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

Answer: C

Preliminary energy-use analysis only: The preliminary energy-use analysis reviews bills and benchmarks the building but includes no site visit; this scope adds a walk-through.

Level 2 energy survey and analysis: Level 2 requires a detailed survey, an end-use breakdown and savings and cost estimates for each measure — more than the manager asked for.

Level 1 walk-through: Correct. BoK 104/113 — ASHRAE 211 Level 1 adds a brief site walk-through to the bill analysis and identifies no-cost/low-cost measures and capital measures that merit further study.

Level 3 detailed analysis of capital-intensive modifications: Level 3 is investment-grade analysis of capital projects with extensive data collection; far beyond a low-cost first look.

ASHRAE 211 levels build on one another: preliminary analysis (bills only) → Level 1 (adds walk-through, low-cost measures) → Level 2 (detailed survey, end-use breakdown, measure costs and savings) → Level 3 (detailed analysis of capital-intensive measures).

Answer: A

1. At the balance point, heating gas use falls to the base load: 520 − 8T = 40 therms/day
2. 8T = 520 − 40 = 480
3. T = 480 ÷ 8 = 60°F

60°F: Correct. BoK 209 — the heating line meets the 40 therm/day base load where 520 − 8T = 40, so T = 480 ÷ 8 = 60°F.

65°F: Solves 520 − 8T = 0 (zero gas use) and ignores the 40 therm/day base load; it also matches the conventional 65°F degree-day base, which is not this building's balance point.

70°F: Adds the base load instead of subtracting it: (520 + 40) ÷ 8 = 70°F.

72°F: Equates the balance point with the thermostat set point; internal and solar gains make the balance point lower than the set point.

The balance point is the outdoor temperature at which weather-dependent use reaches zero (the change point to base load); it sits below the set point because internal gains supply heat.

Answer: A

1. Adjusted baseline = 42,000 + 310 kWh/CDD × 250 CDD = 42,000 + 77,500 = 119,500 kWh
2. Avoided energy = 119,500 kWh − 101,500 kWh = 18,000 kWh

18,000 kWh: Correct. BoK 307 — adjusted baseline = 42,000 + 310 × 250 = 119,500 kWh; 119,500 − 101,500 = 18,000 kWh.

24,000 kWh: Leaves out the 42,000 kWh intercept (base load) and compares 77,500 kWh with the metered use.

42,000 kWh: Reports the intercept (weather-independent base load) as the savings.

119,500 kWh: Reports the weather-adjusted baseline itself instead of the difference from metered use.

Avoided energy = baseline model evaluated at reporting-period conditions − metered use; keep both the intercept and the slope term.

Answer: A

1. Cooling-season lighting savings = 0.5 × 90,000 kWh = 45,000 kWh
2. Heat no longer added = 45,000 kWh × 3,412 Btu/kWh = 153,540,000 Btu
3. Avoided cooling = 153,540,000 Btu ÷ 12,000 Btu/ton-h = 12,795 ton-h
4. Chiller energy saved = 12,795 ton-h × 0.80 kW/ton = 10,236 kWh/yr

10,236 kWh/yr: Correct. BoK 306 — 45,000 kWh × 3,412 Btu/kWh ÷ 12,000 Btu/ton-h = 12,795 ton-h; × 0.80 kW/ton = 10,236 kWh/yr.

12,795 kWh/yr: Stops at ton-hours of avoided cooling and reports them as kWh without applying 0.80 kW/ton.

15,994 kWh/yr: Divides ton-hours by 0.80 kW/ton instead of multiplying (ratio inverted).

20,472 kWh/yr: Applies the cooling effect to all 90,000 kWh instead of the half that occurs in the cooling season.

Lighting savings in conditioned space cut cooling load: kWh × 3,412 ÷ 12,000 gives ton-hours, times kW/ton gives the chiller kWh saved.

Answer: A

1. P/A (6%, 15 yr) = (1 − 1.06⁻¹⁵) ÷ 0.06 = 9.712
2. LCC A = $200,000 + $40,000 × 9.712 = $588,490
3. LCC B = $260,000 + $30,000 × 9.712 = $551,370
4. Difference = $588,490 − $551,370 = $37,120 ≈ $37,100 in favor of Chiller B

$37,100: Correct. BoK 403 — PV of the $10,000/yr operating difference = $10,000 × (1 − 1.06⁻¹⁵) ÷ 0.06 = $97,120; minus the $60,000 extra first cost = $37,100 lower LCC for B.

$60,000: Reports the first-cost difference, which actually favors Chiller A, and ignores operating costs.

$90,000: Uses undiscounted operating savings ($10,000 × 15 = $150,000) minus the $60,000 first-cost premium.

$97,100: Counts the present value of operating savings but forgets to subtract Chiller B's $60,000 higher first cost.

LCC = first cost + present value of recurring costs over the study period; the lower-LCC option is preferred even when its first cost is higher.

Answer: A

1. kW saved = 200 × (62 − 30) W ÷ 1,000 = 6.4 kW
2. kWh saved = 6.4 kW × 4,000 h/yr = 25,600 kWh/yr
3. Energy savings = 25,600 × $0.11 = $2,816/yr
4. Demand savings = 6.4 kW × $14/kW-month × 12 months = $1,075/yr
5. Total = $3,891/yr
6. SPP = $18,000 ÷ $3,891/yr = 4.6 years

4.6 years: Correct. BoK 511 — 6.4 kW saved gives $2,816/yr energy + $1,075/yr demand = $3,891/yr; $18,000 ÷ $3,891 = 4.6 years.

6.2 years: Applies the $14/kW demand charge for one month only instead of 12 months.

6.4 years: Uses energy savings only and omits the $14/kW-month demand savings.

16.7 years: Uses demand savings only and omits the 25,600 kWh/yr energy savings.

Lighting retrofit savings = ΔkW × hours × energy rate + ΔkW × demand rate × months on peak; payback = cost ÷ total annual savings.

Answer: D

1. bhp = 20,000 cfm × 0.5 in. w.g. ÷ (6,356 × 0.65) = 10,000 ÷ 4,131.4 = 2.42 hp
2. Input kW = 2.42 hp × 0.746 kW/hp ÷ 0.90 = 2.006 kW
3. Added energy = 2.006 kW × 5,000 h = 10,032 kWh/yr

6,521 kWh: Omits the 65% fan efficiency, giving air horsepower instead of brake horsepower.

8,126 kWh: Multiplies by the 90% motor efficiency instead of dividing by it.

9,028 kWh: Converts brake hp to kW but omits the motor efficiency, giving motor output instead of electrical input.

10,032 kWh: Correct. BoK 609 — bhp = 20,000 × 0.5 ÷ (6,356 × 0.65) = 2.42 hp; input = 2.42 × 0.746 ÷ 0.90 = 2.006 kW × 5,000 h = 10,032 kWh.

Added fan power = cfm × Δ(in. w.g.) ÷ (6,356 × fan eff) in bhp, then × 0.746 ÷ motor eff for kW input; change filters on pressure drop, not on a calendar alone.

Answer: C

1. Δ efficiency = 0.90 − 0.55 = 0.35 kW/ton
2. Energy saved = 210 tons × 0.35 kW/ton × 2,000 h = 147,000 kWh/yr
3. Energy savings = 147,000 kWh × $0.11/kWh = $16,170/yr
4. Demand reduction = 300 tons × 0.35 kW/ton = 105 kW
5. Demand savings = 105 kW × $14/kW-month × 6 months = $8,820/yr
6. Total = $16,170 + $8,820 = $24,990/yr
7. SPP = $150,000 ÷ $24,990/yr = 6.00 years

4.44 yr: Applies the $14/kW demand savings for 12 months instead of the six cooling months.

4.70 yr: Computes energy savings at the 300-ton full load for all 2,000 hours instead of the 210-ton average load.

6.00 yr: Correct. BoK 612 — energy savings $16,170 plus demand savings $8,820 = $24,990/yr; $150,000 ÷ $24,990 = 6.00 years.

9.28 yr: Counts only the energy savings and omits the demand-charge savings.

Chiller savings = tons × Δ(kW/ton) × hours for energy, plus peak tons × Δ(kW/ton) × $/kW-month × months for demand; payback = cost ÷ total annual savings.

Answer: C

1. Flow reduction = 2.5 − 1.5 = 1.0 gpm
2. Water saved = 250 showers/day × 8 min × 1.0 gpm × 360 days = 720,000 gal/yr
3. Heat saved = 8.34 × 720,000 gal × (105 − 50)°F = 330,264,000 Btu/yr
4. Fuel saved = 330,264,000 ÷ 0.80 = 412,830,000 Btu = 4,128 therms/yr
5. Cost savings = 4,128 therms × $0.90 = $3,715/yr

$2,378: Multiplies the heat saved by the 80% efficiency instead of dividing by it.

$2,972: Leaves out the 80% water heater efficiency, so it counts only the heat delivered to the water.

$3,715: Correct. BoK 705/707 — 720,000 gal/yr saved × 8.34 × 55°F ÷ 0.80 = 4,128 therms × $0.90 = $3,715/yr.

$5,573: Uses the new 1.5 gpm flow instead of the 1.0 gpm reduction, which values the water still used rather than the water saved.

Hot-water savings from low-flow fixtures = gallons saved × 8.34 × ΔT ÷ heater efficiency, then convert to fuel units and cost.

Answer: C

1. Recoverable heat = 90 kW × 0.80 × 3,412 Btu/kWh = 245,664 Btu/h
2. Annual heat delivered = 245,664 Btu/h × 3,000 h = 736,992,000 Btu
3. Gas input avoided = 736,992,000 Btu ÷ 0.80 = 921,240,000 Btu
4. Therms avoided = 921,240,000 ÷ 100,000 = 9,212 therms
5. Savings = 9,212 therms × $1.00/therm = $9,212/yr

$5,896: Multiplies by the furnace efficiency instead of dividing, which understates the fuel avoided.

$7,370: Leaves out the furnace efficiency; each Btu recovered displaces 1 ÷ 0.80 Btu of gas input.

$9,212: Correct. BoK 811 — 90 kW × 0.80 × 3,412 = 245,664 Btu/h; × 3,000 h ÷ 0.80 ÷ 100,000 = 9,212 therms × $1.00 = $9,212/yr.

$11,516: Treats all 90 kW of input as recoverable heat, ignoring the 80% recovery fraction.

Heat recovered from a compressor offsets fuel input = recovered Btu ÷ the displaced heater's efficiency.

Answer: C

1. Baseline = 30 kW × 6,000 h = 180,000 kWh/yr
2. 100%: 30 kW × 1.000 × 2,000 h = 60,000 kWh
3. 80%: 30 kW × 0.8³ = 30 × 0.512 = 15.36 kW × 2,000 h = 30,720 kWh
4. 60%: 30 kW × 0.6³ = 30 × 0.216 = 6.48 kW × 2,000 h = 12,960 kWh
5. VFD total = 60,000 + 30,720 + 12,960 = 103,680 kWh/yr
6. Savings = 180,000 − 103,680 = 76,320 kWh/yr

36,000 kWh: Treats power as proportional to speed (linear), which is the relation for flow, not power.

60,000 kWh: Uses the square of speed; that relation applies to head or pressure, not power.

76,320 kWh: Correct. BoK 806 — power ∝ speed³: 30 kW × (2,000 × 1 + 2,000 × 0.512 + 2,000 × 0.216) = 103,680 kWh; 180,000 − 103,680 = 76,320 kWh saved.

103,680 kWh: This is the energy used with the VFD, not the energy saved.

Affinity laws: flow ∝ speed, head ∝ speed², power ∝ speed³; weight each speed by its hours and compare with the baseline.

Answer: D

1. Annual HDD = 1,000 + 850 + 700 + 400 + 150 + 0 + 0 + 0 + 100 + 400 + 600 + 800 = 5,000 °F·day
2. U existing = 1 ÷ 10 = 0.100; U new = 1 ÷ (10 + 20) = 0.0333 Btu/h·ft²·°F
3. Heat saved = (0.100 − 0.0333) × 10,000 ft² × 5,000 × 24 h/day = 80,000,000 Btu/yr
4. Fuel saved = 80,000,000 ÷ 0.80 = 100,000,000 Btu = 1,000 therms
5. Savings = 1,000 therms × $1.10 = $1,100/yr

$46: Leaves out the 24 h/day factor that converts degree days to degree-hours.

$825: Takes the new U as 1 ÷ 20 (the added insulation alone) instead of 1 ÷ 30 for the whole assembly.

$880: Leaves out the 80% furnace efficiency, so it counts only the heat saved at the roof.

$1,100: Correct. BoK 909 — annual HDD = 5,000; ΔU = 0.100 − 0.0333 = 0.0667; 0.0667 × 10,000 × 5,000 × 24 = 80 MMBtu ÷ 0.80 = 1,000 therms × $1.10 = $1,100/yr.

Annual conduction heat loss = U × A × HDD × 24; to find savings, add R-values first, take U = 1/R, and divide by heating efficiency.

Answer: B

1. 10:00 readings: RAT = 74°F, MAT = 70.4°F, OAT = 56°F
2. OA fraction = (RAT − MAT) ÷ (RAT − OAT) = (74 − 70.4) ÷ (74 − 56) = 3.6 ÷ 18 = 0.20
3. OA flow = 0.20 × 10,000 cfm = 2,000 cfm

486 cfm: Divides (RAT − MAT) by the return temperature itself instead of by (RAT − OAT).

2,000 cfm: Correct. BoK 1005 — OA fraction = (RAT − MAT) ÷ (RAT − OAT) = (74 − 70.4) ÷ (74 − 56) = 3.6 ÷ 18 = 0.20; 0.20 × 10,000 = 2,000 cfm.

2,500 cfm: Divides by (MAT − OAT) instead of (RAT − OAT), which is the ratio of outdoor to return air, not the outdoor fraction.

8,000 cfm: This is the return-air flow; (MAT − OAT) ÷ (RAT − OAT) = 0.80 is the return fraction.

A mixed-air temperature balance gives OA fraction = (RAT − MAT) ÷ (RAT − OAT); a constant 20% with mild outdoor air suggests the economizer is stuck at minimum.

Answer: D

1. Electricity = 500 kW × 7,000 h = 3,500,000 kWh; value = 3,500,000 × $0.12 = $420,000
2. CHP fuel = 5.2 MMBtu/h × 7,000 h = 36,400 MMBtu × $6.00 = $218,400
3. Boiler fuel displaced = 2.4 × 7,000 ÷ 0.80 = 21,000 MMBtu × $6.00 = $126,000
4. O&M = 3,500,000 kWh × $0.02 = $70,000
5. Net = $420,000 − $218,400 + $126,000 − $70,000 = $257,600/yr

$131,600: Leaves out the credit for recovered heat that displaces boiler fuel.

$212,240: Multiplies the recovered heat by the 80% boiler efficiency instead of dividing, undervaluing the heat credit.

$232,400: Credits recovered heat at face value without dividing by the 80% boiler efficiency.

$257,600: Correct. BoK 1106 — $420,000 power value − $218,400 CHP fuel + $126,000 boiler-fuel credit − $70,000 O&M = $257,600.

CHP net savings = displaced grid electricity + displaced boiler fuel (recovered heat ÷ boiler efficiency × fuel price) − CHP fuel − CHP O&M.

Answer: A

1. Annual miles = 2,000 gal × 20.0 mpg = 40,000 mi
2. New fuel use = 40,000 mi ÷ 21.0 mpg = 1,904.8 gal
3. Fuel saved = 2,000 − 1,904.8 = 95.2 gal
4. Savings = 95.2 gal × $3.50/gal = $333/yr

$333: Correct. BoK 1208 — miles = 2,000 × 20 = 40,000; new fuel = 40,000 ÷ 21 = 1,904.8 gal; saved 95.2 gal × $3.50 = $333/yr.

$350: Assumes a 5% mpg gain cuts fuel use by 5% (100 gal); fuel use falls by only 1 − 20/21 = 4.76%.

$368: Divides fuel use by 0.95 (2,000 ÷ 0.95 − 2,000 = 105 gal), applying the percentage the wrong way.

$6,667: Reports the new annual fuel cost (1,904.8 gal × $3.50) instead of the savings.

Fuel use varies with 1/mpg, so a percentage gain in mpg yields a smaller percentage cut in gallons; convert through miles, not by applying the mpg percentage to fuel.