Exercise set

Chemical Process Heat Exchanger Duty, LMTD, NTU, and Fouling Exercises

Worked chemical heat-exchanger exercises for duty, LMTD, area, NTU, fouling, heat flux, heat-balance closure and release gates.

Branch: Chemical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on chemical process heat exchangers as thermal, hydraulic and evidence-controlled assets: duty, LMTD, area, effectiveness-NTU, heat-balance closure, fouling resistance, heat flux, Reynolds and Nusselt numbers, fouled UA, hot-oil film temperature and release gates.

Steam, cooling-water headers, condensate, cooling towers, heat recovery and utility-system capacity are handled in the companion specialist exercise set.

Release Evidence Notes

Heat-exchanger evidence should state process boundary, stream identities, phase state, flow basis, inlet and outlet temperature tags, pressure-drop boundary, exchanger area basis, fouling state, utility condition and operating time window. A duty or UA value is not release evidence unless the hot-side and cold-side measurements close within the accepted uncertainty.

Engineering Boundary Notes

These examples are first-pass design and troubleshooting screens. Real exchanger release should also check phase change, property variation, fouling chemistry, flow maldistribution, pressure drop, vibration, erosion, corrosion, tube integrity, cleaning access, control-valve authority, relief basis and commissioning evidence.

Common Release Mistakes

using clean U as if it represents fouled service;
calculating LMTD without checking terminal temperature compatibility;
diagnosing fouling before checking heat-balance closure;
mixing area bases in overall heat-transfer calculations;
ignoring pressure drop and control range while optimizing area;
treating one commissioning point as proof of all seasons and loads.

Scenario Map

Scenario	Exercises	Main calculation	Release decision
Duty and area	1, 2, 3, 10	Sensible duty, LMTD, required area and correction factor	Size or derate the exchanger.
Fouling and data quality	4, 8, 9, 11, 17	U loss, heat-balance closure, fouled UA, cleaning interval and uncertainty	Clean, retest or hold a fouling diagnosis.
Convective performance	5, 6, 7, 12, 13	Heat flux, Reynolds number, Nusselt number, film coefficient and wall temperature	Check limits, correlations and local hot spots.
Rating and release	14, 15, 16, 18	NTU, heat recovery limit, pressure-drop screen and release gates	Release only when thermal and hydraulic evidence agree.

Validation Package Checklist

stream flow, temperature, pressure, phase and property basis;
exchanger area basis, flow arrangement and correction factor;
clean and fouled U or UA basis;
hot-side and cold-side heat-balance closure;
pressure-drop and control-valve operating boundary;
release action for fouling, data-quality or area-margin failure.

Exercise 1: Process Cooling Duty

A process stream of $2.5\ \text{kg/s}$ is cooled from $85^\circ\text{C}$ to $45^\circ\text{C}$ . Use $c_p=3.6\ \text{kJ/(kg K)}$ . Compute duty.

Solution

\dot{Q}=\dot{m}c_p\Delta T=2.5(3.6)(85-45)=360\ \text{kW}

Engineering Comment

This sensible-heat duty should be checked against phase change, fouling, property range and exchanger approach temperature.

Plausibility Check

A few kilograms per second over forty kelvin with heat capacity near water gives hundreds of kilowatts.

Exercise 2: Log-Mean Temperature Difference

Hot process fluid cools from $120^\circ\text{C}$ to $70^\circ\text{C}$ . Cooling water warms from $30^\circ\text{C}$ to $45^\circ\text{C}$ in counterflow. Compute LMTD.

Solution

\Delta T_1=120-45=75^\circ\text{C}

\Delta T_2=70-30=40^\circ\text{C}

\Delta T_{lm}=\dfrac{75-40}{\ln(75/40)}=55.7^\circ\text{C}

Engineering Comment

LMTD is meaningful only when terminal temperatures match the assumed flow arrangement and no hidden phase-change boundary is missing.

Plausibility Check

The log mean should lie between forty and seventy-five degrees.

Exercise 3: Required Exchanger Area

Heat duty is $520\ \text{kW}$ , overall coefficient is $620\ \text{W/(m}^2\text{K)}$ and corrected LMTD is $48^\circ\text{C}$ . Compute area.

Solution

A=\dfrac{520000}{620(48)}=17.5\ \text{m}^2

Engineering Comment

The selected area should include fouling, pressure drop, cleanability and minimum approach temperature margin.

Plausibility Check

Hundreds of kilowatts divided by tens of thousands of watts per square meter gives tens of square meters.

Exercise 4: Fouling Impact on U

Clean overall coefficient is $U_c=720\ \text{W/(m}^2\text{K)}$ . Fouling resistance is $R_f=0.00025\ \text{m}^2\text{K/W}$ . Estimate fouled U using resistance addition.

Solution

\dfrac{1}{U_f}=\dfrac{1}{720}+0.00025=0.001639

U_f=610\ \text{W/(m}^2\text{K)}

Engineering Comment

Fouling reduces thermal capacity and can also raise pressure drop. Cleaning decisions should use both duty and hydraulic evidence.

Plausibility Check

Adding resistance lowers U from the clean value.

Exercise 5: Heat Flux Check

A jacket removes $240\ \text{kW}$ over wetted area $18\ \text{m}^2$ . Compute heat flux.

Solution

q''=\dfrac{240000}{18}=13.3\ \text{kW/m}^2

Engineering Comment

Heat flux should be checked against local hot spots, wetting, fouling, boiling, viscosity and material limits.

Plausibility Check

Hundreds of kilowatts over tens of square meters gives tens of kilowatts per square meter.

Exercise 6: Tube-Side Reynolds Number

Tube velocity is $1.4\ \text{m/s}$ , diameter is $0.020\ \text{m}$ , density is $995\ \text{kg/m}^3$ and viscosity is $0.0008\ \text{Pa s}$ . Compute Reynolds number.

Solution

Re=\dfrac{\rho vD}{\mu}=\dfrac{995(1.4)(0.020)}{0.0008}=34800

Engineering Comment

The turbulent screen supports higher heat transfer, but erosion, vibration and pressure drop must still be checked.

Plausibility Check

The result is far above the usual turbulent threshold.

Exercise 7: Nusselt Number Estimate

For turbulent tube flow, use $Nu=0.023Re^{0.8}Pr^{0.4}$ with $Re=34800$ and $Pr=5.0$ .

Solution

Nu=0.023(34800)^{0.8}(5.0)^{0.4}=214

Engineering Comment

Correlation use requires checking range, entrance length, property basis, roughness and fouling.

Plausibility Check

Turbulent liquid flow often gives Nusselt numbers in the hundreds.

Exercise 8: Heat-Balance Closure

Hot-side duty is $505\ \text{kW}$ and cold-side duty is $485\ \text{kW}$ . Compute mismatch relative to average duty.

Solution

\bar{Q}=\dfrac{505+485}{2}=495\ \text{kW}

E=\dfrac{505-485}{495}=4.04\%

Engineering Comment

If the closure limit is five percent, the data can support a first-pass UA review. If not, fix measurements before diagnosing fouling.

Plausibility Check

The two duties differ by twenty kilowatts on a roughly five-hundred-kilowatt exchanger.

Exercise 9: Fouled UA Capacity

Area is $22\ \text{m}^2$ and fouled U is $610\ \text{W/(m}^2\text{K)}$ . Compute fouled UA.

Solution

UA=610(22)=13420\ \text{W/K}=13.42\ \text{kW/K}

Engineering Comment

Fouled UA is a useful operating capacity number only if area basis and temperature measurements are consistent.

Plausibility Check

Hundreds of watts per square meter kelvin over tens of square meters gives tens of kilowatts per kelvin.

Exercise 10: LMTD Correction Factor

A multipass exchanger has base LMTD $52^\circ\text{C}$ and correction factor $F=0.86$ . Compute corrected LMTD.

Solution

\Delta T_{corr}=F\Delta T_{lm}=0.86(52)=44.7^\circ\text{C}

Engineering Comment

Correction factors protect against overestimating driving force in nonideal arrangements.

Plausibility Check

The corrected value must be below the uncorrected LMTD.

Exercise 11: Cleaning Interval from Fouling Rate

Fouling resistance increases at $4.0\times10^{-5}\ \text{m}^2\text{K/W per month}$ . The action limit is $3.0\times10^{-4}\ \text{m}^2\text{K/W}$ . Estimate time to action from clean state.

Solution

t=\dfrac{3.0\times10^{-4}}{4.0\times10^{-5}}=7.5\ \text{months}

Engineering Comment

The interval assumes approximately linear fouling under comparable flow, temperature and chemistry.

Plausibility Check

Eight months at the stated rate would exceed the limit slightly.

Exercise 12: Convective Coefficient from Nusselt Number

For a tube, $Nu=214$ , thermal conductivity is $0.62\ \text{W/(m K)}$ and diameter is $0.020\ \text{m}$ . Compute film coefficient.

Solution

h=\dfrac{Nu\,k}{D}=\dfrac{214(0.62)}{0.020}=6634\ \text{W/(m}^2\text{K)}

Engineering Comment

High film coefficient may not govern if fouling, wall conduction or shell-side resistance dominates.

Plausibility Check

Large turbulent Nusselt number over a small tube gives several thousand watts per square meter kelvin.

Exercise 13: Hot-Oil Film Temperature

Bulk hot-oil temperature is $285^\circ\text{C}$ , heat flux is $9000\ \text{W/m}^2$ and film coefficient is $450\ \text{W/(m}^2\text{K)}$ . Estimate wall film temperature.

Solution

\Delta T_f=\dfrac{9000}{450}=20^\circ\text{C}

T_f=285+20=305^\circ\text{C}

Engineering Comment

Film temperature can govern oil degradation even when bulk temperature is acceptable.

Plausibility Check

A moderate heat flux divided by a few hundred watts per square meter kelvin gives tens of degrees.

Exercise 14: Effectiveness-NTU Outlet Temperature

Cold stream heat-capacity rate is $C_c=6\ \text{kW/K}$ , hot stream capacity rate is $C_h=10\ \text{kW/K}$ and effectiveness is $\epsilon=0.62$ . Hot inlet is $120^\circ\text{C}$ and cold inlet is $30^\circ\text{C}$ . Find cold outlet.

Solution

C_{min}=6\ \text{kW/K}

Q=\epsilon C_{min}(T_{h,in}-T_{c,in})=0.62(6)(90)=334.8\ \text{kW}

T_{c,out}=30+\dfrac{334.8}{6}=85.8^\circ\text{C}

Engineering Comment

Effectiveness-NTU is useful for rating when outlet temperatures are unknown, but it depends on the correct flow arrangement and UA.

Plausibility Check

The cold outlet remains below the hot inlet.

Exercise 15: Heat Recovery Limit

A heat-recovery exchanger has $C_{min}=4.5\ \text{kW/K}$ and inlet temperature difference $80^\circ\text{C}$ . What is the maximum possible recovered heat?

Solution

Q_{max}=C_{min}\Delta T=4.5(80)=360\ \text{kW}

Engineering Comment

Actual recovery is lower after effectiveness, approach temperature, fouling and control constraints.

Plausibility Check

Several kilowatts per kelvin over eighty kelvin gives hundreds of kilowatts.

Exercise 16: Pressure-Drop Utilization

Allowable tube-side pressure drop is $70\ \text{kPa}$ . Measured pressure drop is $58\ \text{kPa}$ . Compute utilization.

Solution

U_{\Delta P}=\dfrac{58}{70}=82.9\%

Engineering Comment

Pressure-drop utilization should be trended with fouling. A thermal pass can still fail hydraulically.

Plausibility Check

The measured pressure drop is somewhat below the limit, so utilization is above eighty percent.

Exercise 17: Guarded UA Acceptance

Required UA is $12.5\ \text{kW/K}$ . Measured UA is $13.4\ \text{kW/K}$ with uncertainty $0.6\ \text{kW/K}$ . Does it pass with a guard band?

Solution

UA_g=13.4-0.6=12.8\ \text{kW/K}

Since:

12.8>12.5

it passes.

Engineering Comment

The margin is small. Fouling, seasonal utility temperature or measurement drift could remove it.

Plausibility Check

Subtracting uncertainty leaves only three tenths of a kilowatt per kelvin margin.

Exercise 18: Heat Exchanger Release Gate

A release requires heat-balance closure below $5\%$ , guarded UA above required, pressure-drop utilization below $90\%$ and no unresolved vibration concern. Results are $4.0\%$ , pass, $82.9\%$ and one unresolved vibration concern. Does it release?

Solution

The unresolved vibration concern blocks release:

1>0

The exchanger release fails.

Engineering Comment

Thermal performance cannot override a mechanical integrity concern. Exchanger release needs thermal, hydraulic and mechanical evidence.

Plausibility Check

All numerical thermal and hydraulic screens pass, but the release rule includes an explicit mechanical blocker.

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