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SCHEDULE 6

PEAK FLOW CALCULATION METHOD FOR DRAINAGE BASINS WHOSE AREA IS EQUAL TO OR LESS THAN 60 KM2

The so-called rational method is used to calculate the 10-year interval peak flow. The method was validated for drainage basins whose area is less than 25 km2. Thus, where the area of the drainage basin covers between 25 and 60 km2, 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.

STEPS IN CALCULATION

1. Delimitation of the drainage basin;

2. Calculation of the average slope of the drainage basin;

3. Identification of the use of the territory and of the surface deposits in the drainage basin;

4. Calculation of the total area of the basin, of the proportion of each type of surface deposits per land use type and of the percentage of the basin covered by lakes and bare and semi-bare wetlands;

5. Determination of the watercourse’s length and calculation of the “85-10” slope of the watercourse;

6. Calculation of the weighted runoff coefficient of the drainage basin;

7. Calculation of the drainage basin’s concentration time;

8. Determination of rainfall intensity;

9. Calculation of the correction coefficient for rainfall intensity;

10. Determination of the reduction coefficient for peak flow;

11. Calculation of the 10-year interval peak flow.

EXPLANATION OF THE STEPS TO BE FOLLOWED WITH AN EXAMPLE

Step 1 - Delimitation of the drainage basin

The drainage basin that supplies the watercourse with water at the crossing point is delimited using a topographic map at a scale of 1: 20 000. Figure 1 shows, as an example, the delimitation of a drainage basin under study.

Step 2 - Calculation of the average slope of the drainage basin (Sb)

The average slope is calculated using a grid (1 cm X 1 cm) superimposed on the drainage basin. The number of times each horizontal and vertical line of that grid crosses a contour line must be determined. The length of those lines is also recorded. The calculation made to determine the average slope of the drainage basin under study is given in figure 2.

Step 3 - Identification of the use of the territory and of the surface deposits in the drainage basin

With the help of the surface deposit maps, the forest maps and knowledge of the territory, the use of the lands comprised within the drainage basin must be identified. They may be woodlands, pasturelands or croplands. Then the surface deposits for each land use type must be identified. Bare and semi-bare wetlands must also be located.

Figure 3 identifies the surface deposits and locates the bare and semi-bare wetlands in the drainage basin under study, which is completely wooded.

Step 4 - Calculation of the total area of the basin, of the proportion of each type of surface deposits per land use type and of the percentage of the basin covered by lakes and bare and semi-bare wetlands

In the case of the basin under study, according to figure 3, the results are the following:

Land use type | Identification1 | Area (ha) | Proportion |

Wooded | 2AR | 238 | 57% |

Wooded | 2BEM | 127 | 31% |

Wooded | 2BE | 19 | 5% |

- | Lakes and bare/semi-bare wetlands | 30 | 7% |

- | Total area | 414 | 100% |

1Identification of surface deposits and location of lakes and bare and semi-bare wetlands

Step 5 - Determination of the watercourse’s length (Lc) and calculation of the “85–10” slope of the watercourse (Sc)

The length of the watercourse is measured from the crossing point, following the course of the main watercourse extended to the watershed divide, that is, to the most distant point in the drainage basin determining the longest route a drop of water must travel to reach the crossing point.

The “85-10” slope of the watercourse is defined as the average slope of the section of the watercourse between 2 points located respectively 10% upstream from the crossing point and 15% downstream from the farthest limit of the drainage basin.

Figure 4 locates the line determining the length of the watercourse (Lc) and figure 5 shows the calculation method for the “85-10” slope of the watercourse (Sc) for the drainage basin under study.

Step 6 - Calculation of the weighted runoff coefficient of the drainage basin (Cp)

Firstly, using table 1, the various types of surface deposits in the drainage basin are classified on a hydrological basis.

Table 1: Hydrological classification of surface deposits

Types of surface deposits (designation) | Hydrological classification |

1AB-1BF-1BG-1BI-1BN-1BP-1BPY-1BR-1BT-1P-2-2A-2AE-2AK-2AT-2B-2BD-2BE-2BP-3AC-4GS-5S-6-6A-8AP-8APM-8APY-8AY-8AYP-8CM-8CY-8E-8F-8M-8P-8PM-8Y-9-9A-9R-9S | AB |

1A-1AD-1B-1BC-1BD-1BDY-1BIM-1BIY-2AM-2AR-2AY-2BEM-2BER-2BEY-2BR-3-3A-3AN-3ANY-4P-6S-6SM-6SR-6SY-8A-8AC-8AL-8ALM-8ALY-8AM-8AR-8AS-8ASY-8C-8PY-9SM-9SY-M6S-M8A-M8AP-M8C-M8PY | B |

3AE-3D-3DD-3DE-4-4A-4GSM-4GSR-4GSY-5SM-5SR-5SY-6AM-6AY-6R-8-8G | BC |

1AA-1AAM-1AAR-1ADY-1AM-1AR-1ASY-1AY-1AYR-1M-1Y-2BDY-4AR-4AY-4GA-4GAM-4GAY-4GAR-4GD-5A-5L-5R-5Y-M1- M1A-M1AA-R1-R1A-R1BD-R2A-R2AKR2BE-R3AN-R4-R4GS-R5S-R6-R6S-R8AR8AP-R8C-R8E-R8P-R9S-RS | C |

1AAY-5AM-5AR-5AY-5G-5GR-R-R1AAR4GA-R5A | CD |

7-7E-7L-7R-7T-7TM-7TY-AN-M7T-R7-R7T | n.a. |

Note: Type 7 deposits are classified as bare and semi-bare wetlands.

When the hydrological of surface deposits is completed, the runoff coefficient for each type of deposit is determined using table 2 based on land use and the average slope of the drainage basin.

Table 2: Runoff coefficients(C)

Land use type | Average slope of drainage basin (Sb) | Hydrological classification of surface deposits | ||||

AB | B | BC | C | CD | ||

Croplands | < 3% | 0.30 | 0.36 | 0.41 | 0.47 | 0.51 |

3 to 8% | 0.34 | 0.43 | 0.51 | 0.59 | 0.67 | |

> 8% | 0.43 | 0.51 | 0.61 | 0.67 | 0.73 | |

Pasturelands | < 3% | 0.12 | 0.17 | 0.25 | 0.34 | 0.43 |

3 to 8% | 0.17 | 0.25 | 0.33 | 0.43 | 0.51 | |

> 8% | 0.22 | 0.39 | 0.47 | 0.56 | 0.64 | |

Woodlands | < 3% | 0.09 | 0.15 | 0.21 | 0.29 | 0.37 |

3 to 8% | 0.12 | 0.19 | 0.26 | 0.34 | 0.43 | |

> 8% | 0.18 | 0.26 | 0.34 | 0.43 | 0.51 | |

Lakes and bare/semi-bare wetlands | 0.05 |

Then the weighted runoff coefficient for the drainage basin may be calculated (Cp). In the case of the basin under study, the data and calculations are the following:

Land use type | Identification | Proportion of drainage basin | Hydrological classification | Slope of drainage basin (Sb) | Runoff coefficient (C) |

Woodlands | 2AR | 57% | B | - | 0.26 |

Woodlands | 2BEM | 31% | B | > 8% | 0.26 |

Woodlands | 2BE | 5% | AB | - | 0.18 |

Lakes and bare/semi-bare wetlands | 7% | - | - | 0.05 |

Weighted runoff coefficient

(Cp) = (57% X 0.26)+(31% X 0.26)+(5% X 0.18)+(7% X 0.05)=0.24

Step 7 - Calculation of the drainage basin’s concentration time (tc)

The concentration time of the drainage basin is determined using one of the following 2 formulas:

If Cp < 0.40

tc | = | 3.26 (1,1 - Cp) Lc0.5 |

Sc0.33 |

where:

tc: concentration time (minutes)

Cp: weighted runoff coefficient for the basin

Lc: length of the watercourse (m)

Sc: “85-10” slope of the watercourse (%)

if Cp ≤ 0.20, | Sc minimum to be used = 0.1% |

if 0.20 < Cp < 0.40, | Sc minimum to be used = 0.5% |

tc minimum = 10 minutes

If Cp ≥ 0.40

tc | = | 0.057 Lc |

Sc0,2 Ab0.1 |

where:

tc: concentration time (minutes)

Lc: length of watercourse

Sc: “85-10” slope of watercourse (%)

Ab: area of drainage basin (ha)

tc minimum = 10 minutes

In the case of the basin under study, the Cp is equal to 0.24. Consequently, the first formula must be used.

tc = 3.26 (1.1 - 0.24) x 3,600 0.5 = 136 minutes

1.9 0.33

Step 8 - Determination of rainfall intensity (I)

Rainfall intensity is determined using figures 6 and 7. In figure 6, average total rainfall of a 1-hour duration for the basin under study is indicated by the contour line closest to that basin. Figure 7 indicates the standard deviation for the total rainfall of a 1-hour duration.

The rainfall intensity applicable to the drainage basin is determined as follows:

I = average total rainfall of a 1-hour duration + (1.305 X standard deviation for total rainfall of a 1-hour duration).

In our example, which is located on sheet 21M/6 N.E., the average is 22 mm/hour and the standard deviation is 8 mm/hour. The rainfall intensity applicable to that drainage basin is therefore 32.4 mm/hour, that is, 22 + (1.305 X 8).

Step 9 - Calculation of the correction coefficient for rainfall intensity (Fi)

Depending on the concentration time of the drainage basin, the correction coefficient for rainfall intensity is calculated using one of the following 2 formulas:

Fi | = | 12.25 | for 10 minutes ≤ tc < 60 minutes |

tc0.612 |

Fi | = | 17.07 | for tc ≥ 60 minutes |

tc0.693 |

where:

tc: concentration time (minutes)

In the case of the basin under study, the second formula must be used (tc = 136 minutes).

Fi | = | 17.07 | = 0.567 |

136 0.693 |

Step 10 - Determination of the reduction coefficient for the peak (FL)

The retention zones such as lakes and bare and semi-bare wetlands entail a significant reduction in the peak flow. The reduction coefficient for peak flow is evaluated using the proportion of lakes and bare and semi-bare wetlands calculated at step 4 and figure 8. In the case of the basin under study, that coefficient is 0.69 (curve B, 7% covered by lakes and bare and semi-bare wetlands).

Step 11 - Calculation of the 10-year interval peak flow (Q10)

That flow is calculated using the following formula:

Q10 (m3/s) = | Cp Fi I Ab FL |

360 |

where:

Cp = Weighted runoff coefficient for the drainage basin

Fi = Correction coefficient for rainfall intensity

I = Rainfall intensity (mm/hour)

Ab = Area of the drainage basin (ha)

FL = Reduction coefficient for peak flow

For the basin under study:

Q10 = | 0.24 X 0.567 X 32.4 X 414 X 0.69 |

360 |

Q10 = 3.5 m3/s

A weighted factor of at 5% is then applied to the flow obtained in order to take into account exceptional climatic events.

Ex.: 3.5 m3/s X 1.05 = 3.67 m3/s

Figure 3

Identification of surface deposits in the drainage basin and location of lakes and bare and semi-bare wetlands

Identification of surface deposits in the drainage basin and location of lakes and bare and semi-bare wetlands

SCHEDULE 6

PEAK FLOW CALCULATION METHOD FOR DRAINAGE BASINS WHOSE AREA IS EQUAL TO OR LESS THAN 60 KM2

The so-called rational method is used to calculate the 10-year interval peak flow. The method was validated for drainage basins whose area is less than 25 km2. Thus, where the area of the drainage basin covers between 25 and 60 km2, 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.

STEPS IN CALCULATION

1. Delimitation of the drainage basin;

2. Calculation of the average slope of the drainage basin;

3. Identification of the use of the territory and of the surface deposits in the drainage basin;

4. Calculation of the total area of the basin, of the proportion of each type of surface deposits per land use type and of the percentage of the basin covered by lakes and bare and semi-bare wetlands;

5. Determination of the watercourse’s length and calculation of the “85-10” slope of the watercourse;

6. Calculation of the weighted runoff coefficient of the drainage basin;

7. Calculation of the drainage basin’s concentration time;

8. Determination of rainfall intensity;

9. Calculation of the correction coefficient for rainfall intensity;

10. Determination of the reduction coefficient for peak flow;

11. Calculation of the 10-year interval peak flow.

EXPLANATION OF THE STEPS TO BE FOLLOWED WITH AN EXAMPLE

Step 1 - Delimitation of the drainage basin

The drainage basin that supplies the watercourse with water at the crossing point is delimited using a topographic map at a scale of 1: 20 000. Figure 1 shows, as an example, the delimitation of a drainage basin under study.

Step 2 - Calculation of the average slope of the drainage basin (Sb)

The average slope is calculated using a grid (1 cm X 1 cm) superimposed on the drainage basin. The number of times each horizontal and vertical line of that grid crosses a contour line must be determined. The length of those lines is also recorded. The calculation made to determine the average slope of the drainage basin under study is given in figure 2.

Step 3 - Identification of the use of the territory and of the surface deposits in the drainage basin

With the help of the surface deposit maps, the forest maps and knowledge of the territory, the use of the lands comprised within the drainage basin must be identified. They may be woodlands, pasturelands or croplands. Then the surface deposits for each land use type must be identified. Bare and semi-bare wetlands must also be located.

Figure 3 identifies the surface deposits and locates the bare and semi-bare wetlands in the drainage basin under study, which is completely wooded.

Step 4 - Calculation of the total area of the basin, of the proportion of each type of surface deposits per land use type and of the percentage of the basin covered by lakes and bare and semi-bare wetlands

In the case of the basin under study, according to figure 3, the results are the following:

Land use type | Identification1 | Area (ha) | Proportion |

Wooded | 2AR | 238 | 57% |

Wooded | 2BEM | 127 | 31% |

Wooded | 2BE | 19 | 5% |

- | Lakes and bare/semi-bare wetlands | 30 | 7% |

- | Total area | 414 | 100% |

1Identification of surface deposits and location of lakes and bare and semi-bare wetlands

Step 5 - Determination of the watercourse’s length (Lc) and calculation of the “85–10” slope of the watercourse (Sc)

The length of the watercourse is measured from the crossing point, following the course of the main watercourse extended to the watershed divide, that is, to the most distant point in the drainage basin determining the longest route a drop of water must travel to reach the crossing point.

The “85-10” slope of the watercourse is defined as the average slope of the section of the watercourse between 2 points located respectively 10% upstream from the crossing point and 15% downstream from the farthest limit of the drainage basin.

Figure 4 locates the line determining the length of the watercourse (Lc) and figure 5 shows the calculation method for the “85-10” slope of the watercourse (Sc) for the drainage basin under study.

Step 6 - Calculation of the weighted runoff coefficient of the drainage basin (Cp)

Firstly, using table 1, the various types of surface deposits in the drainage basin are classified on a hydrological basis.

Table 1: Hydrological classification of surface deposits

Types of surface deposits (designation) | Hydrological classification |

1AB-1BF-1BG-1BI-1BN-1BP-1BPY-1BR-1BT-1P-2-2A-2AE-2AK-2AT-2B-2BD-2BE-2BP-3AC-4GS-5S-6-6A-8AP-8APM-8APY-8AY-8AYP-8CM-8CY-8E-8F-8M-8P-8PM-8Y-9-9A-9R-9S | AB |

1A-1AD-1B-1BC-1BD-1BDY-1BIM-1BIY-2AM-2AR-2AY-2BEM-2BER-2BEY-2BR-3-3A-3AN-3ANY-4P-6S-6SM-6SR-6SY-8A-8AC-8AL-8ALM-8ALY-8AM-8AR-8AS-8ASY-8C-8PY-9SM-9SY-M6S-M8A-M8AP-M8C-M8PY | B |

3AE-3D-3DD-3DE-4-4A-4GSM-4GSR-4GSY-5SM-5SR-5SY-6AM-6AY-6R-8-8G | BC |

1AA-1AAM-1AAR-1ADY-1AM-1AR-1ASY-1AY-1AYR-1M-1Y-2BDY-4AR-4AY-4GA-4GAM-4GAY-4GAR-4GD-5A-5L-5R-5Y-M1- M1A-M1AA-R1-R1A-R1BD-R2A-R2AKR2BE-R3AN-R4-R4GS-R5S-R6-R6S-R8AR8AP-R8C-R8E-R8P-R9S-RS | C |

1AAY-5AM-5AR-5AY-5G-5GR-R-R1AAR4GA-R5A | CD |

7-7E-7L-7R-7T-7TM-7TY-AN-M7T-R7-R7T | n.a. |

Note: Type 7 deposits are classified as bare and semi-bare wetlands.

When the hydrological of surface deposits is completed, the runoff coefficient for each type of deposit is determined using table 2 based on land use and the average slope of the drainage basin.

Table 2: Runoff coefficients(C)

Land use type | Average slope of drainage basin (Sb) | Hydrological classification of surface deposits | ||||

AB | B | BC | C | CD | ||

Croplands | < 3% | 0.30 | 0.36 | 0.41 | 0.47 | 0.51 |

3 to 8% | 0.34 | 0.43 | 0.51 | 0.59 | 0.67 | |

> 8% | 0.43 | 0.51 | 0.61 | 0.67 | 0.73 | |

Pasturelands | < 3% | 0.12 | 0.17 | 0.25 | 0.34 | 0.43 |

3 to 8% | 0.17 | 0.25 | 0.33 | 0.43 | 0.51 | |

> 8% | 0.22 | 0.39 | 0.47 | 0.56 | 0.64 | |

Woodlands | < 3% | 0.09 | 0.15 | 0.21 | 0.29 | 0.37 |

3 to 8% | 0.12 | 0.19 | 0.26 | 0.34 | 0.43 | |

> 8% | 0.18 | 0.26 | 0.34 | 0.43 | 0.51 | |

Lakes and bare/semi-bare wetlands | 0.05 |

Then the weighted runoff coefficient for the drainage basin may be calculated (Cp). In the case of the basin under study, the data and calculations are the following:

Land use type | Identification | Proportion of drainage basin | Hydrological classification | Slope of drainage basin (Sb) | Runoff coefficient (C) |

Woodlands | 2AR | 57% | B | - | 0.26 |

Woodlands | 2BEM | 31% | B | > 8% | 0.26 |

Woodlands | 2BE | 5% | AB | - | 0.18 |

Lakes and bare/semi-bare wetlands | 7% | - | - | 0.05 |

Weighted runoff coefficient

(Cp) = (57% X 0.26)+(31% X 0.26)+(5% X 0.18)+(7% X 0.05)=0.24

Step 7 - Calculation of the drainage basin’s concentration time (tc)

The concentration time of the drainage basin is determined using one of the following 2 formulas:

If Cp < 0.40

tc | = | 3.26 (1,1 - Cp) Lc0.5 |

Sc0.33 |

where:

tc: concentration time (minutes)

Cp: weighted runoff coefficient for the basin

Lc: length of the watercourse (m)

Sc: “85-10” slope of the watercourse (%)

if Cp ≤ 0.20, | Sc minimum to be used = 0.1% |

if 0.20 < Cp < 0.40, | Sc minimum to be used = 0.5% |

tc minimum = 10 minutes

If Cp ≥ 0.40

tc | = | 0.057 Lc |

Sc0,2 Ab0.1 |

where:

tc: concentration time (minutes)

Lc: length of watercourse

Sc: “85-10” slope of watercourse (%)

Ab: area of drainage basin (ha)

tc minimum = 10 minutes

In the case of the basin under study, the Cp is equal to 0.24. Consequently, the first formula must be used.

tc = 3.26 (1.1 - 0.24) x 3,600 0.5 = 136 minutes

1.9 0.33

Step 8 - Determination of rainfall intensity (I)

Rainfall intensity is determined using figures 6 and 7. In figure 6, average total rainfall of a 1-hour duration for the basin under study is indicated by the contour line closest to that basin. Figure 7 indicates the standard deviation for the total rainfall of a 1-hour duration.

The rainfall intensity applicable to the drainage basin is determined as follows:

I = average total rainfall of a 1-hour duration + (1.305 X standard deviation for total rainfall of a 1-hour duration).

In our example, which is located on sheet 21M/6 N.E., the average is 22 mm/hour and the standard deviation is 8 mm/hour. The rainfall intensity applicable to that drainage basin is therefore 32.4 mm/hour, that is, 22 + (1.305 X 8).

Step 9 - Calculation of the correction coefficient for rainfall intensity (Fi)

Depending on the concentration time of the drainage basin, the correction coefficient for rainfall intensity is calculated using one of the following 2 formulas:

Fi | = | 12.25 | for 10 minutes ≤ tc < 60 minutes |

tc0.612 |

Fi | = | 17.07 | for tc ≥ 60 minutes |

tc0.693 |

where:

tc: concentration time (minutes)

In the case of the basin under study, the second formula must be used (tc = 136 minutes).

Fi | = | 17.07 | = 0.567 |

136 0.693 |

Step 10 - Determination of the reduction coefficient for the peak (FL)

The retention zones such as lakes and bare and semi-bare wetlands entail a significant reduction in the peak flow. The reduction coefficient for peak flow is evaluated using the proportion of lakes and bare and semi-bare wetlands calculated at step 4 and figure 8. In the case of the basin under study, that coefficient is 0.69 (curve B, 7% covered by lakes and bare and semi-bare wetlands).

Step 11 - Calculation of the 10-year interval peak flow (Q10)

That flow is calculated using the following formula:

Q10 (m3/s) = | Cp Fi I Ab FL |

360 |

where:

Cp = Weighted runoff coefficient for the drainage basin

Fi = Correction coefficient for rainfall intensity

I = Rainfall intensity (mm/hour)

Ab = Area of the drainage basin (ha)

FL = Reduction coefficient for peak flow

For the basin under study:

Q10 = | 0.24 X 0.567 X 32.4 X 414 X 0.69 |

360 |

Q10 = 3.5 m3/s

A weighted factor of at 5% is then applied to the flow obtained in order to take into account exceptional climatic events.

Ex.: 3.5 m3/s X 1.05 = 3.67 m3/s

Figure 3

Identification of surface deposits in the drainage basin and location of lakes and bare and semi-bare wetlands

Identification of surface deposits in the drainage basin and location of lakes and bare and semi-bare wetlands