Hydrology  and  Floodplain  Analysis  Fifth  Edition  

Supplemental  Materials  

Jones  Creek  

Subbasins  of  Jones  Creek  

Example  1:  Calculate  volume  of  the  Direct  Runoff  of  Subbasin  A  and  plot  the  net  rainfall  on  the  resul@ng  hydrograph  

Baseflow for Subbasins A = 100 cfs

Initial infiltration for the first hr = 0.5 in

Constant infiltration = 0.25 in 0

100

200

300

400

500

600

700

0 2 4 6 8 10 12 14

Q (

cfs)

Time (hr)

Storm Hydrograph for Subbasin A

0 0.5

1 1.5

2 2.5

3 3.5

1 2 3

Inte

nsity

(in/

hr)

Time (hr)

Rainfall hyetograph (Subbasin A)

Solu@on  to  Example  1    Separate  out  the  base  flow    

Solu@on  to  Example  1   Calculate  the  volume  of  Direct  Runoff  

DRO = Total Runoff – Base Flow

Total Runoff = 12 hr * 600 cfs * ½ * (3600 s/ hr) * (1 ac-ft/ 43,560 ft3)

= 297.5 ac-ft

Base Flow = (12 hr + 10 hr)/2 * 100 cfs * (3600 s/ hr) * (1 ac-ft/ 43,560 ft3)

= 90.9 ac-ft

DRO = 297.5 ac-ft – 90.9 ac-ft

DRO = 206.6 ac-ft

Solu@on  to  Example  1   Resulting  hydrograph  after  removing  Baseflow  

Solu@on  to  Example  1   Separate  out  the  infiltration  from  the  hyetograph  

Solu@on  to  Example  1   Place  the  rainfall  in  the  upper  left  corner  of  the  resulting  hydrograph  

0  

0.5  

1  

1.5  

2  

2.5  

3  0  

100  

200  

300  

400  

500  

600  

0   2   4   6   8   10   12  

Rainf

all  (in)  

Q  (c

fs)  

Time  (hr)  

Storm  Hydrograph  for  Subbasin  A:  No  Baseflow  

Problem  1:  Calculate  volume  of  the  Direct  Runoff  of  Subbasin  C  and  plot  the  net  rainfall  on  the  resul@ng  hydrograph  

 Use  the  process  in  Example  1  

Base flow for Subbasin C = 200 cfs Initial infiltration for the first hr = 1 in Constant infiltration = 0.5 in

0

0.5

1

1.5

2

2.5

3

1 2 3

Inte

nsity

(in/

hr)

Time (hr)

Rainfall hyetograph (Subbasin C)

Example  2:  a)  Make  sure  the  Unit  hydrograph  of  Subbasin  A  is  a  unit  hydrograph  

 Area  of  Subbasin  A  =  550  ac  

 Assume  no  baseflow  

 Qp  =  220  cfs  

Solu@on  to  Example  2,  Part  a  DRO  =  area  under  the  hydrograph    

DRO  =  220  cfs  *  5  hr  *  ½  

DRO  =  550  cfs-­‐hr  =  550  ac-­‐in    

550  ac-­‐in  /  550  ac  =  1  in  

Example  2:  b)  Given  a  4-­‐period  storm  (1  period  =30  minutes)  and  the  previous  unit  hydrograph  for  Subbasin  A,  create  the  storm  hydrograph  using  the  “add  and  lag”  method  

  Area  of  A  =  550  ac    Assume  no  baseflow  or  

infiltration    Qp  =  220  cfs  

Time (0.5 hr)

Q (cfs)

0 0

1 55

2 110

3 165

4 220

5 183.333

6 146.667

7 110

8 73.333

9 36.667

10 0

0.5 in 0.5 in 1 in 1.5 in

Solu@on  to  Example  2,  Part  b  Time (0.5 hr)

Q (cfs)

0 0

1 55

2 110

3 165

4 220

5 183.333

6 146.667

7 110

8 73.333

9 36.667

10 0

Time (0.5 hr) P1*Un P2*Un P3*Un P4*Un Qn 0 0 0

1 27.5 0 27.5

2 55 55 0 110

3 82.5 110 82.5 0 275

4 110 165 165 27.5 467.5

5 91.6665 220 247.5 55 614.2

6 73.3335 183.333 330 82.5 669.2

7 55 146.667 274.9995 110 586.7

8 36.6665 110 220.0005 91.6665 458.3

9 18.3335 73.333 165 73.3335 330

10 0 36.667 109.9995 55 201.7

11 0 55.0005 36.6665 91.7

12 0 18.3335 18.3

13 0 0

Unit Hydrograph

Solu@on  to  Example  2,  Part  b  Time

(0.5 hr) Qn

(cfs) 0 0

1 27.5

2 110

3 275

4 467.5

5 614.2

6 669.2

7 586.7

8 458.3

9 330

10 201.7

11 91.7

12 18.3

13 0

Problem  2  (informa@on  on  next  slide)    a)  Make  sure  the  Unit  hydrograph  of  Subbasin  C  is  a  unit  hydrograph  (Subbasin  C  area  =  770  ac)  

  b)  Given  a  4-­‐period  storm  (1  period  =30  minutes)  and  the  unit  hydrograph  for  Subbasin  C,  create  the  storm  hydrograph  using  the  “add  and  lag”  method  

Unit  Hydrograph  and  4-­‐period  storm  for  Subbasin  C  

Time (0.5 hr)

Q (cfs)

0 0.0

1 70.0

2 140.0

3 210.0

4 280.0

5 240.0

6 200.0

7 160.0

8 120.0

9 80.0

10 40.0

11 0

Storm  Hydrograph  Worksheet  UH for C

Time (0.5 hr) Qn

0.0 0

70.0 1

140.0 2

210.0 3

280.0 4

240.0 5

200.0 6

160.0 7

120.0 8

80.0 9

40.0 10

0 11

12

13

14

Example  3:  Find  the  @me  to  peak  (tp)  and  the  maximum  flow  (Qp)  by  the  SCS  triangular  unit  hydrograph  method  for  Subbasin  A  (550  ac).  Use  L  =  1  mi,  average  slope  =  25[/mi,  D  =  0.5  hr  and  CN  =  71.  

  tp  =  [L0.8(S  +  1)0.7]/1900y0.5    L  =  length  to  divide  (ft)    y  =  average  watershed  slope  (%)    S  =  (1000/CN)  –  10  in    CN  =  Curve  Number  for  various  soil/land  use  

 Qp  =  [484  *  (Area  in  sq.  mi)]/TR   TR  =  D/2  +  tp  =  rise  time  of  the  hydrograph  

  D  =  rainfall  duration  

Solu@on  to  Example  3  L  =  5280  ft  S  =  (1000/71)  –  10  =  4.085  y  =  25ft/5280ft  =  0.474  %  

tp  =  [5280.8*(5.085)0.7]/1900(.474)0.5  

     tp  =  2.27  hr  

Solu@on  to  Example  3  D  =    1/2  hr  TR  =  D/2  +  tp  

TR  =  1/4  hr  +  2.27  hr  =  2.52  hr  

550  ac(43,560  ft2/  ac)(mi/5280  ft)2  =  0.859  sq.  mi  

Qp  =  [484  *  (.859)]/2.52  hr  Qp  =  165.0  cfs  

Solu@on  to  Example  3  TB  =  time  base  

DRO  =  ½  *  Qp  *  TB  hr  

550  ac-­‐in  =  ½  *  165  cfs  *  TB  hr  

TB  =  6.67  hr  

Note:  B  =  TB  +  TR  

Solu@on  to  Example  3  

Problem  3   Find  the  time  to  peak  (tp)  and  the  maximum  flow  (Qp)  by  the  SCS  triangular  unit  hydrograph  method  for  Subbasin  C  (770  ac).  Use  L  =  1.5  mi,  average  slope  =  30ft/mi,  D  =  0.5  hr  and  CN  =  75.