Energy balance and Bowen's ratio evaporation equation

Energy balance equation

For a water body (lake) over a time period, the conservation of energy gives

$$Q_n = Q_e + Q_h + Q_\theta + Q_v$$

where

• $Q_n$ = net radiation energy received $= Q_s - Q_r + Q_a - Q_{ar} - Q_{br}$, that is, (incoming short-wave minus reflected short-wave) plus (incoming long-wave minus its reflection) minus the long-wave radiation emitted by the water surface, in J/m²/day,
• $Q_e$ = energy used for evaporation $= \rho L E$,
• $Q_h$ = sensible heat transferred to the air by conduction,
• $Q_\theta$ = increase in heat stored in the water body,
• $Q_v$ = net energy advected out of the body by inflow and outflow (zero if neglected).

Here $\rho$ is the density of water, $L$ the latent heat of vaporisation and $E$ the evaporation depth.

Bowen's ratio

Bowen's ratio is the ratio of the heat lost as sensible heat to the heat used for evaporation:

$$\beta = \frac{Q_h}{Q_e} = \gamma\,\frac{T_w - T_a}{e_w - e_a}$$

where $\gamma$ is the psychrometric constant (about 0.66 mb/°C, or 0.49 mm Hg/°C, at sea-level pressure), $T_w$ = water surface temperature, $T_a$ = air temperature, $e_w$ = saturation vapour pressure at $T_w$ and $e_a$ = vapour pressure of the air.

Derivation of the evaporation equation

Substitute $Q_h = \beta Q_e$ in the energy balance:

$$Q_n = Q_e + \beta Q_e + Q_\theta + Q_v \;\Rightarrow\; Q_e(1+\beta) = Q_n - Q_\theta - Q_v$$ $$Q_e = \frac{Q_n - Q_\theta - Q_v}{1+\beta}$$

Since $Q_e = \rho L E_L$:

$$\boxed{E_L = \frac{Q_n - Q_\theta - Q_v}{\rho\,L\,(1+\beta)}}$$

If the storage and advection terms are small over the period, $E_L = \dfrac{Q_n}{\rho L (1+\beta)}$. This gives the evaporation in metres of water per unit time when $Q$ is in J/m² and $\rho L$ in J/m³.