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Math problem set Grade 12 Math

Stonebriar Reservoir: Sediment Inflow and Lost Storage

A grade 12 problem set converting a fictional watershed's sediment yield into lost reservoir storage, then comparing linear and exponential models of capacity loss.

Objectives

  1. Students will convert a watershed sediment yield in tons per acre into acre-feet of reservoir storage lost per year.
  2. Students will construct linear and exponential models of a reservoir's remaining storage.
  3. Students will solve an exponential equation with logarithms for the year a storage threshold is reached.

Vocabulary

Acre-foot
The amount of water it takes to cover one acre of land, about the size of a football field, one foot deep. Western states use acre-feet to measure large amounts of water.
Reservoir sedimentation
The slow filling of a reservoir with mud and sand carried in by the river. Over many years, sediment steals space that was meant to store water.

Also: sediment yield, trap efficiency, bulk density, dead storage

Teacher overview

Every reservoir is filling in from the day it is built, because the river that supplies it also carries sediment that settles once the current slows behind the dam. This set follows the fictional Stonebriar Reservoir, a hydropower and water supply pool on the Merribeck River, from its watershed's sediment yield to the storage it loses each year and the year its turbine intake is threatened. Two errors recur at this grade. The first is converting tons of sediment directly into volume without applying a bulk density, and the second is treating a fixed percent loss per year as identical to a fixed volume loss per year, which produces two noticeably different answers by year 50.

Stonebriar Reservoir: Sediment Inflow and Lost Storage — Grade 12 Math
Name: ____________________________ Date: ______________

Problems

All values below describe the fictional Stonebriar Reservoir and its watershed.

Stonebriar Reservoir holds 42,000 acre-feet when full. Its watershed covers 310 square miles and delivers 0.42 tons of sediment per acre each year. The reservoir traps 92 percent of the sediment that reaches it, and the trapped material settles at a bulk density of 55 pounds per cubic foot. The turbine intake sits above the lowest 3,000 acre-feet of the pool, which is dead storage set aside for sediment. Once sediment fills it, material begins to reach the intake. Use 640 acres per square mile, 2,000 pounds per ton, and 43,560 cubic feet per acre-foot.

  1. Find the watershed area in acres.

  2. Find the tons of sediment the watershed delivers to the reservoir each year.

  3. Find the tons of sediment the reservoir traps each year at 92 percent trap efficiency. Round to the nearest ton.

  4. Convert the trapped sediment to a volume in cubic feet per year, using the bulk density of 55 pounds per cubic foot. Round to the nearest thousand cubic feet.

  5. Convert your problem 4 answer to acre-feet of storage lost per year. Round to the nearest tenth.

  6. Write a linear function S(t) for remaining storage in acre-feet after t years, using the loss rate from problem 5, then find the remaining storage after 50 years.

  7. Using the linear model, how many years pass before the reservoir loses 10 percent of its original 42,000 acre-feet? Round to the nearest tenth of a year.

  8. Engineers argue that trap efficiency falls as the pool shrinks, so they model remaining storage as S(t) = 42,000(0.9985) raised to the power t, a loss of 0.15 percent of remaining storage each year. Find the remaining storage this model predicts after 50 years, to the nearest acre-foot.

  9. Using the exponential model from problem 8, solve for the number of years until only 90 percent of the original storage remains. Round to the nearest tenth of a year.

  10. Return to the linear model. How many years until the trapped sediment equals the 3,000 acre-feet of dead storage below the turbine intake? Round to the nearest tenth.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 198,400 acres.
    310 square miles x 640 acres per square mile = 198,400 acres.

  2. Answer: 83,328 tons.
    198,400 acres x 0.42 tons per acre = 83,328 tons of sediment delivered per year.

  3. Answer: 76,662 tons.
    83,328 x 0.92 = 76,661.76, which rounds to 76,662 tons trapped per year. The remaining 8 percent passes through the dam and continues downstream.

  4. Answer: 2,788,000 cu ft.
    Convert to pounds: 76,662 tons x 2,000 = 153,324,000 pounds. Divide by the bulk density: 153,324,000 / 55 = 2,787,709 cubic feet, which rounds to 2,788,000 cubic feet per year.

  5. Answer: 64.0 acre-ft.
    2,788,000 / 43,560 = 64.0 acre-feet of storage lost per year.

  6. Answer: 38,800 acre-ft.
    The linear model is S(t) = 42,000 - 64t, since the model removes the same 64 acre-feet every year. At t = 50: S(50) = 42,000 - 64(50) = 42,000 - 3,200 = 38,800 acre-feet.

  7. Answer: 65.6 years.
    A 10 percent loss is 0.10 x 42,000 = 4,200 acre-feet. Years = 4,200 / 64 = 65.6 years.

  8. Answer: 38,963 acre-ft.
    S(50) = 42,000 x 0.9985 raised to the 50th power. Since 0.9985 to the 50th power is 0.92769, S(50) = 42,000 x 0.92769 = 38,963 acre-feet, about 163 acre-feet more than the linear model predicts.

  9. Answer: 70.2 years.
    Solve 0.9985 raised to the power t = 0.90. Taking natural logs: t = ln(0.90) / ln(0.9985) = (-0.10536) / (-0.0015011) = 70.2 years. The exponential model buys about 4.6 more years than the linear one, because each year's loss is figured on a shrinking pool.

  10. Answer: 46.9 years.
    Years = 3,000 acre-feet / 64 acre-feet per year = 46.875, which rounds to 46.9 years. Sediment does not settle evenly across the pool, so the practical date for intake problems can arrive sooner than this uniform-fill estimate suggests.

Differentiation

Support

Provide the fictional Stonebriar conversion chain as a labeled ladder reading square miles, acres, tons, pounds, cubic feet, acre-feet, with the factor written on each rung, so problems 1 through 5 become substitution rather than recall. Give problem 6 with the form S(t) = 42,000 - ___ t already printed, and allow a table of values instead of logarithms for problem 9.

Extension

Ask students to use a table or a graph to describe how the gap between the linear and exponential models for the fictional reservoir changes over the first 200 years, and to explain in writing which model a dam owner should use for a 20-year capital plan and which for a 150-year outlook.

ELL supports

Pre-teach "yield," "trap efficiency," "bulk density," and "dead storage" with a cross-section diagram of a reservoir showing sediment building up from the upstream end toward the intake. Provide the two model forms side by side with labels reading "same amount each year" and "same percent each year," and offer a sentence frame for the comparison in problem 9.

Extensions

Have students price the problem. At a fictional dredging cost of $18,500 per acre-foot removed, students calculate what it would cost the fictional Stonebriar Water District to remove one decade of accumulated sediment, then compare that figure with the cost of buying replacement storage elsewhere at a stated price per acre-foot. Ask them to state which assumption in their comparison is least certain and why a dam owner might still defer the work.