Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 13 17 21 26 in hr 150 drain i = Find the peak flowrate, Q sub p, in cubic feet per second. [pause] In this problem, --- pipe tc[min]+20

Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 13 17 21 26 in hr 150 drain i = 3 watersheds drain to a common outlet, and the runoff is conveyed offsite by a drain pipe. pipe tc[min]+20

Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 13 17 21 26 in hr 150 drain i = The area, runoff coefficient and time of concentration for each watershed, are provided in the data table. pipe tc[min]+20

Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 13 17 21 26 in hr 150 drain i = The intensity curve for the given storm is also provided. pipe tc[min]+20

Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = The peak flowrate, Q sub p, in cubic feet per second equals 1.008 --- pipe tc[min]+20 peak flowrate QP ft3 in =1.008 * C * i * A [acres] s hr

Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = times the runoff coefficient, C, --- pipe tc[min]+20 peak flowrate in QP ft3 =1.008 * C * i * A [acres] s hr runoff coefficent

Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = times times the intensity, i, in inches per hour, ---- pipe intensity tc[min]+20 peak flowrate ft3 in QP =1.008 * C * i * A [acres] s hr runoff coefficent

Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = times the area, A, in acres. pipe intensity tc[min]+20 peak flowrate in QP ft3 =1.008 * C * i * A [acres] s hr area runoff coefficent

Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = Considering the precision of the factors, --- pipe intensity tc[min]+20 peak flowrate in QP ft3 =1.008 * C * i * A [acres] s hr area runoff coefficent

Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = the 1.008 terms is usually rounded to 1, and the equation becomes, --- pipe intensity tc[min]+20 peak flowrate in QP ft3 =1.008 * C * i * A [acres] s hr area runoff coefficent

Find: Qp [cfs] QP = C * i tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = Q sub p, equals C times i times A. [pause] Let’s begin with the area, --- pipe intensity tc[min]+20 peak flowrate in QP ft3 = C * i * A [acres] s hr area runoff coefficent

Find: Qp [cfs] A = AA + AB + AC QP = C * i tc Area C [acre] [min] Area 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = A. [pause] The total area equals the sum of watershed areas A, B and C. pipe intensity tc[min]+20 peak flowrate QP ft3 in A = AA + AB + AC = C * i * A [acres] s hr area runoff coefficent

Find: Qp [cfs] A = AA + AB + AC QP = C * i tc Area C [acre] [min] Area 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = After plugging in the values, from the data table, we find --- pipe intensity tc[min]+20 peak flowrate QP ft3 in A = AA + AB + AC = C * i * A [acres] s hr area runoff coefficent

Find: Qp [cfs] A = AA + AB + AC QP = C * i A = 20 [acres] tc Area C [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = the total area, equals, 20 acres. pipe intensity tc[min]+20 peak flowrate QP ft3 in A = AA + AB + AC = C * i * A [acres] s hr A = 20 [acres] area runoff coefficent

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = Next, we’ll solve for the runoff coefficient, C, which equals --- pipe intensity tc[min]+20 peak flowrate in QP ft3 = C * i * A [acres] s hr 20 runoff coefficent

Find: Qp [cfs] Σ C *A C = Σ A QP = C * i 20 tc Area C [acre] [min] B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C Σ C *A C = drain the average of the runoff coefficients from all contributing watersheds, --- Σ A pipe intensity peak flowrate ft3 in QP = C * i * A [acres] s hr 20 runoff coefficent

Find: Qp [cfs] Σ C *A C = Σ A CA*AA+CB*AB+CC*AC C = AA+AB+AC QP tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C Σ C *A C = drain weighted by area. Plugging in the values for C and A, --- Σ A pipe intensity CA*AA+CB*AB+CC*AC peak flowrate C = AA+AB+AC in QP ft3 = C * i * A [acres] s hr 20 runoff coefficent

Find: Qp [cfs] Σ C *A C = Σ A CA*AA+CB*AB+CC*AC C = AA+AB+AC QP tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C Σ C *A C = drain the weight C value equals, --- Σ A pipe intensity CA*AA+CB*AB+CC*AC peak flowrate C = AA+AB+AC in QP ft3 = C * i * A [acres] s hr 20 runoff coefficent

Find: Qp [cfs] Σ C *A C = Σ A CA*AA+CB*AB+CC*AC C = AA+AB+AC QP tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C Σ C *A C = drain 0.563. Σ A pipe intensity CA*AA+CB*AB+CC*AC peak flowrate C = AA+AB+AC ft3 in QP = C * i * A [acres] s hr C =0.563 20 runoff coefficent

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = --- pipe intensity tc[min]+20 peak flowrate in QP ft3 = C * i * A [acres] s hr 20 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in hr 150 drain i = The last variable to solve is i, the intensity. pipe intensity tc[min]+20 peak flowrate in QP ft3 = C * i * A [acres] s hr 20 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 150 drain in i = The storm intensity is a function of the time of concentration, t sub c, The intensity decreases asymptotically, with t sub c. pipe hr intensity tc[min]+20 peak flowrate i QP ft3 in = C * i * A [acres] s hr 20 0.563 tc

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 150 drain in i = For the given equation, we should use the maximum t sub c value, --- pipe hr intensity tc[min]+20 peak flowrate i QP ft3 in = C * i * A [acres] s hr 20 0.563 tc

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C drain in 150 i = for all contributing watersheds. This mean runoff from all areas --- pipe hr intensity tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) QP ft3 in = C * i * A [acres] s hr 20 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 150 drain in i = of all watersheds will contribute to the peak flowrate, Q sub p. [pause] pipe hr intensity tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) QP ft3 in = C * i * A [acres] =max(35, 45, 25) s hr 20 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in 150 drain i = The time of concentration for this problem is 45 minutes. pipe hr intensity tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) in QP ft3 = C * i * A [acres] =max(35, 45, 25) s hr 20 =45 [min] 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in 150 drain i = Plugging 45 minutes into the IDF curve, --- pipe hr intensity tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) in QP ft3 = C * i * A [acres] =max(35, 45, 25) s hr 20 =45 [min] 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in 150 drain 2.308 = the storm intensity equals 2.308, inches per hour. pipe hr intensity tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) in QP ft3 = C * i * A [acres] =max(35, 45, 25) s hr 20 =45 [min] 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in 150 drain 2.308 = Returning to the rational equation, we can compute the peak flowrate, by --- pipe hr 2.308 tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) in QP ft3 = C * i * A [acres] =max(35, 45, 25) s hr 20 =45 [min] 0.563

Find: Qp [cfs] QP = C * i 20 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C in 150 drain 2.308 = plugging in our values for C, i and A, and the peak flowrate equals, --- pipe hr 2.308 tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) in QP ft3 = C * i * A [acres] =max(35, 45, 25) s hr 20 =45 [min] 0.563

Find: Qp [cfs] QP= 25.99 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 150 drain in 2.308 = 25.99 cubic feet per second. pipe hr tc[min]+20 peak flowrate tc=max(tc,A,tc,B,tc,C) ft3 s QP= 25.99 =max(35, 45, 25) =45 [min]

Find: Qp [cfs] 13 17 21 26 QP= 25.99 tc Area C [acre] [min] Area Area B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 13 17 21 26 drain Looking over the possible solutions, --- pipe peak flowrate ft3 s QP= 25.99

Find: Qp [cfs] AnswerD 13 17 21 26 QP= 25.99 tc Area C [acre] [min] B A 7 0.65 35 A B 10 0.40 45 C 3 0.90 25 Area Q C 13 17 21 26 drain the answer is D. pipe peak flowrate ft3 s QP= 25.99 AnswerD

? 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