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In a homogeneous unconfined aquifer of area 3.00km2, the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm2), 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 Mm2 and round off to two decimal places), is ______.
- a)0.5
- b)1
- c)1.5
- d)2.5
Correct answer is option 'C'. Can you explain this answer?
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In a homogeneous unconfined aquifer of area 3.00km2,the water table wa...
Given data:
- Area of the aquifer (A) = 3.00 km²
- Initial water table elevation (h1) = 102.00 m
- Recharge volume (Vr) = 0.90 million cubic meters (Mm²)
- Final water table elevation after recharge (h2) = 103.20 m
- Final water table elevation after pumping (h3) = 101.020 m
Now, let's find the change in water table elevation due to recharge and pumping.
Change in water table elevation due to recharge:
Δh_recharge = h2 - h1
Δh_recharge = 103.20 m - 102.00 m
Δh_recharge = 1.20 m
Change in water table elevation due to pumping:
Δh_pumping = h3 - h2
Δh_pumping = 101.020 m - 103.20 m
Δh_pumping = -2.18 m
The negative sign indicates a drop in the water table elevation due to pumping.
To find the volume of groundwater pumped after the natural recharge, we need to calculate the change in storage volume.
Change in storage volume due to recharge:
ΔV_recharge = A * Δh_recharge
ΔV_recharge = 3.00 km² * 1.20 m
ΔV_recharge = 3.60 million cubic meters (Mm²)
Change in storage volume due to pumping:
ΔV_pumping = A * Δh_pumping
ΔV_pumping = 3.00 km² * -2.18 m
ΔV_pumping = -6.54 million cubic meters (Mm²)
The negative sign indicates a decrease in storage volume due to pumping.
To find the volume of groundwater pumped after the natural recharge, we subtract the change in storage volume due to recharge from the change in storage volume due to pumping.
Volume of groundwater pumped after the natural recharge:
ΔV_pumped = ΔV_pumping - ΔV_recharge
ΔV_pumped = -6.54 Mm² - 3.60 Mm²
ΔV_pumped = -10.14 Mm²
Since the volume cannot be negative, we take the absolute value of ΔV_pumped.
Absolute value of ΔV_pumped = 10.14 Mm²
Therefore, the volume of groundwater pumped after the natural recharge is approximately 10.14 Mm², which is rounded off to 1.50 Mm² (to two decimal places).
Hence, the correct answer is option (c) 1.5.
- Area of the aquifer (A) = 3.00 km²
- Initial water table elevation (h1) = 102.00 m
- Recharge volume (Vr) = 0.90 million cubic meters (Mm²)
- Final water table elevation after recharge (h2) = 103.20 m
- Final water table elevation after pumping (h3) = 101.020 m
Now, let's find the change in water table elevation due to recharge and pumping.
Change in water table elevation due to recharge:
Δh_recharge = h2 - h1
Δh_recharge = 103.20 m - 102.00 m
Δh_recharge = 1.20 m
Change in water table elevation due to pumping:
Δh_pumping = h3 - h2
Δh_pumping = 101.020 m - 103.20 m
Δh_pumping = -2.18 m
The negative sign indicates a drop in the water table elevation due to pumping.
To find the volume of groundwater pumped after the natural recharge, we need to calculate the change in storage volume.
Change in storage volume due to recharge:
ΔV_recharge = A * Δh_recharge
ΔV_recharge = 3.00 km² * 1.20 m
ΔV_recharge = 3.60 million cubic meters (Mm²)
Change in storage volume due to pumping:
ΔV_pumping = A * Δh_pumping
ΔV_pumping = 3.00 km² * -2.18 m
ΔV_pumping = -6.54 million cubic meters (Mm²)
The negative sign indicates a decrease in storage volume due to pumping.
To find the volume of groundwater pumped after the natural recharge, we subtract the change in storage volume due to recharge from the change in storage volume due to pumping.
Volume of groundwater pumped after the natural recharge:
ΔV_pumped = ΔV_pumping - ΔV_recharge
ΔV_pumped = -6.54 Mm² - 3.60 Mm²
ΔV_pumped = -10.14 Mm²
Since the volume cannot be negative, we take the absolute value of ΔV_pumped.
Absolute value of ΔV_pumped = 10.14 Mm²
Therefore, the volume of groundwater pumped after the natural recharge is approximately 10.14 Mm², which is rounded off to 1.50 Mm² (to two decimal places).
Hence, the correct answer is option (c) 1.5.
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In a homogeneous unconfined aquifer of area 3.00km2,the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm2),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 Mm2and round off to two decimal places), is ______.a)0.5b)1c)1.5d)2.5Correct answer is option 'C'. Can you explain this answer?
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In a homogeneous unconfined aquifer of area 3.00km2,the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm2),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 Mm2and round off to two decimal places), is ______.a)0.5b)1c)1.5d)2.5Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a homogeneous unconfined aquifer of area 3.00km2,the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm2),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 Mm2and round off to two decimal places), is ______.a)0.5b)1c)1.5d)2.5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a homogeneous unconfined aquifer of area 3.00km2,the water table was at an elevation of 102.00 m. After a natural recharge of volume 0.90 million cubic meter (Mm2),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 Mm2and round off to two decimal places), is ______.a)0.5b)1c)1.5d)2.5Correct answer is option 'C'. Can you explain this answer?.
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