Answer:
(a). 42.1 ft/s, (b). 3366.66 ft^3/s, (c). 0.235, (d). 18.2 ft, (e). 3.8 ft.
Explanation:
The following parameters are given in the question above and they are;
Induced hydraulic jump, j = 80 ft wide channel, and the water depths on either side of the jump are 1 ft and 10 ft. Let k1 and k2 represent each side of the jump respectively.
(a). The velocity of the faster moving flow can be calculated using the formula below;
k1/k2 = 1/2 [ √ (1 + 8g1^2) - 1 ].
Substituting the values into the equation above a s solving it, we have;
g1 = 7.416.
Hence, g1 = V1/ √(L × k1).
Therefore, making V1 the subject of the formula, we have;
V1 = 7.416× √ ( 32.2 × 1).
V1 = 42.1 ft/s.
(b). R = V1 × j × k1.
R = 42.1 × 80 × 1.
R = 3366.66 ft^3/s.
(c). Recall that R = V2 × A.
Where A = 80 × 10.
Therefore, V2 = 3366.66/ 80 × 10.
V2 = 4.21 ft/s.
Hence,
g2 = V1/ √(L × k2).
g2 = 4.21/ √ (32.2 × 10).
g2 = 0.235.
(d). (k2 - k1)^3/ 4 × k1k2.
= (10 - 1)^3/ 4 × 1 × 10.
= 18.2 ft.
(e).The critical depth;
[ (3366.66/80)^2 / 32.2]^ 1/3.
The The critical depth = 3.80 ft.
Answer:
a) 42.08 ft/sec
b) 3366.33 ft³/sec
c) 0.235
d) 18.225 ft
e) 3.80 ft
Explanation:
Given:
b = 80ft
y1 = 1 ft
y2 = 10ft
a) Let's take the formula:
[tex] \frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1} [/tex]
[tex] 10*2 = \sqrt{1 + 8f^2 - 1[/tex]
1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
[tex] \frac{V_1}{\sqrt{g*y_1}} = 7.416 [/tex]
[tex] V_1 = 7.416 *\sqrt{32.2*1}[/tex]
V1 = 42.08 ft/sec
b) the flow rate will be calculated as
Q = VA
VA = V1 * b *y1
= 42.08 * 80 * 1
= 3366.66 ft³/sec
c) The Froude number of the sub-critical flow.
V2.A2 = 3366.66
Where A2 = 80ft * 10ft
Solving for V2, we have:
[tex] V_2 = \frac{3666.66}{80*10} [/tex]
= 4.208 ft/sec
Froude number, F2 =
[tex] \frac{V_2}{g*y_2} = \frac{4.208}{32.2*10} [/tex]
F2 = 0.235
d) [tex]El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}[/tex]
[tex] El = \frac{(10-1)^3}{4*1*10} [/tex]
[tex] = \frac{9^3}{40} [/tex]
= 18.225ft
e) for critical depth, we use :
[tex] y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3 [/tex]
= 3.80 ft
