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.