S = 10.8sin ( θ ) + 0.03 = 10.8sin ( θ 1 ) + 0.03 S 1 = 10.8sin ( 1.146 ) + 0.03 = 0.2460 Using equation (7.8c) for S 2 S = 16.8sin ( θ ) − 0.50 = 16.8sin ( θ 2 ) − 0.50 S 2 = 16.8sin ( 5.711 ) − 0.50 = 1.172 The ratio of the soil erosion of a 2% slope and a 10% slope with the other factors remaining constant. The soil erosion of the 2% slope to the 10% slope is: A 1 A 2 = RKLCP ∗ 0.2460 RKLCP ∗ 1.172 = 0.2460 1.172 = 0.2099

3) If the soil loss for a given set of conditions is 4.5 Mg/ha for a 60-m length of slope, what soil loss could be expected for a 240-m slope length if the slope is 5%? ( Problem 7.3 of the textbook ) Soil loss = 4.5Mg/ha Length = 60m Slope = 5% A = R× K × L×S×C ×P L = ( l 22 ) b where L = slope length factor, l = slope length b = dimensionless exponent given by: b = sin ( θ ) sin ( θ ) + 0.269 ( sin ( θ ) ) 0.8 + 0.05 where θ = tan − 1 ( s ) L 60 = ¿ slope length constant of 60m L 240 = ¿ slope length constant of 240m θ = tan − 1 ( 0.05 )= 2.86 ° b = sin ( 2.86 ) sin ( 2.86 ) + 0.269 ( sin ( 2.86 ) ) 0.8 + 0.05 = 0.40128 L 60 = ( 60 22 ) 0.40128 = 1.4957 L 240 = ( 240 22 ) 0.40128 = 1.1006 S 60 = S 240 thus using S S = 10.8 sinθ + 0.03 = 0.5689 A 60 = 4.5 = R×K × ( L 60 ) ×S×C×P = RKCP ( 1.4957 )( 0.5689 ) RKCP = 4.5 1.4957 × 0.5689 = 5.2885 A 240 = RKCP ( L 240 ) ( S ) = 5.2885 × 1.1006 × 0.5689 = 3.3113 Mg/ha 4) Given your present location, assume that K = 0.015, l = 91 m, s = 10%, C = 0.2, and that up- and downslope farming is practiced. What is the soil loss? What conservation practice should be adopted if the soil loss is to be reduced to 5 Mg/ha? ( Problem 7.4 of the textbook ) R = 3000 K = 0.015 l = 91m s = 10% C = 0.2 P = 1.0* *Given up-/downslope farming is used A 0 = R × K × L×S×C×P = 3000 × 0.015 ×L×S × 0.2 × 1.0 L 0 = ( l 22 ) b where b = sin ( θ ) sin ( θ ) + 0.269 ( sin ( θ ) ) 0.8 + 0.05 and θ = tan − 1 ( s ) θ = tan − 1 ( s ) = tan − 1 ( 0.10 ) = 5.7106 b = sin ( 5.7106 ) sin ( 5.7106 ) + 0.269 ( sin ( 5.7106 ) ) 0.8 + 0.05 = 0.5183 L 0 = ( 91 22 ) 0.5183 = 2.087 S = 16.8sin ( θ ) − 0.50 = 16.8sin ( 5.7106 ) − 0.50 = 1.1717 A 0 = 3000 × 0.015 × 2.087 × 1.1717 × 0.2 × 1.0 = 22.0080 Mg/ha is the soil loss A desired = 5 = 3000 × 0.015 × 2.087 × 1.1717 × 0.2 ×P => P = 5 3000 × 0.015 × 2.087 × 1.1717 × 0.2 P desired = 0.2272