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DN 200, 5 km, k=0,1 mm

Reservoir, Wasserstand 200 m ü. NN. DN 200, 5 km, k=0,1 mm. A. DN 200, 10 km, k=0,1 mm. C. Mindest-Wasserdruck v on 30 mWs an Entnahmepunkten B und C eingehalten?. DN 100, 1,5 km, k=0,1 mm. Geodät. Höhe: 155 m ü. NN. Wasserentnahme: 15 l/s. B. Geodät. Höhe: 150 m ü. NN.

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DN 200, 5 km, k=0,1 mm

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  1. Reservoir, Wasserstand 200 m ü. NN. DN 200, 5 km, k=0,1 mm A DN 200, 10 km, k=0,1 mm C Mindest-Wasserdruck von 30 mWs an Entnahmepunkten B und C eingehalten? DN 100, 1,5 km, k=0,1 mm Geodät. Höhe: 155 m ü. NN. Wasserentnahme: 15 l/s B Geodät. Höhe: 150 m ü. NN. Wasserentnahme: 5 l/s

  2. Reservoir, Wasserstand 200 m ü. NN. DN 200, 5 km, k=0,1 mm 20 l/s A DN 200, 10 km, k=0,1 mm 15 l/s C DN 100, 1,5 km, k=0,1 mm 5 l/s Geodät. Höhe: 155 m ü. NN. Wasserentnahme: 15 l/s B Geodät. Höhe: 150 m ü. NN. Wasserentnahme: 5 l/s

  3. Druckverluste Reservoir-A: J=2,105 m/km Druckverlust Reservoir-A : 2,105 m/km · 5 km = 10,525 m

  4. Reservoir, Wasserstand 200 m ü. NN. Wasserdrucklinie bei A:200 m – 10,525 m = 189,475 m ü. NN. DN 200, 5 km, k=0,1 mm Druckverlust Δh = 10,525 m A DN 200, 10 km, k=0,1 mm C DN 100, 1,5 km, k=0,1 mm Geodät. Höhe: 155 m ü. NN. Wasserentnahme: 15 l/s B Geodät. Höhe: 150 m ü. NN. Wasserentnahme: 5 l/s

  5. Druckverlust A-B: J=4,974 m/km 4,974 m/km · 1,5 km = 7,461 m Druckverlust A-C: J=1,233 m/km 1,233 m/km · 10 km = 12,33 m

  6. Reservoir, Wasserstand 200 m ü. NN. DN 200, 5 km, k=0,1 mm Drucklinie:189,475 m ü. NN. A Geodät. Höhe: 155 m ü. NN. Wasserentnahme: 15 l/s DN 200, 10 km, k=0,1 mm Δh = 12,33 m C Δh = 7,461 m DN 100, 1,5 km, k=0,1 mm Drucklinie: 189,475 m – 12,33 m = 177,145 m ü.NN. Wasserdruck: 177,145 m – 155 m = 22,145 m Drucklinie: 189,475 m – 7,461 m = 182,014 m ü.NN. B Wasserdruck: 182,014 m – 150 m = 32,014 m Geodät. Höhe: 150 m ü. NN. Wasserentnahme: 5 l/s

  7. J [m/km] 4,509 Lineare Interpolation: DN 200, k=0,1 mm, 23 l/s 20 l/s - 2,105 m/km 30 l/s - 4,509 m/km J = 3/10 · (4,509 m/km - 2,105 m/km) + 2,105 m/km =2,8262 m/km ? 2,105 Q [l/s] 30 20 23

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