Exercise set

Thermal Energy Systems and Heat Exchangers Exercises

Solved thermal systems exercises for heat exchangers, LMTD correction, NTU, storage, exergy, heat recovery, pumping, COP and fouling.

Branch: Energy Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.4.6 · reviewed

These exercises practise thermal energy system calculations from an energy engineering perspective. The goal is not only to compute heat transfer. The goal is to decide whether the thermal boundary, exchanger performance, storage hold time, useful temperature level, auxiliary power, fouling state, bypass behavior, cleaning economics, measurement uncertainty, and operating evidence support an engineering decision.

Assume steady operation, single-phase flow, constant properties, negligible heat loss, and stable controls unless an exercise states otherwise. Real systems also require verified fluid properties, phase-change checks, pressure-drop limits, fouling history, material compatibility, relief and safety review, controls review, commissioning evidence, and measurement uncertainty.

How to Use These Exercises

For each problem:

define the thermal boundary and whether auxiliary loads are inside it;
keep heat, mass flow, temperature, pressure, time, and power units consistent;
state whether the result supports sizing, performance testing, heat recovery, cleaning, operation, or release;
check whether the temperature level makes the energy useful, not only whether the heat rate is large;
identify the measurement needed to validate the calculation.

A common mistake is treating a heat exchanger as an isolated component. Energy systems also care about source temperature, sink demand, timing, pressure drop, pumping or fan energy, controls, fouling, cleaning access, seasonal operation, and whether the recovered heat displaces a real external energy input.

Release Evidence Notes

Thermal-system calculations should be tied to a measured boundary. For each result, record hot and cold stream identity, mass-flow measurement, temperature sensor locations, pressure-drop boundary, auxiliary loads, control state, bypass state, fouling state, heat-loss assumption and time period. A heat duty or UA result without that boundary cannot support release.

Heat-exchanger evidence should separate feasibility from area. LMTD, NTU, pinch, temperature-cross and terminal-approach checks should be completed before blaming exchanger area, fouling or cleaning. If the target outlet temperature violates heat-capacity limits or approach rules, no practical area increase can make that target valid.

Performance evidence should include hydraulics and controls. Heat recovery, fouling tests, cleaning payback, bypass leakage, pumping penalty and COP checks depend on valve position, flow balance, sensor placement, pump or fan power, simultaneity of source and sink demand, turndown, startup and seasonal operation. A thermal pass can still be a net-energy fail if auxiliary power or timing is ignored.

Commissioning evidence should be conservative near release thresholds. Heat-balance mismatch, guarded duty margin, uncertainty, calibration, fouling trend, bypass leakage and pressure-drop increase should be visible before accepting or rejecting performance. If data do not close, the right action is instrumentation review or retest before changing equipment.

The practical release question is whether heat balance, temperature level, pressure drop, control state, fouling condition, auxiliary energy and measurement evidence all support the same operating action. If one layer conflicts, the result should trigger retest, cleaning, bypass repair, control tuning or revised target conditions rather than unconditional release.

Engineering Boundary Notes

These exercises use simplified thermal-system and heat-exchanger screens. They do not replace detailed exchanger rating, phase-change analysis, pressure-drop design, controls review, relief and safety review, material compatibility assessment, fouling program design, commissioning procedures or measurement-system validation. A heat-duty, UA or COP result applies only to the stated streams, flow rates, sensor locations, pressure-drop boundary, control state and time period.

Separate heat rate from useful energy service. Heat duty, exergy, LMTD, NTU, pinch, thermal-storage hold time, pump power, fouling ratio and cleaning payback answer different questions. Release evidence should state whether the decision concerns sizing, performance testing, useful recovery, cleaning, bypass repair, storage availability or operating target.

Common Release Mistakes

blaming exchanger area or fouling before checking heat-capacity limits, terminal approaches and temperature-cross feasibility;
accepting UA from test data when hot-side and cold-side heat balances do not close;
reporting recovered heat without subtracting pump, fan, auxiliary or simultaneity penalties;
using a fouling ratio without matching flow, bypass state, sensor calibration and comparable operating point;
treating large low-temperature waste heat as valuable without checking useful sink temperature and timing;
scheduling cleaning from thermal duty alone while pressure drop, access, downtime and payback are ignored.

Scenario Map

Scenario	Core calculation	Engineering decision
Heat-balance reconciliation	Hot-side and cold-side duties compared by normalized mismatch	Decide whether test data are credible enough for UA or performance claims.
LMTD area sizing	Duty divided by U and log-mean temperature difference	Check sensitivity to small temperature approaches, fouling and arrangement correction.
LMTD correction-factor reserve	Counterflow LMTD, correction factor and guarded installed area	Decide whether a nonideal shell-and-tube arrangement still has enough area margin.
Measured effectiveness	Actual duty divided by maximum possible duty	Interpret performance only at the measured flow arrangement and control state.
Waste-heat exergy	Heat rate multiplied by Carnot quality factor	Prioritize heat sources by temperature level, not heat rate alone.
Heat recovery net benefit	Avoided fuel input minus auxiliary electrical energy	Report net benefit and simultaneity, not only recovered heat.
Pumping penalty	Pressure-drop power divided by recovered heat	Include auxiliary power and pressure-drop trends in heat-recovery decisions.
Heat-pump COP	Useful heat divided by electrical input	Confirm lower supply temperature still satisfies demand at peak conditions.
Fouling test	Current UA divided by clean UA	Trigger cleaning only after checking flow, sensors, bypass and comparable operating point.
Guarded performance test	Measured duty minus uncertainty and limit	Use uncertainty-aware pass/fail margin for release decisions.
Operating action	Duty loss, pressure-drop increase and net recovery	Combine thermal and hydraulic evidence before scheduling cleaning or inspection.
Effectiveness-NTU prediction	UA divided by minimum heat-capacity rate	Predict exchanger duty before outlet temperatures are known.
Cleaning payback	Recovered duty improvement converted to avoided energy value	Decide whether cleaning is justified by recovered service, not only fouling ratio.
Pinch-limited recovery	Minimum approach temperature and auxiliary heat shortfall	Decide how much target heating can really be supplied by recovery.
Bypass leakage diagnosis	Mixed outlet temperature and heat-duty loss	Estimate bypass fraction before blaming exchanger fouling or undersizing.
Temperature-cross feasibility	Terminal approaches and heat-capacity-limited maximum duty	Reject impossible outlet-temperature targets before area sizing, cleaning or release.
Thermal storage standby loss	Exponential temperature decay and useful energy above a minimum temperature	Check whether stored heat remains useful after the required hold time.
Fouling-resistance back calculation	Difference between dirty and clean overall thermal resistance	Compare measured fouling with the design allowance before cleaning or release.

Validation Package Checklist

hot and cold streams, flow rates, sensor locations, pressure-drop boundary and time period are documented;
auxiliary loads, pump or fan power, bypass state, control state, fouling state and heat-loss assumption are recorded;
heat balance, LMTD, NTU, pinch, temperature-cross and terminal-approach checks are completed before area conclusions;
useful temperature level, source-sink simultaneity, storage hold time and external energy displacement are proven;
calibration, uncertainty, heat-balance mismatch, fouling trend, bypass leakage and pressure-drop increase are guarded;
cleaning, repair, control tuning and revised target decisions include net energy, downtime and operating evidence;
final release decision states accept, retest, clean, repair bypass, tune controls, revise target, rerate exchanger or hold.

Exercise 1: Heat-Exchanger Heat Balance Reconciliation

A water-to-water heat exchanger is tested after commissioning. The hot stream data are:

\dot{m}_h=2.8\ \text{kg/s}

C_{p,h}=4.18\ \text{kJ/(kg K)}

T_{h,in}=82^\circ\text{C},\quad T_{h,out}=64^\circ\text{C}

The cold stream data are:

\dot{m}_c=3.4\ \text{kg/s}

C_{p,c}=4.18\ \text{kJ/(kg K)}

T_{c,in}=30^\circ\text{C},\quad T_{c,out}=44^\circ\text{C}

Estimate the hot-side and cold-side heat duties and the normalized heat-balance mismatch.

Solution

Hot-side heat duty:

\dot{Q}_h=\dot{m}_h C_{p,h}(T_{h,in}-T_{h,out})

\dot{Q}_h=2.8(4.18)(82-64)=210.7\ \text{kW}

Cold-side heat duty:

\dot{Q}_c=\dot{m}_c C_{p,c}(T_{c,out}-T_{c,in})

\dot{Q}_c=3.4(4.18)(44-30)=198.9\ \text{kW}

Normalized mismatch:

\displaystyle \epsilon_Q=\frac{\dot{Q}_h-\dot{Q}_c}{(\dot{Q}_h+\dot{Q}_c)/2}

\displaystyle \epsilon_Q=\frac{210.7-198.9}{(210.7+198.9)/2}=0.0576

The heat-balance mismatch is approximately:

5.8\%

Engineering Comment

A 5.8 percent mismatch may be acceptable or unacceptable depending on flow-meter accuracy, temperature-sensor calibration, heat loss, mixing, and property assumptions. The engineer should not use a single side blindly for $UA$ estimation. The test record should state which side is more reliable and whether the mismatch is inside the uncertainty budget.

Plausibility Check

The hot side reports 210.7 kW and the cold side reports 198.9 kW, a difference of 11.8 kW. Normalizing by the average duty gives 5.8 percent mismatch. That is plausible for a field heat-balance test, but it is large enough that flow calibration, temperature sensor placement, mixing and heat loss should be checked before fitting $UA$ from one side only.

Exercise 2: LMTD Area for a Heat Recovery Exchanger

A counterflow heat recovery exchanger must transfer:

\dot{Q}=520\ \text{kW}

The hot stream cools from $150^\circ\text{C}$ to $95^\circ\text{C}$ . The cold stream warms from $55^\circ\text{C}$ to $105^\circ\text{C}$ . Use:

U=640\ \text{W/(m}^2\text{K)}

Estimate the log-mean temperature difference and required area.

Solution

For counterflow:

\Delta T_1=T_{h,in}-T_{c,out}=150-105=45\ \text{K}

\Delta T_2=T_{h,out}-T_{c,in}=95-55=40\ \text{K}

Log-mean temperature difference:

\displaystyle \Delta T_{lm}=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

\displaystyle \Delta T_{lm}=\frac{45-40}{\ln(45/40)}=42.4\ \text{K}

Area:

\displaystyle A=\frac{\dot{Q}}{U\Delta T_{lm}}

Convert duty to watts:

\dot{Q}=520{,}000\ \text{W}

\displaystyle A=\frac{520{,}000}{640(42.4)}=19.2\ \text{m}^2

Engineering Comment

The area estimate is a clean screening result. The small terminal temperature differences mean the design is sensitive to fouling, flow maldistribution, and temperature measurement error. A real review should also check exchanger arrangement correction factor, pressure drop, cleaning access, material compatibility, and whether both streams are simultaneous for enough operating hours.

Plausibility Check

The terminal differences are 45 K and 40 K, so the LMTD of 42.4 K lies between them as expected. Dividing 520 kW by $640\ \text{W/(m}^2\text{K)}$ and 42.4 K gives 19.2 m2. Because the two approaches are small and close together, fouling or measurement error can noticeably change required area and recovered duty.

Exercise 3: Effectiveness from Measured Operating Data

A heat exchanger recovers heat from an exhaust-water loop. The hot stream has:

\dot{m}_h=1.6\ \text{kg/s},\quad C_{p,h}=4.0\ \text{kJ/(kg K)}

T_{h,in}=90^\circ\text{C},\quad T_{h,out}=62^\circ\text{C}

The cold stream has:

\dot{m}_c=2.4\ \text{kg/s},\quad C_{p,c}=4.18\ \text{kJ/(kg K)}

T_{c,in}=28^\circ\text{C}

Estimate exchanger effectiveness using the hot-side measured duty.

Solution

Heat-capacity rates:

C_h=\dot{m}_h C_{p,h}=1.6(4.0)=6.40\ \text{kW/K}

C_c=\dot{m}_c C_{p,c}=2.4(4.18)=10.03\ \text{kW/K}

The minimum heat-capacity rate is:

C_{min}=6.40\ \text{kW/K}

Measured hot-side heat transfer:

\dot{Q}=C_h(T_{h,in}-T_{h,out})

\dot{Q}=6.40(90-62)=179.2\ \text{kW}

Maximum possible heat transfer:

\dot{Q}_{max}=C_{min}(T_{h,in}-T_{c,in})

\dot{Q}_{max}=6.40(90-28)=396.8\ \text{kW}

Effectiveness:

\displaystyle \epsilon=\frac{\dot{Q}}{\dot{Q}_{max}}=\frac{179.2}{396.8}=0.452

The exchanger effectiveness is approximately:

45.2\%

Engineering Comment

The effectiveness is moderate. Before concluding that the exchanger is undersized, check whether the cold stream actually needs more heat, whether controls are bypassing flow, whether the hot stream is fouled, and whether the measured hot outlet temperature is stable. Effectiveness is meaningful only for the stated operating point and flow arrangement.

Plausibility Check

The hot stream is the minimum heat-capacity stream at 6.40 kW/K, so it correctly defines the maximum possible heat transfer. The measured duty is 179.2 kW and the maximum possible duty is 396.8 kW, giving 45.2 percent effectiveness. The value is meaningful only for this operating point; a control bypass or different flow split would change the result.

Exercise 4: Exergy Value of Waste Heat

Two waste-heat streams each contain $300\ \text{kW}$ of heat. Stream A is available at $180^\circ\text{C}$ and stream B is available at $75^\circ\text{C}$ . Use an environment temperature:

T_0=25^\circ\text{C}

Estimate the ideal maximum work potential for each stream using:

\displaystyle \dot{W}_{max}=\dot{Q}\left(1-\frac{T_0}{T}\right)

Temperatures must be absolute.

Solution

Convert temperatures to kelvin:

T_0=25+273=298\ \text{K}

T_A=180+273=453\ \text{K}

T_B=75+273=348\ \text{K}

Stream A:

\displaystyle \dot{W}_{max,A}=300\left(1-\frac{298}{453}\right)=102.6\ \text{kW}

Stream B:

\displaystyle \dot{W}_{max,B}=300\left(1-\frac{298}{348}\right)=43.1\ \text{kW}

Engineering Comment

Both streams have the same heat rate, but stream A has more than twice the ideal work potential because it is hotter. This does not mean stream A will actually produce $102.6\ \text{kW}$ of useful work. Real conversion has losses, finite heat-transfer area, pressure drop, controls, and cost. The result shows why temperature level matters in heat recovery decisions.

Plausibility Check

The 180 degree C stream has an ideal work potential of 102.6 kW, while the 75 degree C stream has 43.1 kW. The ratio is about 2.4 even though both carry 300 kW of heat. This is physically plausible because heat closer to ambient temperature has lower thermodynamic usefulness.

Exercise 5: Heat Recovery Net Benefit with Auxiliary Power

A heat recovery loop delivers:

\dot{Q}_{useful}=680\ \text{kW}

It operates for $3600\ \text{h/year}$ . The recovered heat displaces boiler fuel with efficiency:

\eta_{boiler}=0.84

The heat recovery loop adds $24\ \text{kW}$ of pump and fan power while operating. Estimate annual avoided fuel energy, annual auxiliary electrical energy, and net useful energy benefit on an energy basis.

Solution

Avoided fuel input:

\displaystyle E_{fuel,avoided}=\frac{\dot{Q}_{useful}t}{\eta_{boiler}}

\displaystyle E_{fuel,avoided}=\frac{680(3600)}{0.84}=2{,}914{,}286\ \text{kWh/year}

E_{fuel,avoided}=2914\ \text{MWh/year}

Auxiliary electrical energy:

E_{aux}=24(3600)=86{,}400\ \text{kWh/year}=86.4\ \text{MWh/year}

Net energy benefit by simple subtraction:

E_{net}=2914-86.4=2828\ \text{MWh/year}

Engineering Comment

The simple energy balance looks attractive, but fuels and electricity may have different cost, emissions, demand charges, and grid impacts. The engineering review should also check whether the heat demand is simultaneous, whether pressure drop affects process capacity, whether fouling will reduce delivered heat, and whether backup heat is needed when the recovery loop is offline.

Plausibility Check

The useful recovered heat is 680 kW for 3600 h, or 2448 MWh of delivered heat. Because boiler efficiency is 84 percent, avoided fuel input is larger at 2914 MWh/year. The auxiliary energy is 86.4 MWh/year, leaving 2828 MWh/year by simple subtraction. The result is favorable, but it mixes fuel and electricity on an energy basis and still needs cost, emissions and simultaneity checks.

Exercise 6: Pumping Power Penalty

A heat exchanger added for heat recovery increases loop pressure drop by:

\Delta P=85\ \text{kPa}

The volumetric flow rate is:

Q=0.042\ \text{m}^3/\text{s}

Pump efficiency is:

\eta_p=0.68

The exchanger recovers $410\ \text{kW}$ of useful heat. Estimate the additional hydraulic pumping power and the pumping penalty as a fraction of recovered heat.

Solution

Pumping power:

\displaystyle P_p=\frac{\Delta P Q}{\eta_p}

Convert pressure:

\Delta P=85{,}000\ \text{Pa}

\displaystyle P_p=\frac{85{,}000(0.042)}{0.68}=5250\ \text{W}=5.25\ \text{kW}

Pumping penalty fraction:

\displaystyle f_p=\frac{P_p}{\dot{Q}_{recovered}}=\frac{5.25}{410}=0.0128

The pumping penalty is:

1.28\%

Engineering Comment

The auxiliary penalty is small at this point, but it can grow quickly with higher flow, fouling, undersized piping, partially closed valves, or poor control. A heat recovery project should report net benefit, not only recovered heat. The pressure-drop measurement is also a useful fouling indicator over time.

Plausibility Check

The added pumping power is $85{,}000\times0.042/0.68=5.25$ kW. Compared with 410 kW of recovered heat, the penalty is 1.28 percent. That is small enough for this screen, but pressure drop scales strongly with flow and can rise with fouling, so it should be trended rather than ignored.

Exercise 7: Heat-Pump COP and Temperature Lift

A heat pump supplies $900\ \text{kW}$ of useful heat to a district heating loop. Electrical input is $260\ \text{kW}$ . After lowering the supply temperature and improving the heat-source temperature, useful heat remains $900\ \text{kW}$ but electrical input falls to $220\ \text{kW}$ .

Calculate the COP before and after the change, and the electrical demand reduction.

Solution

Heat-pump coefficient of performance:

\displaystyle COP=\frac{\dot{Q}_{useful}}{P_{input}}

Initial COP:

\displaystyle COP_0=\frac{900}{260}=3.46

Improved COP:

\displaystyle COP_1=\frac{900}{220}=4.09

Electrical demand reduction:

\Delta P=260-220=40\ \text{kW}

Relative reduction:

\displaystyle r=\frac{40}{260}=0.154=15.4\%

Engineering Comment

The COP improvement is caused by reducing temperature lift and improving source conditions, not by changing the heat demand. The review should confirm that the lower supply temperature still satisfies the load at peak conditions, that flow rates and emitters can deliver heat, and that auxiliary pumps, defrost, standby power, and part-load operation are included in the reported boundary.

Plausibility Check

COP improves from 3.46 to 4.09 while useful heat stays at 900 kW. Electrical demand falls by 40 kW, or 15.4 percent. This is plausible when temperature lift is reduced, but the lower supply temperature must still satisfy the real heat emitters and peak-load condition before the operating change is accepted.

Exercise 8: Fouling Performance Test

A clean heat exchanger had:

UA_{clean}=42\ \text{kW/K}

During an end-of-run performance test, measured data indicate:

UA_{test}=31\ \text{kW/K}

The operating procedure requires cleaning if effective $UA$ falls below $80\%$ of clean performance. Determine the performance ratio and cleaning decision.

Solution

Performance ratio:

\displaystyle R_{UA}=\frac{UA_{test}}{UA_{clean}}

\displaystyle R_{UA}=\frac{31}{42}=0.738

As a percent:

R_{UA}=73.8\%

Cleaning trigger:

73.8\%<80\%

The exchanger meets the cleaning trigger.

Engineering Comment

The calculation supports cleaning, but the engineer should first verify flow rates, fluid properties, sensor calibration, bypass position, and whether the test was at a comparable operating point. A low apparent $UA$ can come from fouling, air binding, control valve position, flow maldistribution, incorrect temperature measurement, or heat loss outside the selected boundary.

Plausibility Check

The end-of-run $UA$ is 31 kW/K versus 42 kW/K clean, so performance is 73.8 percent of clean. That is 6.2 percentage points below the 80 percent cleaning trigger. The trigger is numerically met, but the apparent loss should be confirmed against comparable flow, temperature, bypass and sensor conditions.

Exercise 9: Guarded Performance-Test Margin

A thermal energy system must deliver at least:

\dot{Q}_{limit}=240\ \text{kW}

A field test estimates:

\dot{Q}_{measured}=252\ \text{kW}

The expanded uncertainty of the heat-duty estimate is:

U_Q=9\ \text{kW}

Use a conservative decision rule:

M_Q=\dot{Q}_{measured}-U_Q-\dot{Q}_{limit}

Determine whether the system passes with guarded margin.

Solution

Guarded margin:

M_Q=252-9-240=3\ \text{kW}

Since:

M_Q\geq 0

the system passes the selected guarded decision rule.

Engineering Comment

The pass margin is small. The test should preserve raw flow and temperature data, calibration records, steady-state evidence, fluid-property assumptions, heat-loss boundary, and uncertainty budget. If the system is expected to operate at higher ambient temperature, fouled condition, or lower flow, this single test may not cover the release envelope.

Plausibility Check

The nominal measured margin is 12 kW above the 240 kW limit, but the expanded uncertainty consumes 9 kW of that margin. The guarded margin is therefore only 3 kW. This is a pass under the stated rule, but it is narrow and should not be extrapolated to fouled, hot-ambient or low-flow operation without additional evidence.

Exercise 10: Choosing an Operating Action

A heat recovery exchanger normally delivers $500\ \text{kW}$ at clean operation. Trend data show:

Condition	Recovered heat	Pressure drop
clean baseline	$500\ \text{kW}$	$42\ \text{kPa}$
current operation	$420\ \text{kW}$	$78\ \text{kPa}$

The added pump power from the pressure-drop increase is estimated at $3.6\ \text{kW}$ . Calculate duty loss, pressure-drop increase, and net recovered heat after subtracting added pump power. State the likely operating action.

Solution

Duty loss:

\Delta \dot{Q}=500-420=80\ \text{kW}

Relative duty loss:

\displaystyle r_Q=\frac{80}{500}=0.16=16\%

Pressure-drop increase:

\Delta(\Delta P)=78-42=36\ \text{kPa}

Relative pressure-drop increase:

\displaystyle r_{\Delta P}=\frac{36}{42}=0.857=85.7\%

Net recovered heat after subtracting added pump power:

\dot{Q}_{net}=420-3.6=416.4\ \text{kW}

Engineering Comment

The exchanger is still recovering useful heat, but the combination of 16 percent duty loss and 85.7 percent pressure-drop increase is a strong fouling signal. The likely action is to schedule cleaning or inspection, verify measurements, and check filtration or water chemistry. If the exchanger serves a critical load, the operating decision may also include bypass review, backup heating readiness, and a retest after cleaning.

Plausibility Check

Current duty is 420 kW, so the lost duty from the 500 kW baseline is 80 kW or 16 percent. Pressure drop rose by 36 kPa from a 42 kPa baseline, an 85.7 percent increase, while added pump power is 3.6 kW. Net recovered heat is still 416.4 kW, but the combined duty loss and hydraulic penalty support cleaning or inspection rather than continued normal operation.

Exercise 11: Counterflow Effectiveness-NTU Prediction

A counterflow exchanger is being screened before outlet temperatures are known. The hot-stream heat-capacity rate is:

C_h=7.0\ \text{kW/K}

The cold-stream heat-capacity rate is:

C_c=4.0\ \text{kW/K}

The inlet temperatures are:

T_{h,in}=110^\circ\text{C},\quad T_{c,in}=30^\circ\text{C}

The estimated overall conductance is:

UA=18\ \text{kW/K}

Use the counterflow relation:

\displaystyle \epsilon=\frac{1-\exp[-NTU(1-C_r)]}{1-C_r\exp[-NTU(1-C_r)]}

where:

\displaystyle NTU=\frac{UA}{C_{min}},\quad C_r=\frac{C_{min}}{C_{max}}

Estimate effectiveness, heat duty and outlet temperatures.

Solution

Minimum and maximum heat-capacity rates:

C_{min}=4.0\ \text{kW/K}

C_{max}=7.0\ \text{kW/K}

Capacity ratio:

\displaystyle C_r=\frac{4.0}{7.0}=0.571

Number of transfer units:

\displaystyle NTU=\frac{18}{4.0}=4.5

Effectiveness:

\displaystyle \epsilon=\frac{1-\exp[-4.5(1-0.571)]}{1-0.571\exp[-4.5(1-0.571)]}=0.932

Maximum possible heat transfer:

\dot{Q}_{max}=C_{min}(T_{h,in}-T_{c,in})

\dot{Q}_{max}=4.0(110-30)=320\ \text{kW}

Predicted heat duty:

\dot{Q}=\epsilon\dot{Q}_{max}=0.932(320)=298\ \text{kW}

Hot outlet temperature:

\displaystyle T_{h,out}=110-\frac{298}{7.0}=67.4^\circ\text{C}

Cold outlet temperature:

\displaystyle T_{c,out}=30+\frac{298}{4.0}=104.6^\circ\text{C}

Engineering Comment

The NTU method is useful when outlet temperatures are not yet known, but it depends on a credible $UA$ , flow arrangement and heat-capacity rates. A final design still needs pressure drop, fouling margin, bypass behavior, control range, material compatibility, approach temperature, safety review and commissioning evidence.

Plausibility Check

The cold side has the smaller heat-capacity rate, so it limits the maximum heat transfer. The predicted duty of $298\ \text{kW}$ is below the $320\ \text{kW}$ thermodynamic maximum, and the cold outlet reaches $104.6^\circ\text{C}$ , still below the hot inlet of $110^\circ\text{C}$ . That ordering is physically plausible for a counterflow exchanger with high effectiveness.

Exercise 12: Fouling Cleaning Payback from Recovered Duty

A fouled heat exchanger currently recovers:

\dot{Q}_{current}=360\ \text{kW}

After cleaning, the expected recovered heat is:

\dot{Q}_{cleaned}=450\ \text{kW}

The exchanger operates:

t=2600\ \text{h/year}

The avoided energy value is:

c_E=0.08\ \text{currency units/kWh}

The cleaning outage and service cost is:

C_{clean}=16{,}500\ \text{currency units}

Estimate the annual recovered-energy value and simple payback.

Solution

Recovered-duty improvement:

\Delta \dot{Q}=450-360=90\ \text{kW}

Annual additional recovered energy:

E_{add}=90(2600)=234{,}000\ \text{kWh/year}

Annual value:

V_{annual}=234{,}000(0.08)=18{,}720\ \text{currency units/year}

Simple payback:

\displaystyle P_b=\frac{16{,}500}{18{,}720}=0.881\ \text{year}

In months:

P_b=0.881(12)=10.6\ \text{months}

Engineering Comment

Cleaning is justified by recovered service only if the added heat displaces real purchased energy during simultaneous operating hours. The decision should also include outage timing, lost production, cleaning risk, pressure-drop recovery, fouling recurrence rate, emissions value, maintenance access and post-clean validation. A short payback can still be rejected if cleaning creates unacceptable downtime or safety risk.

Plausibility Check

The cleaned exchanger is expected to recover $90\ \text{kW}$ more heat. Over $2600$ hours, that is $234\ \text{MWh}$ of additional recovered energy. At $0.08$ per kWh, the annual value is $18{,}720$ , slightly higher than the $16{,}500$ cleaning cost, so a payback just under one year is plausible.

Exercise 13: Pinch-Limited Heat Recovery and Auxiliary Heat

A counterflow heat recovery exchanger uses a hot process stream with heat-capacity rate:

C_h=8.0\ \text{kW/K}

and inlet temperature:

T_{h,in}=120^\circ\text{C}

The cold stream has heat-capacity rate:

C_c=5.0\ \text{kW/K}

and inlet temperature:

T_{c,in}=45^\circ\text{C}

The process wants the cold stream outlet to reach:

T_{c,target}=112^\circ\text{C}

The heat exchanger design rule requires minimum terminal approach:

\Delta T_{min}=12\ \text{K}

The unit operates:

t=3000\ \text{h/year}

and any shortfall is supplied by a boiler with efficiency:

\eta_b=0.88

Find the target heating duty, whether the target violates the hot-end approach, the maximum cold outlet temperature allowed by the approach rule, the recovered duty, the auxiliary heat shortfall and the annual auxiliary fuel input.

Solution

Target heating duty:

\dot{Q}_{target}=C_c(T_{c,target}-T_{c,in})

\dot{Q}_{target}=5.0(112-45)=335\ \text{kW}

Hot outlet if that target duty were recovered:

\displaystyle T_{h,out,target}=T_{h,in}-\frac{\dot{Q}_{target}}{C_h}

\displaystyle T_{h,out,target}=120-\frac{335}{8.0}=78.1^\circ\text{C}

Hot-end terminal approach at the target cold outlet:

\Delta T_{hot\ end}=T_{h,in}-T_{c,target}

\Delta T_{hot\ end}=120-112=8\ \text{K}

The target violates the $12\ \text{K}$ minimum approach rule.

Maximum cold outlet allowed by the hot-end approach is:

T_{c,out,max}=T_{h,in}-\Delta T_{min}

T_{c,out,max}=120-12=108^\circ\text{C}

Maximum recovered duty:

\dot{Q}_{rec}=C_c(T_{c,out,max}-T_{c,in})

\dot{Q}_{rec}=5.0(108-45)=315\ \text{kW}

Corresponding hot outlet:

\displaystyle T_{h,out}=120-\frac{315}{8.0}=80.6^\circ\text{C}

Cold-end approach:

\Delta T_{cold\ end}=T_{h,out}-T_{c,in}=80.6-45=35.6\ \text{K}

The cold-end approach is not limiting.

Auxiliary heat shortfall:

\dot{Q}_{aux}=335-315=20\ \text{kW}

Annual auxiliary useful heat:

E_{aux,useful}=20(3000)=60{,}000\ \text{kWh/year}

Annual auxiliary fuel input:

\displaystyle E_{fuel,aux}=\frac{60{,}000}{0.88}=68{,}182\ \text{kWh/year}

or:

68.2\ \text{MWh/year}

Engineering Comment

The exchanger may recover most of the duty, but the final temperature lift is blocked by the minimum approach. Reporting only the recovered $315\ \text{kW}$ would hide the fact that another $20\ \text{kW}$ of useful heat is still needed at a higher temperature level. A release review should state whether the shortfall is acceptable, supplied by auxiliary heat, solved by a hotter source, or handled by a different process integration arrangement.

Plausibility Check

The requested cold outlet of $112^\circ\text{C}$ is only $8\ \text{K}$ below the hot inlet, so it is reasonable that a $12\ \text{K}$ approach rule rejects it. Limiting the outlet to $108^\circ\text{C}$ reduces duty by $5(4)=20\ \text{kW}$ , exactly matching the auxiliary heat shortfall.

Exercise 14: Bypass Leakage from Mixed Outlet Temperature

A heat-recovery exchanger is commissioned with a cold-stream total flow of:

\dot{m}_c=4.0\ \text{kg/s}

and:

C_{p,c}=4.18\ \text{kJ/(kg K)}

The cold-stream inlet temperature is:

T_{c,in}=30^\circ\text{C}

The clean all-flow model predicts the exchanger cold outlet, before any downstream mixing, as:

T_{c,hx,out}=68^\circ\text{C}

During the test, the measured downstream mixed cold outlet is only:

T_{c,mix}=61^\circ\text{C}

Assume some cold flow leaks through a bypass valve, remains at the inlet temperature, and mixes downstream with exchanger outlet flow. The commissioning limit for closed-bypass leakage is:

\beta_{max}=0.05

where $\beta$ is the bypassed fraction of total cold flow. Estimate the bypass fraction, expected clean duty, measured useful duty, lost useful duty and commissioning action.

Solution

For downstream mixing:

T_{c,mix}=(1-\beta)T_{c,hx,out}+\beta T_{c,in}

Solve for bypass fraction:

\displaystyle \beta=\frac{T_{c,hx,out}-T_{c,mix}}{T_{c,hx,out}-T_{c,in}}

\displaystyle \beta=\frac{68-61}{68-30}=0.184

So the estimated bypass leakage is:

18.4\%

Expected all-flow clean duty:

\dot{Q}_{clean}=\dot{m}_c C_{p,c}(T_{c,hx,out}-T_{c,in})

\dot{Q}_{clean}=4.0(4.18)(68-30)=635.4\ \text{kW}

Measured useful duty at the mixed outlet:

\dot{Q}_{mix}=\dot{m}_c C_{p,c}(T_{c,mix}-T_{c,in})

\dot{Q}_{mix}=4.0(4.18)(61-30)=518.3\ \text{kW}

Lost useful duty:

\Delta \dot{Q}=635.4-518.3=117.1\ \text{kW}

The estimated bypass fraction is much larger than the $5\%$ closed-bypass limit, so the test should not be used to declare the exchanger undersized or fouled. The justified action is to inspect and repair the bypass valve or control logic, verify valve isolation, repeat the heat balance, and then reassess exchanger performance.

Engineering Comment

Bypass leakage can make a good exchanger look weak because the downstream sensor sees mixed water, not the true exchanger outlet. A commissioning report should separate core exchanger performance from valve position, bypass piping, sensor location, actuator command, control sequence and mixing point. Otherwise the team may clean, oversize or replace an exchanger when the real fault is hydraulic short-circuiting.

Plausibility Check

The measured outlet is $7\ \text{K}$ below the expected exchanger outlet but still $31\ \text{K}$ above the inlet, so the bypassed fraction should be significant but less than half. The computed $18.4\%$ matches that intuition. The useful-duty loss also equals the missing $7\ \text{K}$ multiplied by $4.0(4.18)$ , giving about $117\ \text{kW}$ , so the heat balance is internally consistent.

Exercise 15: Temperature-Cross Feasibility Screen

A heat recovery project proposes to heat a cold process stream using a warm effluent stream. The hot stream has heat-capacity rate:

C_h=3.2\ \text{kW/K}

and inlet temperature:

T_{h,in}=86^\circ\text{C}

The cold stream has heat-capacity rate:

C_c=5.0\ \text{kW/K}

and inlet temperature:

T_{c,in}=24^\circ\text{C}

The proposed recovered-heat target is:

T_{c,target}=76^\circ\text{C}

The design basis requires both terminal approaches to remain at least:

\Delta T_{min}=8\ \text{K}

Check whether the target is thermally feasible. If not, estimate the maximum feasible recovered duty, the maximum cold outlet temperature, and the remaining useful heat shortfall.

Solution

Target recovered duty:

\dot{Q}_{target}=C_c(T_{c,target}-T_{c,in})

\dot{Q}_{target}=5.0(76-24)=260\ \text{kW}

Hot outlet temperature implied by that target:

\displaystyle T_{h,out,target}=T_{h,in}-\frac{\dot{Q}_{target}}{C_h}

\displaystyle T_{h,out,target}=86-\frac{260}{3.2}=4.8^\circ\text{C}

Hot-end terminal approach at the proposed target:

\Delta T_{hot\ end}=T_{h,in}-T_{c,target}

\Delta T_{hot\ end}=86-76=10\ \text{K}

The hot-end approach passes the $8\ \text{K}$ rule. The cold-end approach implied by the same target is:

\Delta T_{cold\ end}=T_{h,out,target}-T_{c,in}

\Delta T_{cold\ end}=4.8-24=-19.2\ \text{K}

That is a temperature cross, so the target is not feasible for this stream pair.

The cold-end approach rule requires:

T_{h,out,min}=T_{c,in}+\Delta T_{min}

T_{h,out,min}=24+8=32^\circ\text{C}

Maximum feasible recovered duty from the hot stream is therefore:

\dot{Q}_{max}=C_h(T_{h,in}-T_{h,out,min})

\dot{Q}_{max}=3.2(86-32)=172.8\ \text{kW}

Corresponding maximum cold outlet temperature:

\displaystyle T_{c,out,max}=T_{c,in}+\frac{\dot{Q}_{max}}{C_c}

\displaystyle T_{c,out,max}=24+\frac{172.8}{5.0}=58.6^\circ\text{C}

Remaining useful heat shortfall:

\dot{Q}_{shortfall}=\dot{Q}_{target}-\dot{Q}_{max}

\dot{Q}_{shortfall}=260-172.8=87.2\ \text{kW}

The recovery target should not be released as an exchanger area problem. It needs a lower cold-outlet target, a hotter source, a different integration arrangement, additional auxiliary heating, or a changed process requirement.

Engineering Comment

This is a useful release gate because the hot-end approach alone looks acceptable. The infeasibility appears only after the heat-capacity rates are applied and the hot outlet is calculated. A project team that skips this check may keep increasing exchanger area even though no finite area can make the hot stream leave below the cold inlet while preserving a positive terminal approach.

Plausibility Check

The target asks the cold stream to gain $52\ \text{K}$ at $5.0\ \text{kW/K}$ , or $260\ \text{kW}$ . The hot stream can only cool from $86^\circ\text{C}$ to $32^\circ\text{C}$ before the required cold-end approach is reached, giving $172.8\ \text{kW}$ . The maximum cold outlet of $58.6^\circ\text{C}$ is far below the requested $76^\circ\text{C}$ , so the $87.2\ \text{kW}$ shortfall is a thermodynamic feasibility issue, not a minor sizing margin.

Exercise 16: Thermal Storage Standby Loss and Useful Hold Time

A hot-water thermal storage tank is charged at the end of a heat-recovery period. The stored water mass is:

m=18{,}000\ \text{kg}

Use:

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

The initial uniform tank temperature is:

T_0=82^\circ\text{C}

Ambient temperature is:

T_a=20^\circ\text{C}

The tank heat-loss coefficient is:

UA=0.28\ \text{kW/K}

The downstream process can use stored heat only while tank temperature is above:

T_{min}=68^\circ\text{C}

The required hold time is:

t=12\ \text{h}

The process needs at least:

E_{req}=160\ \text{kWh}

of useful stored heat above $68^\circ\text{C}$ after the hold period. Estimate the tank thermal capacitance, time constant, tank temperature after 12 hours, heat lost during standby, useful energy remaining above the process threshold, and maximum hold time before the useful-energy requirement is just met.

Solution

Tank thermal capacitance:

C_{tank}=mC_p

C_{tank}=18{,}000(4.18)=75{,}240\ \text{kJ/K}

Convert to kWh per kelvin:

\displaystyle C_{tank}=\frac{75{,}240}{3600}=20.9\ \text{kWh/K}

The first-order thermal time constant is:

\displaystyle \tau=\frac{C_{tank}}{UA}

\displaystyle \tau=\frac{20.9}{0.28}=74.6\ \text{h}

For standby cooling toward ambient:

T(t)=T_a+(T_0-T_a)e^{-t/\tau}

After 12 hours:

T(12)=20+(82-20)e^{-12/74.6}

T(12)=72.8^\circ\text{C}

Temperature drop during standby:

\Delta T_{loss}=82-72.8=9.2\ \text{K}

Heat lost from the tank:

E_{loss}=C_{tank}\Delta T_{loss}

E_{loss}=20.9(9.2)=192\ \text{kWh}

Useful stored heat remaining above the process threshold:

E_{useful,12}=C_{tank}(T(12)-T_{min})

E_{useful,12}=20.9(72.8-68)=100\ \text{kWh}

The tank does not meet the $160\ \text{kWh}$ useful-energy requirement after 12 hours.

Tank temperature required to retain $160\ \text{kWh}$ above the useful threshold:

\displaystyle T_{req}=T_{min}+\frac{E_{req}}{C_{tank}}

\displaystyle T_{req}=68+\frac{160}{20.9}=75.7^\circ\text{C}

Solve for the hold time when $T(t)=T_{req}$ :

\displaystyle t_{max}=-\tau\ln\left(\frac{T_{req}-T_a}{T_0-T_a}\right)

\displaystyle t_{max}=-74.6\ln\left(\frac{75.7-20}{82-20}\right)=8.0\ \text{h}

The release decision is to limit the hold time to about 8 hours, improve insulation, raise the initial storage temperature if safe, increase storage mass, reduce standby duration, or provide auxiliary heat before the process demand begins.

Engineering Comment

Thermal storage capacity is useful only above the temperature required by the downstream service. A large tank can still fail if standby loss, stratification, minimum process temperature, sensor location or schedule mismatch erodes useful temperature before demand arrives. A release package should preserve tank mass, usable volume, stratification assumption, insulation condition, ambient case, heat-loss coefficient, charge temperature, minimum useful temperature, hold-time requirement, safety limits and auxiliary-heat fallback.

Plausibility Check

The tank stores about $20.9\ \text{kWh}$ per kelvin, so losing about $9.2\ \text{K}$ over 12 hours corresponds to roughly $192\ \text{kWh}$ of standby loss. The final tank temperature remains above $68^\circ\text{C}$ , but only by about $4.8\ \text{K}$ , leaving about $100\ \text{kWh}$ of useful heat. That makes the failure against a $160\ \text{kWh}$ requirement plausible even though the tank is still warm.

Exercise 17: Fouling Resistance from Dirty Overall Coefficient

A heat-recovery exchanger was commissioned with a clean overall heat-transfer coefficient:

U_{clean}=850\ \text{W/(m}^2\text{K)}

After one operating season, a comparable test estimates:

U_{dirty}=620\ \text{W/(m}^2\text{K)}

Use the simplified relation:

\displaystyle \frac{1}{U_{dirty}}=\frac{1}{U_{clean}}+R_f

The design fouling allowance is:

R_{f,allow}=0.00035\ \text{m}^2\text{K/W}

The exchanger area is:

A=42\ \text{m}^2

During the reviewed operating case, the log-mean temperature difference is:

LMTD=32\ \text{K}

The process needs:

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

Estimate fouling resistance, compare it with the design allowance, calculate clean and dirty duty at the same LMTD, and state the release decision.

Solution

Fouling resistance from the clean and dirty coefficients:

\displaystyle R_f=\frac{1}{U_{dirty}}-\frac{1}{U_{clean}}

\displaystyle R_f=\frac{1}{620}-\frac{1}{850}

R_f=0.001613-0.001176=0.000436\ \text{m}^2\text{K/W}

Allowance margin:

M_R=R_{f,allow}-R_f

M_R=0.000350-0.000436=-0.000086\ \text{m}^2\text{K/W}

Relative exceedance of the allowance:

\displaystyle E_R=100\frac{0.000436-0.000350}{0.000350}

E_R=24.7\%

Clean duty at the same area and LMTD:

\dot{Q}_{clean}=U_{clean}A(LMTD)

\dot{Q}_{clean}=850(42)(32)=1{,}142{,}400\ \text{W}

\dot{Q}_{clean}=1142\ \text{kW}

Dirty duty:

\dot{Q}_{dirty}=620(42)(32)=833{,}280\ \text{W}

\dot{Q}_{dirty}=833\ \text{kW}

Duty loss relative to clean:

\Delta \dot{Q}=1142-833=309\ \text{kW}

Dirty-duty margin against the process requirement:

M_Q=833-900=-67\ \text{kW}

The dirty test fails both the fouling-allowance screen and the required-duty screen. The release action should be cleaning, flow and sensor verification, bypass check, or operating derating before relying on this exchanger for the $900\ \text{kW}$ duty.

Engineering Comment

Back-calculating fouling resistance is more informative than saying the exchanger is “dirty.” It turns a measured $U$ loss into a thermal-resistance term that can be compared with the design fouling allowance. The result should still be tied to comparable flow rates, fluid properties, phase state, LMTD, bypass position, pressure drop, sensor placement and heat-balance closure. A bad $U_{dirty}$ estimate from poor measurements can look like fouling even when the real fault is flow maldistribution or instrumentation error.

Plausibility Check

The dirty coefficient is about $73\%$ of the clean value, so a large duty loss is expected at the same area and LMTD. The computed fouling resistance of about $0.00044\ \text{m}^2\text{K/W}$ is slightly above the $0.00035$ allowance, and the dirty duty of $833\ \text{kW}$ is below the $900\ \text{kW}$ process need. Those two failures support the same decision rather than conflicting.

Exercise 18: LMTD Correction Factor and Installed-Area Reserve

A heat-recovery project proposes a shell-and-tube exchanger for:

\dot{Q}=740\ \text{kW}

The estimated overall heat-transfer coefficient is:

U=520\ \text{W/(m}^2\text{K)}=0.520\ \text{kW/(m}^2\text{K)}

The expected terminal temperatures are:

Stream	Inlet	Outlet
Hot	$150^\circ\text{C}$	$95^\circ\text{C}$
Cold	$35^\circ\text{C}$	$82^\circ\text{C}$

The available installed area is:

A_{installed}=34\ \text{m}^2

For the selected one-shell-pass arrangement, the LMTD correction factor from the exchanger method is:

F=0.78

The project requires a 15 percent area reserve after applying the correction factor. Calculate the ideal counterflow LMTD, the uncorrected area, the corrected area, the guarded area, and the installed-area margin. Then check sensitivity if the verified correction factor is only:

F=0.72

Solution

For counterflow terminal differences:

\Delta T_1=T_{h,in}-T_{c,out}=150-82=68\ \text{K}

\Delta T_2=T_{h,out}-T_{c,in}=95-35=60\ \text{K}

The ideal counterflow LMTD is:

\displaystyle \Delta T_{lm}=\frac{\Delta T_1-\Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

\displaystyle \Delta T_{lm}=\frac{68-60}{\ln(68/60)}=\frac{8}{0.1252}=63.9\ \text{K}

If the exchanger were treated as ideal counterflow, the required area would be:

\displaystyle A_{ideal}=\frac{\dot{Q}}{U\Delta T_{lm}}

\displaystyle A_{ideal}=\frac{740}{0.520(63.9)}=22.3\ \text{m}^2

For the nonideal arrangement, apply the correction factor:

\displaystyle A_{corr}=\frac{\dot{Q}}{UF\Delta T_{lm}}=\frac{A_{ideal}}{F}

\displaystyle A_{corr}=\frac{22.3}{0.78}=28.6\ \text{m}^2

Add the required 15 percent area reserve:

A_{guard}=1.15A_{corr}=1.15(28.6)=32.9\ \text{m}^2

The installed-area margin is:

M_A=A_{installed}-A_{guard}=34.0-32.9=1.1\ \text{m}^2

Relative to installed area:

\displaystyle M_{A,\%}=\frac{1.1}{34.0}(100)=3.2\%

This is a conditional pass, not a generous margin. If the correction factor is verified as only $0.72$ :

\displaystyle A_{corr,0.72}=\frac{22.3}{0.72}=31.0\ \text{m}^2

and:

A_{guard,0.72}=1.15(31.0)=35.7\ \text{m}^2

The revised margin is:

M_{A,0.72}=34.0-35.7=-1.7\ \text{m}^2

So a slightly poorer arrangement correction changes the decision from conditional pass to fail.

Engineering Comment

The correction factor prevents an ideal counterflow calculation from overstating performance for multipass, crossflow or otherwise nonideal exchangers. In this example, ignoring $F$ would report an apparent area requirement of only $22.3\ \text{m}^2$ and a large false margin. The corrected and guarded requirement is $32.9\ \text{m}^2$ , leaving only about $1.1\ \text{m}^2$ . A release package should preserve the exchanger arrangement, correction-factor source, terminal temperatures, heat-balance closure, pressure-drop limit, fouling allowance, bypass state and sensor uncertainty.

Plausibility Check

The terminal differences are close, so the LMTD of $63.9\ \text{K}$ correctly falls between $60$ and $68\ \text{K}$ . Applying $F=0.78$ increases area by about $1/0.78=1.28$ , so the jump from $22.3$ to $28.6\ \text{m}^2$ is expected. Adding 15 percent reserve brings the requirement close to the installed $34\ \text{m}^2$ , making the sensitivity to $F=0.72$ credible.

Review Checklist

For each thermal energy exercise or field calculation, ask:

Does the heat balance close within measurement uncertainty?
Are the hot and cold stream temperature approaches physically realistic and above any required pinch limit?
Is recovered heat available at a useful temperature and at the same time as demand?
Are pump, fan, control, standby, cleaning, and downtime penalties visible?
Does exergy or temperature level change the priority of heat sources?
Is the method appropriate: LMTD when outlet temperatures are known, NTU when they are being predicted?
Is any multipass, crossflow or shell-and-tube correction factor included before area margin is claimed?
Does the test represent clean, fouled, startup, turndown, peak-load, or degraded operation?
Are uncertainty, calibration, and decision rules explicit?
Could low outlet temperature or low duty be caused by bypass leakage, mixing, valve position, or sensor location rather than exchanger area alone?
Does any target outlet temperature create a temperature cross after heat-capacity rates and terminal approaches are checked?
Does thermal storage retain enough useful energy above the service temperature after standby loss and hold time?
Is fouling resistance compared with a design allowance at a comparable flow, LMTD, pressure drop and measurement boundary?
Does the result imply sizing, cleaning, control tuning, monitoring, or release action?

Common Mistakes

Treating recovered heat duty as useful energy without checking source temperature, sink demand timing and auxiliary power.
Blaming exchanger area when the target outlet temperature violates pinch, heat-capacity-rate or terminal-approach constraints.
Using LMTD sizing before confirming the proposed hot and cold outlet temperatures are thermodynamically feasible.
Ignoring the LMTD correction factor for multipass, crossflow or shell-and-tube arrangements and overstating installed-area margin.
Comparing clean and fouled tests taken at different flow rates, control states, valve positions or sensor locations.
Comparing dirty $U$ with clean $U$ without converting the difference into fouling resistance or checking the design allowance.
Ignoring pressure-drop and pump or fan power when reporting heat-recovery savings.
Treating low outlet temperature as fouling while bypass leakage, mixing, valve position or sensor placement remains unresolved.
Releasing a heat-pump COP from mild conditions while peak lift, defrost, part load or source-temperature limits control performance.
Counting stored heat without checking standby loss, useful temperature threshold, stratification, hold time and auxiliary fallback.
Scheduling cleaning from UA ratio alone without energy value, downtime, pressure drop, fouling trend and access constraints.
Accepting commissioning data when hot-side and cold-side heat balances do not close within uncertainty.
Changing control sequences after performance testing without repeating the evidence at the same thermal boundary.

Thermal energy engineering is strongest when the calculation leads to a decision that can be validated by measured flow, temperature, pressure, power, fouling, and operating data.

REF