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Es recalcula la pèrdua <strong>de</strong> càrrega ∆h:<br />
1,90<br />
−3<br />
0,016<br />
∆h0 − E = 4,08 ⋅10<br />
⋅ 0,32 ⋅ ⋅ 73 m = 3,44 m<br />
4,90<br />
0,0968<br />
v = 2,17 m<br />
0 F<br />
- Tram EF<br />
∆h − = 9,94 m<br />
=<br />
+<br />
∆h0− F ∆ho−E<br />
∆hE−F<br />
9,94 m = 3,44 m + E F<br />
∆h − ; E F<br />
1,90<br />
−3<br />
Q<br />
∆h = 4,08 ⋅10<br />
⋅K<br />
⋅ ⋅L<br />
⋅F<br />
4,90<br />
D<br />
Longitud ramal = L = 38 m<br />
Coeficient <strong>de</strong> <strong>Fi</strong>sher = F = 0,528<br />
⎛ 0,016 ⎞<br />
⎜ ⎟<br />
−3<br />
3<br />
6,50 = 4,08 ⋅10<br />
⋅0,32<br />
⋅<br />
⎝ ⎠<br />
4,90<br />
D<br />
D =<br />
1,<br />
93 ⋅10<br />
4,9 −7<br />
=<br />
0,043 m<br />
=<br />
1,90<br />
43 mm<br />
PE d’alta <strong>de</strong>nsitat:<br />
Ø exterior = 63 mm<br />
Espessor = 5,8 mm<br />
Ø interior = 51,4 mm<br />
Es recalcula la pèrdua <strong>de</strong> càrrega ∆h:<br />
1,90<br />
⎛ 0,016 ⎞<br />
⎜ ⎟<br />
−3<br />
3<br />
∆h = 4,08 ⋅10<br />
⋅0,32<br />
⋅<br />
⎝ ⎠<br />
4,90<br />
0,0514<br />
Comprovació <strong>de</strong> la velocitat <strong>de</strong> l’aigua:<br />
∆h − = 6,50 m<br />
⋅38<br />
⋅0,528<br />
⋅ 64 m ⋅0,528<br />
=<br />
4,39 m<br />
Annexos Memòria<br />
60