In the first case, 1 in 1,760 = 3 in mile.
1 in 5,280 = 1 in mile.
h = r ∕ 2 = ¾.
v = 55√(h × 2f)
= 55√(¾ × 6) = 165/√ = 118.
In second case v = 55√(¾ × 2) = 55√(3 ∕ 2) = 67·9.
Thus the velocity of the two streams would be as 118: 67·9.
Supposing a sewer to have a gradient of 1 in 300, how much would the velocity and discharge be increased by altering the gradient to 1 in 100?
- 1 in 300 = 17·6 in mile.
- 1 in 100 = 52·8 in mile.
As h is not given, we must assume it = ¼;, as it does in circular sewers running full or half full.
- v = 55 √(h x 2f)
- = 55 √(35·2 ∕ 4) = 163 feet per minute.
- v1 = 55 √(104 ∕ 4) = 281 feet per minute.
The increase in discharge may be similarly calculated.
Describe the relation existing in a sewer between gradient, volume, velocity, and size.
By the formula v = 55 √(h. f.)