where I and O are the corresponding rates of inflow and outflow. An alternative statement of this equation is that the rate of change of storage within the reach at any instant is given by:

This, the continuity equation, forms the basis of all the storage routing methods. The routing problem consists of finding O as a function of time, given I as a function of time, and having information or making assumptions about S. Equation (14.2) cannot be solved directly.

Any procedure for routing a hydrograph generally has to adopt a numerical approximation such as the finite difference technique. Choosing a suitable time interval for the routing period, At, the continuity equation can be represented in a finite difference form as:

The subscripts 1 and 2 refer to the start and end of any At time step. The routing time step has to be chosen small enough such that the assumption of a linear change of flow rates within the time step is acceptable (as a working guide, At should be less than one-sixth of the time of rise of the inflow hydrograph).

At the beginning of a time step, all the variables in (14.3) are known except O2 and S2. Thus with two unknowns, a second equation is needed to solve for O2 at the end of a time step. A second equation is obtained by relating S to O alone, or to I and O together. The two equations are then used recursively to find sequential values of O through the necessary number of At intervals until the outflow hydrograph can be fully defined. It is the nature of the second equation for the storage relationship that distinguishes two methods of storage routing.

For a reservoir or a river reach with a determinate control, such as a weir or overspill crest, upstream of which a nearly level pool is formed, the temporary storage can be evaluated from the topographical dimensions of the 'reservoir', assuming a horizontal water surface (Fig. 14.6). For a level pool, the temporary storage, S, is directly and uniquely related to the head, H, of water over the crest of the control. The discharge from the 'pool' is also directly and uniquely related to H. Hence S is indirectly but uniquely a function of O.

Temporary

Stored Peak O O

Temporary

It is convenient to rearrange (14.3) to move the unknowns S2 and O2 to one side of the equation and to adjust the Oi term to produce:

Since S is a function of O, [(S/At) + (O/2)] is also a specific function of O (for a given At), as in Fig. 14.7. Replacing [(S/At) + (O/2)] by G, for simplification, (14.5) can be written:

where Im = (Ii + Ā”2)/2. Fig 14.7 defines the relationship between O and G = (S/At + O/2), and this curve needs to be determined by using the common variable H to fix values of S and O and then G, for a specific At.

Table 14.1 Order of calculations for level-pool routing

Gi Go

Gi G3

Equation (14.5) and the auxiliary curve of Fig. 14.7 now provide an elegant and rapid step-by-step solution. At the beginning of a step, Gi and Oi are known from the previous step (or from conditions prior to the flood for the first step). Im is also known from the given inflow hydrograph. Thus all three known terms in (14.5) immediately lead to G2 at the end of the time step, and then to O2 from Fig. 14.7.

The recursive calculation of values for the outflow hydrograph is easily carried out in a table or, if the relationship of Fig. 14.7 is known in a functional form (such as for flow over a weir), in a spreadsheet. The calculation starts with all the inflows known (column A in Table 14.1). The mean inflows for a time step At given by Im = (I1 +12)/2 are then entered in column B. The initial outflow O1 is known from the discharge before the flood (column C). Thus the first value in column E, G1, may be determined from Fig. 14.7. The mean inflow during the first time interval less the initial outflow gives the first entry in column D and this is added to the first column E value to give the second entry in column E, G2, (14.5). Then the second value of the outflow hydrograph O2 can be read off from Fig. 14.7 (or calculated from a known functional relationship). This now becomes the O1 value for the next interval. The sequence of calculations is continued until the outflow hydrograph is completed or until the required outflow discharges are known.

A useful check on the validity of any level pool routing calculation is that the peak of the outflow hydrograph should occur at the intersection of the inflow and outflow hydrographs on the same plot (Fig. 14.6). At that point, I = 0, so dS/dt = 0, i.e. storage is a maximum and therefore O is a maximum. Thereafter, the temporary storage is depleted.

Example: Discharge from a reservoir is over a spillway with discharge characteristic Q(m3s-1) = 110 H1-5m3s-1, where H m is the head over the spillway crest. The reservoir surface area is 7.5 km2 at spillway crest level and increases linearly by 1.5 km2 per metre rise of water level above crest level. The design storm inflow, assumed to start with the reservoir just full, is given by a triangular hydrograph, base length 36 h and a peak flow of 360 m3 s-1 occurring 12 h after start of inflow. Estimate the peak outflow over the spillway and its time of occurrence relative to the start of the inflow. Solution. A level water surface in the reservoir is assumed. Temporary storage above crest level is given by integrating the change in surface area with increasing depth in

Table 14.2 Derivation of O and G for the auxiliary curve of O versus G

10.26 10.38 10.47 10.55 11.53 11.78 12.03 12.27 12.51 12.75

213 432 653 878 1199 1470 1752 2042 2339 2652

the reservoir as:

Outflows over the crest are given by O = 110H15 m3 s 1. Taking a value of At = 2

The derivation of O and G values are shown in Table 14.2. O is plotted against G and a curve drawn through the points (Fig. 14.8). (Note that the relationship is specifically for At = 2 h.)

The inflows into the reservoir are evaluated for each 2h period from the beginning of the storm inflow and the outflows are computed recursively using the storage equation (equation 14.5 and the auxiliary curve). The results are given in Table 14.3 and plotted on Fig. 14.9. The peak outflow over the spillway is 180 m3 sā1 and it occurs 24 h after the commencement of the inflow. The approach to the peak outflow becomes obvious during the computations as the net inflow rate (Im ā O) declines, and after the maximum O is reached, (Im ā O) becomes negative.

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