CEM Practice Questions (2026) — Free Sample Questions with Answers

Answer: B

1. Givens: purchased electricity 4,200,000 kWh at 0.85 lb CO₂e/kWh; on-site gas 60,000 therms at 11.7 lb CO₂e/therm.
2. Classify: purchased electricity = Scope 2 (indirect); gas burned in owned boilers = Scope 1 (direct), so it is excluded.
3. Scope 2 = 4,200,000 kWh × 0.85 lb/kWh = 3,570,000 lb CO₂e.
4. Convert: 3,570,000 lb ÷ 2,204.6 lb/metric ton = 1,619 metric tons CO₂e.

318 metric tons CO₂e: This is the natural gas combustion total, which is Scope 1 (direct emissions from owned boilers), not Scope 2.

1,619 metric tons CO₂e: Correct. Scope 2 covers indirect emissions from purchased electricity: 4,200,000 × 0.85 = 3,570,000 lb ÷ 2,204.6 = 1,619 metric tons.

1,785 metric tons CO₂e: Divided by 2,000 lb (short tons) instead of 2,204.6 lb per metric ton.

1,938 metric tons CO₂e: Added the Scope 1 natural gas emissions to the purchased-electricity emissions.

GHG Protocol scopes: Scope 1 is direct combustion on owned equipment, Scope 2 is purchased electricity, steam or chilled water, Scope 3 is other value-chain emissions; emissions = activity × emission factor.

Answer: C

1. Givens: annual peak 1,200 kW; ratchet 80%; demand charge $12.00/kW-month.
2. Ratchet floor = 0.80 × 1,200 kW = 960 kW. Billing demand = max(actual, 960).
3. Excess over actual: Oct 60, Nov 140, Dec 200, Jan 260, Feb 220, Mar 0 (980 > 960) → total 880 kW-months.
4. Added cost = 880 kW-months × $12.00 = $10,560.

$3,120: Calculated the ratchet penalty for January only.

$10,320: Counted March (980 kW, above the 960 kW ratchet) as a negative 20 kW instead of zero penalty.

$10,560: Correct. Ratchet demand = 0.80 × 1,200 = 960 kW; excess billed kW = 60 + 140 + 200 + 260 + 220 + 0 = 880 kW × $12 = $10,560.

$27,600: Applied a 100% ratchet (1,200 kW) instead of 80%.

$69,360: Reported the total demand charges billed for the six months, not the extra cost caused by the ratchet.

A demand ratchet sets a floor on billing demand for up to 11 months after a peak, so one summer peak can raise demand charges all winter; peak control pays off year-round.

Answer: C

1. Givens: 620,000 kWh; 18,500 therms; 48,000 ft²; 1 kWh = 3,412 Btu; 1 therm = 100,000 Btu
2. Electricity: 620,000 × 3,412 = 2,115,440,000 Btu
3. Gas: 18,500 × 100,000 = 1,850,000,000 Btu
4. Total: 3,965,440,000 Btu/yr
5. EUI = 3,965,440,000 ÷ 48,000 = 82,613 Btu/ft²-yr ≈ 82.6 kBtu/ft²-yr

38.5 kBtu/ft²-yr: Counted only the natural gas.

44.1 kBtu/ft²-yr: Counted only the electricity.

82.6 kBtu/ft²-yr: Correct. Site EUI = (620,000 × 3,412 + 18,500 × 100,000) Btu ÷ 48,000 ft² = 3,965,440,000 ÷ 48,000 ≈ 82,600 Btu/ft²-yr = 82.6 kBtu/ft²-yr.

167.7 kBtu/ft²-yr: Converted kWh at a 10,000 Btu/kWh power-plant heat rate instead of the site value of 3,412 Btu/kWh.

429.5 kBtu/ft²-yr: Treated therms as dekatherms (1,000,000 Btu each) instead of 100,000 Btu.

Site EUI converts every fuel to Btu at site values (3,412 Btu/kWh, 100,000 Btu/therm), sums them and divides by floor area.

Answer: E

Sling psychrometer: A psychrometer measures air dry-bulb and wet-bulb temperatures for humidity, not trap performance.

Infrared thermometer: The stem already shows that surface temperatures cannot separate flash steam from blow-through on this trap.

Combustion analyzer: A combustion analyzer measures flue gases at the boiler, not trap operation.

Pitot tube and manometer: A pitot tube measures air velocity pressure in ducts, not steam flow through a trap.

Ultrasonic tester: Correct. An ultrasonic tester lets the auditor hear the trap's operating sound: a normal bucket trap cycles open and shut, while a failed trap gives a continuous rushing sound of blow-through.

When temperature readings are ambiguous, use ultrasonic testing to tell a cycling steam trap from one that is blowing through.

Answer: A

1. Givens: investment $200,000; energy savings $28,000/yr; added maintenance $3,000/yr; n = 15; i = 6%; P/A = 9.7122
2. Net annual savings = 28,000 − 3,000 = $25,000
3. PW of net savings = 25,000 × 9.7122 = $242,806
4. SIR = 242,806 ÷ 200,000 = 1.21 (> 1.0, so cost-effective)

1.21: Correct. SIR = PW of net savings ÷ investment = (28,000 − 3,000) × 9.7122 ÷ 200,000 = 242,806 ÷ 200,000 ≈ 1.21.

1.36: Ignored the added maintenance cost (28,000 × 9.7122 ÷ 200,000).

1.88: Ignored discounting ((25,000 × 15) ÷ 200,000).

2.10: Ignored both the maintenance cost and discounting ((28,000 × 15) ÷ 200,000).

SIR = present worth of savings net of operating and maintenance cost changes ÷ present worth of investment; SIR > 1 is cost-effective.

Answer: D

1. Shaft load: 75 hp × 0.746 kW/hp × 0.80 = 44.76 kW.
2. ΔkW = 44.76 × (1/0.910 − 1/0.950) = 44.76 × (1.0989 − 1.0526) = 44.76 × 0.04627 = 2.071 kW.
3. Energy savings: 2.071 kW × 5,200 h = 10,769 kWh × $0.072 = $775.
4. Demand savings: 2.071 kW × $13/kW-month × 12 = $323.
5. Total ≈ $775 + $323 = $1,098 per year.

$775: Counted only the energy savings and omitted 12 months of demand charges.

$802: Applied the demand charge for one month instead of 12.

$950: Multiplied by efficiency instead of dividing (kW = hp × 0.746 × LF × η).

$1,098: Correct. ΔkW = 75 × 0.746 × 0.80 × (1/0.91 − 1/0.95) = 2.07 kW; savings = 2.07 × 5,200 × $0.072 + 2.07 × 12 × $13 ≈ $775 + $323 = $1,098.

$1,373: Ignored the 0.80 load factor and assumed full load.

Motor input kW = hp × 0.746 × LF ÷ η, so savings use the difference of reciprocal efficiencies; include both energy and 12 months of coincident demand.

Answer: D

1. Givens: E = 45 fc; A = 60 × 40 = 2,400 ft²; 6,000 lm per luminaire; CU = 0.65; LLF = 0.85.
2. Lumen method: N = (E × A) ÷ (lumens per luminaire × CU × LLF).
3. N = (45 × 2,400) ÷ (6,000 × 0.65 × 0.85) = 108,000 ÷ 3,315 = 32.6.
4. Round up to 33 luminaires to meet the maintained target.

10: Multiplied by CU and LLF instead of dividing by them.

22: Omitted the coefficient of utilization, assuming all lamp lumens reach the work plane.

28: Omitted the light loss factor, giving initial rather than maintained footcandles.

33: Correct. N = E × A ÷ (lumens × CU × LLF) = 45 × 2,400 ÷ (6,000 × 0.65 × 0.85) = 32.6, rounded up to 33.

The lumen method divides the required work-plane lumens (fc × ft²) by the useful maintained lumens per luminaire (initial lumens × CU × LLF).

Answer: D

1. Given: A = 12,000 ft²; ΔU = 0.25 − 0.08 = 0.17 Btu/h·ft²·°F; 6,200 HDD; η = 0.78; $1.05/therm.
2. ΔQ = ΔU × A × HDD × 24 = 0.17 × 12,000 × 6,200 × 24 = 303,552,000 Btu/yr.
3. Fuel saved = 303,552,000 ÷ 0.78 = 389,169,000 Btu = 3,892 therms.
4. Savings = 3,892 × $1.05 = $4,086/yr.

$170: Omitted the 24 h/day that converts degree-days to degree-hours.

$2,486: Multiplied by the furnace efficiency instead of dividing.

$3,187: Left out the furnace efficiency.

$4,086: Correct. ΔQ = 0.17 × 12,000 × 6,200 × 24 = 303.6 × 10⁶ Btu; ÷ 0.78 = 389.2 × 10⁶ Btu = 3,892 therms × $1.05 = $4,086.

$6,009: Used the existing U-factor (0.25), which is the current wall loss, not the savings.

Seasonal conduction Q = U·A·DD·24; use the change in U for savings and divide by heating-system efficiency to get fuel.

Answer: E

1. Given: 250 tons; COP = 0.68; pump work ignored.
2. Evaporator load = 250 × 12,000 = 3,000,000 Btu/h.
3. Generator input = 3,000,000 ÷ 0.68 = 4,411,765 Btu/h.
4. Energy balance: heat rejected = 3.00 + 4.41 = 7.41 MMBtu/h.

1.41 MMBtu/h: Subtracted the cooling output from the generator input; both heat streams leave through the tower.

3.00 MMBtu/h: This is only the evaporator (cooling) load; it omits the generator heat input.

4.41 MMBtu/h: This is only the generator heat input; it omits the heat absorbed in the evaporator.

5.04 MMBtu/h: Multiplied the cooling load by COP instead of dividing to get the generator input.

7.41 MMBtu/h: Correct. Heat rejected = evaporator load + generator input = 3.00 + 3.00/0.68 = 3.00 + 4.41 = 7.41 MMBtu/h.

For an absorption chiller, heat rejected ≈ cooling load + generator heat; with a COP well below 1, its tower must be much larger than an electric chiller’s.

Answer: C

1. Givens: UA = 8,000 Btu/h·°F; ΔT reduction = 70 − 60 = 10°F; 10 h/night × 150 nights = 1,500 h; η = 0.80; $1.20/therm.
2. Heat loss avoided = UA × ΔT × h = 8,000 × 10 × 1,500 = 120,000,000 Btu/yr.
3. Fuel saved = 120,000,000 Btu ÷ 0.80 = 150,000,000 Btu = 1,500 therms (1 therm = 100,000 Btu).
4. Cost saved = 1,500 therms × $1.20 = $1,800/yr.

$180: Leaves out the 10 hours per night, treating Btu/h as Btu per night.

$1,440: Ignores boiler efficiency; the reduced heat load must be divided by 0.80 to get the fuel saved.

$1,800: Correct. Heat saved = UA × ΔT × hours = 8,000 × 10 × 10 × 150 = 120,000,000 Btu; fuel = 120,000,000 ÷ 0.80 = 1,500 therms × $1.20 = $1,800.

$4,320: Applies the 10°F setback for 24 hours a day instead of the 10 setback hours each night.

Setback savings equal UA × the temperature reduction × setback hours, divided by heating-plant efficiency to convert the avoided load to fuel.

Answer: A

Full storage: Correct. Full (load-shifting) storage moves the entire on-peak cooling load to off-peak chiller operation, so no chiller runs during the peak window.

Partial storage, load leveling: Load leveling runs the chiller at a constant output all day, with the tank covering only the load above that output.

Partial storage, demand limiting: Demand limiting keeps the chiller running on peak at a reduced, capped output; here the chillers are off.

Chiller-priority operation: Chiller priority meets load with the chiller first and uses storage only for the excess, the opposite of this case.

Full storage shifts all on-peak cooling to off-peak; partial storage (load leveling or demand limiting) keeps a chiller running on peak and needs a smaller tank.

Answer: B

1. Givens: 8,000 lb/h; hf(150 psig) = 338 Btu/lb; hf(10 psig) = 208 Btu/lb; hfg(10 psig) = 952 Btu/lb.
2. Flash fraction = (hf,high − hf,low) / hfg,low = (338 − 208) / 952 = 0.1366.
3. Flash steam = 8,000 × 0.1366 = 1,092 lb/h.

897 lb/h: Divided by hg at 10 psig (208 + 952 = 1,160) instead of hfg; the flashed water only needs its latent heat.

1,092 lb/h: Correct. Flash fraction = (338 − 208)/952 = 13.7%; 8,000 × 0.137 = 1,092 lb/h.

1,214 lb/h: Used hfg at the high pressure (857); evaporation happens at the flash-tank pressure, so use hfg at 10 psig.

2,840 lb/h: Divided the full high-pressure hf (338) by hfg; only the enthalpy drop (338 − 208) is available to flash water.

5,000 lb/h: Divided the enthalpy drop by hf at 10 psig instead of hfg at 10 psig.

Percent flash = (hf at the upstream pressure − hf at the flash pressure) ÷ hfg at the flash pressure.

Answer: D

1. Givens: heat rate 11,000 Btu/kWh; recovered heat 5,000 Btu/kWh; boiler η = 0.80; O&M $0.020/kWh; grid $0.110/kWh.
2. Boiler gas displaced = 5,000 ÷ 0.80 = 6,250 Btu per kWh generated.
3. Net CHP fuel = 11,000 − 6,250 = 4,750 Btu/kWh = 0.00475 MMBtu/kWh.
4. Break-even: 0.00475 × P + 0.020 = 0.110 → P = 0.090 ÷ 0.00475 = $18.95/MMBtu.

$8.18/MMBtu: Left out the boiler-fuel credit: 0.090 ÷ 0.011 MMBtu/kWh; recovered heat displaces boiler gas and lowers the net fuel per kWh.

$10.00/MMBtu: Left out both the heat credit and the O&M cost: 0.110 ÷ 0.011 MMBtu/kWh.

$15.00/MMBtu: Credited the 5,000 Btu of recovered heat as 5,000 Btu of gas; at 80% boiler efficiency it displaces 6,250 Btu of gas.

$18.95/MMBtu: Correct. Net fuel = 11,000 − 5,000/0.80 = 4,750 Btu/kWh; break-even when 0.00475 × P + 0.020 = 0.110, so P = 0.090 ÷ 0.00475 = $18.95/MMBtu.

A CHP screen charges the unit its full fuel, credits the boiler fuel its recovered heat displaces (heat ÷ boiler efficiency) and adds O&M; CHP pays while that net cost per kWh is below the avoided grid price.

Answer: D

1. Givens: 40,000 cfm; TP = 6.0 in. w.g.; fan η = 0.70; motor η = 0.93; 1 hp = 0.746 kW.
2. bhp = 40,000 × 6.0 / (6,356 × 0.70) = 240,000 / 4,449 = 53.9 hp.
3. Shaft kW = 53.9 × 0.746 = 40.2 kW.
4. Electrical input = 40.2 / 0.93 = 43.3 kW.

28.2 kW: Used air horsepower (no fan efficiency) and left out motor efficiency.

37.4 kW: Multiplied by motor efficiency instead of dividing by it.

40.2 kW: Converted brake horsepower to kW but left out the 93% motor efficiency.

43.3 kW: Correct. bhp = 40,000 × 6.0 ÷ (6,356 × 0.70) = 53.9 hp; input = 53.9 × 0.746 ÷ 0.93 = 43.3 kW.

53.9 kW: Reported brake horsepower as kW without converting (× 0.746) or dividing by motor efficiency.

Fan input kW = cfm × TP ÷ (6,356 × η_fan) × 0.746 ÷ η_motor; divide by each efficiency in the chain.

Answer: B

1. Givens: stack 360°F → 460°F; fuel 24,000 MMBtu/yr; $7.50/MMBtu; rule 1% fuel per 40°F.
2. Stack-temperature rise = 460 − 360 = 100°F.
3. Fuel penalty = 100 ÷ 40 × 1% = 2.5%.
4. Fuel saved = 24,000 × 0.025 = 600 MMBtu/yr.
5. Cost saved = 600 × $7.50 = $4,500/yr.

$3,690: Applied the 2.5% to the boiler's useful output (24,000 × 0.82) instead of to the fuel burned.

$4,500: Correct. Rise = 460 − 360 = 100°F → 100/40 × 1% = 2.5%; 24,000 MMBtu × 0.025 = 600 MMBtu × $7.50 = $4,500.

$5,488: Divided the fuel savings by the 82% efficiency; the rule already applies to fuel input, so efficiency is not applied again.

$20,700: Divided the full 460°F stack temperature by 40 (11.5%) instead of the 100°F rise above the clean baseline.

A rising net stack temperature at the same firing rate signals fouling; estimate the penalty from the rise above the clean baseline (about 1% per 40°F) and apply it to fuel input.

Answer: C

Option A, key parameter measurement: Option A measures only the key parameter and stipulates the rest; here neither fan power nor hours can be reasonably stipulated.

Option D, calibrated simulation: Option D suits cases with no baseline data or many interacting measures; direct submetering of the fans is simpler and more accurate.

Option B, all parameter measurement: Correct. Option B isolates the retrofit and continuously measures all parameters (fan kW over time), which suits loads and hours that both vary.

Option C, whole facility: Savings of about 4% are too small to separate reliably from normal variation in the whole-building meter; Option C generally needs larger savings.

IPMVP: A = retrofit isolation with key-parameter measurement; B = retrofit isolation measuring all parameters; C = whole-facility meter (savings large relative to use); D = calibrated simulation.