Heat exchangers, coupled transport and performance

Use an enthalpy balance before a temperature formula

A heat exchanger transfers energy between streams through combined fluid-side convection and wall conduction. In steady operation with negligible external heat exchange, shaft work and kinetic/potential changes, the hot stream’s enthalpy decrease equals the cold stream’s increase. If each fluid remains in one phase with approximately constant cp, Q̇=Ch(Th,in−Th,out)=Cc(Tc,out−Tc,in).

Here C=ṁcp is a heat-capacity rate in W/K, not a body capacitance in J/K. It combines mass flow rate and specific heat. Equal mass flow rates need not give equal C if the fluids have different cp. If phase change occurs, use appropriate enthalpy differences rather than forcing a single constant-cp temperature formula.

The local temperature difference drives transfer

In parallel flow, both streams enter the same end. In counterflow, they enter opposite ends, so a position graph must show the cold stream read opposite to its direction of travel. A counterflow cold outlet can be warmer than the hot outlet without any local heat flow from colder to hotter fluid; they occupy opposite ends.

For a simple steady arrangement with approximately constant overall coefficient U, Q̇=UA ΔTlm. The log-mean temperature difference is(ΔT1−ΔT2)/ln(ΔT1/ΔT2), using the correct terminal differences for that flow arrangement. When those differences are equal, use their common value rather than treating0/0 as a failure. U combines wall and film resistances and must be tied to the area convention.

Separate effectiveness from other costs

For a sensible two-stream exchanger, define Q̇max=Cmin(Th,in−Tc,in), with Cmin the smaller capacity rate. Effectiveness ε=Q̇/Q̇max measures how much of that inlet-state-limited transfer is achieved. It is not a fraction of heat converted to work and does not replace an energy-conservation check.

NTU=UA/Cmin compares conductance with stream capacity rate. Performance also depends on the capacity-rate ratio and flow arrangement; NTU alone is insufficient. Increasing flow can alter both convection and pumping demand. Fouling, contact paths, pressure loss and property changes may reduce or change performance. Equal hot/cold energy estimates are a useful consistency check, while a discrepancy could indicate measurement error, storage or an omitted outward loss.

Pair the ends before taking a logarithmic mean

The symbol ln means natural logarithm, the inverse of the exponential already used for transient temperatures. If exp(a)=r for a positive dimensionless ratio r, then ln(r)=a; in particular ln(1)=0. Thus ln(55/40)≈0.31845 because exp(0.31845)≈1.375. A ratio of temperature differences is dimensionless; do not take a logarithm of an unreferenced number of kelvin.

For counterflow, pair quantities at the same physical end: ΔT1=Th,in−Tc,out and ΔT2=Th,out−Tc,in. Under the stated constant-U, constant-capacity-rate, no-loss model, incremental transfer is proportional to the local difference while each stream changes temperature. Combining those local stream balances produces the logarithmic mean, rather than giving both terminal values equal area weight. The result lies between two positive terminal differences. Using only their arithmetic mean is generally a different approximation.

For a new case keep Ch=100 W/K and hot 90→60 °C, giving 3000 W, but set Cc=200 W/K and the cold inlet to 20 °C. The cold rise is 3000/200=15 K, so its outlet is 35 °C. The same-end pairs are recorded below. This is a separate constructed exchanger state, not the earlier equal-capacity case or its straight-profile diagram.

Check a consistent equal-capacity counterflow case

Take constructed Ch=Cc=100 W/K. Hot fluid cools90→60 °C, so Q̇h=100×30=3000 W. Cold fluid warms20→50 °C and gives Q̇c=100×30=3000 W. Both terminal counterflow differences are40 K, so ΔTlm=40 K and the implied UA=3000/40=75 W/K.

The inlet-limited maximum is 100×(90−20)=7000 W. Effectiveness is 3000/7000=3/7≈42.9%, and NTU=75/100=0.75. Equal capacity rates make the local difference constant in this uniform-conductance model; they do not make a finite exchanger100% effective.

For the new unequal-capacity case, ΔTlm=(55−40)/ln(55/40)≈15/0.31845≈47.10 K, so the implied UA=3000/47.1026≈63.69 W/K. Both streams still give 3000 W. Effectiveness is 3000/[100×(90−20)]≈42.9%, but the required conductance differs from the earlier 75 W/K equal-capacity state. Equal transfer and inlet conditions do not by themselves fix conductance when the other capacity rate changes.

The arithmetic mean of these terminal differences is 47.5 K. Substituting it would infer UA=3000/47.5≈63.16 W/K, rather than 63.69 W/K in the logarithmic model. The small discrepancy for these particular values does not make the formulas identical. Pairing 90−20 and 60−35 would be a more fundamental mistake: those compare temperatures at opposite physical ends. The existing straight profiles apply only to the equal-capacity example; the new local difference varies along the exchanger.

Try the reasoning

1. Why do both plotted temperatures decrease from left to right in the counterflow example?

   Show the answer and reasoning

   The horizontal coordinate follows the hot stream. The cold stream flows the other way, warming from right to left.

2. Does equal hot/cold capacity rate mean perfect heat exchange?

   Show the answer and reasoning

   No. Finite conductance and inlet conditions still limit transfer.

3. Can a constant-cp temperature balance cover condensation unchanged?

   Show the answer and reasoning

   Not generally. Phase-change enthalpy must be included.

4. In another supplied counterflow state, keep the hot 90→60 °C stream at Ch=100 W/K and cold inlet 20 °C, but set Cc=150 W/K. Find the cold outlet, both terminal differences, logarithmic mean and implied UA.

   Show the answer and reasoning

   The transfer is 3000 W, so the cold rise is 3000/150=20 K and outlet 40 °C. Same-end differences are 90−40=50 K and 60−20=40 K. ΔTlm=10/ln(50/40)≈44.81 K and UA≈3000/44.8142=66.94 W/K. This is another supplied state, not a prediction that the original unchanged exchanger will necessarily produce these endpoints.

5. A calculation substitutes ln(55)/ln(40) for ln(55/40) in the new case. Why is that wrong?

   Show the answer and reasoning

   A logarithm of a ratio is a difference of logarithms on a compatible dimensionless reference, not their quotient: ln(55/40)=ln(1.375)≈0.31845. Use the ratio of the same-end temperature differences directly; the logarithm is not an extra temperature averaging operation.

What to carry forward

Balance both streams, define the arrangement and temperature differences, then evaluate effectiveness with a compatible boundary.