---
title: "Energy balance and Bowen's ratio evaporation equation"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:bowen-ratio-evap"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/bowen-ratio-evap
site: BCT Engineering
---

# Energy balance and Bowen's ratio evaporation equation

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2082 Bhadra, 5 marks; 2073 Shrawan, 4 marks (asked 2 times)
- Video: [Surface Energy Balance and Bowen Ratio](https://www.youtube.com/watch?v=MIyTEWz3dpk) — Mobushir Khan
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/bowen-ratio-evap
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

Explain the energy balance equation and derive the evaporation equation using Bowen's ratio.

## Short answer

From energy balance Qn = Qe+Qh+Qθ+Qv and Bowen ratio β = γ(Tw-Ta)/(ew-ea), evaporation is EL = (Qn - Qθ - Qv) / [ρL(1+β)].

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Bowen's ratio is:
   A. Qe/Qh
   B. Qh/Qe
   C. Qn/Qe
   D. Q_theta/Qn
   Answer: B. It is the ratio of sensible heat loss to evaporation heat loss. (Key point: Bowen ratio)

2. Substituting Qh = beta Qe into the balance gives:
   A. Qe = (Qn - Q_theta - Qv)/(1 + beta)
   B. Qe = (Qn - Q_theta - Qv) x (1 + beta)
   C. Qe = Qn/beta
   D. Qe = (Qn + Q_theta)/(1 - beta)
   Answer: A. Qe(1+beta) = Qn - Q_theta - Qv. (Key point: Derivation step)

3. Qe is related to evaporation depth E by:
   A. Qe = L E / rho
   B. Qe = rho L E
   C. Qe = rho E / L
   D. Qe = E/(rho L)
   Answer: B. Qe = rho L E, so E = Qe/(rho L). (Key point: Qe link)

4. If storage and advection are small, evaporation is:
   A. Qn rho L (1+beta)
   B. Qn / (rho L (1 + beta))
   C. Qn beta / (rho L)
   D. Qn / (rho L beta)
   Answer: B. With Q_theta = Qv = 0, E = Qn/(rho L (1+beta)). (Key point: Simplification)

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