If n is given and is not equal to .015 the solution is not so simple. In Fig. 15 and 16 the diagram is so drawn that the position of the diameter scales for all values of n is fixed on the vertical “diameter” line. The scales of diameter change for each value of n. These scales of diameter are shown for each value of n from .010 to .020 on vertical lines to the left of the “diameter” line. For use, the proper diameter scale for any given value of n must be projected horizontally upon the vertical “diameter” line. The velocity can be determined on Fig. 15 and 16, only when the diameter has been determined and then only when the diameter scale for n equal .015 is used, since the only scale shown for velocity is for n = .015.
For example, in Case 1, Example 3 there are given n = .020, Q, and d. Find the intersection of the vertical line for n = .020 with the sloping diameter line for d = 6 inches. Project the intersection horizontally to the right to the vertical “diameter” line. Place a straight-edge at this point and at Q = 1.0 on the quantity scale. The required value of S is read at the intersection of the straight-edge and the slope scale and is equal to 0.13. The intersection of the straight-edge in this position with the velocity scale is not the required value of the velocity since the velocity scale is made out for n = .015 and not .020. It is necessary to change the position of the straight-edge so that it may lie on Q equal 1.0 and on d equal 6 inches for n equal .015. The value of V is shown in this position as 5 feet per second.
The reverse process for Fig. 15 and 16 is illustrated by Case 4, Example 2 in which n = .011 and Q and V are also given. When Q and V are given the value of d is fixed independent of all other factors. Therefore the value of d can be read from the scale with n = .015 and is found to be 12 inches. Now find the value of d = 12 inches on the scale for n = .011 and project on to the “diameter” line. Place the straight-edge at this point and at Q = 2. The required slope is read as .0022.
Fig. 17 is prepared for the solution of problems in which n = .015 only. For problems in which n has some other value it is necessary to transform the data to equivalent conditions in which n = .015. This is done by means of the conversion factors shown in Fig. 18. The given slope or velocity is multiplied by the proper factor to convert from or to the value of n = .015.
For example in Case 1, Example 4 there are given n = .020, Q, and d. With Q and d given the value of V can be read from Fig. 17 without conversion. The corresponding value of S for n = .015 is .0065. It is now necessary to use the transformation diagram Fig. 18. The hydraulic radius of the given pipe is one foot. On Fig. 18 at the intersection of the slope line for R = 1.0 foot and n = .020 the value of the factor is read as 1.92. Since the given n is for rougher material than that represented by n = .015 the required slope must be greater than for n = .015 to give the same velocity. It is therefore necessary to multiply .0065 × 1.92 and the required slope is .0125.
In Case 6, Example 1 there are given n = .018, d, and S. The remaining factors are to be solved by Fig. 17. Solve first as though n = .015 in order to find an approximate value of d or R. In this case it is evident that d is greater than 57 inches. The value of R is therefore about 1.25. Referring to Fig. 18 the conversion factor for the slope for n = .018 is about 1.52. Since the given slope for n = .018 is .001, for an equal velocity and for n = .015 the slope should be less. Therefore in reading Fig. 17 it is necessary to use a slope of .001
1.52 = .00066. The diameter is found to be about 80 inches. Since this is nearer to the correct diameter the value of the conversion factor must be corrected for this approximation. The hydraulic radius for an 80 inch pipe is 1.67 feet, and the conversion factor from Fig. 18 is about 1.48. The slope for n = .015 should be therefore .001
1.48 = .000675 and from Fig. 17 the required diameter and quantity are read as 80 inches and 185 second-feet, respectively.
If n is not given but must be solved for, the solution on Fig. 15 and 16 is relatively simple. The desired value of n is read at the intersection of the sloping diameter line representing the known diameter and the horizontal projection of the intersection of the straight-edge with the vertical “diameter” line.
For example in Case 7, Example 1 there are given Q, d, and S. Lay the straight-edge on the given values of Q = 3 and S = .002. At the point where the straight-edge crosses the vertical “diameter” line project a horizontal line to the sloping diameter line for d = 18 inches. The vertical line passing through this point represents a value of n = .019. In order to find the value of V lay the straight-edge on Q = 3 and d = 18 inches for n = .015. The value of V is read as 1.7.
A slightly different condition is illustrated in the solution of Case 8, Example 1 in which Q, V and S are given. Determine first the value of d as though n = .015. Then proceed to determine n as in the preceding examples.
The solution for an unknown value of n on Fig. 17 is not so simple. It must be determined by working backwards from the conversion factor.