in which a and are constants for different classes of material. For cast-iron pipe a is 0.00007726 and is 0.00000647. This formula is seldom used in sewerage design.

Exponential formulas have been developed as the result of experiments which have demonstrated that V does not vary as the one-half power of R and S but that the relation should be expressed as,

in which p and q are constants and C is a factor dependent on the character of the material. The various formulas coming under this classification have been given the names of the experimenters proposing them. Examples of these formulas are: Flamant’s, in English units, for new cast-iron pipe, which is,

and LampÉ’s for the same material which is,

These formulas are useful only for the material to which they apply, but they can be used for conduits of any shape. A. V. Saph and E. W. Schoder have shown^[31] that the general formula for all materials lies between the limits,

Hazen and Williams’ formula is in the form,

in which C is a factor dependent on the character of the material of the conduit. The values of C as given by Hazen and Williams are,

This formula is of as general application as Kutter’s formula and is easier of solution, but being more recently in the field and because of the ease of the solution of Kutter’s formula by diagrams it is not in such general use. Exponential formulas are used more in waterworks than in sewerage practice.

Manning’s formula is in the form,

in which n is the same as for Kutter’s formula. Charts for the solution of Manning’s formula are given in Eng. News-Record, Vol. 85, 1920, p. 837.

36. Solution of Formulas.—The solution of even the simplest of these formulas, such as Flamant’s, is laborious because of the exponents involved. Darcy’s and Kutter’s formulas are even more cumbersome because of the character of the coefficient. The labor involved in the solution of these formulas has resulted in the development of a number of diagrams and other short cuts. Since each formula involves three or more variables it cannot be represented by a single straight line on rectangular coordinate paper. The simplest form of diagram for the solution of three or more variables is the nomograph, an example of which is shown in Fig. 13 for the solution of Flamant’s formula. A straight-edge placed on any two points of the scales of two different vertical lines will cross the other line at a point on the scale corresponding to its correct value in the formula. Such a diagram is in common use for the solution of problems for the flow of water in cast-iron pipe.

Fig. 14 has been prepared to simplify the solution of Hazen and Williams’ formula. The scales of slope for different classes of material are shown on vertical lines to the left of the slope line. For use these scales must be projected horizontally on the slope line. The scales for other factors are shown on independent reference lines.

For example let it be required to find the loss of head in a 12 inch pipe carrying 1 cubic foot per second when the coefficient of roughness is 100. A straight-edge placed at 1.0 cubic feet per second on the quantity scale, and 12 inches on the diameter scale crosses the slope line at .00092 opposite the slope scale for c = 100. It crosses the velocity line at 1.31 feet per second.

Kutter’s formula is the most commonly used for sewer design and has been generally accepted as a standard in spite of its cumbersomeness. Fig. 15 is a graphical solution of Kutter’s formula for small pipes, and Fig. 16 for larger pipes. The diagrams are drawn on the nomographic principle and give solutions for a wide range of materials, but they are specially prepared for the solution of problems in which n = .015. In their preparation the effect of the slope on the coefficient has been neglected. Fig. 17 is drawn on ordinary rectangular coordinate paper and can be used only for the solution of problems in which n = .015. Both diagrams are given for practice in the use of the different types.

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.

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.

For example in Case 7, Example 2 there are given Q, d, and S. The value of V is read directly as though n = .015 as 7 feet per second. The value of S read for n = .015 is .0075. But the given slope is .005. Since the given slope is flatter than that for n = .015 the conversion factor is less than unity and is therefore .005
.0075 = 0.67. With this value of the conversion factor and the value of R given as 0.75 the value of n is read from Fig. 18 as slightly greater than .012.

38. Flow in Circular Pipes Partly Full.—The preceding examples have involved the flow in circular pipes completely filled. The same methods of solution can be used for pipes flowing partly full except that the hydraulic radius of the wetted section is used instead of the diameter of the pipe. Diagrams are used to save labor in finding the hydraulic radius and the other hydraulic elements of conduits flowing partly full.

The hydraulic elements of a conduit for any depth of flow are: (a) The hydraulic radius, (b) the area, (c) the velocity of flow, and (d) the quantity or rate of discharge. The velocity and quantity when partly full as expressed in terms of the velocity and quantity when full as calculated by Kutter’s formula will vary slightly with different diameters, slopes and coefficients of roughness. The other elements are constant for all conditions for the same type of cross-section. The hydraulic elements for all depths of a circular section for two different diameters and slopes are shown in Fig. 19. The differences between the velocity and quantity under the different conditions are shown to be slight, and in practice allowance is seldom made for this discrepancy.

In the solution of a problem involving part full flow in a circular conduit the method followed is to solve the problem as though it were for full flow conditions and then to convert to partial flow conditions by means of Fig. 19, or to convert from partial flow conditions to full flow conditions and solve as in the preceding section.

For example let it be required to determine the quantity of flow in a 12–inch diameter pipe with n = .015 when on a slope of .005 and the depth of flow is 3 inches. First find the quantity for full flow. From Fig. 15 this is 2.0 cubic feet per second. The depth of flow of 3 inches is one-fourth or 0.25 of the full depth of 12 inches. From Fig. 19, running horizontally on the 0.25 depth line to meet the quantity curve, the proportionate quantity at this depth is found to be on the 0.13 vertical line, and the quantity of flow is therefore 2 × 0.13 = 0.26 cubic feet per second.

Another problem, involving the reversal of this process is illustrated by the following example:

Let it be required to determine the diameter and full capacity of a vitrified pipe sewer on a grade of 0.002 if the velocity of flow is 3.0 feet per second when the sewer is discharging at 30 per cent of its full capacity, the depth of flow being 12 inches. From Fig. 19 the depth of flow when the sewer is carrying 30 per cent of its full capacity is 0.38 of its full depth. Since the partial depth is 12 inches the full diameter is 12
.038 = 31.6 inches. The velocity of flow at 38 per cent depth is 86 per cent of the full velocity. Since the velocity given is 3.0 feet per second, the full velocity is 3.0
.86 = 3.5 feet per second. With a full velocity of 3.5 feet per second and a diameter of 31.6 inches from Fig. 16 the full capacity of the sewer is 18 cubic feet per second.

39. Sections Other than Circular.—The ordinary shape used for small sewers is circular. The difficulty of constructing large sewers in a circular shape, special conditions of construction such as small head room, soft foundations, etc., or widely fluctuating conditions of flow have led to the development of other shapes. For conduits flowing full at all times a circular section will carry more water with the same loss of head than any other section under the same conditions. In any section the smaller the flow the slower the velocity, an undesirable condition. The ideal section for fluctuating flows would be one that would give the same velocity of flow for all quantities. Such a section is yet to be developed. Sections have been developed that will give relatively higher velocities for small quantities of flow than are given by a circular section. The best known of these sections is the egg shape, the proportions and hydraulic elements of which are shown in Fig. 20. Other shapes that have the same property, but which were not developed for the same purpose are the rectangular, the U-shape, and the section with a cunette. The egg-shaped section has been more widely used than any other special section. It is, however, more difficult and expensive to build under certain conditions, and has a smaller capacity when full than a circular sewer of the same area of cross-section. Various sections are illustrated in Fig. 22 and 23.

The U-shaped section is suitable where the cover is small, or close under obstructions where a flat top is desirable and the fluctuations of flow are so great as to make advantageous a special shape to increase the velocity of low flows. The proportions of a U-shaped section are shown in Fig. 23 (6). Other sections used for the same purpose are the semicircular and special forms of the rectangular section.

The proportions and the hydraulic elements of the square-shaped section are shown in Fig. 21. This is useful under low heads where a flat roof is required to carry heavy loads, and the fluctuations of flow are not large.

Sections with cunettes have not been standardized. A cunette is a small channel in the bottom of a sewer to concentrate the low flows, as shown in Fig. 22 (7). A cunette can be used in any shape of sewer.

Sections developed mainly because of the greater ease of construction under certain conditions are the basket handle, the gothic, the catenary, and the horse shoe. Some of these shapes are shown in Fig. 22 and 23. They are suitable for large sewers on soft foundations, where it is desirable to build the sewer in three portions, such as invert, side walls, and arch. They are also suitable for construction in tunnels where the shape of the sewer conforms to the shape of the timbering, or in open cut work where the shape of the forms are easier to support.

Problems of flow in all sections can be solved by determining the hydraulic radius involved, and substituting directly in the desired formula, or by the use of one of the diagrams after converting to the equivalent circular diameter. The determination of the hydraulic radius of these special sections is laborious, and hence other less difficult methods are followed. Problems are more commonly solved by converting the given data into an equivalent circular sewer, solving for the elements of this circular sewer and then reconverting into the original terms, or by working in the other direction. The hydraulic elements of various sections when full are given in Table 18.

5. Soft Foundation. 6. Wet ground.