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MechanicalHeat TransfermediumCourseworkPE / Advanced

Log Mean Temperature Difference, Counterflow

In a counterflow heat exchanger:

  • The hot fluid enters at 150∘C150^\circ\text{C} and leaves at 90∘C90^\circ\text{C}
  • The cold fluid enters at 30∘C30^\circ\text{C} and leaves at 70∘C70^\circ\text{C}

Find the log mean temperature difference (LMTD).

Given
Th,in150 degCHot inlet
Th,out90 degCHot outlet
Tc,in30 degCCold inlet
Tc,out70 degCCold outlet
Hint 1

Draw the temperature profile along the exchanger. In counterflow, which end does the hot inlet sit at?

Hint 2

The hot inlet faces the cold outlet: ΔT₁ = 150 − 70 = 80.

Hint 3

LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), with a natural logarithm.

Worked solution — try the problem first

LMTD=ΔT1−ΔT2ln⁡(ΔT1/ΔT2)\text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}

Pair the ends correctly. In counterflow the fluids move in opposite directions, so the hot inlet faces the cold outlet:

ΔT1=Th,in−Tc,out=150−70=80∘C\Delta T_1 = T_{h,in} - T_{c,out} = 150 - 70 = 80^\circ\text{C} ΔT2=Th,out−Tc,in=90−30=60∘C\Delta T_2 = T_{h,out} - T_{c,in} = 90 - 30 = 60^\circ\text{C}

This end-pairing is the whole difference between counterflow and parallel flow, and it is where most errors originate.

LMTD=80−60ln⁡(80/60)=20ln⁡(1.3333)=200.28768=69.52∘C\text{LMTD} = \frac{80 - 60}{\ln(80/60)} = \frac{20}{\ln(1.3333)} = \frac{20}{0.28768} = 69.52^\circ\text{C}

LMTD=69.5∘C\boxed{\text{LMTD} = 69.5^\circ\text{C}}

Sanity check: the LMTD must lie between ΔT2=60\Delta T_2 = 60 and ΔT1=80\Delta T_1 = 80, and slightly below their arithmetic mean of 70. It does — 69.5. Any answer outside 60-80 is wrong on inspection.

Concepts:Heat exchangersLMTDCounterflow vs parallel flow

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