← Back to Blog Chart thumbnail: the two entering-air enthalpy conventions, saturated at the wet bulb 85.05 kJ/kg against real moist air 84.76 kJ/kg

The Merkel Number Explained: KaV/L and the Entering-Air Convention

Every cooling tower duty calculation runs through one group of symbols, written KaV/L and called the Merkel number. It is the number a test produces, the number a fill’s characteristic is expressed in, and the number that two parties argue about when a freshly rebuilt tower still does not hit its guarantee. Surprisingly little of that argument is about the measurement itself — a lot of it is about a convention that the Merkel integral leaves open.

What KaV/L is

KaV/L combines the fill’s mass-transfer behaviour with how hard the tower is working:

  • Ka is the volumetric mass-transfer coefficient of the fill — how effectively a cubic metre of fill moves water vapour into the passing air;
  • V is the fill volume;
  • L is the water mass flow.

The product KaV per unit water flow is a dimensionless duty; it is the same for two towers only when they are compared at the same water-to-air ratio. The air side enters through L/G, where G is the dry-air mass flow. This is why fill data is always quoted as KaV/L against L/G: change either side and the number means something different.

Where the integral comes from

The Merkel formulation treats heat and mass transfer between water and air as a single enthalpy exchange. The enthalpy potential at any point in the fill is the difference between the air saturation enthalpy at the local water temperature, hs(Tw), and the enthalpy of the air actually at that point, ha(Tw). Adding up the reciprocal of that potential over the whole water temperature range gives the required duty:

KaV/L = ∫ cp,w dTw / (hs(Tw) − ha(Tw))

Read it as a cost integral, not a heat balance: the numerator is the heat each slice of water carries, and the denominator is how hard the air is being pushed to absorb it. Where the driving potential is wide, the fill does its work cheaply; where the water is already close to the incoming air’s capability, the denominator becomes small and the required duty grows quickly. That is the origin of the practical fact that a tower approaching the wet bulb needs disproportionate fill volume for the last degree.

The operating line

The air’s enthalpy is not measured through the fill — it is modelled as rising linearly with the water it passes:

ha(Tw) = hin + (L/G) cp,w (Tw − Tc)

hin is the enthalpy of the air entering the fill, L/G is the water-to-air mass-flow ratio, and cp,w is the specific heat of water. The slope of the operating line is (L/G) cp,w: with a large L/G, the air rises steeply in enthalpy and the driving potential collapses over a short distance. This is the single most sensitive assumption in the calculation, and it is why an accurate water flow measurement and an accurate air flow estimate both matter more than they look.

The convention nobody writes down

hin is where the ambiguity lives. Two conventions are in common use, and they give different answers:

ConventionWhat hin is taken asWhere it is used
Real moist-air enthalpyEnthalpy of the true moist-air state at the measured dry-bulb and wet-bulb temperaturesPhysically consistent modelling — the operating-line derivation assumes this
Saturated air at the entering wet bulbEnthalpy of saturated air at the entering wet-bulb temperature aloneCross-checks against published demand curves and worksheets that treat air enthalpy as a function of wet bulb only

The two are not the same quantity. Saturated air at a given wet bulb carries more enthalpy than an unsaturated mixture at the same wet bulb and a higher dry bulb, because the saturation state contains the water vapour the mixture does not — and because the enthalpy of moist air includes the vapour’s contribution, not just its temperature.

Merkel number: the saturation-enthalpy curve with two entering-air conventions — saturated at the 27 °C wet bulb and the real moist-air state at 33 °C dry bulb — showing an offset of 0.29 kJ/kg (illustrative)Illustrative Merkel demand chart showing the two entering-air enthalpy conventions. The rising curve is the saturation enthalpy of air at the local water temperature, hs(Tw). The two straight lines are the operating line ha(Tw) = hin + (L/G) cp,w (Tw − Tc) with hin taken as saturated air at the entering wet bulb (85.05 kJ/kg) and as the real moist-air state at 33 °C dry bulb / 27 °C wet bulb (84.76 kJ/kg) — an offset of 0.29 kJ/kg. Reference condition: cold water 32 °C, hot water 42 °C, L/G 1.50, cp,w 4.18 kJ/kg·K. Over the same range the saturated-at-wet-bulb convention needs KaV/L 1.52 against 1.50 for the real moist-air convention, 1.06 % more — the post's "about 1.1 % higher", computed here. Values are computed from the formulas, not measured. (chart-merkel-conventions)Air conventions — Illustrative — not a measurementTwo entering-air enthalpy conventions32343638404280100120140160180Water temperature Tw (°C)Air enthalpy (kJ/kg dry air)85.0584.76saturated at the wet bulbreal moist airReference: water 32–42 °C · entering air 33 °C db / 27 °C wb · L/G 1.50Saturated-at-wet-bulb vs real moist-air entering enthalpyhin saturated at 27 °C wb (kJ/kg)85.05hin real moist air 33/27 (kJ/kg)84.76offset Δhin (kJ/kg)0.29KaV/L: saturated ÷ real air1.52 ÷ 1.50Computed from the formula — not field data.
Illustrative, computed from the properties — not a measurement. The rising curve is the saturated-air enthalpy hs(Tw); the two straight lines are the operating line with hin taken as saturated air at the entering wet bulb (85.05 kJ/kg) and as the real moist-air state at 33 °C dry bulb / 27 °C wet bulb (84.76 kJ/kg) — an offset of 0.29 kJ/kg. A higher hin flattens the enthalpy gap across the whole fill, so the same cooling range needs more fill: KaV/L 1.518 against 1.502, 1.07% more — the “about 1.1%” comparison above, computed. Every chart in this series has its own page, with one shared conditions bar: Tower Lab.

When we compared the two conventions inside our own implementation at a reference condition — 42 °C hot water, 32 °C cold water, 33 °C dry bulb, 27 °C wet bulb, L/G = 1.5 — the saturated-at-wet-bulb assumption returned a Merkel number about 1.1% higher than the real-moist-air treatment. The direction is structural and worth remembering: taking the entering air as saturated raises hin, flattens the enthalpy gap, and the tower needs slightly more fill to do the same job.

That figure is not a universal constant. It belongs to that reference point: at a small approach or a low range the difference changes size, and at some conditions it changes sign in practical terms because the dry-bulb offset works the other way. The lesson to carry into a test report is simpler than the number: state which convention was used.

Illustrative arithmetic

The following is an illustrative example, not a measured result.

Suppose a real-moist-air calculation returns KaV/L = 1.500 at the reference condition above. Applying the roughly 1.1% offset observed with the other convention gives:

1.500 × 1.011 ≈ 1.517

Nothing about the tower changed — only the treatment of the entering air did. That is the size of error a report can acquire by silently switching conventions, and it is large enough to consume a plausible performance shortfall.

Two ways to evaluate the integral

The integral has no convenient closed form against real property data, so it is evaluated numerically. Two quadrature choices dominate in cooling tower work:

  • Composite Simpson’s rule with a configurable even number of segments — the general-purpose choice, with error falling as the segment width to the fourth power.
  • Four-point equal-weight Tchebycheff quadrature — deliberately chosen when a result has to be comparable with the tabulated worksheets and demand curves that use it. It is less accurate than a fine Simpson grid, and that is exactly why it is used for cross-checking: matching a published curve’s integration scheme removes one source of difference before the engineering comparison starts.

A note on what the integral cannot do: if hs(Tw) − ha(Tw) reaches zero or goes negative anywhere in the range, the state is physically infeasible — the water cannot be cooled below what the air can absorb — and the integral has no defined value. Good implementations raise that as a domain error instead of integrating through it, and root-solvers bracket the cold-water temperature between the wet bulb and the hot-water temperature rather than taking unconstrained steps near the pinch, where the function becomes extremely steep.

What the Merkel formulation leaves out

It is a simplification, and a useful one, but the omissions matter at the margins:

  • Evaporation is not counted as a mass loss. The water flow entering the fill is not the water flow leaving it, and the Merkel model treats them as the same stream. The L/G used in a Merkel calculation is therefore a convention, not a direct measurement.
  • A Lewis factor of one is assumed, which equates the heat and mass transfer coefficients. The more complete Poppe formulation handles the Lewis factor and evaporation separately and is used when the loss of accuracy matters more than the extra computation.
  • Uniform flow is assumed in both directions, which is rarely exactly true inside a real cell.

None of these make the Merkel number the wrong tool. They make it a definition that must travel with its assumptions, which is why a Merkel number without its L/G, entering-air convention and property source is not a comparable result.

Read more

Sources

Frequently Asked Questions

What is the Merkel number and what are its units?

The Merkel number is the group written KaV/L: the fill's mass-transfer coefficient Ka, multiplied by the fill volume V, divided by the water mass flow L. It is dimensionless — often expressed in kg of water per kg of dry air — and it is the required duty of the tower, obtained by integrating the enthalpy driving potential across the water temperature range.

Why does the entering-air enthalpy convention change my KaV/L?

The Merkel integral is driven by the difference between saturated-air enthalpy at the local water temperature and the air's actual enthalpy. If the entering air is treated as saturated at the entering wet bulb, its enthalpy is higher than a real mixed-air state at the same wet bulb, so the driving force shrinks and the required KaV/L rises. That is why the two conventions do not agree.

How do I compare my result with a published demand curve?

Match the convention the published curve uses before comparing anything. If a curve was produced with entering air taken as saturated at the wet bulb, use the same assumption; otherwise the difference in convention will appear as a performance error that is not real.

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