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    HVAC / cooling coil

    Cooling coil selection and part-load curve

    A coil is not a fixed capacity: it is a surface temperature plus a bypass factor. This selector places the apparatus dew point, finds the leaving state on the coil line, converts it into sensible and latent capacity, and traces what happens to that capacity as the airflow falls.

    CoilWetContact factor74.3%Condensate11.2 L/h
    Total capacity
    23.29 kW
    15.47 sensible · 7.82 latent
    Sensible heat ratio
    0.664
    Wet coil, latent load present
    Apparatus dew point
    8.5 °C
    Water 7.0 °C + 1.5 K approach
    Coil rows
    4
    NTU 1.36 · BF 0.257
    Psychrometric state points
    The coil line runs from the entering air to the apparatus dew point; the leaving state sits on it at the bypass factor.
    23.3 kW
    Psychrometric state points
    StateDry bulb (°C)RH (%)W (g/kg)h (kJ/kg)
    Entering air26.05010.4752.9
    Apparatus dew point8.51006.8825.9
    Leaving air13.0847.8032.8

    Dry-air mass flow 1.160 kg/s · chilled water 1.01 L/s at 5.5 K rise.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Part-load capacity curve
    Capacity falls more slowly than airflow because the bypass factor drops as NTU rises.
    Part-load capacity curve0.0005.2210.4315.6520.8726.0910%10%20%20%30%30%40%40%50%50%60%60%70%70%80%80%90%90%100%100%Coil capacity (kW)Linear reference
    Sensible and latent split at part load
    The leaving air moves toward the ADP, so the latent share grows as the load falls.
    Sensible and latent split at part load0.0003.476.9310.4013.8617.3310%10%20%20%30%30%40%40%50%50%60%60%70%70%80%80%90%90%100%100%SensibleLatent
    Canonical equations
    The exact relationships implemented by this engine.

    ADP = t_water,in + approach · The effective surface temperature of the coil.

    BF = (t_leaving − ADP) / (t_entering − ADP) · Fraction of air that misses the surface.

    Q = ṁ_da · (h₁ − h₂) · Sensible plus latent, from the enthalpy difference.

    NTU = −ln BF, rows = ⌈NTU / NTU_row⌉ · Each row adds a fixed amount of NTU.

    engineering methodology

    A clear path from inputs to results.

    A cooling coil is a surface temperature plus a bypass factor, not a fixed capacity. This selector places the apparatus dew point, finds the leaving state on the coil line, converts it into sensible and latent capacity, sizes the rows and the water flow, and traces what happens to all of it as the airflow falls.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the airflow and the entering dry-bulb and relative humidity.
    2. Set the leaving dry-bulb the space needs.
    3. Enter the chilled-water temperature and the approach to the apparatus dew point.
    4. Set the NTU per row and the water temperature rise.
    5. Read the capacity, the SHR, the rows and the part-load curve.

    The formula

    ADP=tw+Δtapp  ;  BF=t2−ADPt1−ADP  ;  Q=m˙da(h1−h2)\mathrm{ADP} = t_w + \Delta t_{app} \;;\; \mathrm{BF} = \frac{t_2 - \mathrm{ADP}}{t_1 - \mathrm{ADP}} \;;\; Q = \dot m_{da}(h_1 - h_2)

    Reference: ASHRAE Handbook — Fundamentals, psychrometrics and cooling-coil apparatus dew point; Carrier psychrometric chart construction.

    Worked example

    1 m³/s from 26 °C / 50% to 13 °C on 7 °C water

    Inputs
    • Airflow 1 m³/s at 26 °C dry bulb, 50% relative humidity.
    • Leaving dry bulb 13 °C.
    • Water in at 7 °C, approach 1.5 K, 0.35 NTU per row, 5.5 K water rise.
    Calculation path
    1. 1

      ADP = 7 + 1.5 = 8.5 °C, below the entering dew point, so the coil runs wet.

    2. 2

      BF = (13 − 8.5) / (26 − 8.5) = 0.257.

    3. 3

      W₂ = 0.00688 + 0.257 × (0.01047 − 0.00688) = 0.00780 kg/kg, about 7.8 g/kg.

    4. 4

      Enthalpy falls from 52.8 to 32.8 kJ/kg, so Q = 1.188 × 20.0 = 23.3 kW.

    5. 5

      NTU = −ln 0.257 = 1.36, so four rows at 0.35 NTU each.

    6. 6

      Condensate 11.2 L/h; water flow 23.3 / (4.186 × 5.5) = 1.01 L/s.

    Result

    23.3 kW · SHR 0.664 · ADP 8.5 °C · 4 rows · 1.01 L/s.

    Interpreting the result

    The sensible heat ratio is set by the coil line, so it moves with the ADP and the entering air, not with the airflow. That is why a coil cannot be re-rated for a different entering condition by scaling capacity: a colder surface dehumidifies more, a warmer one less. The part-load curve is concave because NTU rises as the airflow falls, so capacity drops more slowly than flow.

    Assumptions & limitations

    • Chilled-water coil with a constant ADP at part load; a DX coil would drop its ADP and remove relatively more moisture.
    • Bypass factor follows BF = e^−NTU with a fixed NTU per row.
    • One-dimensional coil line from the entering state to the ADP; no carry-over or reheat.
    • Standard atmospheric pressure, dry-air mass flow from the entering density.

    Frequently asked questions

    Why is my coil dry?

    The apparatus dew point is above the entering dew point, so the surface never reaches saturation and no moisture condenses. Lower the water temperature or the approach.

    Why does capacity fall more slowly than airflow?

    At lower airflow the same coil surface has more NTU per unit of air, so the bypass factor drops and each kilogram of air leaves closer to the ADP.

    What does the bypass factor mean physically?

    It is the share of the air that passes through the coil without touching the surface. A BF of 0.257 means about a quarter of the air leaves essentially untreated.

    How many rows should I choose?

    Rows follow from the bypass factor the duty needs. More rows buy a lower BF, which buys deeper dehumidification at the cost of fan and water pressure.

    Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.