Exercise set

Building Energy Systems and HVAC Performance Exercises

Worked HVAC exercises for loads, ventilation, latent cooling, filters, low delta-T, tower approach, heat-pump balance point and COP.

These exercises practise building energy and HVAC performance calculations from an engineering decision perspective. The aim is to connect each number to a boundary, service requirement, operating mode, measured evidence, and action: resize, retune, repair, recommission, monitor, or reject an energy claim.

Assume simplified steady operation unless an exercise states otherwise. Real building decisions also require climate data, humidity control, indoor-air-quality requirements, pressure relationships, envelope details, equipment curves, control sequences, safety interlocks, maintenance access, sensor calibration, and measured trend data.

Release Evidence Notes

Use the worked answers as commissioning or operating evidence only when the weather period, occupancy schedule, control sequence, sensor calibration, trend interval, equipment state, ventilation basis, humidity condition and utility boundary all match the decision being made. An energy-saving calculation is weak if it does not prove that comfort, pressure, humidity, outdoor-air and indoor-air-quality constraints remain protected.

The strongest release records separate load calculation, control sequence, measured trend, service-quality guard and energy boundary. Each gate should state whether the result supports tuning, repair, rebalance, demand response, energy reporting or rejection, and it should identify the measured evidence needed when a nominal saving conflicts with comfort, ventilation or plant reliability.

How to Use These Exercises

For each calculation:

  1. define the building, zone, plant, air-handling unit, or loop boundary;
  2. state whether the result supports peak load, annual energy, fault diagnosis, commissioning, or demand response;
  3. keep airflow, hydronic flow, heat, power, time, weather, and uncertainty units consistent;
  4. check that energy savings do not reduce comfort, ventilation, humidity control, pressure control, or safety margin;
  5. convert the result into an engineering action.

The most common mistake is treating HVAC energy as a utility number only. A credible building calculation also states the service delivered: indoor conditions, air quality, pressure relationships, schedule, occupancy, and recovery time.

Engineering Boundary Notes

These exercises use simplified building-energy and HVAC performance screens. They do not replace load calculations, psychrometric design, ventilation-code review, control-sequence validation, testing and balancing, commissioning, M&V plans, safety interlock review or indoor-air-quality assessment. A calculated energy or load margin applies only to the stated zone, plant, loop, schedule, weather period, sensor calibration and service requirement.

Separate energy, comfort, ventilation, humidity, pressure and reliability boundaries before release. Fan power, chiller kW per ton, heat-pump COP, economizer savings, delta-T recovery and demand-response reduction can all pass numerically while the building fails comfort, IAQ, laboratory airflow, humidity control, recovery time or plant reliability.

Common Release Mistakes

  • reporting energy savings without proving comfort, outdoor air, pressure and humidity service were maintained;
  • comparing utility meters before weather, occupancy, schedule and service changes are normalized;
  • accepting coil, chiller or AHU performance from one side of a heat balance without sensor calibration and uncertainty;
  • treating low delta-T, filter pressure drop or tower approach as energy issues only instead of fault evidence;
  • using heat-pump COP without auxiliary heat, defrost, demand limit and cold-weather capacity evidence;
  • releasing control changes without trend data, alarm review, rebound check and operator override state.

Scenario Map

ExerciseBuilding or HVAC scenarioDecision supported
1Winter perimeter zone with envelope and infiltration lossesScreen peak heating load and decide whether air leakage matters
2Humid outdoor-air cooling conditionEstimate ventilation load without separating sensible and latent terms
3Mild-weather air-side economizer operationDiagnose whether mixed-air data supports free cooling
4Static-pressure reset on a variable-speed fanEstimate fan-energy saving and service-quality checks
5Hydronic coil commissioning heat balanceDecide whether water-side and air-side measurements close well enough
6Heat-pump retrofit with auxiliary heatChoose the correct energy boundary for COP reporting
7Ventilation heat recovery in heating modeEstimate recovery effectiveness and useful recovered heat
8Demand-response event with rebound energyConfirm net energy reduction and guarded comfort margin
9Laboratory airflow commissioningApply uncertainty before accepting a minimum-flow requirement
10Weather-normalized retrofit M&VReport simple and guarded savings without reducing service quality
11Chiller plant at a measured cooling loadSeparate chiller-only COP from total plant kW per ton
12VAV terminal with simultaneous cooling and reheatQuantify avoidable reheat energy before changing controls
13Humid cooling coil with dehumidificationSeparate sensible load, latent load, condensate rate and coil release margin
14Demand-controlled ventilation during partial occupancyEstimate outdoor-air load reduction while guarding minimum flow and CO2 service evidence
15Low chilled-water delta-T in a hydronic cooling loopQuantify overpumping, pump-power penalty and corrective savings
16Loaded AHU filter with differential-pressure alarmQuantify fan power penalty and guarded replacement threshold
17Heat-pump balance point during cold weatherCheck auxiliary heat and demand-limit risk when heat-pump capacity derates
18Cooling-tower approach degradationQuantify condenser-water temperature penalty, plant kW per ton and annual energy impact

Validation Package Checklist

  • zone, building, plant, AHU, loop, schedule, weather period and utility boundary are documented;
  • airflow, hydronic flow, temperature, humidity, pressure, power and sensor calibration evidence are recorded;
  • comfort, ventilation, IAQ, pressure relationship, humidity control and recovery-time limits are protected;
  • control sequence, setpoints, overrides, alarm state, maintenance condition and trend interval match the calculation;
  • weather normalization, occupancy normalization, uncertainty and rebound effects are included for savings claims;
  • equipment curves, fouling, filter state, low delta-T, tower approach, auxiliary heat and derating are guarded where relevant;
  • final release decision states accept, retune, recommission, repair, monitor, restrict demand response, revise M&V or hold.

Exercise 1: Envelope and Infiltration Heating Load

A perimeter office zone has an envelope conductance:

H_{env}=2.4\ \text{kW/K}

The indoor setpoint is 21^\circ\text{C} and the outdoor design condition is -6^\circ\text{C}. The zone volume is 2400\ \text{m}^3 and infiltration is estimated at 0.35\ \text{h}^{-1}. Use:

\rho=1.2\ \text{kg/m}^3,\quad C_p=1.01\ \text{kJ/(kg K)}

Estimate the envelope load, infiltration sensible load, and total screened heating load.

Solution

Temperature difference:

\Delta T=21-(-6)=27\ \text{K}

Envelope load:

\dot{Q}_{env}=H_{env}\Delta T=2.4(27)=64.8\ \text{kW}

Infiltration flow:

\displaystyle Q_{inf}=\frac{ACH\cdot V}{3600}=\frac{0.35(2400)}{3600}=0.233\ \text{m}^3/\text{s}

Infiltration sensible load:

\dot{Q}_{inf}=\rho C_p Q_{inf}\Delta T
\dot{Q}_{inf}=1.2(1.01)(0.233)(27)=7.62\ \text{kW}

Total screened heating load:

\dot{Q}_{total}=64.8+7.62=72.4\ \text{kW}

Engineering Comment

The infiltration contribution is about 10.5\% of the screened heating load. That is large enough to matter for comfort, peak heat, and energy use, but it should be verified with air-leakage testing or calibrated trend data before equipment capacity is changed.

Plausibility Check

The envelope term dominates:

\displaystyle \frac{64.8}{72.4}=0.895

while infiltration contributes:

\displaystyle \frac{7.62}{72.4}=0.105

or about 10.5 percent of the screened load. That fraction is too large to ignore, but small enough that a wrong air-change estimate would not explain the entire heating requirement by itself.

Exercise 2: Ventilation Cooling Load with Enthalpy

An air-handling unit supplies 1.8\ \text{m}^3/\text{s} of outdoor air during a humid cooling condition. Outdoor-air enthalpy is 72\ \text{kJ/kg} and indoor return-air enthalpy is 50\ \text{kJ/kg}. Use:

\rho=1.18\ \text{kg/m}^3

Estimate the total outdoor-air cooling load.

Solution

Outdoor-air mass flow:

\dot{m}_{OA}=\rho Q_v=1.18(1.8)=2.124\ \text{kg/s}

Total load from enthalpy difference:

\dot{Q}_{OA}=\dot{m}_{OA}(h_{out}-h_{in})
\dot{Q}_{OA}=2.124(72-50)=46.7\ \text{kW}

Engineering Comment

This load includes sensible and latent effects because enthalpy is used. Reducing outdoor air would reduce load, but it may violate ventilation, pressure, or air-quality requirements. The better engineering question is whether the required outdoor air is correct, measured, heat-recovered, and controlled without sensor error.

Plausibility Check

The enthalpy difference is:

72-50=22\ \text{kJ/kg}

and the outdoor-air mass flow is just over 2 kg/s:

1.18(1.8)=2.124\ \text{kg/s}

Multiplying gives about 47 kW, so the result is credible. The load scales directly with outdoor-air flow, which is why flow measurement and ventilation requirement are release-critical.

Exercise 3: Mixed-Air Economizer Diagnosis

During mild weather an air-handling unit has:

T_{RA}=23^\circ\text{C}
T_{OA}=8^\circ\text{C}
T_{mix}=18.5^\circ\text{C}

The supply-air target is 13^\circ\text{C}. Calculate the actual outdoor-air fraction, the outdoor-air fraction required for free cooling to the supply target, and the diagnostic implication.

Solution

Actual outdoor-air fraction:

\displaystyle f_{OA}=\frac{T_{RA}-T_{mix}}{T_{RA}-T_{OA}}
\displaystyle f_{OA}=\frac{23-18.5}{23-8}=\frac{4.5}{15}=0.300

Required fraction:

\displaystyle f_{OA,req}=\frac{T_{RA}-T_{SA,set}}{T_{RA}-T_{OA}}
\displaystyle f_{OA,req}=\frac{23-13}{23-8}=\frac{10}{15}=0.667

Engineering Comment

The unit is using about 30\% outdoor air when about 67\% is needed for free cooling. If humidity, freeze protection, pressure control, and smoke mode do not explain the low fraction, inspect damper position, actuator linkage, sensor placement, and the economizer sequence. The displayed damper command alone is not enough evidence.

Plausibility Check

The actual outdoor-air fraction is less than half of the required fraction:

\displaystyle \frac{0.300}{0.667}=0.45

The fraction gap is:

0.667-0.300=0.367

That is a large diagnostic difference, not a rounding issue. The next step should compare measured damper position, mixed-air temperature, outdoor-air temperature and return-air temperature in the same time window.

Exercise 4: Fan Speed Reset Energy Estimate

A variable-speed supply fan draws 22\ \text{kW} at full speed. Static-pressure reset allows the fan speed to fall to 78\% of full speed for 2100\ \text{h/year}. Use the ideal fan affinity relation:

\displaystyle \frac{P_2}{P_1}=\left(\frac{N_2}{N_1}\right)^3

Estimate the reduced fan power and annual energy reduction during those hours.

Solution

Reduced power:

P_2=22(0.78)^3=10.45\ \text{kW}

Power reduction:

\Delta P=22-10.45=11.55\ \text{kW}

Annual energy reduction:

\Delta E=11.55(2100)=24{,}255\ \text{kWh/year}

Engineering Comment

The affinity-law saving is a screening estimate. The real result depends on minimum ventilation, terminal unit behaviour, duct leakage, filter loading, fan and drive efficiency, and whether zones still meet temperature and pressure requirements. Trend data should confirm service quality as well as energy reduction.

Plausibility Check

The cubic speed relation makes the power fraction:

0.78^3=0.475

so fan power should fall to roughly 48 percent of full power:

\displaystyle \frac{10.45}{22}=0.475

The annual saving is large because the reduced condition lasts 2100 hours. This estimate should be rejected if static pressure reset causes low-flow complaints, pressure failures or simultaneous reheat.

Exercise 5: Hydronic Coil Duty and Heat-Balance Mismatch

A heating coil is checked during commissioning. Water-side measurements show:

\dot{m}_w=1.15\ \text{kg/s}
C_{p,w}=4.18\ \text{kJ/(kg K)}
T_{w,in}=58^\circ\text{C},\quad T_{w,out}=46^\circ\text{C}

Air-side measurements estimate:

\dot{Q}_{air}=53\ \text{kW}

Calculate water-side heat duty and the normalized heat-balance mismatch.

Solution

Water-side duty:

\dot{Q}_w=\dot{m}_w C_p(T_{w,in}-T_{w,out})
\dot{Q}_w=1.15(4.18)(58-46)=57.7\ \text{kW}

Normalized mismatch:

\displaystyle \epsilon_Q=\frac{\dot{Q}_w-\dot{Q}_{air}}{(\dot{Q}_w+\dot{Q}_{air})/2}
\displaystyle \epsilon_Q=\frac{57.7-53.0}{(57.7+53.0)/2}=0.0849

The mismatch is about:

8.5\%

Engineering Comment

An 8.5 percent mismatch may be reasonable for field measurements, but it is large enough to review sensor calibration, air mixing, water flow measurement, coil bypass, and condensate or heat-loss assumptions. Commissioning should define an acceptable mismatch before using the result as acceptance evidence.

Plausibility Check

The water temperature drop is:

58-46=12\ \text{K}

and the water-side heat duty is about 4.7 kW higher than the air-side estimate:

57.7-53.0=4.7\ \text{kW}

Normalizing by the average heat rate gives 8.5 percent. That is a commissioning-quality question, not a design-load question.

Exercise 6: Heat-Pump System COP Boundary

A heat pump delivers 120\ \text{kW} of useful heating. The compressor draws 31\ \text{kW}, fans draw 5\ \text{kW}, and pumps draw 4\ \text{kW}. During cold weather, electric auxiliary heat adds 18\ \text{kW} of heat and draws 18\ \text{kW}.

Calculate compressor-only COP, system COP excluding auxiliary heat, and total heating system COP including auxiliary heat.

Solution

Compressor-only COP:

\displaystyle COP_{comp}=\frac{120}{31}=3.87

System COP excluding auxiliary heat:

\displaystyle COP_{sys,no\ aux}=\frac{120}{31+5+4}=\frac{120}{40}=3.00

Total useful heat:

\dot{Q}_{total}=120+18=138\ \text{kW}

Total electrical input:

P_{total}=31+5+4+18=58\ \text{kW}

Total system COP:

\displaystyle COP_{total}=\frac{138}{58}=2.38

Engineering Comment

The compressor-only value overstates building performance because fans, pumps, and auxiliary heat consume electricity. The total boundary is the correct value for building energy analysis. A heat-pump retrofit should also check supply temperature, defrost, part-load cycling, and whether envelope improvements could reduce auxiliary heat.

Plausibility Check

Adding fans and pumps lowers COP from:

3.87

to:

3.00

Adding auxiliary resistance heat lowers the total boundary again:

\displaystyle \frac{2.38}{3.00}=0.79

so the total system COP is about 21 percent lower than the heat-pump-only system boundary. A performance claim should state which boundary it uses.

Exercise 7: Heat Recovery Ventilation Benefit

A ventilation heat recovery unit handles 2.2\ \text{m}^3/\text{s} of outdoor air. Outdoor air is 2^\circ\text{C}, return air is 22^\circ\text{C}, and supply air after heat recovery is 14^\circ\text{C}. Use:

\rho=1.2\ \text{kg/m}^3,\quad C_p=1.01\ \text{kJ/(kg K)}

Calculate heating effectiveness and recovered sensible heat.

Solution

Heating effectiveness:

\displaystyle \epsilon_{HR}=\frac{T_{supply,after}-T_{outdoor}}{T_{return}-T_{outdoor}}
\displaystyle \epsilon_{HR}=\frac{14-2}{22-2}=0.60

Recovered heat:

\dot{Q}_{HR}=\rho C_p Q_v(T_{supply,after}-T_{outdoor})
\dot{Q}_{HR}=1.2(1.01)(2.2)(14-2)=32.0\ \text{kW}

Engineering Comment

The unit recovers about 32\ \text{kW} under this condition. The review should also check added fan power, frost protection, cross-leakage, bypass control, filter loading, maintenance access, and whether recovery is active during the hours that matter for heating energy.

Plausibility Check

The supply air after recovery has gained:

14-2=12\ \text{K}

out of a possible:

22-2=20\ \text{K}

so the effectiveness is 60 percent. With about 2.64 kg/s of air, a 12 K sensible rise gives about 32 kW, matching the heat-recovery estimate.

Exercise 8: Demand-Response Net Reduction and Comfort Margin

A building reduces HVAC electrical demand by 280\ \text{kW} for 2\ \text{h} during a demand-response event. Recovery adds 90\ \text{kW} for 3\ \text{h}. The zone temperature limit is 26.0^\circ\text{C}, measured peak zone temperature is 25.4^\circ\text{C}, and expanded sensor uncertainty is 0.3\ \text{K}.

Find net electrical energy reduction and guarded temperature margin.

Solution

Event reduction:

E_{event}=280(2)=560\ \text{kWh}

Recovery energy:

E_{recovery}=90(3)=270\ \text{kWh}

Net energy reduction:

E_{net}=560-270=290\ \text{kWh}

Guarded temperature margin:

M_T=T_{limit}-(T_{measured}+U_T)
M_T=26.0-(25.4+0.3)=0.3\ \text{K}

Engineering Comment

The event achieved a net energy reduction and remained inside the guarded temperature limit, but the margin is small. Demand-response acceptance should also check humidity, ventilation, complaints, critical zones, and recovery time. A building can reduce peak power and still fail service if recovery or comfort is not controlled.

Plausibility Check

Recovery energy consumes almost half of the event reduction:

\displaystyle \frac{270}{560}=0.48

leaving:

290\ \text{kWh}

net. The guarded temperature margin is only 0.3 K, the same size as the expanded uncertainty. That means one additional sensor or normalization error could change the acceptance decision.

Exercise 9: Guarded Airflow Commissioning Check

A laboratory support space requires at least:

Q_{req}=1.55\ \text{m}^3/\text{s}

The measured supply airflow is:

Q_{meas}=1.63\ \text{m}^3/\text{s}

Expanded measurement uncertainty is:

U_Q=0.06\ \text{m}^3/\text{s}

Use a guarded rule:

M_Q=Q_{meas}-U_Q-Q_{req}

Determine whether the airflow passes.

Solution

Guarded margin:

M_Q=1.63-0.06-1.55=0.02\ \text{m}^3/\text{s}

Since:

M_Q\geq 0

the airflow passes under the selected guarded decision rule.

Engineering Comment

The margin is narrow. The commissioning record should preserve the measurement method, instrument calibration, damper positions, filter state, fan speed, door positions, and pressure relationships. A later filter change, control reset, or balancing change could remove the margin.

Plausibility Check

The unguarded measurement is:

1.63-1.55=0.08\ \text{m}^3/\text{s}

above the requirement. After uncertainty is subtracted, the accepted margin is only:

0.02\ \text{m}^3/\text{s}

or about:

\displaystyle \frac{0.02}{1.55}=1.3\%

of the required airflow. This passes the rule, but it is not a robust margin.

Exercise 10: Normalized Baseline Energy Saving

A building retrofit is evaluated over two comparable weather-normalized periods. The normalized baseline electricity use for HVAC is:

E_{baseline}=186{,}000\ \text{kWh}

Measured post-retrofit HVAC electricity use is:

E_{measured}=159{,}500\ \text{kWh}

Estimated expanded uncertainty in the savings estimate is:

U_E=4{,}800\ \text{kWh}

Calculate simple savings and guarded savings.

Solution

Simple savings:

E_{savings}=E_{baseline}-E_{measured}
E_{savings}=186{,}000-159{,}500=26{,}500\ \text{kWh}

Guarded savings:

E_{savings,guarded}=E_{baseline}-E_{measured}-U_E
E_{savings,guarded}=26{,}500-4{,}800=21{,}700\ \text{kWh}

Engineering Comment

The retrofit still shows positive guarded savings. The result is credible only if the normalization preserved the same service: occupancy, operating hours, indoor conditions, ventilation, humidity control, and maintenance state. Apparent savings can come from milder use, reduced service, sensor drift, or unreported overrides.

Plausibility Check

Simple savings are:

\displaystyle \frac{26{,}500}{186{,}000}=14.2\%

of the normalized baseline. Guarded savings are:

\displaystyle \frac{21{,}700}{186{,}000}=11.7\%

The uncertainty consumes about 18 percent of the simple savings, so the result remains positive but should be reported with the guarded value and the service-normalization assumptions.

Exercise 11: Chiller Plant COP and kW per Ton

A chilled-water plant is measured at a cooling load of:

\dot{Q}_{cool}=620\ \text{kW}

The chiller compressor draws:

P_{chiller}=160\ \text{kW}

The chilled-water pumps draw 18\ \text{kW}, condenser-water pumps draw 14\ \text{kW}, and cooling-tower fans draw 12\ \text{kW}.

Calculate chiller-only COP, total plant COP, cooling tons and total plant \text{kW/ton}.

Solution

Chiller-only COP:

\displaystyle COP_{chiller}=\frac{620}{160}=3.88

Total plant input:

P_{plant}=160+18+14+12=204\ \text{kW}

Total plant COP:

\displaystyle COP_{plant}=\frac{620}{204}=3.04

Cooling tons:

\displaystyle RT=\frac{620}{3.517}=176.3\ \text{tons}

Total plant power per ton:

\displaystyle \frac{\text{kW}}{\text{ton}}=\frac{204}{176.3}=1.16\ \text{kW/ton}

Engineering Comment

Chiller-only COP is useful for equipment diagnosis, but plant COP and kW per ton are better for building energy performance because pumps and tower fans are part of the delivered cooling service. A commissioning review should also check chilled-water temperature reset, condenser-water temperature, tower approach, pump differential pressure, flow balance, part-load staging, sensor calibration and whether the measured load is representative.

Plausibility Check

The auxiliaries add:

18+14+12=44\ \text{kW}

to the chiller input, so plant COP must be lower than chiller-only COP. The 620\ \text{kW} cooling load is about 176 tons, and 204\ \text{kW} over that load gives a value a little above 1\ \text{kW/ton}, which is consistent with the calculated 1.16\ \text{kW/ton}.

Exercise 12: Simultaneous Cooling and Reheat Penalty

A VAV terminal receives supply air at:

T_{SA}=13^\circ\text{C}

During a lightly loaded period, the terminal reheats the air to:

T_{discharge}=18^\circ\text{C}

The airflow is:

Q_v=1.4\ \text{m}^3/\text{s}

Use:

\rho=1.2\ \text{kg/m}^3,\quad C_p=1.01\ \text{kJ/(kg K)}

The condition lasts 1800\ \text{h/year} and the hot-water plant efficiency is 0.86. Estimate reheat load, annual delivered reheat energy and fuel input.

Solution

Temperature rise across the reheat coil:

\Delta T=18-13=5\ \text{K}

Air mass flow:

\dot{m}=\rho Q_v=1.2(1.4)=1.68\ \text{kg/s}

Reheat load:

\dot{Q}_{reheat}=\dot{m}C_p\Delta T
\dot{Q}_{reheat}=1.68(1.01)(5)=8.484\ \text{kW}

Annual delivered reheat energy:

E_{reheat}=8.484(1800)=15{,}271\ \text{kWh/year}

Fuel input to the hot-water plant:

\displaystyle E_{fuel}=\frac{15{,}271}{0.86}=17{,}757\ \text{kWh/year}

Engineering Comment

Reheat may be necessary for humidity control, minimum ventilation, pressurization or laboratory safety, but it can also indicate a control fault. The response should check minimum airflow settings, supply-air temperature reset, zone load, simultaneous heating and cooling, humidity constraints, pressure requirements, occupancy schedule and whether reducing reheat would create comfort or air-quality failures.

Plausibility Check

The airflow is 1.68\ \text{kg/s} and the reheat rise is only 5\ \text{K}, so an 8.5\ \text{kW} load is plausible. Running that condition for 1800 hours creates about 15.3\ \text{MWh} of delivered heat. Dividing by 0.86 raises the required fuel input because the hot-water plant is not perfectly efficient.

Exercise 13: Latent Cooling Load and Condensate Rate

A cooling coil is commissioned during a humid operating condition. Mixed air entering the coil is:

T_{in}=27^\circ\text{C}

with humidity ratio:

\omega_{in}=0.014\ \text{kg water/kg dry air}

Leaving air is:

T_{out}=13^\circ\text{C}

with humidity ratio:

\omega_{out}=0.009\ \text{kg water/kg dry air}

Airflow is:

Q_v=2.4\ \text{m}^3/\text{s}

Use:

\rho=1.18\ \text{kg/m}^3,\quad C_p=1.02\ \text{kJ/(kg K)},\quad h_{fg}=2450\ \text{kJ/kg}

The coil schedule lists total cooling capacity:

\dot{Q}_{coil}=70\ \text{kW}

and the condensate drain acceptance limit is:

\dot{m}_{cond,limit}=55\ \text{kg/h}

Estimate sensible cooling load, moisture removal rate, latent load, total load, sensible heat ratio and release decision.

Solution

Approximate dry-air mass flow:

\dot{m}_{air}=\rho Q_v=1.18(2.4)=2.832\ \text{kg/s}

Sensible cooling load:

\dot{Q}_{sens}=\dot{m}_{air}C_p(T_{in}-T_{out})
\dot{Q}_{sens}=2.832(1.02)(27-13)=40.4\ \text{kW}

Moisture removed per second:

\dot{m}_{cond}=\dot{m}_{air}(\omega_{in}-\omega_{out})
\dot{m}_{cond}=2.832(0.014-0.009)=0.01416\ \text{kg/s}

Convert condensate rate to hourly units:

\dot{m}_{cond,h}=0.01416(3600)=51.0\ \text{kg/h}

Latent cooling load:

\dot{Q}_{lat}=\dot{m}_{cond}h_{fg}
\dot{Q}_{lat}=0.01416(2450)=34.7\ \text{kW}

Total cooling load:

\dot{Q}_{total}=40.4+34.7=75.1\ \text{kW}

Sensible heat ratio:

\displaystyle SHR=\frac{\dot{Q}_{sens}}{\dot{Q}_{total}}
\displaystyle SHR=\frac{40.4}{75.1}=0.538

Coil capacity margin:

M_{coil}=70.0-75.1=-5.1\ \text{kW}

Condensate drain margin:

M_{cond}=55.0-51.0=4.0\ \text{kg/h}

The condensate drain passes this simplified screen, but the coil capacity does not pass the humid-condition load.

Engineering Comment

The latent load is almost as large as the sensible load, so a dry-bulb-only cooling check would miss the problem. The engineering response should review entering-air humidity, outdoor-air fraction, coil leaving-air target, chilled-water temperature, coil cleanliness, face velocity, drain slope and whether the scheduled capacity was selected for this humidity condition. Reducing outdoor air or dehumidification load must not violate ventilation, pressure or indoor-air-quality requirements.

Plausibility Check

The coil removes 0.005\ \text{kg water/kg dry air} from about 2.8\ \text{kg/s} of air, so a condensate rate near 51\ \text{kg/h} is credible. The sensible heat ratio is only 0.538, meaning nearly half the total cooling is latent. That explains why a 70\ \text{kW} coil can fail even though the sensible load is only about 40\ \text{kW}.

Exercise 14: Demand-Controlled Ventilation Energy and IAQ Guard

A conference zone normally operates with design outdoor-air flow:

Q_{OA,base}=0.62\ \text{m}^3/\text{s}

During partial occupancy, demand-controlled ventilation reduces measured outdoor-air flow to:

Q_{OA,DCV}=0.36\ \text{m}^3/\text{s}

The occupied minimum ventilation requirement for the current schedule is:

Q_{OA,min}=0.30\ \text{m}^3/\text{s}

Expanded airflow measurement uncertainty is:

U_Q=0.04\ \text{m}^3/\text{s}

Outdoor-air enthalpy is h_{out}=68\ \text{kJ/kg}, return-air enthalpy is h_{return}=50\ \text{kJ/kg}, and air density is \rho=1.18\ \text{kg/m}^3. The condition lasts 1100\ \text{h/year} and the cooling plant COP is 3.1.

The zone CO2 trend shows C_{meas}=880\ \text{ppm} with expanded sensor uncertainty U_C=70\ \text{ppm}. The service limit is C_{limit}=1000\ \text{ppm}.

Estimate outdoor-air cooling-load reduction, annual cooling-electricity saving, guarded ventilation margin and guarded CO2 margin.

Solution

Outdoor-air flow reduction:

\Delta Q_{OA}=Q_{OA,base}-Q_{OA,DCV}
\Delta Q_{OA}=0.62-0.36=0.26\ \text{m}^3/\text{s}

Mass-flow reduction:

\Delta\dot{m}_{OA}=\rho\Delta Q_{OA}
\Delta\dot{m}_{OA}=1.18(0.26)=0.3068\ \text{kg/s}

Enthalpy difference:

\Delta h=h_{out}-h_{return}=68-50=18\ \text{kJ/kg}

Cooling-load reduction:

\Delta\dot{Q}_{cool}=\Delta\dot{m}_{OA}\Delta h
\Delta\dot{Q}_{cool}=0.3068(18)=5.52\ \text{kW}

Cooling-electric demand reduction:

\displaystyle \Delta P_{elec}=\frac{\Delta\dot{Q}_{cool}}{COP}
\displaystyle \Delta P_{elec}=\frac{5.52}{3.1}=1.78\ \text{kW}

Annual cooling-electricity saving:

\Delta E\approx1.78(1100)=1960\ \text{kWh/year}

Guarded ventilation flow:

Q_{guard}=Q_{OA,DCV}-U_Q
Q_{guard}=0.36-0.04=0.32\ \text{m}^3/\text{s}

Guarded ventilation margin:

M_Q=Q_{guard}-Q_{OA,min}=0.32-0.30=0.02\ \text{m}^3/\text{s}

Guarded CO2 margin:

M_C=C_{limit}-(C_{meas}+U_C)
M_C=1000-(880+70)=50\ \text{ppm}

The DCV sequence produces a cooling-energy reduction while passing the selected guarded ventilation and CO2 checks.

Engineering Comment

The result is acceptable only because service evidence is preserved. The guarded flow margin is narrow, so the commissioning record should verify sensor calibration, CO2 sensor placement, minimum-damper position, occupancy schedule, pressure relationships, economizer lockouts, trend interval and alarm response. If the CO2 margin or guarded outdoor-air margin disappears, the energy saving is not a valid improvement.

Plausibility Check

The outdoor-air reduction is 0.26\ \text{m}^3/\text{s}, less than half the original design flow, so a single-digit kilowatt load reduction is plausible. The annual electricity saving is modest because the cooling plant COP divides the thermal reduction. The 0.02\ \text{m}^3/\text{s} guarded ventilation margin is small, which matches the need for trend-based commissioning evidence.

Exercise 15: Low Chilled-Water Delta-T and Pumping Penalty

A building chilled-water loop is carrying a measured cooling load of:

\dot{Q}_{cool}=900\ \text{kW}

The design chilled-water temperature rise is:

\Delta T_{design}=6.0\ \text{K}

but trend data show a measured operating temperature rise of only:

\Delta T_{low}=3.8\ \text{K}

Use:

C_p=4.18\ \text{kJ/(kg K)}

Assume water density is close enough to 1000\ \text{kg/m}^3 that 1\ \text{kg/s} is approximately 1\ \text{L/s}. At design flow, the secondary pump draws:

P_{pump,design}=18\ \text{kW}

Use the screening pump-affinity estimate:

\displaystyle \frac{P_{pump,low}}{P_{pump,design}}\approx\left(\frac{\dot{m}_{low}}{\dot{m}_{design}}\right)^3

The condition occurs for 2200\ \text{h/year}. Estimate design flow, low-delta-T flow, flow ratio, low-delta-T pump power and annual extra pump energy. Then estimate the pump energy if corrective work raises the chilled-water temperature rise to:

\Delta T_{corrected}=5.5\ \text{K}

Solution

For a hydronic cooling loop:

\dot{Q}=\dot{m}C_p\Delta T

Design mass flow is:

\displaystyle \dot{m}_{design}=\frac{\dot{Q}_{cool}}{C_p\Delta T_{design}}
\displaystyle \dot{m}_{design}=\frac{900}{4.18(6.0)}=35.9\ \text{kg/s}

This is approximately:

35.9\ \text{L/s}

Low-delta-T mass flow is:

\displaystyle \dot{m}_{low}=\frac{900}{4.18(3.8)}=56.7\ \text{kg/s}

This is approximately:

56.7\ \text{L/s}

Flow ratio:

\displaystyle R_Q=\frac{\dot{m}_{low}}{\dot{m}_{design}}
\displaystyle R_Q=\frac{56.7}{35.9}=1.58

Low-delta-T pump power estimate:

P_{pump,low}=18(1.58)^3
P_{pump,low}=70.9\ \text{kW}

Extra pump power:

\Delta P_{pump}=70.9-18.0=52.9\ \text{kW}

Annual extra pump energy:

\Delta E_{pump}=52.9(2200)=116{,}000\ \text{kWh/year}

If corrective work raises the temperature rise to 5.5\ \text{K}:

\displaystyle \dot{m}_{corrected}=\frac{900}{4.18(5.5)}=39.1\ \text{kg/s}

Corrected flow ratio relative to design:

\displaystyle R_{Q,corrected}=\frac{39.1}{35.9}=1.09

Corrected pump power estimate:

P_{pump,corrected}=18(1.09)^3=23.4\ \text{kW}

Annual pump-energy reduction from the low-delta-T condition to the corrected condition is:

\Delta E_{saved}=(70.9-23.4)(2200)=104{,}000\ \text{kWh/year}

Engineering Comment

Low chilled-water delta-T means the same cooling load is carried with too much water flow. The symptom can come from coil fouling, control-valve leakage, three-way bypasses, incorrect differential-pressure reset, poor valve authority, low supply-air temperature strategy, sensor error or simultaneous load diversity. The calculation does not identify the root cause by itself, but it quantifies why the plant should not accept the condition as harmless.

The cubic pump-power estimate is a screen, not a certified measurement. Real pump power depends on the pump curve, control head, bypass flow, valve positions, minimum-flow logic, differential-pressure sensor location and pump efficiency. The commissioning response should compare this estimate with trend data for flow, differential pressure, pump kW, valve position, supply temperature and return temperature.

Plausibility Check

The cooling load is unchanged, so flow must rise when temperature rise falls. Dropping from 6.0\ \text{K} to 3.8\ \text{K} gives a flow ratio of:

\displaystyle \frac{6.0}{3.8}=1.58

which matches the calculated overpumping ratio. Because pump power scales roughly with the cube of similar flow changes, a 58 percent flow increase can create an approximately fourfold pump-power estimate. That explains why correcting delta-T from 3.8\ \text{K} to 5.5\ \text{K} produces a large pump-energy reduction even though the cooling load stays at 900\ \text{kW}.

Exercise 16: Dirty-Filter Pressure Drop and Fan Power Penalty

An air-handling unit operates at an occupied supply airflow of:

Q_v=4.0\ \text{m}^3/\text{s}

The clean-filter differential pressure is:

\Delta p_{clean}=120\ \text{Pa}

Trend data show a loaded-filter differential pressure of:

\Delta p_{loaded}=260\ \text{Pa}

Use a total fan and drive efficiency of:

\eta_{fan}=0.58

The unit operates in this occupied mode for:

t_{occ}=3200\ \text{h/year}

The maintenance rule requires replacement or investigation when the guarded filter pressure drop exceeds:

\Delta p_{limit}=250\ \text{Pa}

The expanded uncertainty of the differential-pressure measurement is:

U_{\Delta p}=15\ \text{Pa}

Estimate the fan-power component associated with the clean filter, the loaded filter, the extra fan power, annual extra fan energy, and the guarded maintenance decision.

Solution

Filter pressure-drop power at the clean condition:

\displaystyle P_{filter,clean}=\frac{Q_v\Delta p_{clean}}{\eta_{fan}}
\displaystyle P_{filter,clean}=\frac{4.0(120)}{0.58}=828\ \text{W}=0.828\ \text{kW}

Filter pressure-drop power at the loaded condition:

\displaystyle P_{filter,loaded}=\frac{Q_v\Delta p_{loaded}}{\eta_{fan}}
\displaystyle P_{filter,loaded}=\frac{4.0(260)}{0.58}=1793\ \text{W}=1.79\ \text{kW}

Extra fan power associated with the loaded filter:

\Delta P_{filter}=P_{filter,loaded}-P_{filter,clean}
\Delta P_{filter}=1.79-0.828=0.966\ \text{kW}

Annual extra fan energy:

\Delta E_{fan}=\Delta P_{filter}t_{occ}
\Delta E_{fan}=0.966(3200)=3090\ \text{kWh/year}

Guarded filter pressure drop:

\Delta p_{guard}=\Delta p_{loaded}+U_{\Delta p}
\Delta p_{guard}=260+15=275\ \text{Pa}

Guarded margin against the maintenance limit:

M_{\Delta p}=\Delta p_{limit}-\Delta p_{guard}
M_{\Delta p}=250-275=-25\ \text{Pa}

The loaded filter fails the guarded maintenance screen. If replacement returns the filter pressure drop to the clean value at the same airflow, the avoidable fan-energy penalty is about 3090\ \text{kWh/year} for this occupied mode.

Engineering Comment

The calculated energy penalty is only the filter component of fan work, not total AHU power. A dirty filter can also reduce airflow, increase static-pressure setpoint demand, push VAV boxes open, increase noise, mask coil fouling, or trigger comfort and ventilation failures. The response should check airflow trend, fan speed, static pressure, filter bank bypass, sensor tubing, alarm threshold, replacement interval and whether the same airflow was actually maintained.

Plausibility Check

The loaded filter pressure drop is:

260-120=140\ \text{Pa}

above the clean condition. At 4.0\ \text{m}^3/\text{s}, that extra pressure drop is roughly:

4.0(140)=560\ \text{W}

of air power before efficiency. Dividing by 0.58 gives just under 1.0\ \text{kW} of extra fan input, so an annual penalty near 3.1\ \text{MWh} over 3200 hours is plausible. The guarded pressure drop is also above the 250\ \text{Pa} limit, so the maintenance decision is not based on energy alone.

Exercise 17: Heat-Pump Balance Point and Auxiliary-Heat Demand Gate

A building heat-pump retrofit is reviewed for a cold-weather operating point. The indoor heating setpoint is:

T_i=21^\circ\text{C}

The outdoor air temperature is:

T_o=-4^\circ\text{C}

The building heat-loss coefficient is:

H=3.2\ \text{kW/K}

Internal gains during occupied operation are:

Q_{int}=14\ \text{kW}

Use the building load model:

Q_{load}=H(T_i-T_o)-Q_{int}

The heat-pump heating-capacity curve is approximated by:

Q_{HP}=68-1.2(7-T_o)

where Q_{HP} is in kW and T_o is in degrees Celsius. At the reviewed condition, the heat-pump COP is:

COP_{HP}=2.3

Any heating load above heat-pump capacity is supplied by electric auxiliary heat. The demand-response limit for heating equipment during this period is:

P_{limit}=34\ \text{kW}

Calculate building heating load, available heat-pump capacity, auxiliary heat, total electric power and demand-limit margin. Then find the outdoor balance point where heat-pump capacity equals building load. Finally, check whether a verified envelope and control action that reduces load by:

5\ \text{kW}

would pass the demand limit at the same outdoor condition.

Solution

Building heating load at -4^\circ\text{C}:

Q_{load}=3.2(21-(-4))-14
Q_{load}=3.2(25)-14=66.0\ \text{kW}

Heat-pump capacity at the same outdoor temperature:

Q_{HP}=68-1.2(7-(-4))
Q_{HP}=68-1.2(11)=54.8\ \text{kW}

Auxiliary electric heat required:

Q_{aux}=Q_{load}-Q_{HP}
Q_{aux}=66.0-54.8=11.2\ \text{kW}

Heat-pump electric input:

\displaystyle P_{HP}=\frac{Q_{HP}}{COP_{HP}}=\frac{54.8}{2.3}=23.8\ \text{kW}

Auxiliary heat has a resistance-heating COP of about 1 for this screen, so:

P_{aux}=11.2\ \text{kW}

Total heating electric demand:

P_{total}=23.8+11.2=35.0\ \text{kW}

Demand-limit margin:

M_P=P_{limit}-P_{total}=34.0-35.0=-1.0\ \text{kW}

The reviewed cold-weather condition fails the demand-response power limit.

To find the balance point, set building load equal to heat-pump capacity:

3.2(21-T_o)-14=68-1.2(7-T_o)

Expand both sides:

53.2-3.2T_o=59.6+1.2T_o

Solve:

-6.4=4.4T_o
T_{balance}=-1.45^\circ\text{C}

Because:

-4^\circ\text{C}<-1.45^\circ\text{C}

the reviewed condition is below the heat-pump balance point, so auxiliary heat is expected.

With a verified 5\ \text{kW} load reduction:

Q_{load,red}=66.0-5.0=61.0\ \text{kW}

Reduced auxiliary heat:

Q_{aux,red}=61.0-54.8=6.2\ \text{kW}

Reduced total electric demand:

P_{total,red}=23.8+6.2=30.0\ \text{kW}

Reduced-load demand margin:

M_{P,red}=34.0-30.0=4.0\ \text{kW}

The verified load-reduction action passes the demand limit at the same outdoor condition.

Engineering Comment

The balance point is where a heat pump stops covering the full building load without auxiliary heat. It depends on the building load curve, internal gains, actual delivered capacity, defrost, supply temperature, airflow, water temperature, controls and sensor calibration. A release decision should verify that auxiliary heat staging, demand-response limits, comfort recovery and freeze protection all work together during the same cold-weather period.

Plausibility Check

At -4^\circ\text{C}, the building load is 66.0\ \text{kW} while the heat pump can deliver 54.8\ \text{kW}, so about 11\ \text{kW} of auxiliary heat is plausible. The balance point of about -1.45^\circ\text{C} is warmer than the reviewed condition, so auxiliary heat should appear. Reducing load by 5\ \text{kW} lowers auxiliary heat one-for-one and moves total electric demand below the 34\ \text{kW} limit.

Exercise 18: Cooling-Tower Approach and Chiller-Lift Penalty

A water-cooled chiller plant is measured during a warm operating period. The cooling load is:

\dot Q_{cool}=700\ \text{kW}

The outdoor wet-bulb temperature is:

T_{wb}=24^\circ\text{C}

The target condenser-water supply temperature is:

T_{cws,target}=29^\circ\text{C}

The measured condenser-water supply temperature is:

T_{cws,meas}=33^\circ\text{C}

At the target condenser-water condition, the chiller input power is expected to be:

P_{ch,target}=155\ \text{kW}

For this operating range, estimate chiller input-power penalty as:

2.4\% \text{ per K}

of condenser-water supply temperature above target. The target tower fan power is:

P_{tower,target}=12\ \text{kW}

The measured tower fan power is:

P_{tower,meas}=18\ \text{kW}

Chilled-water pump power is:

P_{chw}=16\ \text{kW}

and condenser-water pump power is:

P_{cw}=14\ \text{kW}

The operating mode runs for:

t=1800\ \text{h/year}

Calculate tower approach, chiller power penalty, total plant kW per ton and annual extra energy. Then check a corrective action that restores condenser-water supply to:

T_{cws,corr}=30^\circ\text{C}

and tower fan power to:

P_{tower,corr}=13\ \text{kW}

Solution

Target tower approach is:

A_{target}=T_{cws,target}-T_{wb}
A_{target}=29-24=5\ \text{K}

Measured tower approach is:

A_{meas}=T_{cws,meas}-T_{wb}
A_{meas}=33-24=9\ \text{K}

Approach degradation:

\Delta A=9-5=4\ \text{K}

The measured condenser-water supply temperature is 4\ \text{K} above target. Chiller input-power increase is:

\Delta f_{ch}=0.024(4)=0.096

or 9.6\%.

Measured chiller power estimate:

P_{ch,meas}=P_{ch,target}(1+\Delta f_{ch})
P_{ch,meas}=155(1.096)=169.9\ \text{kW}

Chiller power penalty:

\Delta P_{ch}=169.9-155=14.9\ \text{kW}

Target plant power at the same cooling load is:

P_{plant,target}=P_{ch,target}+P_{tower,target}+P_{chw}+P_{cw}
P_{plant,target}=155+12+16+14=197\ \text{kW}

Measured plant power estimate is:

P_{plant,meas}=169.9+18+16+14=217.9\ \text{kW}

Plant power penalty:

\Delta P_{plant}=217.9-197=20.9\ \text{kW}

Convert cooling load to tons:

\displaystyle \text{tons}=\frac{700}{3.517}=199.0\ \text{tons}

Target plant performance:

\displaystyle \frac{\text{kW}}{\text{ton}}_{target}=\frac{197}{199.0}=0.99\ \text{kW/ton}

Measured plant performance:

\displaystyle \frac{\text{kW}}{\text{ton}}_{meas}=\frac{217.9}{199.0}=1.09\ \text{kW/ton}

Annual extra energy in this operating mode:

\Delta E=\Delta P_{plant}t
\Delta E=20.9(1800)=37{,}600\ \text{kWh/year}

For the corrective action, condenser-water supply remains:

T_{cws,corr}-T_{cws,target}=30-29=1\ \text{K}

above target. Chiller power increase after correction:

\Delta f_{ch,corr}=0.024(1)=0.024

Corrected chiller power:

P_{ch,corr}=155(1.024)=158.7\ \text{kW}

Corrected plant power:

P_{plant,corr}=158.7+13+16+14=201.7\ \text{kW}

Power reduction from measured to corrected operation:

\Delta P_{saved}=217.9-201.7=16.2\ \text{kW}

Annual energy reduction:

\Delta E_{saved}=16.2(1800)=29{,}200\ \text{kWh/year}

The corrective action does not restore the ideal target condition, but it removes most of the measured penalty.

Engineering Comment

Cooling-tower approach is a plant diagnostic, not only a weather observation. A high approach can come from fouled fill, low airflow, fan control limits, poor water distribution, scaling, blocked strainers, incorrect condenser-water setpoint, sensor offset or an overloaded tower cell. The chiller then sees warmer condenser water, higher lift and higher compressor power even if the chilled-water load is unchanged.

The release record should preserve wet-bulb measurement, condenser-water supply and return temperatures, tower fan speed, basin condition, water treatment, cell staging, condenser flow, chiller lift, compressor power, pump power and whether the same load and weather interval are compared. A tower fix is credible only when the measured approach, chiller power and plant kW per ton move in the same direction.

Plausibility Check

The approach rises from 5\ \text{K} to 9\ \text{K}, so a 4\ \text{K} condenser-water penalty is visible and operationally meaningful. At 2.4\% per kelvin, the chiller power penalty is just under 10\%, or about 15\ \text{kW} on a 155\ \text{kW} chiller input. Adding the extra tower fan power brings the plant penalty to about 21\ \text{kW}, which is consistent with the shift from roughly 0.99 to 1.09\ \text{kW/ton}.

Review Checklist

Before accepting a building energy result, ask:

  1. Is the boundary a zone, AHU, plant, building, or campus?
  2. Are weather, occupancy, schedule, and service requirements stated?
  3. Are ventilation, humidity, pressure, and comfort constraints preserved?
  4. Are demand-controlled ventilation savings guarded by minimum outdoor-air flow and indoor-air-quality evidence?
  5. Are sensible and latent loads separated when humidity or dehumidification matters?
  6. Are hydronic delta-T, flow, pump power and valve/control evidence checked before accepting plant savings?
  7. Are filter pressure drop, fan speed, airflow and static-pressure evidence checked before accepting maintenance or fan-energy claims?
  8. Are fan, pump, auxiliary heat, standby, and recovery effects in the right boundary?
  9. Are chiller-only COP, plant COP, kW per ton, reheat and auxiliary loads reported with clear boundaries?
  10. Are cooling-tower approach, condenser-water temperature, tower fan power and chiller lift checked before accepting plant performance?
  11. Are heat-pump capacity curves, balance point, auxiliary heat staging and demand limits checked at cold-weather conditions?
  12. Is uncertainty included when the result is close to a limit?
  13. Does the result imply a clear action: tune, repair, clean, rebalance, resize, monitor, or reject?

Common Mistakes

  • Claiming energy savings without preserving ventilation, humidity, pressure, comfort and indoor-air-quality constraints.
  • Mixing zone, AHU, plant and campus boundaries when calculating COP, fan power, pump power or recovered heat.
  • Treating design loads as operating evidence without weather normalization, occupancy schedule and trend verification.
  • Counting economizer, DCV or demand-response savings while ignoring sensor calibration, damper position, minimum outdoor air and override behavior.
  • Combining sensible and latent loads when humidity, condensate or dehumidification capacity controls the decision.
  • Accepting low chilled-water delta-T as harmless because cooling load is met, while pump energy, valve authority and plant capacity are degraded.
  • Treating dirty filters as only a maintenance issue while ignoring fan power, airflow, static pressure, alarm uncertainty and filter bypass.
  • Reporting chiller-only COP as plant performance while excluding pumps, towers, reheat, standby and auxiliary loads.
  • Treating high condenser-water temperature as a weather fact while tower approach, fan control, fouling, water treatment and sensor calibration remain unchecked.
  • Reporting heat-pump COP without checking balance point, auxiliary heat lockout, defrost, comfort recovery and demand-limit impact.
  • Using a short trend window to justify a permanent control change without checking startup, shutdown, peak, part-load and fault modes.

Building energy calculations are strongest when they connect energy use to measurable service quality.

REF

See also