37. =Use of Diagrams.=—There are five factors in Kutter’s formula: _n_, _Q_, _V_, _d_ (or _R_), and _S_. If any three of these are given the other two can be determined, except when the three given are _Q_, _V_, and _d_. These three are related in the form _Q_ = _AV_, which is independent of slope or the character of the material. There are only nine different combinations possible with these five factors, which will be met in the solution of Kutter’s formula. The solution of the problems by means of the diagrams is simple when the data given include _n_ = .015. For other given values of _n_ the solution is more complicated. Results of the solution of types of each of the nine problems are given in Table 17 and the explanatory text below.

_If n is given and is equal to .015_, the solution is simple.

Sewerage and Sewage Treatment — Part 42

For example in Table 17 _case 1, example 1_; to be solved on Fig. 15. Place a straight-edge at 1.0 on the _Q_ line and at 6 inches on the diameter line for _n_ = .015. The slope and the velocity will be found at the intersection of the straight-edge with these respective scales.

Sewerage and Sewage Treatment — Part 42

All problems in which _n_ is given as .015 and the solution for which falls within the limits of Fig. 15 or 16 should be solved by placing a straight-edge on the two known scales and reading the two unknown results at the intersection of the straight-edge and the remaining scales.

Sewerage and Sewage Treatment — Part 42

For example in _case 1_, _example 2_ find the intersection of the horizontal line representing _Q_ = 100 with the sloping diameter line representing _d_ = 48 inches. The vertical slope line passing through this point represents _S_ = .0065 and the sloping velocity line passing through this point represents 8.5 feet per second.

In general problems in which _n_ = .015, can be solved on Fig. 17 by finding the intersection of the two lines representing the given data, and reading the values of the remaining variables represented by the other two lines passing through this point.

SOLUTIONS OF PROBLEMS BY KUTTER’S FORMULA

─────┬───────┬────────────────────────────┬──────────────────────────── Case │Example│ Given │ Found ─────┼───────┼─────┬─────┬────┬────┬──────┼─────┬─────┬────┬────┬────── │ │ _n_ │ _Q_ │_V_ │_d_ │ _S_ │ _n_ │ _Q_ │_V_ │_d_ │ _S_ ─────┼───────┼─────┼─────┼────┼────┼──────┼─────┼─────┼────┼────┼────── 1 │ 1 │0.015│ 1.0│ 2.5│ 6│ │ │ │5.0 │ │0.0575 1 │ 2 │ .015│100.0│ │ │ │ │ │8.5 │ │.0065 1 │ 3 │ .020│ 1.0│ │ 6│ │ │ │5.0 │ │.13 1 │ 4 │ .020│100.0│ │ 48│ │ │ │8.5 │ │.0125 2 │ 1 │ .015│ 5.0│ │ │0.0003│ │ │1.2 │28 │ 2 │ 2 │ .010│ 5.0│ │ │.0003 │ │ │1.7 │23.5│ 3 │ 1 │ .015│ │ │ 18│.002 │ │ 4.0│2.25│ │ 3 │ 2 │ .018│ │ │ 18│.0008 │ │ 2.0│1.1 │ │ 4 │ 1 │ .015│ 2.0│ 2.5│ │ │ │ │ │12 │.00475 4 │ 2 │ .011│ 2.0│ 2.5│ │ │ │ │ │12 │.0022 5 │ 1 │ .015│ │ 5.0│ 36│ │ │ 35.0│ │ │.0038 6 │ 1 │ .018│ │ 5.0│ │.001 │ │185.0│ │80 │ 7 │ 1 │ │ 3.0│ │ 18│.002 │0.019│ │1.7 │ │ 7 │ 2 │ │ 50.0│ │ 36│.005 │ .012│ │7.0 │ │ 8 │ 1 │ │ 6.0│ 2.5│ │.003 │ .018│ │ │21 │ 9 │ 1 │ │ │ 4.2│ 66│.00059│ .011│100.0│ │ │ ─────┴───────┴─────┴─────┴────┴────┴──────┴─────┴─────┴────┴────┴──────