MECH3610 · Advanced Thermofluids
Heat Exchangers: LMTD & Effectiveness-NTU
Heat exchangers are analysed by two complementary methods: the log-mean-temperature-difference (LMTD) method when the inlet and outlet temperatures are known, and the effectiveness-NTU method when outlet temperatures are not. This chapter covers parallel- vs counter-flow profiles, the stream energy balances, the F correction factor and the effectiveness relations. It underpins the Week-4 Heat-Exchangers lab (feeding the 20% report) and appears on the exams.
What this chapter covers
- 01Stream energy balances Q = m_dot_h*c_p,h*(T_h,i - T_h,o) = m_dot_c*c_p,c*(T_c,o - T_c,i)
- 02Rate equation Q = U*A*deltaT_lm and the overall coefficient U from a resistance network
- 03Counter-flow vs parallel-flow temperature profiles and their end differences deltaT1, deltaT2
- 04Log-mean temperature difference deltaT_lm = (deltaT1 - deltaT2)/ln(deltaT1/deltaT2)
- 05Shell-and-tube / cross-flow correction factor F: deltaT_lm = F*deltaT_lm,counterflow
- 06Heat-capacity rates C = m_dot*c_p, C_min and C_max, and C_r = C_min/C_max
- 07Effectiveness eps = Q/Q_max with Q_max = C_min*(T_h,i - T_c,i) and NTU = U*A/C_min
- 08Choosing LMTD (outlets known) vs effectiveness-NTU (outlets unknown, sizing/rating)
Counter-flow heat exchanger area by the LMTD method
- +1Heat duty from the hot-side energy balance Q = m_dot_h*c_p,h*(T_h,i - T_h,o) = 0.5*2100*(120 - 60) = 63000 W = 63 kW.
- +1For counter-flow the ends are deltaT1 = T_h,i - T_c,o = 120 - 50 = 70 C and deltaT2 = T_h,o - T_c,i = 60 - 20 = 40 C (hot inlet paired with cold outlet).
- +1Log-mean temperature difference deltaT_lm = (deltaT1 - deltaT2)/ln(deltaT1/deltaT2) = (70 - 40)/ln(70/40) = 30/ln(1.75) = 30/0.560 = 53.6 C.
- +1Required area A = Q/(U*deltaT_lm) = 63000/(350*53.6) = 63000/18760 = 3.36 m2.
Key terms
- LMTD method
- Sizes or rates an exchanger from Q = U*A*deltaT_lm using the log-mean temperature difference; best when all four inlet and outlet temperatures are known.
- Log-mean temperature difference
- deltaT_lm = (deltaT1 - deltaT2)/ln(deltaT1/deltaT2), the correct average driving difference for an exchanger because the local difference varies exponentially along the length.
- Correction factor (F)
- A chart-read factor (0 < F <= 1) that adjusts the counter-flow LMTD for shell-and-tube or cross-flow geometries: deltaT_lm = F*deltaT_lm,counterflow. F near 1 means the unit performs close to counter-flow.
- Heat-capacity rate (C)
- C = m_dot*c_p [W/K] for a stream. The smaller of the two, C_min, limits the maximum possible heat transfer; C_r = C_min/C_max is the capacity ratio.
- Effectiveness (eps)
- eps = Q/Q_max, the actual heat transfer divided by the thermodynamic maximum Q_max = C_min*(T_h,i - T_c,i). It depends only on NTU and C_r for a given flow arrangement.
- Number of transfer units (NTU)
- NTU = U*A/C_min, a dimensionless size of the exchanger. With C_r it gives the effectiveness directly, which is why the effectiveness-NTU method suits problems with unknown outlet temperatures.
Heat Exchangers: LMTD & Effectiveness-NTU FAQ
When do I use LMTD versus effectiveness-NTU?
Use LMTD when you know all the inlet and outlet temperatures (a straightforward sizing problem). Use effectiveness-NTU when one or both outlet temperatures are unknown — for example, rating an existing exchanger of known area — because LMTD would need iteration but eps-NTU gives the answer directly from NTU and C_r.
How do I pair the end temperature differences?
It depends on the arrangement. In counter-flow the streams run opposite ways, so the hot inlet meets the cold outlet: deltaT1 = T_h,i - T_c,o and deltaT2 = T_h,o - T_c,i. In parallel flow both fluids enter together: deltaT1 = T_h,i - T_c,i and deltaT2 = T_h,o - T_c,o. Counter-flow gives a larger mean difference for the same temperatures.
Why is counter-flow better than parallel flow?
Counter-flow keeps a more uniform temperature difference along the exchanger and can raise the cold outlet above the hot outlet, so it achieves a larger LMTD and higher effectiveness for the same area and flow rates. Parallel flow's difference collapses as the streams approach a common temperature.
How is this examined in MECH3610?
Through the Week-4 Heat-Exchangers lab, which feeds the 20% laboratory report, and as LMTD/effectiveness-NTU calculation questions on the exams. Marks go to the correct energy balance, end-difference pairing and method choice. Confirm the lab and exam details on Moodle.
Exam move
Learn the two methods as a fork: known outlets -> LMTD (Q = U*A*deltaT_lm); unknown outlets -> effectiveness-NTU (find NTU = U*A/C_min and C_r, read eps, then Q = eps*C_min*(T_h,i - T_c,i)). Drill the counter-flow vs parallel-flow end-difference pairing until it is automatic, and always start from a stream energy balance to get Q or a missing temperature. Sketch the temperature-length profile for each arrangement so the pairing is visible. For the lab, connect the measured stream temperatures to U through Q = U*A*deltaT_lm. Confirm the lab and exam format on Moodle.
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