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Calcimator

Heat Exchanger Sizing Calculator

Size a counterflow heat exchanger using the LMTD method. Calculate heat duty, log mean temperature difference, and required surface area.

About this calculator

This calculator sizes a counterflow liquid-to-liquid heat exchanger using the classic LMTD (log mean temperature difference) method. It first computes the heat duty, Q = ṁ·Cp·ΔT, converting your hot-side flow rate from GPM to a mass flow rate using water's density (8.33 lb/gal) and multiplying by the hot fluid's temperature drop and specific heat — so for anything other than water, adjusting the specific heat input is essential since the density conversion itself is fixed to water. The LMTD is then calculated from the two temperature differences at each end of a counterflow arrangement (hot-in vs. cold-out, and hot-out vs. cold-in), using the standard logarithmic-mean formula, with a fallback to the simple arithmetic difference in the degenerate case where both end differences are nearly equal.

Required surface area follows directly from A = Q / (U × LMTD): a lower overall heat transfer coefficient (U) — say, water-to-oil versus water-to-water — demands proportionally more surface area for the same duty. Effectiveness is reported as actual heat transferred over the theoretical maximum possible (assuming the hot stream is the limiting, minimum-capacity side), giving a quick sense of how hard the exchanger is working relative to the ideal. Two assumptions are worth flagging: the model assumes true counterflow (parallel flow or crossflow would need a correction factor applied to the LMTD), and it assumes the hot side has the smaller thermal capacity rate — if your cold side actually has less capacity, the effectiveness number will be off.

Inputs

°F
°F
°F
°F
GPM
BTU/lb·°F
BTU/hr·ft²·°F

Results

Heat Duty

3,998,400 BTU/hr

Required Surface Area

340.4 ft²

≈ 8 king-size beds

LMTD78.3 °F
Effectiveness57.1%
Mass Flow Rate49,980 lb/hr
How to Use This Calculator
  1. Enter the Hot Fluid Inlet and Outlet temperatures in °F to define the hot-side duty.
  2. Enter the Cold Fluid Inlet and Outlet temperatures in °F — verify the cold outlet is lower than the hot inlet for a valid counterflow arrangement.
  3. Enter the Hot Side Flow Rate in GPM and the Specific Heat of the fluid (water = 1.0 BTU/lb·°F).
  4. Set the Overall U-Value in BTU/hr·ft²·°F: water-to-water ≈ 150–300, steam-to-water ≈ 200–500.
  5. Review the Heat Duty in BTU/hr and the LMTD in °F — the LMTD drives the required surface area.
  6. Use the Required Surface Area in ft² to select or specify a heat exchanger model from vendor catalogs.

How the result changes with Hot Fluid Inlet Temp

Hot Fluid Inlet TempHeat DutyRequired Surface Area
100-999,600 BTU/hr0 ft²
1501,499,400 BTU/hr182.2 ft²
3008,996,400 BTU/hr515.8 ft²
50018,992,400 BTU/hr706.5 ft²

What each input means

Hot Fluid Inlet Temp
Temperature of the hot fluid entering the heat exchanger.
Hot Fluid Outlet Temp
Desired temperature of the hot fluid leaving the heat exchanger.
Cold Fluid Inlet Temp
Temperature of the cold fluid entering the heat exchanger.
Cold Fluid Outlet Temp
Expected or desired temperature of the cold fluid leaving the heat exchanger.
Hot Side Flow Rate
Volumetric flow rate of the hot fluid (water assumed). For other fluids, adjust specific heat.
Specific Heat (Cp)
Specific heat of the fluid. Water = 1.0; ethylene glycol (50%) ≈ 0.81; oil ≈ 0.45-0.55.
Overall U-Value
Overall heat transfer coefficient. Water-to-water ≈ 150-300; water-to-oil ≈ 20-60; steam-to-water ≈ 200-500.

How this is calculated

Worked example, using the default values

  1. Identify Input Parameters
    4 parameters
    Hot Fluid Inlet Temp = 200, Hot Fluid Outlet Temp = 120, Cold Fluid Inlet Temp = 60, Cold Fluid Outlet Temp = 100 = 7 input(s) provided
  2. Calculate Heat Duty
    Heat Duty
    3998400 = 3998400
  3. Calculate Required Surface Area
    Required Surface Area
    340.4 = 340.4
  4. Calculate LMTD
    LMTD
    78.3 = 78.3
  5. Calculate Effectiveness
    Effectiveness
    57.1 = 57.1

Engine last updated . Checked against 2 independently-derived tests — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.

Frequently Asked Questions

Why does required surface area go up so much when I switch from water-to-water to water-to-oil?

Required area is A = Q/(U×LMTD), so it's inversely proportional to the overall heat transfer coefficient (U). Water-to-oil systems have a much lower U — roughly 20-60 BTU/hr·ft²·°F versus 150-300 for water-to-water — because oil's poor thermal conductivity and higher viscosity resist heat transfer, so for the same duty and temperature driving force, a water-to-oil exchanger needs proportionally more surface area.

Why do I still need to change the Specific Heat input if I'm not using water?

The mass flow rate conversion from GPM is hardcoded to water's density (8.33 lb/gal), so the calculator always assumes a water-like fluid for that step, but heat duty (Q = ṁ·Cp·ΔT) still depends on the fluid's actual specific heat. If you're sizing for glycol or oil, enter that fluid's specific heat (glycol ≈ 0.81, oil ≈ 0.45-0.55) even though the density behind the flow conversion isn't adjusted.

What does the Effectiveness output tell me, and when is it not reliable?

Effectiveness is the actual heat transferred divided by the theoretical maximum possible with infinite area, assuming the hot stream is the side with the smaller thermal capacity rate (mass flow × specific heat). If your cold side actually has the smaller capacity rate instead, that assumption breaks and the number won't reflect the true thermodynamic limit — you'd need to identify the real minimum-capacity side yourself.

Why does the LMTD calculation fall back to a plain difference in some cases?

The standard log-mean formula divides by the natural log of the ratio of the two end-temperature differences, which is undefined when those differences are equal (the ratio is 1, and ln(1) = 0). The calculator detects that near-equal case and falls back to the simple arithmetic difference, since in the limit the log-mean and arithmetic mean converge to the same value anyway.

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