Find: Pe [in] N P=6 [in] Ia=0.20*S Land use % Area

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Presentation transcript:

Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C Industrial D 4.0 4.2 4.4 4.6 Find the excess precipitation, P sub e, in inches. [pause] In this problem, --- N

Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C Industrial D 4.0 4.2 4.4 4.6 a lot is divided into 4 distinct land uses, whose percent area and soil type are provided. N

Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C Industrial D 4.0 4.2 4.4 4.6 The total rainfall is 6 inches, and the initial abstraction, is 20% of the maximum retention. [pause] N

Find: Pe [in] N P=6 [in] Ia=0.20*S 4.0 4.2 4.4 4.6 Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C Industrial D 4.0 4.2 4.4 4.6 The SCS rainfall-runoff equation --- N

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C Industrial D (P-Ia)2 for excess precipitation, is, --- N Pe= P-Ia+S

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C Industrial D runoff depth (P-Ia)2 the excess precipitation, equals the quantity, --- N Pe= P-Ia+S

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 the total rainfall depth, minus the initial abstraction, squared, --- N Pe= P-Ia+S rainfall depth

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 divided by the rainfall depth, minus the initial abstraction, --- N Pe= P-Ia+S rainfall depth

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 plus the maximum retention. [pause] N Pe= P-Ia+S rainfall maximum depth retention

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 The problem statement provides the rainfall depth, as P = 6 inches, --- N Pe= P-Ia+S rainfall maximum depth retention

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 and an initial abstraction of 0.2 times the maximum retention. N Pe= P-Ia+S rainfall maximum depth retention

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 Substituting this value in 0.20 times S, in for the initial abstraction, --- N Pe= P-Ia+S rainfall maximum depth retention

Find: Pe [in] N P=6 [in] Ia=0.20*S (P-Ia)2 Pe= P-Ia+S (P-0.2*S)2 Pe= Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 The equation reduces to a function of the rainfall depth, P, and the maximum retention, S. N Pe= P-Ia+S (P-0.2*S)2 Pe= P+0.8*S

Find: Pe [in] P=6 [in] Ia=0.20*S (P-Ia)2 1,000 S= -10 Pe= CN P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 The maximum retention equals 1,000 divided by the curve number, minus 10, --- 1,000 S= -10 Pe= CN P-Ia+S (P-0.2*S)2 Pe= P+0.8*S

Find: Pe [in] P=6 [in] Ia=0.20*S (P-Ia)2 1,000 S= -10 Pe= CN P-Ia+S Land use % Area Soil Type Residential 40% 25% 20% 15% C P=6 [in] Open space D Ia=0.20*S Commercial C initial Industrial D abstraction runoff depth (P-Ia)2 and the curve number is a function of the land uses, percent areas, and soil types. [pause] Curve numbers based on various land uses, 1,000 S= -10 Pe= CN P-Ia+S (P-0.2*S)2 Pe= P+0.8*S

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] Land use % Area Soil Type CN Residential 40% 25% 20% 15% C Open space D Commercial C Industrial D for hydrologic soil types A through D, are provided in tables, and generally range between 40 and 98. 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] A (Sand)  Low CN Find: Pe [in] Land use % Area Soil Type CN Residential 40% 25% 20% 15% C Open space D Commercial C Industrial D Soil type A refers to sandy soils which have a high infiltration potential, and low curve number. 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C Open space D Commercial C Industrial D 1,000 Soil type D refers to clayey soils which have low infiltration potential, and higher curve numbers. Soil types B and C are in between. (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C Open space D Commercial C Industrial D 1,000 For a medium density residential land use, with a soil type C, --- (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 The curve number equals 83. [pause] The other three curve numbers are looked up as well --- 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in] A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 [pause]. 1,000 (P-0.2*S)2 S= -10 Pe= CN P+0.8*S P=6 [in]

Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S P=6 [in] A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 The curve number to use in the retention equation, should be a composite curve number which account for all four land uses. 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S P=6 [in]

Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp= A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 This value should be the weighted average of all curve numbers, based on the percent area of each land use. 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S n Σ CNcomp= (% Area)i * CNi P=6 [in] i=1

Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp= A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 Plugging in the values for percent areas and curve numbers, the composite curve number computes to --- 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S n Σ CNcomp= (% Area)i * CNi P=6 [in] i=1

Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp= A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 85.95. [pause] Substituting this value in, --- 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S n Σ CNcomp= (% Area)i * CNi = 85.95 P=6 [in] i=1

Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S CNcomp= A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 the maximum retention becomes, --- 1,000 (P-0.2*S)2 S= -10 Pe= CNcomp P+0.8*S n Σ CNcomp= (% Area)i * CNi = 85.95 P=6 [in] i=1

Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 1.634 inches. [pause] Finally, the retention depth --- 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp P+0.8*S n Σ CNcomp= (% Area)i * CNi = 85.95 P=6 [in] i=1

Σ Find: Pe [in] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 and rainfall depth are substituted in for variables P, and S, and the excess rainfall equals, --- 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp P+0.8*S n Σ CNcomp= (% Area)i * CNi = 85.95 P=6 [in] i=1

Σ Find: Pe [in] Pe= 4.40 [in] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 4.40 inches. [pause] 1,000 (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp P+0.8*S n Pe= 4.40 [in] Σ CNcomp= (% Area)i * CNi i=1

Σ Find: Pe [in] Pe= 4.40 [in] 4.0 4.2 4.4 4.6 1,000 (P-0.2*S)2 S= A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 4.0 4.2 4.4 4.6 1,000 When reviewing the possible solutions, --- (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp P+0.8*S n Pe= 4.40 [in] Σ CNcomp= (% Area)i * CNi i=1

Σ Find: Pe [in] AnswerC Pe= 4.40 [in] 4.0 4.2 4.4 4.6 1,000 A (Sand)  Low CN Find: Pe [in] D (Clay)  High CN Land use % Area Soil Type CN Residential 40% 25% 20% 15% C 83 Open space D 80 Commercial C 94 Industrial D 93 4.0 4.2 4.4 4.6 1,000 the answer is C. (P-0.2*S)2 S= -10 = 1.634 [in] Pe= CNcomp P+0.8*S n Pe= 4.40 [in] Σ CNcomp= (% Area)i * CNi AnswerC i=1

? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1 Find: σ’v at d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ γ d d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4