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Problem:
Water is conducted from source to water distribution reservoir as seen in the below figure. The
length is L, and max water elevations are A and B. Max capacity of the water transmission line is
0.185 m3
/s. The yield of the source is 0.1 m3
/s in summer, and 0.2 m3
/s in winter. In these two cases
(summer and winter) how would the hydraulic grade lines shape up, draw them. Gate valve is
complately open in both cases. If gate valve is gradually closed, how would be the shape of the new
hydraulic grade line in summer?
Qwinter = 0.2 m³/s therefore, the max capacity is the 0.185 m³/s
0,185= V winter∗π∗(0.45)²/4
It is suitible for DN 450 Pipe
V winter = 1.163 m/s
J winter= λ∗v²/(D∗2g) = 0.03∗(1.163)²/(0. 45∗2∗9.81) = 0.0045
Qsummer = 0.1 m³/s < 0.185 m³/s
0,1= V summer∗π∗(0.45)²/4
V summer = 0.628 m/s
J summer= λ∗v²/(D∗2g) = 0.03∗(0.628)²/(0.45∗2∗9.81) = 0.0013
For Summer:
For Winter:
For Gradually Closed in Summer:

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ENVIRONMENTAL Ce322 hw

  • 1. Problem: Water is conducted from source to water distribution reservoir as seen in the below figure. The length is L, and max water elevations are A and B. Max capacity of the water transmission line is 0.185 m3 /s. The yield of the source is 0.1 m3 /s in summer, and 0.2 m3 /s in winter. In these two cases (summer and winter) how would the hydraulic grade lines shape up, draw them. Gate valve is complately open in both cases. If gate valve is gradually closed, how would be the shape of the new hydraulic grade line in summer? Qwinter = 0.2 m³/s therefore, the max capacity is the 0.185 m³/s 0,185= V winter∗π∗(0.45)²/4 It is suitible for DN 450 Pipe V winter = 1.163 m/s J winter= λ∗v²/(D∗2g) = 0.03∗(1.163)²/(0. 45∗2∗9.81) = 0.0045 Qsummer = 0.1 m³/s < 0.185 m³/s 0,1= V summer∗π∗(0.45)²/4 V summer = 0.628 m/s J summer= λ∗v²/(D∗2g) = 0.03∗(0.628)²/(0.45∗2∗9.81) = 0.0013
  • 2. For Summer: For Winter: For Gradually Closed in Summer: