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:
1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
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:
= 4.208 ft/sec
Froude number, F2 =
F2 = 0.235
d) 
= 18.225ft
e) for critical depth, we use :
![y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3](https://tex.z-dn.net/?f=%20y_c%20%3D%20%5B%5Cfrac%7B%28%5Cfrac%7B3366.66%7D%7B80%7D%29%5E2%7D%7B32.2%7D%5D%5E1%5E%2F%5E3%20)
= 3.80 ft
Answer:
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if I am right then make me brainliest
Answer:
I believe reverse boost valve.
Answer:
Explanation:
I'm not 100% this is what you want, but here it is:
2
3
13
8
11
A
13
