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Engineers Canada Agricultural / Biosystems / Bioresource / Food Engineering examination syllabus Agricultural Waste Management & Environmental Engineering: practice questions

Engineers Canada Agricultural / Biosystems / Bioresource / Food Engineering examination syllabus · Canada · Provincial and territorial regulators (syllabus by Engineers Canada)

Original practice questions written to the published syllabus · not past papers
10worked here, free
97written to this exam in this subject
9syllabus units

What this subject covers

Handling residues, manure and wastewater so that farms and plants do not pollute soil, water and air.

In the exams: NCEES PE Natural Resources & Environmental Systems and nutrient management; PRC ABE waste management; electives in the ICAR 6th Deans' Committee B.Tech

Where the 97 questions fall

How our questions for Engineers Canada Agricultural / Biosystems / Bioresource / Food Engineering examination syllabus spread across the units of Agricultural Waste Management & Environmental Engineering. Difficulty: 23 easy, 47 medium, 27 hard.

Basics of wastewater treatment18
Composting14
Wastewater characteristics: BOD and COD13
Crop-residue management13
Environmental impact assessment12
Air quality and odour10
Anaerobic lagoons7
Manure handling systems5
Nutrient management plans5

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Wastewater characteristics: BOD and COD · hard

Lactose, the main sugar in whey, oxidises as C12H22O11 + 12 O2 → 12 CO2 + 11 H2O. Its theoretical oxygen demand (molar mass 342 g/mol) is:

  1. 0.89 g O2 per g lactose
  2. 1.12 g O2 per g lactose
  3. 1.54 g O2 per g lactose
  4. 1.07 g O2 per g lactose
Answer and working

B. 1.12 g O2 per g lactose

12 × 32 = 384 g O2 per 342 g lactose = 1.12 g/g. 1.07 is the glucose value; 0.89 inverts the ratio; 1.54 is grams of CO2 produced per gram of lactose, not oxygen consumed.

2 · Wastewater characteristics: BOD and COD · medium

A food plant sends 750 kg of BOD5 per day to treatment. Taking the domestic contribution as 60 g BOD5 per person per day (this figure differs between countries and texts), its population equivalent is:

  1. 45,000 persons
  2. 125,000 persons
  3. 1,250 persons
  4. 12,500 persons
Answer and working

D. 12,500 persons

PE = 750,000 g/d ÷ 60 g/person/d = 12,500 persons. 1,250 and 125,000 are factor-of-ten slips in converting kg to g; 45,000 multiplies 750 by 60 instead of dividing.

3 · Basics of wastewater treatment · hard

An activated sludge plant treating sugar-rich food wastewater develops an SVI above 250 mL/g with a cloudy, slow-settling sludge. Which combination is most likely to cause this?

  1. Very high dissolved oxygen together with a short sludge age of one day
  2. Over-wasting of sludge that lowered the MLSS well below the design value
  3. Excess nitrogen and phosphorus added at well above the BOD:N:P need
  4. Filamentous growth favoured by low dissolved oxygen and nutrient shortage
Answer and working

D. Filamentous growth favoured by low dissolved oxygen and nutrient shortage

Filamentous bulking is linked to low DO, low F/M, nutrient deficiency and readily degradable sugars, all common in food wastewaters. Surplus nutrients and high DO discourage filaments; low MLSS alone does not raise SVI.

4 · Composting · easy

A compost mix of poultry manure and little bedding has a starting C:N of 10:1. The most likely problem is:

  1. Loss of nitrogen as ammonia and strong odour
  2. Temperatures that never rise above ambient
  3. Excess carbon locked up as stable humus
  4. Very slow decomposition from nitrogen shortage
Answer and working

A. Loss of nitrogen as ammonia and strong odour

With too little carbon relative to nitrogen, microbes cannot use all the nitrogen, which is lost as ammonia, causing odour and lower fertiliser value. Slow decomposition from nitrogen shortage is the problem of high-C:N mixes such as straw alone.

5 · Composting · medium

Which observation best indicates that compost has become stable and mature?

  1. A C:N ratio that has risen well above that of the starting mix
  2. Low respiration rate and no reheating after the pile is turned
  3. A core temperature that rises again to 60 °C after turning
  4. A strong ammonia smell when the pile is opened and turned
Answer and working

B. Low respiration rate and no reheating after the pile is turned

Mature compost has low oxygen uptake or CO2 evolution, does not self-heat after turning, has no ammonia odour and normally a lower C:N than the starting mix. Reheating and ammonia show active decomposition still under way.

6 · Manure handling systems · easy

In standard methods for manure analysis, total solids and volatile solids are determined by heating the sample at, respectively:

  1. 60 °C for one hour, then 900 °C for six hours
  2. 550 °C first, then 103–105 °C to constant mass
  3. 20 °C for five days, then 150 °C under reflux
  4. 103–105 °C to constant mass, then 550 °C
Answer and working

D. 103–105 °C to constant mass, then 550 °C

TS is the residue after drying at 103–105 °C; VS is the part of TS lost on ignition at about 550 °C, and the ash remaining is fixed solids. 20 °C for five days and 150 °C reflux describe BOD and COD conditions.

7 · Basics of wastewater treatment · hard

For an activated sludge plant treating dairy wastewater, the operator increases the sludge wasting rate so that SRT falls from 20 days to 3 days. The most likely consequence is:

  1. Loss of nitrification, since nitrifiers wash out at short SRT
  2. Better nitrification, since nitrifiers grow faster than heterotrophs
  3. Lower F/M, since more biomass is kept in the system
  4. Less surplus sludge, since shorter SRT reduces cell yield
Answer and working

A. Loss of nitrification, since nitrifiers wash out at short SRT

Nitrifiers grow slowly; at short SRT they are wasted faster than they grow. Shorter SRT also raises net sludge yield and raises F/M because less biomass is held.

8 · Composting · hard

In forced-aeration composting, the airflow needed to remove heat and hold temperature is far greater than that needed to supply oxygen for decomposition. The design consequence is that:

  1. Moisture is gained as air passes, so the pile needs no water management
  2. Aeration fans are sized from the stoichiometric oxygen demand of the feedstock alone
  3. Aeration is controlled by temperature feedback rather than oxygen demand alone
  4. Aeration should be stopped once the pile reaches 70 °C to save energy
Answer and working

C. Aeration is controlled by temperature feedback rather than oxygen demand alone

Heat removal dominates airflow requirements, so fans are sized and controlled for temperature, and the large airflow dries the pile, making moisture addition necessary. Stopping aeration at high temperature lets the pile overheat further.

9 · Manure handling systems · hard

A slurry has 8% total solids, 2.0 kg/m³ of P and 4.0 kg/m³ of ammonium-N. A separator captures 60% of solids, 50% of P and 10% of ammonium in the solid fraction. For the liquid fraction, the ratio of ammonium-N to P compared with the raw slurry is:

  1. 0.6 times higher
  2. 1.0 times higher
  3. 1.8 times higher
  4. 3.6 times higher
Answer and working

C. 1.8 times higher

Raw NH4-N:P = 4.0/2.0 = 2.0. Liquid keeps 90% of ammonium (3.6) and 50% of P (1.0), ratio 3.6; 3.6/2.0 = 1.8 times higher. 3.6 is the liquid ratio itself, not the change; 1.0 assumes no change; separation shifts P to solids more than ammonium, so the liquid becomes relatively richer in N.

10 · Wastewater characteristics: BOD and COD · hard

A seeded BOD test gives these data: sample bottle DO drop 5.0 mg/L with 10 mL sample and 2 mL seed in 300 mL; seed control DO drop 4.0 mg/L with 10 mL seed in 300 mL. The BOD5 of the sample is:

  1. -450 mg/L
  2. 150 mg/L
  3. 30 mg/L
  4. 126 mg/L
Answer and working

D. 126 mg/L

Seed correction = 4.0 × (2/10) = 0.8 mg/L. BOD5 = (5.0 − 0.8)/(10/300) = 4.2 × 30 = 126 mg/L. 150 ignores the seed; 30 subtracts the whole seed-control drop; −450 inverts the seed ratio, an impossible negative result.

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Agricultural Waste Management & Environmental EngineeringThe full study manual, free Engineers Canada Agricultural / Biosystems / Bioresource / Food Engineering examination syllabusFormat, sections and how it has changed

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