NPSH 2 ft 3 n 120 64 35 s ft 71 72 100 30 68 80 8 in 20 I0 HP 15 71 HP 40 6 in 1
ID: 1714452 • Letter: N
Question
NPSH 2 ft 3 n 120 64 35 s ft 71 72 100 30 68 80 8 in 20 I0 HP 15 71 HP 40 6 in 10 S HP 3 HP 2 HP 100 150 200 00 350 00 450 Capacity (gal/min) 1000 1200 1600 1800 Capacity (Umin) FIGURE 13.25 Performance for 3 × 4-10 centrifugal pump at 1750 rpm. (Source: Goulds Pumps, Inc., Seneca Falls, NY) 7. (1 point) Critical depth can be calculated by Manning's equation. True /Falsel 8. (1 point) For a given cross-section and flowrate Minimum specific energy, Es,min= 1.5 * Normal depth. [True/False] 9. ( point) For steep slope, Ye>Y; where Ye critical depth and Yn normal depth. 10. (2 points) Flow rate remains fixed over space in a steady flow. [True/Ealsel 11. (1 point) If upstream Froude number, Fri 5, it is a strong jump. [True /Falsel 12. (1 point) Head loss (SI unit) between point a and point b in a culvert True /Falsel n2Lv2 13. 3 points) Determine barrel length and head for an outlet control circular C.M. culvert. Assume n = 0.020, discharge = 60 cfs and diameter = 3 ft.Explanation / Answer
7. False. v=1/n*r3/2*s1/2
8. false. Emin = 3/2*Y critical
9. True
10.False. For steady flow, Q remains constant over time for a specific point.
11. False. Fr. number need to be >9.
12. True.
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