=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. [Illustration: 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: