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Chemical Engineering DISCUSSION
I am sizing an exchanger to cool oil from 150 °C to 90 °C using water that heats from 30 °C to 70 °C. The duty is 200 kW and I am assuming an overall coefficient of 400 W/m²·K. With the same four temperatures I get a log mean temperature difference of 69.5 K for counter-current flow and 55.8 K for co-current flow.
Why does the flow direction change the mean temperature difference when the inlet and outlet temperatures are identical, and how much does it matter for the area?
The LMTD is built from the temperature differences at the two ends of the exchanger: LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2). Which temperatures face each other at each end depends on the flow direction.
Counter-current: hot inlet meets cold outlet, 150 - 70 = 80 K, and hot outlet meets cold inlet, 90 - 30 = 60 K. LMTD = 20 / ln(80 / 60) = 69.5 K. Co-current: both inlets together, 150 - 30 = 120 K, and both outlets together, 90 - 70 = 20 K. LMTD = 100 / ln(6) = 55.8 K.
Area follows from A = Q / (U × LMTD): 200,000 / (400 × 69.5) = 7.2 m² for counter-current and 200,000 / (400 × 55.8) = 9.0 m² for co-current, 25 percent more surface for the same duty.
The physical reason is how evenly the driving force is spread. In co-current flow the first part of the exchanger has a huge temperature difference and transfers heat very fast, but the two streams converge, and the last part works with only 20 K. Much of the area is doing little. In counter-current flow the difference stays between 60 and 80 K along the whole length, so every square metre is used well.
There is also a hard limit. In co-current flow the cold stream can never leave hotter than the hot stream leaves, because both approach a common temperature. In counter-current flow it can: heating the water to 100 °C while the oil leaves at 90 °C is possible in counter-current and impossible in co-current, however large the exchanger.
Most industrial exchangers are neither. A shell-and-tube unit with one shell pass and two tube passes has one tube pass running with the shell fluid and one against it. The mean temperature difference is then F × LMTD for counter-current flow, where the correction factor F is read from charts, or computed from a formula, using two ratios. Here R = (150 - 90) / (70 - 30) = 1.5 and P = (70 - 30) / (150 - 30) = 0.33.
For a one-shell, two-tube-pass exchanger those values give F of about 0.91, so the effective difference is 63 K and the area about 7.9 m². Designers look for F above about 0.75 to 0.8; below that the configuration is too close to a temperature cross, and shells in series or a true counter-current type such as a plate or double-pipe exchanger is used.
Co-current flow still has its uses. Because the hottest fluid meets the coldest at the inlet, the wall temperature is more moderate along the exchanger, which can protect a heat-sensitive product from overheating at the hot end or stop a viscous fluid from congealing at the cold end. It is also used when a stream must be quenched quickly.
When one stream is at constant temperature, direction makes no difference at all. With condensing steam at 120 °C heating water from 30 to 70 °C, the end differences are 90 K and 50 K either way, the LMTD is the same, and F is 1 for any pass arrangement.
Some calculation checks. Do the energy balance before the LMTD: the water flow needed is Q / (c_p × ΔT) = 200 / (4.18 × 40) = 1.2 kg/s, and the four temperatures must be consistent with both flow rates. If the two end differences are equal, the formula gives zero divided by zero; the mean difference is then simply that common value.
The LMTD assumes a constant overall coefficient and constant heat capacities with no phase change part-way. If a stream desuperheats, condenses and subcools in one unit, split it into zones and treat each separately. And when the outlet temperatures are what you are trying to find, for an existing exchanger, the effectiveness-NTU method avoids the trial and error that the LMTD method needs.
