Cu07821 ppt9 recapitulation

Watermanagement

“Recapitulation”

Goal Course Watermanagement

• Knowledge of the working of the water system, how it works, what has an influence on what, what are you aiming at, which processes take place

• Policy

Goals

• Optimal circumstances for farmers and nature, optimal water table and gain

• Prevention of water trouble, fulfill on the rainfall event NBW, 1:25 year shower, Stowa

• Prevention of water shortage? No guidelines exist, even for a dry period, maintain the water level?

• Water framework directive, European directive, ecological quality. Is about the quality of the water system itself.

• Prevention of salination?

Model Sobek

• Area 1 500 hagsl +0,8 mDepth drainage 1 mSummer level (weir) -0,3 m , freeboard 1,1 m

• Area 2 500 hagsl +0,1 mDepth drainage 0,7 mSummer level (pump) -0,7 m , freeboard 0,8 m

De Zeeuw Hellinga

Higher factor, faster discharge of water to the ditch

• Infiltration 99 mm/hour !!• Depression 5 mm• μ=0,060 storage in soil• Drainage De Zeeuw Hellinga• No seepage• Pump 11 mm/day = 1,27 m3/s • Precipitation 96 hour 100mm, 4 days• Calculated steps 5 min, for 1 month

Used Rainfall 1:25 year 96 hours, type long, lengthened to 1 month

Oversight Summer

Initial GWL =

Open water level

Questions

• What is the percentage surface water?• Surface water max 11,2 ha = 1,1 %• μ=0,060 storage soil• How much m3 of water can be stored in a

layer of 10 cm soil, in an area of 1000 ha.• 1000 x 10.000 x 0,1 x 0,06 = 60.000 m3• What is the amount of precipitation in m3?• 1000 x 10.000 x 100/1000 = 1.000.000 m3• How much water will be discharged (max)?• 11/1000 x 4 x 1000 x 10.000 =440.000 m3

More questions

• Give an estimate of the storage in the surface water.

• Average height (1,1 + 0,8)/2 = 0,95 m

• Assume rectangular ditch• 11,2 x 10.000 x 0,95 = 106.400 m3

And more questions!

• Suppose the entire area of 1000 ha is flat, will it submerge? (simpel approach to the problem)

• Rainfall 1.000.000 m3• Discharge pump 440.000 m3• Storage surface water 106.400 m3• Storage in soil 0,95 x 10 x 60.000 = 570.000 m3• Total of storage + discharge =1.116.400 m3• Answer is no, differences in height are needed, so

the water can flow to the lowest points.

Calibration!!!!

• Which information do you need for a good calibration?

• Precipitation• Variances in water level• GWL at the start of the rain shower• Discharge of the pump• Model that can be modified

Summer, maximum wl

Surface runoff

Gebied 1= 0

Groundwater discharge

Infiltration

Storage on land (mm)

Groundwaterlevel

Summer, larger culvert

Winter

• Area 1gsl +0,8 mDepth drainage 1 mWinter level (weir) -0,5 m , dry height 1,3 mGWL is 0,5 m below ground surface level 

• Area 2gsl +0,1 mDepth drainage 0,7 mWinter level (pump) -0,9 m , dry height 1,0 mGWL is 0,5 m below ground surface level

Winter, maximum wl

10 ha open water extra downstreams

Winter pump capacity 2 m3/s

Water level lowered 0,2m + 5000 m2 storage

Small culvert round 600 mm

Other possibilities

• Heighten the land• Discharge partly to another area.• Less drainage?

Effects of bad maintenance

• Culverts closed with dirt• Pump capacity is lower• Collapsed banks, lower storage and lower

discharge• Plants growing in ditch (problem?)