Versions Compared

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.

...

The denominator of equation (26) (current live volume contributing to high flow loss) is then:

Equation 29 Image Added

Equation (26) can now be recast in terms of the total volume in the division:

Equation 30 Image Added

Rearranged:

Equation 31 Image Added

...

In this case, the index flow rate is the same as the inflow rate, and an owner’s share of inflow will determine their share of active storage.  The proportional routing formula (equation (19)) can be rewritten and applied to inflow as shown in equation (32), below and then rearranged to determine the owner’s share of storage.  

Equation 32Image Added
Equation 33Image Added

The owner’s share of proportional lateral flux is then found by substituting into equation (20) and rearranging.

Equation 34Image Added

Each owner’s storage Storage(o,t) - StoragedsMAX(o) is compared to StorageHFTRatioHFT(o) to determine whether a high flow loss applies.  If it does, equation (24) is used to determine this value.  Mass balance is applied to determine each owner’s share of outflow.

Determining whether an owner has a high flow loss when x ≠ 1

If the division’s Muskingum weighting x ≠ 1 both inflow and outflow impact storage, equation (23) is used to determine each owner’s share of outflow O(o).  From this a trial estimate of each owner’s storage (Storage(o,t)) is calculated by assuming LossHF = 0.  It is then possible to determine if owners have a share of the high flow loss or not, i.e. whether Storage(o,t) - StoragedsMAX(o) is less or more than their share of the high flow threshold: StorageHFTRatioHFT(o).

Case of Muskingum weighting x ≠ 1, owner without a high flow loss

This solution is applied in divisions where Muskingum weighting x ≠ 1 to owners that have Storage(o,t) ≤ StoragedsMAX(o) + StorageHFTRatioHFT(o). To solve mass balance, the outflow volume (O(o)) is recast in terms of the division’s live storages. Recalling the index flow rate q‾(o) from equation (17) and rearranging for O(o):

Equation 35