1. Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.6535 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

2. Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.6535 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

3. Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.6535 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

4. Find: Qp [cfs] 13 17 21 26 tc Area C [acre] [min] Area Area B A 7 0.6535 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

5. Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A7
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
pipe
tc[min]+20
peak flowrate QP
ft3 in
=1.008 * C * i
* A [acres] s hr

6. Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A7
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

7. Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A7
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

8. Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A7
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

9. Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A7
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

10. Find: Qp [cfs] QP =1.008 * C * i tc Area C [acre] [min] Area Area B A7
0.65 35 A B 10
0.40 45 C 3
0.90 25
Area Q C in hr
150
drain i =
the 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

11. Find: Qp [cfs] QP = C * i tc Area C [acre] [min] Area Area B A 7 0.6535 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

12. Find: Qp [cfs] A = AA + AB + AC QP = C * i tc Area C [acre] [min] Area7
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

13. Find: Qp [cfs] A = AA + AB + AC QP = C * i tc Area C [acre] [min] Area7
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

14. 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

15. 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

16. 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

17. Find: Qp [cfs] Σ C *A C = Σ A CA*AA+CB*AB+CC*AC C = AA+AB+AC QPtc
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

18. Find: Qp [cfs] Σ C *A C = Σ A CA*AA+CB*AB+CC*AC C = AA+AB+AC QPtc
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

19. Find: Qp [cfs] Σ C *A C = Σ A CA*AA+CB*AB+CC*AC C = AA+AB+AC QPtc
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

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. 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

31. Find: Qp [cfs] QP= 25.99 tc Area C [acre] [min] Area Area B A 7 0.6535 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]

32. Find: Qp [cfs] 13 17 21 26 QP= 25.99 tc Area C [acre] [min] Area AreaB 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

33. 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

34. ? 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