Fig. 18.—Conversion Factors for Kutter’s Formula.
In Figs. 15 and 16 the diameter scales are varied for different values of the roughness coefficient n. The velocity scale is shown only for a value of n = .015. The velocity for other values of n can be determined by the method given in the following paragraphs.
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.
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.
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.
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.
| TABLE 17 | |||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| 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 | |||||