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Heat exchanger area
Enter the duty, the four temperatures and an overall coefficient. The calculator returns the log mean temperature difference and the surface area, and tells you if the temperatures cross.
Your figures
Counter-current gets more out of the same area and is the default for a reason.
Plate water-water 3000–5000, shell & tube water-oil 300–600, steam-liquid 800–1500.
Result
Area
- Surface area
- 4.8 m²
- LMTD
- 64.9 K
- Approach, hot end
- 70 K
- Approach, cold end
- 60 K
Q ÷ (U × LMTD) = 250 kW ÷ (800 × 64.9)
Counter-current
A close approach buys area steeply — halving the approach roughly doubles the exchanger.
Check
- Hot side drop
- 60 K
- Cold side rise
- 50 K
- Hot flow needed
- 3,583 kg/h
- Cold flow needed
- 4,300 kg/h
If the hot side is water — Q = m cp ΔT at 4.186 kJ/kg·K
Same basis
Questions
What does it mean if the temperatures cross?
That no exchanger of any size can do what you have asked. In counter-current flow the hot outlet can approach the cold inlet but cannot go below it. If the calculator reports a cross, one of the four temperatures is wrong — usually the hot outlet set too low.
Counter-current or co-current?
Counter-current wherever you have the choice. It gets a larger mean temperature difference out of the same four temperatures, so it needs less area for the same duty, and it can take the hot stream below the cold outlet temperature — which co-current cannot do at all.
What coefficient should I use?
It depends entirely on the fluids and the exchanger type. Plate exchangers on water duties run 3,000 to 5,000 W per square metre per kelvin; shell and tube on water to oil is nearer 300 to 600; steam to a liquid is 800 to 1,500. Fouling allowance reduces all of them, sometimes by half.
Why does a close approach get expensive so fast?
Because the LMTD shrinks toward zero as the approach closes, and area is duty divided by LMTD. Halving the approach roughly doubles the exchanger. The last few degrees of heat recovery are always the dearest ones to buy.