Hydrological Losses

Penman's method combines the energy balance (net radiation) and the aerodynamic (mass transfer) methods to estimate evaporation or potential evapotranspiration (PET) from standard weather data. It needs no surface temperature.

Procedure

1. Collect data: mean monthly air temperature $T$, relative humidity RH, observed sunshine hours $n$, wind speed $u_2$ at 2 m, latitude, and the reflection coefficient $r$ of the surface.
2. Find from tables: saturation vapour pressure $e_w$ at $T$ (mm Hg), the slope $A = de_w/dT$ at $T$, the extraterrestrial radiation $H_a$ (mm of evaporable water/day) for the latitude and month, and the possible sunshine hours $N$.
3. Actual vapour pressure: $e_a = RH \times e_w$.
4. Net radiation (mm/day):

with $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$ and $T_a$ in kelvin.

5. Aerodynamic term:

with $u_2$ in km/day.

6. Compute $PET$ using the equation above with $\gamma = 0.49$ mm Hg/$^\circ$C (psychrometric constant).
7. For the monthly value multiply by the number of days. For open-water evaporation, use $r = 0.05$; for crops use $r$ of the crop (0.15-0.25) and multiply by a crop coefficient if actual crop ET is required.

Energy balance equation

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

where

• $Q_n$ = net radiation energy received = $Q_s - Q_r - Q_b + Q_a - Q_{ar} - Q_{bs}$ (incoming short-wave minus reflected, plus net long-wave) 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 sensible heat loss to evaporation heat loss:

with $\gamma \approx 0.61\,(P/1000)$ mb/°C (0.49 mm Hg/°C), $T_w$ = water surface temperature, $T_a$ = air temperature, $e_w$ = saturation vapour pressure at $T_w$ and $e_a$ = vapour pressure of air.

Derivation of the evaporation equation

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

Since $Q_e = \rho L E_L$:

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.

The energy budget method applies the law of conservation of energy to a lake over a period (usually a day or a week) to find the energy used in evaporation.

Energy balance

In short, $Q_n = Q_e + Q_h + Q_\theta + Q_v$, where

Evaporation

Sensible heat is found by the Bowen ratio $\beta = Q_h/Q_e = \gamma (T_w - T_a)/(e_w - e_a)$. Substituting:

Data needed

• Radiation: from pyranometer or computed from sunshine hours.
• Water temperatures at various depths (for $Q_\theta$), and air temperature and humidity (for $\beta$).
• Inflow, outflow and their temperatures (for $Q_v$).

Merits and demerits

• Gives accurate evaporation, and can be used for any time period; basis for calibrating other methods.
• Needs costly and detailed instruments, and measurement of each term is difficult; $Q_v$ and $Q_\theta$ are hard to find; so it is used for research on large lakes rather than routine estimation.

Effective (excess) rainfall in each hour of a sub-basin is $P_t - \phi$ (if positive; otherwise zero). The sub-basin values are then area-weighted for the whole catchment.

Step 1: Effective rainfall (mm) in each sub-basin

Example (15 km², $\phi = 10$): 16-10 = 6, 48-10 = 38, 22-10 = 12, 10-10 = 0.

Step 2: Area-weighted effective rainfall for 80 km²

Check: $15\times6 + 25\times1 + 0 + 0 = 115$ (1st hour); $15\times38 + 25\times27 + 35\times19 + 5\times26 = 2040$ (2nd hour); $15\times12 + 25\times5 + 5\times2 = 315$ (3rd hour).

Step 3: Runoff from the catchment

Total effective rainfall $= 1.4375 + 25.5 + 3.9375 + 0 = 30.875$ mm.

Average runoff rate over the 4 h $= 2.47\times10^6/(4\times3600) = 171.5$ m³/s.

Answer: Runoff depth = 30.875 mm, volume = 2.47 × 10⁶ m³; hourly effective rainfall = 1.44, 25.50, 3.94 and 0 mm.

Given: $A = 500$ ha $= 5\times10^6$ m²; runoff volume $= 0.361\ \text{Mm}^3 = 3.61\times10^5$ m³; baseflow negligible; 18 h of rain.

Step 1: Runoff depth and rainfall

Total rainfall $P = 12.6$ cm. Rainfall in each 2-h block (cm): 0.8, 1.8, 0.2, 1.3, 3.2, 3.5, 1.0, 0.6, 0.2.

Step 2: $\phi$-index by trial

Find $\phi$ (cm per 2 h) so that the sum of $(P_i - \phi)$ over the blocks with $P_i > \phi$ is 7.22 cm. Try $\phi = 0.73$ cm per 2 h; the blocks 0.8, 1.8, 1.3, 3.2, 3.5, 1.0 exceed it (0.2, 0.2 and 0.6 do not):

Step 3: Duration of rainfall excess

Rainfall excess occurs in 6 blocks of 2 h, so the duration of rainfall excess $t_r = 6\times2 = 12$ h.

Step 4: W-index

Other losses (depression storage, interception) $= 0.1\times10^6/5\times10^6 = 0.02$ m $= 2.0$ cm.

Answer: $\phi$-index = 0.365 cm/h; duration of rainfall excess = 12 h; W-index = 0.282 cm/h.

Penman equation (potential evapotranspiration):

with

where $H_n$ is the net radiation in mm of evaporable water per day, $E_a$ the aerodynamic term in mm/day, $u_2$ the wind speed at 2 m in km/day, and $e_w, e_a$ in mm of Hg.

Data and constants used

• $T = 10.0\,^\circ$C $\Rightarrow T_a = 283.0$ K; RH $= 60\%$; $n = 9$ h; $N = 12.9$ h; $u_2 = 120$ km/day
• $H_a = 11$ mm/day; $r = 0.25$; $\varphi = 28.50^\circ$; $e_w = 9.2$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• $H_a = 11$ is taken as mm of evaporable water per day, wind $5\ \text{km/h} = 120$ km/day, and albedo is used as $r$. $b = 0.52$.

Step 1: Actual vapour pressure

Step 2: Net radiation $H_n$

$a = 0.29\cos(28.50^\circ) = 0.2549$, and $n/N = 0.6977$

Step 3: Aerodynamic term

Step 4: PET

Answer: PET = 1.98 mm/day, which is about 61 mm for the 31 days of March.

Each interval is 20 min = 1/3 h, so depth in an interval = rate $\times \frac{1}{3}$.

The initial loss (0.8 mm) is absorbed in the first interval, where the rain is only 1.0 mm and is already at the $\phi$ rate, so it does not change the excess rainfall.

Total rainfall

Net runoff

W-index

Answer: Total rainfall = 10.0 mm; net runoff = 4.2 mm; W-index = 2.14 mm/h.

Penman's method combines the energy-balance and aerodynamic (mass-transfer) approaches:

where $A$ = slope of the saturation vapour pressure curve, $\gamma$ = psychrometric constant, $H_n$ = net radiation (mm of water/day) and $E_a$ = drying power of air (mm/day).

Step 1: Vapour pressure

Step 2: Net radiation $H_n$

Sunshine ratio $n/N = 7/11.9 = 0.588$.

Short-wave part:

Long-wave part, with $T_a = 12.5 + 273 = 285.5$ K:

Step 3: Drying power of air

Step 4: PET

For February (28 days): $1.555\times28 \approx 43.6$ mm.

Answer: PET = 1.56 mm/day, about 43.6 mm for February.

(The latitude and longitude are not needed because the extraterrestrial radiation is already given.)

Each interval is 20 min = 1/3 h. Depth = rate $/3$.

Total rainfall

Net runoff

Excess rainfall before the initial loss $= 0.833+1.833+4.167+3.167+2.167 = 12.167$ mm. The initial loss (0.5 mm) is taken from the start of the storm, so:

W-index

The whole storm duration (2 h) is used for $t$. The initial loss is deducted from the first interval before the $\phi$ loss is applied.

Answer: Total rainfall = 17.0 mm; net runoff = 11.67 mm; W-index = 2.42 mm/h.

Runoff depth

Rainfall in each 2-h interval

Total rain $P = 12.6$ cm, so total loss $= 12.6 - 6.8 = 5.8$ cm.

Trial for $\phi$

Let the loss per 2-h interval be $\phi \times 2$. Try intervals with rain above the loss: 2-4, 6-8, 8-10, 10-12, 12-14 (5 intervals). Their rain $= 1.8 + 1.3 + 3.2 + 3.5 + 1.0 = 10.8$ cm.

Check: loss per interval $= 0.8$ cm. The five intervals have rain $> 0.8$ cm (the smallest is 1.0), and the other intervals (0.8, 0.2, 0.6, 0.2) have rain $\le 0.8$ cm, so they give no excess. The assumption is correct.

Duration of rainfall excess

Five intervals of 2 h each produce excess: $5\times2 = 10$ h (periods 2-4 h and 6-14 h).

Answer: $\phi$-index = 0.40 cm/h; duration of rainfall excess = 10 h.

• Solar radiation: It is the source of energy that drives the hydrologic cycle. It evaporates water from oceans and land, heats the air to create winds and convective storms, and melts snow and glaciers (important for Himalayan rivers). The rate of evaporation and transpiration mainly depends on the radiation received.
• Relative humidity: It is the ratio of actual to saturation vapour pressure of the air. It governs how much more vapour the air can take: low humidity gives a large vapour pressure deficit and high evaporation, while at 100% humidity evaporation stops. It also controls cloud formation, condensation and the amount of precipitation.
• Evapotranspiration: The combined loss of water by evaporation from soil and water surfaces and by transpiration from plants. It returns about two-thirds of land precipitation to the atmosphere. It is the largest loss in the water balance, decides crop water requirement, reservoir loss and catchment yield, and links the surface water with the atmosphere in the cycle.

Direct runoff depth: Total direct runoff volume $= \sum Q\,\Delta t = (30+60+45+30+15)\times3600 = 648\,000$ m³.

Note on area: with $A = 0.2$ km² the runoff depth ($648000/2\times10^5 = 3.24$ m) is far greater than the rainfall (5.18 cm), which is impossible. The area must be 200 km² (0.2 × 10³ km²); this is assumed.

Total rainfall $= 1.05+1.28+0.80+0.75+0.70+0.60 = 5.18$ cm, so the total loss $= 5.18 - 0.324 = 4.856$ cm.

Finding $\phi$

Runoff occurs only when rainfall rate exceeds $\phi$. Try the hours with rain greater than $\phi$: 1.05 and 1.28 cm/h (the others are below $\phi$ and infiltrate fully).

Check: $1.05, 1.28 > 1.003$ and $0.80, 0.75, 0.70, 0.60 < 1.003$. Valid.

Cumulative infiltration (loss) curve

Infiltration in each hour $= \min(\text{rain rate}, \phi)$:

The last ordinate (4.856 cm) equals the total loss, confirming the result.

Answer: $\phi$-index = 1.003 cm/h; cumulative infiltration ordinates = 1.003, 2.006, 2.806, 3.556, 4.256, 4.856 and 4.856 cm at 1 to 7 h.

Rainfall in each 2-hour period is the difference of cumulative rainfall; the intensity is depth/2 h. Loss rate $\phi = 0.40$ cm/h.

Excess rainfall hyetograph: bars of excess depth for 2-h periods:

 cm
 1.6 |        ##
 1.4 |     ## ##
 1.0 |     ## ##       ##
 0.6 |     ## ## ##    ##
 0.1 |     ## ## ## ## ##
     +----------------------
       0-2 2-4 4-6 6-8 8-10 10-12 12-14 h

(Bar heights, 2-4 h: 1.40; 4-6 h: 1.60; 6-8 h: 0.60; 8-10 h: 0.10; 10-12 h: 0.90.)

Volume of direct runoff:

Answer: Total excess rainfall = 4.60 cm; volume of direct runoff = 2.3 × 10⁵ m³ (0.23 million m³).

Evaporation from a lake or reservoir is estimated by the following methods.

1. Evaporation pans: Standard pans (US Class A land pan, ISI pan, sunken Colorado pan) are placed near the lake and the fall of water level is measured. Lake evaporation $= C_p \times$ pan evaporation, where the pan coefficient $C_p$ is 0.6-0.8 (0.7 for Class A). Simple and widely used, but the pan coefficient is only an approximation.

2. Empirical formulae: Evaporation is related to vapour pressure deficit and wind speed.

   • Meyer's formula: $E_L = K_M (e_w - e_a)\left(1 + \dfrac{u_9}{16}\right)$ (mm/day).
   • Rohwer's formula: $E_L = 0.771(1.465 - 0.000732 P_a)(0.44 + 0.0733 u_0)(e_w - e_a)$. They are easy to use but need calibration for the local area.

3. Water budget method: Applies continuity to the lake: $E_L = P + V_{is} + V_{ig} - V_{os} - V_{og} - \Delta S - T_L$. It is simple in principle, but seepage and the other terms are hard to measure, so errors are large for short periods.

4. Energy budget method: Applies conservation of energy with Bowen's ratio: $E_L = \dfrac{Q_n - Q_\theta - Q_v}{\rho L(1+\beta)}$. Accurate but needs detailed instruments.

5. Mass transfer (aerodynamic) method: Uses theory of turbulent transfer of vapour: $E_L = B(e_w - e_a)$ with $B$ a wind-speed function.

6. Penman (combination) method: Combines energy balance and aerodynamic terms: $E = \dfrac{A H_n + E_a \gamma}{A+\gamma}$ with $r = 0.05$ for water. Gives good results from routine weather data.

7. Analytical/remote sensing methods: Used for large lakes and where ground data are scarce.

Lake evaporation $= C_p \times$ pan evaporation, and the volume lost $=$ depth $\times$ mean water-spread area.

Step 1: Annual pan evaporation

$16.7+14.3+17.8+25.0+28.6+21.4+16.7+16.7+16.7+16.7+21.4+16.7 = 228.7$ cm

Step 2: Lake evaporation depth

Step 3: Mean water-spread area

Step 4: Volume of evaporation loss

Answer: Evaporation depth = 160.1 cm; volume lost = 4.28 × 10⁶ m³ (about 4.28 million m³) in the year.

Meyer's formula:

where $u_9$ is the wind speed in km/h at 9 m above the ground. The given wind is at 2 m, so it is converted by the 1/7 power law.

Step 1: Wind speed at 9 m

Step 2: Vapour pressures

$e_w = 23.76$ mm Hg; $e_a = RH\times e_w = 0.70\times23.76 = 16.632$ mm Hg; $e_w - e_a = 7.128$ mm Hg

Step 3: Evaporation rate

Step 4: Volume in March (30 days)

Depth $= 4.952\times30 = 148.6$ mm $= 0.1486$ m

Answer: Evaporation = 4.95 mm/day, so the volume evaporated in March is about 297,117 m³ (≈ 2.97 × 10⁵ m³).

(If the 2 m wind is used directly without correction, the answer is 2.69 × 10⁵ m³.)

The runoff is the rainfall minus infiltration and evaporation: $R = P - F - E$.

Given: $P = 10$ cm in 24 h; Horton: $f_0 = 0.8$, $f_c = 0.3$ cm/h, $K = 0.5$ h⁻¹; pan drop $= 0.5$ cm, $C_p = 0.7$; $A = 1$ km².

Step 1: Total infiltration in 24 h

Taking infiltration at capacity rate throughout, Horton's equation $f_p = f_c + (f_0 - f_c)e^{-Kt}$ integrates to

Step 2: Evaporation loss

Step 3: Runoff depth and volume

Answer: Runoff = 1.45 cm, i.e. 1.45 × 10⁴ m³ (14,500 m³) from the 1 km² catchment.

Penman equation (potential evapotranspiration):

with

where $H_n$ is the net radiation in mm of evaporable water per day, $E_a$ the aerodynamic term in mm/day, $u_2$ the wind speed at 2 m in km/day, and $e_w, e_a$ in mm of Hg.

Data and constants used

• $T = 20\,^\circ$C $\Rightarrow T_a = 293$ K; RH $= 75\%$; $n = 10$ h; $N = 13.5$ h; $u_2 = 192$ km/day
• $H_a = 14.4$ mm/day; $r = 0.028$; $\varphi = 27.00^\circ$; $e_w = 11$ mm Hg; $A = 1.42$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• $H_a = 14.4$ is taken as mm of evaporable water per day. $\gamma = 0.49$ is assumed (not given). $e_w$ and $A$ are taken as given.

Step 1: Actual vapour pressure

Step 2: Net radiation $H_n$

$a = 0.29\cos(27.00^\circ) = 0.2584$, and $n/N = 0.7407$

Step 3: Aerodynamic term

Step 4: PET

Answer: PET = 4.74 mm/day, about 147 mm for the 31 days of May.

Method

The $\phi$-index is the constant rate of loss such that the rainfall in excess of it equals the direct runoff. Losses = $15.0 - 8.7 = 6.3$ cm over 8 h.

Trial

Assume all hours exceed $\phi$: $15 - 8\phi = 8.7 \Rightarrow \phi = 0.7875$ cm/h. But the 1st hour (0.6) is below $\phi$, so exclude it (loss in that hour is 0.6 cm). Try 7 hours: $(15 - 0.6) - 7\phi = 8.7$, so $\phi = 0.8143$ cm/h. The 8th hour (0.75) is also below $\phi$, so exclude it too. Try 6 hours:

Check

The excess equals the 8.7 cm runoff, and the 6 hours with excess have rainfall > 0.825 cm/h, so the trial is valid.

Answer: $\phi$-index = 0.825 cm/h.

Water budget method

Continuity equation applied to the lake over a period:

so

where $P$ = precipitation, $V_{is}, V_{os}$ = surface inflow and outflow, $V_{ig}, V_{og}$ = groundwater inflow and outflow, $T_L$ = transpiration loss, $\Delta S$ = change in storage (all in m³). It is simple, but seepage, groundwater flow and measurement errors make it inaccurate for short periods; it suits long periods.

Energy budget method

Conservation of energy for the lake: $Q_n = Q_e + Q_h + Q_\theta + Q_v$. Using Bowen's ratio $\beta = Q_h/Q_e$:

$Q_n$ is net radiation, $Q_\theta$ stored heat, $Q_v$ advected energy, $\rho$ density and $L$ latent heat. It gives accurate results for any period but needs costly instruments and many measurements.

Penman equation (potential evapotranspiration):

with

where $H_n$ is the net radiation in mm of evaporable water per day, $E_a$ the aerodynamic term in mm/day, $u_2$ the wind speed at 2 m in km/day, and $e_w, e_a$ in mm of Hg.

Data and constants used

• $T = 22.5\,^\circ$C $\Rightarrow T_a = 295.5$ K; RH $= 75\%$; $n = 10$ h; $N = 12.7$ h; $u_2 = 80$ km/day
• $H_a = 14.9$ mm/day; $r = 0.2$; $\varphi = 26.82^\circ$; $e_w = 20.4$ mm Hg; $A = 1.24$ mm Hg/$^\circ$C; $\gamma = 0.49$ mm Hg/$^\circ$C; $\sigma = 2.01\times10^{-9}$ mm/day/K$^4$
• Latitude $26^\circ49' = 26.82^\circ$. Elevation (250 m) is not needed because $\gamma$ is given. The closed green crop is represented by $r = 0.20$ (given).

Step 1: Actual vapour pressure

Step 2: Net radiation $H_n$

$a = 0.29\cos(26.82^\circ) = 0.2588$, and $n/N = 0.7874$

Step 3: Aerodynamic term

Step 4: PET

Answer: PET = 4.69 mm/day, so about 141 mm for April (30 days).

Step 1: Rainfall in each 0.5-h interval (cm)

Total rainfall $P = 8.2$ cm; direct runoff $R = 3.8$ cm; total loss $= 4.4$ cm over 5 h.

Step 2: Trial for $\phi$ (let $\phi'$ be the loss per 0.5 h) Assume only the 0.4 cm interval is below $\phi'$ (the other 9 intervals exceed it):

Step 3: Check The first nine intervals (0 to 4.5 h) each have rain of at least 0.5 cm, which is more than 0.444 cm, and the last interval (0.4 cm) is below it. Sum of excess $= 7.8 - 9\times0.444 = 3.8$ cm. Valid.

Step 4: Duration of rainfall excess Excess occurs in 9 intervals of 0.5 h, from 0 to 4.5 h:

Answer: $\phi$-index = 0.889 cm/h (about 0.89 cm/h); duration of rainfall excess = 4.5 h.

Horton's equation

The infiltration capacity (rate) at time $t$ is