Log Mean Temperature Difference (Parallel Flow)

ΔTlm=ΔT1−ΔT2ln⁡(ΔT1/ΔT2)\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}

Worked example: Same streams in parallel flow → 55.81 K (vs 69.52 counter) — press Try an example to run it live, then adjust anything.

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Log Mean Temperature Difference (Parallel Flow) explained

Th,inTh,outTc,inTc,outΔTlm

In parallel (co-current) flow both fluids enter at the same end, so the pairing changes: ΔT₁ = Th,in − Tc,in at the inlet end and ΔT₂ = Th,out − Tc,out at the outlet end. The same 150 °C oil and 30 °C water, delivered to 90 °C and 70 °C, now give terminals of 120 K and 20 K and a log mean of only 55.8 K — a fifth less driving force than the counterflow arrangement, from identical fluids at identical temperatures. Same duty, same U, and you need 25% more surface. That is why counterflow is the default and parallel flow needs a reason.

It does have reasons. Parallel flow puts the biggest ΔT where the cold fluid is coldest, which brings a viscous fluid up to temperature fast, and it holds the hot-end wall temperature lower, which matters when a product scorches, a coating cures or a thermally sensitive fluid must never see a hot tube. The hard limit is thermodynamic: the two outlet temperatures can approach each other but can never cross, so a parallel-flow unit can never heat the cold stream above the hot stream's exit. If your process needs a cross, no amount of surface in a co-current unit will deliver it.

Log Mean Temperature Difference (Parallel Flow) formula

ΔTlm=ΔT1−ΔT2ln⁡(ΔT1/ΔT2)\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}
Where
  • ΔTlm\Delta T_{lm}= Log mean temperature difference (C°)
  • Th,inT_{h,in}= Hot stream inlet (°C)
  • Th,outT_{h,out}= Hot stream outlet (°C)
  • Tc,inT_{c,in}= Cold stream inlet (°C)
  • Tc,outT_{c,out}= Cold stream outlet (°C)