Every second, water you cannot see leaves the ground you are standing on — from lakes, from soil, and through up to a hundred thousand microscopic gates on every square centimetre of leaf. Over most of the land surface this invisible flux is the largest term in the water balance, and it is also the one we measure worst. This companion teaches you to quantify it three ways for open water, to see how a plant does it, and to model both with one equation. Click a part of the landscape below (or a chapter card) to explore, experiment, and test yourself.
Companion to the Part 03 lecture slides · Evaristo Critical Zone Hydrology Lab, University of Georgia. Scores are self-assessment only — nothing is recorded, and progress resets if you reload.
Rain announces itself. Streams can be gauged. But the flux that returns most of the rain to the sky leaves no puddle and makes no sound — and the first thing to understand about it is that there are two of it: a physical one and a biological one.
Evaporation is physics: liquid water at a free surface — a lake, a pan, a wet soil — gains enough energy to leave as vapor. Transpiration is biology: water absorbed by roots moves up through the plant and leaves as vapor through the stomata of its leaves. Both need energy, usually from the sun, and both respond to temperature, humidity, wind, and radiation. Transpiration adds two more levers — the plant type and the soil moisture in its root zone — because a plant can choose to close its gates. Together they are evapotranspiration (ET): in most catchments the largest term of the water balance, often exceeding runoff and recharge.
The lecture summarizes the contrast in a table. Here it is as a game — sort each statement to the process it describes, then check.
You have seen the picture since grade school: two jars of water in the sun, one sealed, one open, and the open one slowly emptying. The physics behind it carries the whole chapter. Molecules at the surface are always leaving; molecules in the vapor are always returning. In the sealed jar the vapor builds until the return traffic equals the departures — the air is saturated, and the vapor pressure above the water is the saturation vapor pressure es(T), a ceiling that rises steeply with temperature. In the open jar the vapor escapes, the air above the water never saturates, and net evaporation continues — faster when the air is warm, dry, and moving. Play with the temperature, the wind, and the lid, and watch the two-way traffic.
Since 1850 the global surface temperature has climbed to about 1 °C above the twentieth-century average while atmospheric CO2 rose from roughly 285 to over 415 ppm. A warmer atmosphere should be thirstier, and evaporation from open water should have risen. Yet across the Northern Hemisphere, half a century of evaporation pans said the opposite: pan evaporation fell year after year. Two suspects were named. The humidity suspect: wetter surroundings make the air over the pan damper. The sunlight suspect: less solar energy is reaching the ground. Roderick and Farquhar (2002) worked the case with two clues you can reproduce below: the vapor pressure deficit had not changed in fifty years, and the diurnal temperature range had shrunk — night-time minima rising twice as fast as daytime maxima. Work both tabs, then watch the video.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.
How do you measure something you cannot see? You account for it. Put a control volume around a pan of water, write conservation of mass and of energy across its surface, and evaporation falls out as the only place the missing water and the missing energy could have gone.
Draw a dashed cylinder around an evaporation pan (Chow et al., 1988). Inside are liquid water (density ρw) and water vapor (density ρa). The pan’s sides are impermeable, so the only way liquid mass leaves is as vapor — and the only way energy leaves with it is as latent heat. Write continuity for the liquid, continuity for the vapor, then the heat-energy equation from Notes01, and strip away every term that is zero for a pan. Fourteen equations later you hold the energy balance equation for evaporation. Step through the derivation below; each step lights up the piece of the pan it is about.
Energy can free the molecules, but something must carry them away. The aerodynamic method looks at the air’s ability to transport vapor: a humidity gradient near the surface supplies the push, wind supplies the pull. The ratio of the vapor flux to the momentum flux (Thornthwaite and Holzman, 1939) leads to a working equation with a vapor transfer coefficient B that grows with wind speed and with surface roughness:
The saturation vapor pressure is the ceiling from the two-jar review, and it rises exponentially with temperature: a gentle slope in the cold, a steep climb above 20 °C. Its slope, Δ = deas/dT, is the single most important number in the rest of this Part — it decides how readily added energy becomes added evaporation. Drag the temperature below and watch the tangent line; then set the humidity to see the deficit the wind has to work with.
When energy is not limiting, use the aerodynamic method; when vapor transport is not limiting, use the energy balance. Usually neither is comfortably true, so Penman combined them — weighting the two estimates with Δ and the psychrometric constant γ (≈ 66.8 Pa °C−1) so that no surface temperature is needed:
The combination method is best for small areas with detailed climatology — net radiation, temperature, humidity, wind, pressure. Over very large areas energy governs, and Priestley and Taylor (1972) noticed that the aerodynamic term runs at about 30 % of the energy term, which collapses the equation to one that needs only radiation and temperature:
A hydrologist should be able to hear a situation and name the method. Six situations below — tag each, then check.
The lecture’s standing problem: average net radiation 185 W m−2, air temperature 28.5 °C, relative humidity 55 %, wind 2.7 m s−1 at 2 m, water density 996.3 kg m−3, roughness height 0.03 cm. Find the open-water evaporation in mm d−1 by the energy balance, aerodynamic, combined, and Priestley–Taylor methods, then ask: are the four answers comparable, and why? Solve along — nine intermediate values, each checked as you go — and the live calculator unlocks at the end so you can change the weather and see which method flinches.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.
A plant lives by taking CO2 from the air, and CO2 can only enter its tissue dissolved in water. So the leaf opens a wet cavity to the sky — and every open stoma leaks vapor. Transpiration is not a leak the plant failed to fix. It is the price of breathing.
Three steps (Hewlett, 1982): absorption of soil water by roots; translocation, in liquid form, through the vascular system of roots, stem, and branches to the leaves and on through the leaf’s veins to the walls of tiny stomatal cavities, where evaporation takes place; and transpiration proper, as the vapor in the cavity moves out through the stoma. Click any part of the leaf to learn what it does, play the water’s journey from root hair to sky, and then dial the stomatal density — 10,000 to 100,000 gates per square centimetre — to see how crowded a leaf surface is.
Every stoma is flanked by two guard cells that swell and shrink to open and close it. They respond to five signals — remember L-I-G-H-T: Light intensity, Internal (ambient) CO2, the vapor-pressure Gradient between leaf and air, Heat (leaf temperature), and Turgidity (leaf water content). Dingman turns four of them into factor functions that reduce a maximum leaf conductance C*leaf:
Run the control room. Each slider is one environmental signal, using the same factor functions as the class spreadsheet; the stoma redraws with the resulting conductance, and the canopy conductance you will need in Chapter 4 updates alongside.
You cannot put a rain gauge on a tree, but you can time its sap. Sap-flow methods apply heat at one level on the trunk — continuously or in pulses — and watch how fast it travels upward with the sap stream. In the heat-pulse configuration (Dingman; Smith & Allen, 1996) a heater probe sits between two thermistors, one xu below and one xd above. Heat spreads by conduction in both directions but is carried by the sap in only one; the moment the two sensors read the same temperature, te, tells you how far the sap has shifted the pulse:
Converting velocity to a flux needs the cross-sectional area of conducting sapwood — the ring between the cambium (radius R) and the heartwood boundary (radius h) — which varies with species and site. Set a sap velocity, fire a pulse, read te off the traces, and scale the tree up.
Heavy water is a label. Rain, soil water, plant water, and vapor each carry a distinct ratio of 2H and 18O (Notes04), and because root uptake does not fractionate, a plant’s xylem water carries the signature of the water its roots took. Inside the Biosphere 2 Tropical Rain Forest — 1,936 m2 under glass, 92 plant species, rain on command — Evaristo et al. (2019) imposed a 68-day drought, then broke it with 66 mm of rain labeled at +152 ‰ δ2H over four events, followed by background rain at −60 ‰. For nine months they followed the label into seepage, soil water, and five tree species. Mix the sources in the first tab, then read the ages in the second.
Multiply one tree’s thirst by a plateau. China’s Loess Plateau went from roughly one-third vegetated in 1999 to two-thirds by 2019 — and the new forests are dying back in dry spells. The clue is a ghost from the atomic age: tritium from 1960s bomb tests, carried down by that decade’s rain, now sits 7–14 m deep, because water creeps downward only 1–2 m a year. The planted trees root 5–15 m down. They are drinking rain that fell before their planters were born — paleo-water — and running up a hydrological debt. Run the bank account below: a deep-soil reservoir that recharges slowly and a plantation that withdraws faster, with the adaptive strategies from the video as switches.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.
Penman (1948) welded the energy balance to the aerodynamic method. Monteith (1965) added the plant: one more resistance, in series with the air’s. The result is the most widely used equation in evapotranspiration — and once you can read its four parts, it is not intimidating at all.
The numerator holds the two drivers: the energy available for ET and the drying power of the air. The denominator holds the two brakes: the energy cost of vaporizing water and the overall resistance to vapor transfer, which combines the atmosphere’s and the canopy’s. Hover or tap each part of the equation to read its role, then try the four “what if” buttons — each kills one term and tells you what happens to ET.
Open-water evaporation crosses one resistance: the atmosphere’s, 1/Cat. Transpiration is a two-step process — from the stomatal cavity to the leaf surface, then from the leaf surface into the air — so the same driving force Δev operates across two resistances in series: 1/Cleaf and 1/Cat. Series resistances add, so whichever conductance is smaller sets the pace. The atmospheric conductance comes from the wind profile over the canopy:
Wire the circuit: set the wind and canopy height for Cat, set the canopy conductance, and watch the current — the vapor flux — respond. The factor (1 + Cat/Ccan) in the Penman–Monteith denominator is this circuit in one number.
This is PenmanMonteith.xlsx rebuilt cell for cell (modified from Dingman, 3rd ed.): twelve inputs, the intermediate values, and the ET rate — 0.583 mm d−1 for the default overcast day over a 16.5-m forest. The assignment asks you to explore the effect of any two inputs on ET and describe the sensitivity with graphs. Move a slider to change one input; pick a variable to sweep it across its range; and read the tornado chart, which nudges every input ±20 % and ranks what matters.
Calculate ET for a corn field on a typical July day: net radiation K + L = 400 W m−2, air temperature 25 °C, relative humidity 60 %, wind 3 m s−1 at 2 m, canopy height 2 m, LAI = 4, pressure 101.3 kPa. The temperature-dependent parameters are given: Δ = 0.189 kPa °C−1, γ = 0.0665 kPa °C−1, ρa = 1.18 kg m−3, ρw = 997 kg m−3, λv = 2.45 × 106 J kg−1, ca = 1013 J kg−1 °C−1, e*a = 3.17 kPa. Work the four steps; each one checks your number before the next opens.
Scored, with explanations after grading. Retake as many times as you like — questions reshuffle.