Science & engineering

Counterflow or parallel heat exchanger NTU and LMTD calculator

An exchanger can be rated from inlet temperatures, stream heat-capacity rates and overall conductance without first knowing outlet temperatures.

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Explore the result

An exchanger can be rated from inlet temperatures, stream heat-capacity rates and overall conductance without first knowing outlet temperatures. The effectiveness–NTU method predicts those outlets under a stated ideal flow arrangement.

Worked example comparison
ExampleOutputWhat to notice
Worked exampleIdeal sensible heat transfer rate4,000 WFor inlet temperatures one hundred and twenty degrees, equal capacity rates one hundred watts per kelvin and UA one hundred, counterflow NTU is one and effectiveness is one half. Heat transfer is four thousand watts and both outlet temperatures are sixty degrees. The equal-capacity counterflow case uses its analytical limiting expression instead of dividing by a vanishing capacity-ratio difference.
Compare a changed assumptionIdeal sensible heat transfer rate3,458.658867 WKeep conductance and capacities fixed but switch arrangement; parallel-flow effectiveness is lower for this example.

Check your understanding

Can counterflow cold outlet exceed hot outlet?

Yes, because those outlets occur at opposite ends. The model still respects the local thermal driving differences.

Predict the comparison before running it

Keep conductance and capacities fixed but switch arrangement; parallel-flow effectiveness is lower for this example.

Behind the numbers

How this calculation works

An exchanger can be rated from inlet temperatures, stream heat-capacity rates and overall conductance without first knowing outlet temperatures. The effectiveness–NTU method predicts those outlets under a stated ideal flow arrangement. This page compares single-pass counterflow and parallel flow with constant capacity rates. It also reports the logarithmic mean temperature approach equivalent to the predicted heat transfer and entered UA.

THE FORMULANTU=UA/Cmin; Q=effectiveness Cmin(Th,in−Tc,in); equivalent LMTD=Q/UA

A worked example

For inlet temperatures one hundred and twenty degrees, equal capacity rates one hundred watts per kelvin and UA one hundred, counterflow NTU is one and effectiveness is one half. Heat transfer is four thousand watts and both outlet temperatures are sixty degrees. The equal-capacity counterflow case uses its analytical limiting expression instead of dividing by a vanishing capacity-ratio difference.

Good to know

The reported mean approach is Q divided by UA; at zero UA its limiting inlet approach is shown while heat transfer is zero. This avoids reporting an undefined zero-over-zero result as a measured driving difference. Multipass shell geometry, fouling, heat loss to surroundings and pressure drop are excluded. The calculation is an ideal rating comparison rather than approval of a specific exchanger or operating condition.

What this calculation answers

An exchanger can be rated from inlet temperatures, stream heat-capacity rates and overall conductance without first knowing outlet temperatures. The effectiveness–NTU method predicts those outlets under a stated ideal flow arrangement.

This page compares single-pass counterflow and parallel flow with constant capacity rates. It also reports the logarithmic mean temperature approach equivalent to the predicted heat transfer and entered UA.

The Heat energy calculator relates an amount of heat to temperature change. Exchanger rating additionally needs two flowing capacity rates, conductance and stream arrangement.

Use the right input conventions

Heat-capacity rate is mass flow times specific heat, in watts per kelvin; it is not specific heat alone. UA is overall heat-transfer coefficient times area in the same watts-per-kelvin unit.

Enter hot inlet above cold inlet and use positive stream capacity rates. Both streams exchange sensible heat only, so condensation, evaporation and strongly changing specific heat are outside the model.

Interpret the result

Effectiveness is heat transfer divided by the maximum possible value based on the smaller capacity rate and inlet temperature gap. NTU is conductance divided by that smaller capacity rate.

Hot outlet follows its heat loss and cold outlet follows the same heat gain. Counterflow permits a cold outlet above the hot outlet without violating local temperature ordering because the two outlets are at opposite physical ends.

Work through the example

For inlet temperatures one hundred and twenty degrees, equal capacity rates one hundred watts per kelvin and UA one hundred, counterflow NTU is one and effectiveness is one half.

Heat transfer is four thousand watts and both outlet temperatures are sixty degrees. The equal-capacity counterflow case uses its analytical limiting expression instead of dividing by a vanishing capacity-ratio difference.

Where the model stops

The reported mean approach is Q divided by UA; at zero UA its limiting inlet approach is shown while heat transfer is zero. This avoids reporting an undefined zero-over-zero result as a measured driving difference.

Multipass shell geometry, fouling, heat loss to surroundings and pressure drop are excluded. The calculation is an ideal rating comparison rather than approval of a specific exchanger or operating condition.

A few questions, answered

Can counterflow cold outlet exceed hot outlet?

Yes, because those outlets occur at opposite ends. The model still respects the local thermal driving differences.

What changes in the comparison?

Keep conductance and capacities fixed but switch arrangement; parallel-flow effectiveness is lower for this example.

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