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This heat exchanger design sample shows how a student might structure the calculation for the shell-and-tube heat exchanger HE2 design project, following Option 1 (HE2 designed alone). The brief’s flow-diagram values are not reproduced on the question page, so every input below is illustrative and must be replaced with your own figures. For the reasoning behind each step, read how to approach the heat exchanger design project.
Planning table
| Part of the brief | Suggested share of answer | What markers look for |
|---|---|---|
| Inputs and energy balance | about 10% | Duty fixed from the oil side; units and signs consistent |
| LMTD and F-factor | about 10% | Correct R and P; F above about 0.8 |
| Assumed U, area, tube layout | about 20% | Standard tube, triangular pitch, tube count, passes |
| Tube-side and shell-side coefficients | about 30% | Re and Pr checked, correct correlations, velocity in range |
| Recalculated U and error check | about 15% | Fouling and wall included; error between 0% and +20% |
| Iteration table and summary | about 15% | Every item the brief asks for, stated once, clearly |
Percentages are suggestions only; check your brief for the marking scheme.
1. Illustrative inputs and assumptions
| Item (illustrative figures) | Value |
|---|---|
| Oil mass flow, inlet and outlet temperature | 10.0 kg/s, 90 °C to 50 °C |
| Water mass flow, inlet temperature | 12.0 kg/s, 20 °C |
| Oil properties (brief) | cp 2,500 J/kg·K; ρ 840 kg/m³; μ 0.0012 Pa·s; k 0.135 W/m·K |
| Water properties (brief) | cp 4,200 J/kg·K; ρ 995 kg/m³; μ 0.0008 Pa·s; k 0.59 W/m·K |
| Fouling resistance, each side (illustrative) | 0.0002 m²·K/W |
| Tube wall conductivity (brief) | 50 W/m·K |
Water is placed on the tube side because it is the fouling-prone, higher-pressure stream and the velocity can be controlled by the pass count. The oil is on the shell side (Kern, 1950; Towler and Sinnott, 2022).
2. Energy balance
The duty is fixed from the oil side: Q = ṁocp,o(Tin − Tout) = 10.0 × 2,500 × 40 = 1.000 × 106 W.
The water outlet follows from the same duty: ΔTw = Q / (ṁwcp,w) = 106 / (12.0 × 4,200) = 19.84 K, so the water leaves at 39.84 °C.
3. LMTD and correction factor
For counter-current flow, ΔT1 = 90 − 39.84 = 50.16 K and ΔT2 = 50 − 20 = 30.00 K. The log-mean difference is (50.16 − 30.00) / ln(50.16 / 30.00) = 39.22 K.
For one shell pass and an even number of tube passes, R = (90 − 50) / 19.84 = 2.016 and P = 19.84 / (90 − 20) = 0.2834. The standard 1–2 expression (Bergman et al., 2017) gives F = 0.906, so the corrected mean difference is 0.906 × 39.22 = 35.52 K. An F value this far above 0.8 means the temperature profile is practical.
4. Iteration on the overall coefficient
The design used 19.05 mm OD tubes with 15.75 mm ID (1.65 mm wall), 4.88 m per pass, triangular pitch pt = 1.25 × OD = 23.8 mm, and four tube passes. Four passes were chosen because two passes gave a water velocity of about 0.8 m/s, below the 1–2.5 m/s guidance on the question page. Check the tube dimensions against the standard-tube table in your brief.
4.1 Trial 1: assumed U = 400 W/m²·K
Required area A = Q / (U F ΔTlm) = 106 / (400 × 35.52) = 70.4 m². Area per tube = π × 0.01905 × 4.88 = 0.2921 m², so 241 tubes are needed, rounded up to 244 (a multiple of four passes), giving 61 tubes per pass.
Tube side. Velocity u = (12.0 / 995) / (61 × π × 0.01575² / 4) = 1.01 m/s. Re = ρud/μ = 19,879 and Pr = cpμ/k = 5.69. Dittus–Boelter for a heated fluid, Nu = 0.023 Re0.8 Pr0.4 (Dittus and Boelter, 1985), gives Nu = 126.7, so hi = 126.7 × 0.59 / 0.01575 = 4,745 W/m²·K.
Shell side. The bundle diameter is Db = Do(Nt/K1)1/n1, with K1 = 0.175 and n1 = 2.285 for four passes on a 1.25 triangular pitch (Towler and Sinnott, 2022; confirm against your table). That gives Db = 0.453 m. A clearance of 50 mm for a split-ring floating head (read from the brief’s chart or formula; 50 mm is assumed here) gives a shell diameter Ds = 0.503 m.
Taking baffle spacing B = 0.4 Ds = 0.201 m, the equivalent diameter is de = 1.10 (pt² − 0.917 Do²) / Do = 13.5 mm. The cross-flow area is As = (pt − Do) Ds B / pt = 0.0202 m², so Gs = 494.3 kg/m²·s and the shell-side velocity is 0.59 m/s.
Res = Gsde/μ = 5,572 and Pr = 22.2. The Kern-type correlation Nu = 0.36 Re0.55 Pr1/3 (Kern, 1950; Towler and Sinnott, 2022), with the viscosity-ratio term of Sieder and Tate (1936) taken as 1, gives Nu = 116.3 and ho = 116.3 × 0.135 / 0.0135 = 1,161 W/m²·K.
The overall coefficient, referred to the outside area, is the sum of resistances: 1/U = 1/ho + Rf,o + Do ln(Do/Di) / 2kw + Rf,iDo/Di + Do/(Dihi). The calculated U is 627 W/m²·K, an error of +56.8% against the assumed 400. That is outside 0 to +20%, so the trial is rejected.
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4.2 Iteration table
Each trial repeats the steps above with a new assumed U. The values below were calculated from the same equations (illustrative figures).
| Trial | U assumed (W/m²·K) | Area (m²) | Tubes (per pass) | Velocity (m/s) | hi | Ds (m) | ho | U calculated | Error |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 400 | 70.4 | 244 (61) | 1.01 | 4,745 | 0.503 | 1,161 | 627 | +56.8% |
| 2 | 500 | 56.3 | 196 (49) | 1.26 | 5,653 | 0.462 | 1,276 | 677 | +35.5% |
| 3 | 600 | 46.9 | 164 (41) | 1.51 | 6,520 | 0.431 | 1,377 | 719 | +19.9% |
| 4 | 650 | 43.3 | 152 (38) | 1.63 | 6,929 | 0.418 | 1,422 | 737 | +13.4% |
Trial 3 passes by only 0.1 percentage points, which is too fragile if any input shifts slightly, so trial 4 is adopted. The coefficients rise with each trial because fewer tubes raise both velocities.
5. Final design check (trial 4)
With 152 tubes (38 per pass), the tube-side velocity is 1.63 m/s, Re = 31,911, Pr = 5.69 and hi = 6,929 W/m²·K. On the shell side Db = 0.368 m, Ds = 0.418 m, B = 0.167 m, As = 0.01399 m², Gs = 714.9 kg/m²·s, Res = 8,058 and ho = 1,422 W/m²·K.
| Resistance, referred to outside area (illustrative figures) | Value (m²·K/W) |
|---|---|
| Shell-side film, 1/ho | 7.03 × 10-4 |
| Shell-side fouling | 2.00 × 10-4 |
| Tube wall | 0.36 × 10-4 |
| Tube-side fouling (scaled by Do/Di) | 2.42 × 10-4 |
| Tube-side film (scaled by Do/Di) | 1.75 × 10-4 |
| Total | 13.56 × 10-4 |
This gives U = 1 / 0.001356 = 737 W/m²·K and ε = (737 − 650) / 650 × 100 = +13.4%, which lies within the accepted 0 to +20% window. The shell-side film is the controlling resistance at about 52% of the total, which is typical when oil is the shell fluid.
The installed area is 152 × π × 0.01905 × 4.88 = 44.4 m². The area needed at the calculated U is 106 / (737 × 35.52) = 38.2 m², a margin of 16%. The shell is slender (4.88 m long for a 0.418 m diameter, a ratio of about 11.7), so a student could discuss a shorter tube length with more tubes, in line with TEMA (2019) practice.
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6. Summary of the design (Option 1, illustrative)
| Item | Result (illustrative figures) |
|---|---|
| Configuration | 1 shell pass, 4 tube passes, water in tubes, oil in shell |
| Duty | 1.000 MW; water outlet 39.8 °C |
| Tubes | 19.05 mm OD, 15.75 mm ID, 4.88 m per pass, 152 tubes (38 per pass) |
| Shell diameter and baffle spacing | 0.418 m; 0.167 m |
| Overall U (OD basis) | 737 W/m²·K (assumed 650, error +13.4%) |
| Heat transfer area (OD basis) | 44.4 m² |
For Option 2, the HE1 energy balance and the mixing condition (the two water outlets combining to the target temperature) would be added before step 2, giving m1, m2, m3 and T1, T2, T3. The rest of the method is unchanged.
When your own figures are ready, you can send your brief through the assignment submission page, or see the wider engineering assignment support available. Reference lists can also be checked with the citation and referencing check.
References
Bergman, T.L., Lavine, A.S., Incropera, F.P. and DeWitt, D.P. (2017) Fundamentals of heat and mass transfer. 8th edn. Hoboken, NJ: John Wiley & Sons.
Dittus, F.W. and Boelter, L.M.K. (1985) ‘Heat transfer in automobile radiators of the tubular type’, International Communications in Heat and Mass Transfer, 12(1), pp. 3–22.
Kern, D.Q. (1950) Process heat transfer. New York: McGraw-Hill.
Sieder, E.N. and Tate, G.E. (1936) ‘Heat transfer and pressure drop of liquids in tubes’, Industrial & Engineering Chemistry, 28(12), pp. 1429–1435.
Tubular Exchanger Manufacturers Association (TEMA) (2019) Standards of the Tubular Exchanger Manufacturers Association. 10th edn. Tarrytown, NY: TEMA.
Towler, G. and Sinnott, R. (2022) Chemical engineering design: principles, practice and economics of plant and process design. 3rd edn. Oxford: Butterworth-Heinemann.

Singapore Assignment Help Editorial Team
This model answer was prepared by our editorial team in response to a real assignment brief we received. It shows one way to approach the task and is meant as a study guide, not for submission.