Determine the time to infilling of an estuary/lagoon system that will receive a
ID: 111768 • Letter: D
Question
Determine the time to infilling of an estuary/lagoon system that will receive a sediment volume 1,000,000 cubic meter per year continuously. The system in question has an area of approximately 200 Km^2, and it's comprised of 50 % marsh, and 50 % water. The basin experiences a subsidence rate of ~4 mm/yr, and eustatic sea level is rising at a rate of ~3 mm/yr. Assuming marsh inundation is negligible, and all sediment is available for estuarine sedimentation, estimate the depth of the lagoon after 50 years. Consider using a step-wise approach to the problem, and perform sediment continuity (mass balance) every year. The average depth for all open water areas is 1.5 m. Assuming uniform deposition throughout the estuary, and a sediment retention of ~60 %, a uniform consolidation rate of newly deposited sediment of 15% the annual deposited material, estimate the estuarine depth after 50 years.Explanation / Answer
step-1:
Area of the basin= 200km2 =200*10^6m2. Average depth=1.5m.
Every year depth of the basin increases=(3mm+4mm)= 7mm due to both subsidence and sea level rising.
So, after 50 years total depth increase will be= 7*50=350mm=(.35)m.
so, after 50 years total volume of the basin = (200*10^6)*(1.5+0.35)m3=37*10^7m3 .
step-2:
Now the total volume of sediment deposited per year=10^6m3 .
the volume of sediments retains per year= 10^6*60%= 6*10^5m3 .
there is 15% consolidation rate, so, the volume of sediments after consolidation=(6*10^5)*85%= 48*10^4m3/year.
So, in 50 years deposited sediment volume will be= (48*10^4)*50=24*10^6m3.
step-3: now after 50 years of deposition the volume of basin remain unfilled by sediments= (37*10^7)-(24*10^6)= 346*10^6m3.
So, after 50 years the depth of the estuary will be= (346*10^6)m3/(200*10^6)m3= 1.73m (answer)
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