Shell and tube heat exchanger design starts with a heat balance, a feasible temperature profile, and a preliminary heat-transfer area. When the inlet and outlet temperatures are known, the LMTD method calculates area from A = Q/(U × F × LMTD). The correction factor F accounts for the selected flow arrangement’s departure from ideal counterflow.
This guide develops a sensible-heat oil cooler example with one shell pass and two tube passes. The calculated duty is 660 kW, the counterflow LMTD is 63.83 K, and F is 0.920. With an illustrative design overall coefficient of 350 W/(m²·K), the preliminary outside tube area is 32.10 m².
That area is a thermal starting point. A complete exchanger design must also establish film coefficients, pressure drops, fouling, tube layout, vibration resistance, materials, thermal expansion, and pressure-boundary integrity.
What the LMTD method calculates
The log mean temperature difference represents the effective temperature driving force for heat transfer under the model’s assumptions. For counterflow:
ΔT₁ = Th,in − Tc,out
ΔT₂ = Th,out − Tc,in
LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)
Use the natural logarithm. Celsius temperature differences and kelvin differences have the same numerical magnitude. If both positive terminal differences are equal, LMTD equals that common difference; the apparent 0/0 in the formula is a removable mathematical limit.
The basic thermal relation is Q = UAΔTm. For a corrected counterflow calculation, ΔTm = F × LMTD. The ASHRAE heat-transfer chapter explains mean temperature difference and overall heat-transfer relationships.
Use our LMTD calculator to compare terminal-temperature calculations. Select the applicable arrangement and inspect the correction-factor assumptions before using the area result.
Collect a consistent design basis first.
| Design information | Why it matters |
|---|---|
| Fluid identity, composition, and phase | Determines properties, compatibility, and the appropriate thermal model |
| Inlet temperatures and required outlet temperatures | Defines duty and temperature approaches |
| Mass flows and operating cases | Establishes energy balance, velocities, and turndown performance |
| Density, viscosity, heat capacity, conductivity | Needed for film coefficients and hydraulic calculations |
| Allowable pressure drop on each side | Constrains passes, tube size, baffles, and velocity |
| Fouling basis and cleaning method | Affects design U and maintainable construction |
| Operating and design pressures/temperatures | Needed for mechanical design and abnormal-condition review |
| Materials, corrosion, and leakage consequences | Influences tube material and construction selection |
Do not independently specify every temperature and both flow rates without checking the heat balance. The hot-side heat release must agree with the cold-side heat gain, allowing for any explicitly modeled heat loss. If measured duties disagree, investigate instrumentation, property data, and operating stability before resizing the exchanger.
For approximately constant heat capacity and no phase change:
Q = ṁhcp,h(Th,in − Th,out) = ṁccp,c(Tc,out − Tc,in)
For significant property variation or phase change, calculate duty from enthalpy differences and divide the exchanger into suitable zones. A single average heat capacity can conceal a limiting local temperature approach.
Why shell-and-tube exchangers need an LMTD correction factor
A multipass exchanger does not generally maintain ideal counterflow everywhere. Tube passes and shell-side flow create a different temperature distribution, so using the counterflow LMTD without correction can understate the area requirement.
F depends on temperatures and arrangement; it is not a fouling factor or discretionary safety factor. Use an equation, chart, or validated model that matches the shell configuration and pass arrangement. University teaching material on LMTD correction factors describes this approach and its assumptions, including approximately uniform U and equal heat-transfer area per pass.
For the idealized one-shell-pass, two-tube-pass sensible-heat arrangement used here, define:
R = (Th,in − Th,out)/(Tc,out − Tc,in)
P = (Tc,out − Tc,in)/(Th,in − Tc,in)
With S = √(R² + 1), for R ≠ 1:
F = [S/(R − 1)] × ln[(1 − P)/(1 − PR)] ÷ ln{[2 − P(R + 1 − S)]/[2 − P(R + 1 + S)]}
Check the logarithm arguments before evaluating this expression. It is not a universal formula for every shell type. R = 1 requires the limiting expression or a validated alternative; near that value, direct substitution can also be numerically unstable.
A very low F signals an unfavorable temperature arrangement and sensitivity to operating changes. Review another configuration, utility condition, or number of shells rather than forcing the equation to return a number. Any minimum acceptable F should come from the project’s design basis and engineering review, not an invented universal code limit.
Worked example: preliminary design of an oil cooler
Assume an oil stream is cooled from 140°C to 80°C using water heated from 30°C to 60°C. Both streams remain liquid. Heat loss to the surroundings is neglected, heat capacities are treated as constant, and the proposed thermal arrangement has one shell pass and two tube passes.
| Parameter | Illustrative value |
|---|---|
| Oil mass flow | 5.0 kg/s |
| Oil mean heat capacity | 2,200 J/(kg·K) |
| Oil inlet/outlet | 140°C / 80°C |
| Water inlet/outlet | 30°C / 60°C |
| Water mean heat capacity | 4,180 J/(kg·K) |
| Assumed design U, outside-area basis | 350 W/(m²·K) |
| Trial tube outside diameter | 19.05 mm |
| Trial effective straight-tube length | 4.5 m |
The oil properties and U are explicit assumptions. They are not verified properties of a specified commercial oil or a rated exchanger. The water outlet temperature is also a process assumption; confirm utility-system constraints and fouling implications for an actual plant.
Step 1: Calculate heat duty
Q = 5.0 × 2200 × (140 − 80) = 660,000 W = 660 kW
The required water flow is:
ṁwater = 660000/[4180 × (60 − 30)] = 5.263 kg/s
This equals approximately 18,947 kg/h. Calculating the water-side duty with the unrounded flow returns 660 kW, so the specified thermal conditions balance.
Step 2: Calculate counterflow LMTD
ΔT₁ = 140 − 60 = 80 K
ΔT₂ = 80 − 30 = 50 K
LMTD = (80 − 50)/ln(80/50) = 63.83 K
The arithmetic mean is 65 K. It is close in this particular case, but substituting it is unnecessary and introduces avoidable error. For more unequal terminal differences, the discrepancy can be larger.
Step 3: Determine the correction factor
R = (140 − 80)/(60 − 30) = 2.000
P = (60 − 30)/(140 − 30) = 0.27273
Substituting these values into the one-shell-pass expression gives F = 0.92045. Therefore:
Corrected mean temperature difference = 0.92045 × 63.8293 = 58.75 K
Keep full numerical precision in the spreadsheet and round only the displayed results.
Step 4: Estimate required heat-transfer area
Ao = 660000/(350 × 58.7517) = 32.10 m²
Using F = 1 would give 29.54 m². Applying the actual example correction increases the estimated area by approximately 8.64% relative to that uncorrected result.
The subscript o indicates outside tube area. Since U was assumed on that basis, area must also be based on tube outside diameter. Mixing an inside-area U with outside area silently changes the thermal prediction.
Step 5: Translate area into a trial tube count
For straight tubes with effective heat-transfer length L:
Aper tube = πdoL = π × 0.01905 × 4.5 = 0.26931 m²
Nrequired = 32.0963/0.26931 = 119.18 tubes
A preliminary round-up to 120 straight tubes gives 32.32 m². With two equal tube passes, this means 60 tubes carrying the flow in parallel per pass, not 120 tubes per pass.
This is only an area-based count. A real tubesheet layout must accommodate pitch, pass-partition lanes, clearances, sealing, and the selected construction. The 120-tube arrangement has less than 1% area above this preliminary estimate and is not a final design margin or vendor selection.
Selecting U and accounting for fouling
The assumed U should eventually be replaced by a calculated or vendor-rated value. For cylindrical tubes on an outside-area basis:
1/Uo = 1/ho + Rf,o + [doln(do/di)]/(2kw) + (do/di)(Rf,i + 1/hi)
Here h represents film coefficients, Rf fouling resistance, kw tube-wall conductivity, and d tube diameter. Inside and outside resistances must be converted to the same area basis. Film coefficients depend on geometry, velocity, properties, and flow regime.
If the specified design U already includes fouling, do not add the same fouling resistance again. Likewise, distinguish fouling allowance, duty contingency, and any additional area allowance; stacking undocumented margins can obscure the design basis. Our heat-transfer fundamentals guide explains the underlying conduction and convection mechanisms.
| Assumed design U, W/(m²·K) | Required area at the same duty and temperatures |
|---|---|
| 250 | 44.93 m² |
| 350 | 32.10 m² |
| 450 | 24.96 m² |
This sensitivity table holds F and terminal temperatures fixed. It shows why an optimistic U can dominate preliminary sizing. It does not establish an expected U range for every oil cooler.
Reproduce the LMTD calculation in Excel
Enter the following inputs in column B, with labels and units beside them. The formulas use English Excel names and comma separators.
| Cell | Input | Value |
|---|---|---|
| B2 | Hot inlet, °C | 140 |
| B3 | Hot outlet, °C | 80 |
| B4 | Cold inlet, °C | 30 |
| B5 | Cold outlet, °C | 60 |
| B6 | Hot mass flow, kg/s | 5 |
| B7 | Hot heat capacity, J/(kg·K) | 2200 |
| B8 | Cold heat capacity, J/(kg·K) | 4180 |
| B9 | Design U, W/(m²·K) | 350 |
| Cell | Output | Formula |
|---|---|---|
| B12 | Duty, W | =B6*B7*(B2-B3) |
| B13 | Cold mass flow, kg/s | =B12/(B8*(B5-B4)) |
| B14 | ΔT₁, K | =B2-B5 |
| B15 | ΔT₂, K | =B3-B4 |
| B16 | Counterflow LMTD, K | =IF(MIN(B14,B15)<=0,NA(),IF(ABS(B14-B15)<0.000001,(B14+B15)/2,(B14-B15)/LN(B14/B15))) |
| B17 | R | =(B2-B3)/(B5-B4) |
| B18 | P | =(B5-B4)/(B2-B4) |
| B19 | S | =SQRT(B17^2+1) |
| B20 | Correction factor F | =IF(ABS(B17-1)<0.000001,NA(),IFERROR((B19/(B17-1))*LN((1-B18)/(1-B18*B17))/LN((2-B18*(B17+1-B19))/(2-B18*(B17+1+B19))),NA())) |
| B21 | Corrected temperature difference, K | =IF(AND(B20>0,B20<=1),B20*B16,NA()) |
| B22 | Outside area, m² | =B12/(B9*B21) |
Validate positive mass flow, heat capacities, and U; require hot inlet above hot outlet, cold outlet above cold inlet, and both terminal differences positive. Check that 0 < P < 1, 1 − PR > 0, and both expressions inside the final logarithm’s ratio are positive for this model. Invalid cases must stop the calculation rather than receive an assumed F.
The R = 1 flag is intentional. It requests a limiting-expression calculation instead of hiding a division-by-zero problem. A small F or an error does not justify setting F to 1. This article supplies formulas for a worksheet; it does not include a newly created downloadable Excel file.
Complete the hydraulic and mechanical design
After selecting trial geometry, calculate tube-side velocity using the number of tubes per pass and actual tube inside diameter. Then evaluate tube-side pressure drop, including entrances, exits, and return regions. Calculate shell-side performance with a method appropriate to its baffles, leakage paths, bypass streams, and flow arrangement.
Changing tube passes can increase velocity and improve film transfer while increasing pressure drop. Adding tubes to gain area can lower velocity and reduce U. Consequently, thermal area and hydraulic performance require iteration rather than one independent calculation each.
Check cleaning access, differential expansion, tube-to-tubesheet joints, nozzle loads, and flow-induced vibration. Decide fluid allocation using pressure, fouling, corrosion, viscosity, maintainability, and leakage consequences together. Avoid a blanket rule that one fluid always belongs on the tube side.
The TEMA Standards page identifies the 2026 edition and its exchanger-specific design provisions. ASME BPVC Section VIII, Division 1 addresses pressure-vessel construction where applicable. Use the editions and requirements adopted by the project and jurisdiction. An LMTD worksheet does not demonstrate compliance with either document.
Temperature feasibility and operating checks
A cold outlet hotter than the hot outlet can be possible in counterflow; those two outlets are at different ends. It does not by itself prove an impossible temperature cross. However, positive counterflow terminal differences alone do not establish feasibility for a particular multipass shell arrangement. Check its correction-factor domain and detailed thermal rating.
When an approach temperature tends toward zero, the driving force can become limiting and required area can grow sharply. Do not use absolute values inside the LMTD formula to conceal negative terminal differences. Review the specified temperatures and chosen arrangement instead.
For unknown outlet temperatures in an existing exchanger, an effectiveness–NTU method or iterative rating model may be more convenient. For boiling, condensation, mixtures with temperature glide, or strong property variation, use appropriate zones and enthalpy calculations. The single-phase correction-factor worksheet here should not be extended automatically to those services.
Check summer cooling-water temperature, minimum flow, fouled operation, and start-up conditions. If operating performance declines, reconcile hot- and cold-side heat balances before blaming fouling. Reduced utility flow, instrument error, bypass leakage, and changed process properties can also reduce apparent performance.
Frequently asked questions
1. What is the main equation for shell and tube heat exchanger design?
For preliminary LMTD sizing, A = Q/(U × F × LMTD). Use a consistent area basis for U and A, and determine F for the actual flow arrangement.
2. Why is a correction factor required?
Multipass flow differs from ideal counterflow. F converts the counterflow LMTD into the model’s effective temperature difference; it is not a safety or fouling allowance.
3. What happens when the two terminal differences are equal?
LMTD equals their common positive value. Use the mathematical limit in Excel rather than dividing zero by zero.
4. Can I assume U = 350 W/(m²·K)?
Only as an explicitly justified preliminary assumption. The example uses that value for demonstration; actual U depends on both fluids, geometry, velocity, wall resistance, and fouling.
5. Is a 120-tube exchanger sufficient for the worked example?
It supplies approximately 32.32 m² of nominal effective outside area at the assumed dimensions. Hydraulic rating, achievable U, layout, mechanical requirements, and project allowances remain unresolved.
6. Is F = 0.8 always the minimum permitted value?
No universal acceptance threshold is established here. Apply the project criterion and evaluate thermal feasibility, sensitivity, and the manufacturer’s rating for the selected arrangement.
7. Is LMTD enough to design a condenser?
LMTD can be useful, but duty should use enthalpy change and the model must account for relevant desuperheating, condensation, subcooling, pressure loss, and mixture effects.
Use LMTD as the start of an iterative design
The example’s 660 kW duty requires a preliminary 32.10 m² at U = 350 W/(m²·K), F = 0.920, and LMTD = 63.83 K. Carry that result into a geometry-based thermal and hydraulic rating, then complete mechanical checks and document the governing operating cases. Find more practical engineering guides and tools at GrowMechanical.
