87
11. A 60-m-long siphon spillway discharges water at the rate of 0.32 m3/sec. The siphon crown
is 1.2 m above the water surface elevation of the reservoir and a length of 10 m away from
the siphon entrance. If the siphon diameter is 30 cm, the friction factor is 0.02, and the head
loss coefficients are 0.2 (inlet) and 1.0 (exit), determine the elevation difference between the
siphon crown and the downstream pool. What is the pressure at the crown in kN/m2?
Ans. Balancing energy from reservoir to downstream pool:
L
h
g
VP
z
g
VP
z 22
2
22
2
2
11
1
; hL = hent + hf + hexit;
V = Q/A = (0.32)/[π(0.15)2] = 4.53 m/s
z1 + 0 + 0 = z2 + 0 + 0 + (0.2+0.02(60/0.3)+1.0)[4.532/2g]
z1 – z2 = 5.44 m; thus hc – h2 = 1.2 + 5.44 = 6.64 m (i.e., the elev. difference between
crown and downstream pool). Balancing energy from reservoir to siphon crown:
L
C
Ch
g
V
P
z
g
VP
z 22
2
2
11
1
; hL = hf + he
0 + 0 + 0 = 1.2 + Pc/γ + [4.532/2g] + (0.2+0.02(10/0.3))[4.532/2g];
Pc/γ = -3.15 m. Pc = (-3.15m)(9.79 kN/m3) = -30.8 kN/m2
12. Stage recorders register headwater and tailwater depths (4.05 m and 3.98 m) of a culvert
during a flood event. The 2 m by 2 m concrete culvert (square-edged entrance) is 15 m long
and has a slope of 3.0%. Based on this information, determine the flood flow rate that passed
through the culvert. The stream depths are measured above the culvert inverts (bottom).
Ans. Since the outlet is submerged, the hydraulic operation category is (a) and the energy
balance is: Hup + SoL = Hdown + hL; where H = stream depths
4.05 + (0.03)(15) = 3.98 + hL; hL = 0.52 m
The head loss from hydraulic category (a) is:
hL = [Ke + {n2L/Rh4/3}(2g) + 1]{Q2/(2g∙A2)};
where Rh = A/P = (4 m2)/(8 m) = 0.5 m. Thus,
0.52 = [0.5+{0.0132(15)/(0.5)4/3}(2g)+1]{Q2/(2g(42)}; Q = 10.0 m3/sec