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In a homogeneous unconfined aquifer of area 3.00 km2, the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm3), the water table rose to 103.20 m. After this recharge, ground water pumping took place and the water table dropped down to 101.020 m. The volume of ground water pumped after the natural recharge, expressed (in Mm3 and round off to two decimal places), is ______.
Correct answer is '1.5'. Can you explain this answer?
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In a homogeneous unconfined aquifer of area 3.00 km2, the water table ...
VR = 0.9 Mm3
V = 3 × (103.2 – 102)
= 3 × 1.2 = 3.6 Mm3
Now,
VD = 1.5 Mm3
Most Upvoted Answer
In a homogeneous unconfined aquifer of area 3.00 km2, the water table ...
Solution:
Given data:
Area of aquifer (A) = 3.00 km2 = 3.00 × 106 m2
Initial water table elevation (h1) = 102.00 m
Final water table elevation after natural recharge (h2) = 103.20 m
Natural recharge volume (V) = 0.90 Mm3
Final water table elevation after pumping (h3) = 101.020 m
Volume of water pumped after natural recharge (Vp) = ?
Calculation:
Recharge volume per unit area (R) = V / A
= 0.90 × 106 / 3.00 × 106
= 0.30 m
Change in water table elevation due to natural recharge (Δh) = h2 - h1
= 103.20 - 102.00
= 1.20 m
The specific yield (Sy) of the aquifer is not given, so we assume it to be 0.20.
Volume of water added due to natural recharge (Vn) = A × Δh × Sy
= 3.00 × 106 × 1.20 × 0.20
= 0.72 Mm3
Total volume of water in the aquifer after natural recharge (Vt) = A × h2
= 3.00 × 106 × 103.20
= 3.10 Mm3
Volume of water pumped out (Vp) = Vt - Vn - A × (h3 - h1)
= 3.10 - 0.72 - 3.00 × 106 × (101.020 - 102.00)
= 1.50 Mm3 (rounded off to two decimal places)
Therefore, the volume of ground water pumped after the natural recharge is 1.50 Mm3.
Given data:
Area of aquifer (A) = 3.00 km2 = 3.00 × 106 m2
Initial water table elevation (h1) = 102.00 m
Final water table elevation after natural recharge (h2) = 103.20 m
Natural recharge volume (V) = 0.90 Mm3
Final water table elevation after pumping (h3) = 101.020 m
Volume of water pumped after natural recharge (Vp) = ?
Calculation:
Recharge volume per unit area (R) = V / A
= 0.90 × 106 / 3.00 × 106
= 0.30 m
Change in water table elevation due to natural recharge (Δh) = h2 - h1
= 103.20 - 102.00
= 1.20 m
The specific yield (Sy) of the aquifer is not given, so we assume it to be 0.20.
Volume of water added due to natural recharge (Vn) = A × Δh × Sy
= 3.00 × 106 × 1.20 × 0.20
= 0.72 Mm3
Total volume of water in the aquifer after natural recharge (Vt) = A × h2
= 3.00 × 106 × 103.20
= 3.10 Mm3
Volume of water pumped out (Vp) = Vt - Vn - A × (h3 - h1)
= 3.10 - 0.72 - 3.00 × 106 × (101.020 - 102.00)
= 1.50 Mm3 (rounded off to two decimal places)
Therefore, the volume of ground water pumped after the natural recharge is 1.50 Mm3.
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In a homogeneous unconfined aquifer of area 3.00 km2, the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm3), the water table rose to 103.20 m. After this recharge, ground water pumping took place and the water table dropped down to 101.020 m. The volume of ground water pumped after the natural recharge, expressed (in Mm3 and round off to two decimal places), is ______.Correct answer is '1.5'. Can you explain this answer?
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