## 2 X 9807

example 3 Calculate the minimum total system head using conditions in Example 2 and a seal well, as shown in Figure 27. Use 2.8 ft (0.853 m) for the frictional head loss hfs-2).

The maximum siphon height Zs in Example 2 was found to be 32.63 ft (9.95 m). Therefore from Eq. 14 the total system head after priming is

TH = Z + hf(1 — 2) In USCS units hfi1—2) = hf1—S) + hfS—2) = 3 + 2.8 = 5.8 ft

Z = Z1 — ZS = 40 — 32.63 = 7.37 ft In SI units hA 1—2) = hf1—S) + h(S—2) = 0.91 + 0.853 = 1.763 ft Z = Z1 — ZS = 12.2 — 9.95 = 2.25 m

Note that the seal well elevation is above discharge level. Therefore in USCS units TH = 7.37 + 5.8 = 13.17 ft in SI units TH = 2.25 + 1.763 = 4.01 m example 4 The dimensions of the down leg in Example 3 require a minimum velocity of 5 ft/s (1.52 m/s) flowing full to purge air from the system and start the siphon. Calculate the system head the pump must overcome to prime the siphon.

V(pipe ID in inches)2 5 x 482

T oT 3,u V( pipe ID in cm )2 1.52 X 121.92 In SI units m3/h =-3-54-=-3-54-= 6400

The critical depth is found from Table 1:

ft3/s 62.8

 ft3's 62.8 = . d2 = 45'2 = ■ Dcrit

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