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Quiz & test questions Grade 12 Engineering & Design

Ashenholt Digester: Biosolids Loading and Energy Recovery Exam

A grade 12 exam on anaerobic digestion at a fictional facility, from solids loading and volatile solids destruction to biogas, electricity, and detention time.

Objectives

  1. Students will calculate dry and volatile solids loading from sludge volume and percent solids.
  2. Students will trace volatile solids destruction into biogas production and generated electricity.
  3. Students will compute digester detention time and solve for a volume that meets a target.

Vocabulary

Detention time
Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.
Biosolids
Biosolids are the nutrient-rich solid material left over after sewage sludge has been treated and made safe. Many farms use treated biosolids as fertilizer for their fields.

Also: anaerobic digestion, volatile solids, biogas, removal efficiency

Teacher overview

This 50-minute exam is worth 38 points and covers anaerobic digestion and energy recovery at the fictional Ashenholt Water Resource Recovery Facility. It mixes one multiple-choice item, nine computation items, and two short-answer items, following one sludge stream from the thickener through the digester to the engine-generator. The misconception this exam targets is that sludge is mostly solids. A stream of 14,000 gallons per day at 4.2 percent solids is more than 95 percent water by weight, which is why the loading calculation has to start from gallons and percent rather than from a volume that students picture as a pile of material.

Ashenholt Digester: Biosolids Loading and Energy Recovery Exam — Grade 12 Engineering & Design
Name: ____________________________ Date: ______________

Assessment

The fictional Ashenholt Water Resource Recovery Facility digests sludge anaerobically and burns the resulting gas in an engine-generator. Use 8.34 pounds per gallon of water throughout. Removal efficiency equals influent minus effluent, divided by influent, times 100.

  1. (2 points) In anaerobic digestion, the organisms that produce methane
    • A) require dissolved oxygen above 5 mg/L in the digester
    • B) work without oxygen and convert organic acids into methane and carbon dioxide
    • C) are the same organisms that convert ammonia into nitrate
    • D) grow only when the digester is held below 10 degrees Celsius
  2. (4 points) Ashenholt pumps 14,000 gallons per day of primary sludge at 4.2 percent solids. Find the pounds of dry solids fed to the digester each day.
  3. (4 points) Volatile solids make up 72 percent of the dry solids. Using a dry solids load of 4,903.9 pounds per day, find the pounds of volatile solids fed each day.
  4. (4 points) The digester destroys 58 percent of the volatile solids it receives. Using a volatile solids load of 3,530.8 pounds per day, find the pounds destroyed each day.
  5. (4 points) The digester produces 15.0 cubic feet of gas for every pound of volatile solids destroyed. Using 2,047.9 pounds destroyed per day, find the daily gas production in cubic feet.
  6. (4 points) The engine-generator produces 5.0 kWh for every 100 cubic feet of digester gas burned. Using 30,718.5 cubic feet per day, find the daily electricity production.
  7. (3 points) Electricity is worth $0.11 per kWh. Using 1,535.9 kWh per day over 365 days, find the annual value of the electricity produced.
  8. (3 points) The digester holds 320,000 gallons and receives 14,000 gallons per day of primary sludge plus 6,000 gallons per day of thickened waste activated sludge. Find the detention time in days.
  9. (3 points) Ashenholt wants 20 days of detention at the same combined feed of 20,000 gallons per day. Write an equation relating volume, feed rate, and detention time, then solve it for the required digester volume in gallons.
  10. (3 points) The plant's influent BOD is 240 mg/L and its effluent BOD is 18 mg/L. Find the BOD removal efficiency as a percent.
  11. (2 points) Secondary treatment standards require a monthly average effluent BOD of 30 mg/L or less. State whether an effluent BOD of 18 mg/L meets that standard and give the reason.
  12. (2 points) Explain in one or two sentences why destroying volatile solids in the digester lowers what Ashenholt pays to haul biosolids away.

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: B) work without oxygen and convert organic acids into methane and carbon dioxide. Methane-forming organisms are strict anaerobes, so choice A describes conditions that would kill them. Choice C describes nitrifying bacteria, which are aerobic, and choice D reverses the temperature relationship, since digesters are heated rather than chilled.
  2. Answer: 4,903.9 lb/day. At the fictional facility, 14,000 gallons times 8.34 pounds per gallon equals 116,760 pounds of sludge, and 116,760 times 0.042 equals 4,903.92 pounds of dry solids per day.
  3. Answer: 3,530.8 lb/day. 4,903.9 pounds per day times 0.72 equals 3,530.81 pounds of volatile solids per day.
  4. Answer: 2,047.9 lb/day. 3,530.8 pounds per day times 0.58 equals 2,047.86 pounds destroyed per day.
  5. Answer: 30,718.5 cubic feet per day. 2,047.9 pounds destroyed times 15.0 cubic feet per pound equals 30,718.5 cubic feet of gas per day.
  6. Answer: 1,535.9 kWh per day. 30,718.5 divided by 100 equals 307.185 units of gas, and 307.185 times 5.0 kWh equals 1,535.925, which rounds to 1,535.9 kWh per day.
  7. Answer: $61,666.39 per year. 1,535.9 kWh per day times 365 days equals 560,603.5 kWh per year, and 560,603.5 times $0.11 equals $61,666.385, which rounds to $61,666.39.
  8. Answer: 16 days. The combined feed is 14,000 plus 6,000, or 20,000 gallons per day. Then 320,000 gallons divided by 20,000 gallons per day equals 16 days of detention.
  9. Answer: 400,000 gallons. Writing volume equals detention time times feed rate gives V equals 20 days times 20,000 gallons per day, so V equals 400,000 gallons.
  10. Answer: 92.5 percent. 240 minus 18 equals 222 mg/L removed, and 222 divided by 240 equals 0.925, which is 92.5 percent removal.
  11. Answer: Yes. An effluent BOD of 18 mg/L sits below the 30 mg/L monthly average limit for secondary treatment, so the plant is in compliance on that parameter.
  12. Answer: Fewer pounds of solids leave the plant. Accept any answer noting that hauling is paid by weight or by volume of biosolids, and destroying volatile solids converts part of that mass into gas, so less material remains to be dewatered, trucked, and land applied.

Differentiation

Support

Provide item 2 with the two multiplications already staged as 14,000 times 8.34, then times 0.042, so students begin from a set-up rather than from a formula. Supply a worked model on a different sludge stream, such as 8,000 gallons per day at 3.0 percent solids, before the exam begins.

Extension

Ask these students to close the loop on cost at the fictional facility. Have them find how many pounds of solids per day remain after digestion, then estimate the annual hauling cost at $0.06 per pound and compare that figure against the annual electricity value from item 7.

ELL supports

Provide a flow diagram labeling thickener, digester, gas holder, and engine-generator, with an arrow for solids and a separate arrow for gas, so each computation item maps to one arrow. Offer sentence frames for items 11 and 12: "The effluent is ___ mg/L, which is ___ the 30 mg/L limit, so the plant ___," and "Destroying volatile solids leaves ___ pounds to haul, which lowers ___."

Extensions

Suppose the fictional Ashenholt facility is offered a food-waste receiving station that would add 1,200 pounds per day of volatile solids to the digester feed. Ask students to recompute the daily gas production and electricity value at the same destruction rate and gas yield, then to name two operating limits the plant would have to check before accepting the additional load.