{"id":95628,"date":"2018-06-04T13:53:46","date_gmt":"2018-06-04T17:53:46","guid":{"rendered":"https:\/\/mrnf.gouv.qc.ca\/RADF\/schedule-7\/"},"modified":"2026-03-04T13:21:35","modified_gmt":"2026-03-04T18:21:35","slug":"schedule-7","status":"publish","type":"cpt_radf","link":"https:\/\/mrnf.gouv.qc.ca\/en\/RADF\/guide\/schedules\/schedule-7\/","title":{"rendered":"Schedule 7 – Calculation method for the peak flow of a drainage basin of an area greater than 60 km2"},"content":{"rendered":"

[et_pb_section fb_built=”1″ module_class=”bg-radf” _builder_version=”4.16″ global_colors_info=”{}”][et_pb_row module_class=”wrapper-radf” _builder_version=”4.16″ global_colors_info=”{}”][et_pb_column type=”4_4″ _builder_version=”4.16″ custom_padding=”|||” global_colors_info=”{}” custom_padding__hover=”|||”][et_pb_text _builder_version=”4.27.4″ global_colors_info=”{}”]<\/p>\n

Schedule 7<\/h1>\n

Calculation method for the peak flow of a drainage basin of an area greater than 60 km2<\/sup><\/h2>\n

The HP-40 statistical method is used to calculate the 20-year interval maximum daily flow. The method was validated for drainage basins whose area is greater than 150 km2<\/sup>. Thus, where the area of the basin covers between 60 and 150 km2<\/sup>, the result must be validated in the field by looking for signs indicating the water level reached by the floods of previous years or by establishing a relationship with basins that were measured on the same territory or near it.
\"\"<\/p>\n

\n

1<\/h2>\n

Additional information<\/h4>\n

Validation of peak flow calculation when the result of this calculation imposes the installation of a conduit with a diameter much larger than the average width of the watercourse.<\/strong><\/p>\n

The methods for calculating peak flows in Schedules 6 and 7 are based on several assumptions, which may make them less suitable for some drainage basins. Sometimes significant differences can be observed between calculated peak flows and flows recorded in the field.<\/p>\n

Where the results of flow calculations require the installation of conduits with diameters far greater than the average width of the watercourse observed in the field, it is possible, based on evidence indicating the water level reached by flooding in previous years or by establishing a relationship with basins measured on or near the same territory, to notify the\u00a0management unit in charge \"\"<\/a> (in French) prior to the installation of a conduit with a diameter less than that determined by the calculation. Supporting documentation submitted by the professional in charge must be submitted (including, but not limited to, calculation of peak flow, average watercourse width, free flow of fish to be insured or not, result of field surveys reflecting choice, identification of inconsistency in flow calculation, etc.).<\/p>\n

In this situation, the diameter of the conduit should be consistent with the average width of the watercourse to be crossed. Under no circumstances should the conduit reduce the width of the watercourse by more than 20%, measured at the upper bank limit. The height of water in the conduit must be less than or equal to 85% of the clearance height after it is buried.<\/p>\n<\/div>\n

Steps in calculation<\/span><\/p>\n

    \n
  1. Delimitation of the drainage basin with a topographic map at a scale of 1:20 000;<\/li>\n
  2. Calculation of the drainage basin’s area;<\/li>\n
  3. Calculation of the “85-10” slope of the watercourse;<\/li>\n
  4. Calculation of the proportion of the basin covered by lakes and bare and semi-bare wetlands;<\/li>\n
  5. Calculation of the 20-year interval maximum daily flow.<\/li>\n<\/ol>\n

    The delimitation of a drainage basin is shown as an example in step 1 of Schedule 6. The calculation method for the “85-10” slope of the watercourse is the same as that used for drainage basins of 60 km2<\/sup> or less (Schedule 6<\/a> \u2014 step 5). The 20-year interval maximum daily flow (Q1.20) is determined using the following formula:<\/p>\n

    Q1,20<\/span>(m\u00b3\/s) = (0,7882 (Ab<\/span>\/100) 0,93<\/span> (Sc) 0,30<\/span>)\/St<\/span> 0,24<\/p>\n

    where:<\/p>\n

    Ab<\/span> = area of the drainage basin (ha)<\/p>\n

    Sc<\/span> = “85-10” slope of the watercourse (%)<\/p>\n

    St<\/span> = percentage of the area of the drainage basin covered by lakes and bare and semi-bare wetlands (%)<\/p>\n

    Exemple :<\/p>\n

    Ab<\/span> = 75 km\u00b2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Q1,20<\/span> = (0,7882 (75) 0,93<\/span> (1) 0,30<\/span>)\/ (5)\u00a00,24<\/span>= 29,7 m\u00b3\/s<\/p>\n

    where:<\/p>\n

    Sc<\/span> = 1%<\/p>\n

    St<\/span> = 5%<\/p>\n

    A weighted factor of at least 5% is then applied to the flow obtained in order to take into account exceptional climatic events.<\/p>\n

    i.e. : 29,7 m\u00b3\/s X 1,05 = 31,2 m\u00b3\/s<\/p>\n

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