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.
Fig. 19.—Hydraulic Elements of Circular Sections.
| d = 12′ 0″ | s = .0004 | n = .015 |
| d = 1′ 0″ | s = .01 | n = .013 |
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.