Transpiration
Transpiration (abbreviated
E) is the loss of water from plants by evaporation. It is an extremely
important process - for example, in untouched areas of the Amazon, nearly
half (50%) of the water that falls as rain was originally transpired by
the plants in the forest! In some locations, it may be 85%!!
Over the lifetime of
a corn plant, nearly 99% of the water taken up by the roots is transpired
away. And a typical full grown corn plant may take up 3-4 liters
of water per day, so this is not a trivial event. In fact, the global
hydrologic cycle is intimately associated with the activity of plants.
Thus, it is imperative that we study how water is lost from plants, and
the forces governing it.
Not all plants are profligate water losers - some cacti may lose only a few ml of water over a period of months or even years! How can this be? This is one aspect of the transpiration process we will delve into.
We can calculate a transpiration ratio as:
Kgs of H2O lost by E / Kgs of dry material produced
This tells us how much water is necessary to produce a certain amount of biomass. The inverse of this is a water use efficiency measure, or, how much biomass can be produced per unit water transpired. Certain plants are more efficient than others, as we will find out.
For corn, the E ratio is about 350. Assuming we produce 3600 Kg of corn, this would require 1.26 x 106 Kg of water, which translates into a 30 cm rainfall (12"). In other words, in an area that receives only 110 cm of water per year (the Piedmont area of NC), raising a corn crop would require nearly 25% of the annual rainfall. But if we assume corn only grows for 6 months, and the rain is evenly spread out all year (which it is), then the corn actually requires 55% of the rain that falls during the growing season. Now you see why some farmers have to irrigate in the east. For your information, sunflowers are even more water demanding than corn!
Stomatal and Cuticular E
When plants moved from
water to land, they faced enormous obstacles, most notably, how to prevent
dessication. As you saw from calculating water potentials of the
atmosphere, the drying power of the air is very great. To solve this
dilemma, plants evolved a cuticle - a waxy covering that slows down (it
is not waterproof, as some believe!) evaporation from plant surfaces.
This waxy cuticle is built up of several layers of waxy material called
cutin. This material is a long-chain fatty acid (typically 16-18
carbons long) and hydroxylated. Ester bonds form between the hydroxyl
and carboxyl groups, interlocking the acids into a polymeric network.
Other waxes (with more carbons, sometimes nearly 40) form a matrix in which the cutin is embedded. This can include saturated hydrocarbons, alcohols, aldehydes, and ketones. These waxes are all very hydrophobic, so they repel water very well. This prevents plants from becoming soggy when it rains. They also prevent excessive water loss from plant surfaces.
But such a layer presents a dilemma for plants. If they are to take up carbon dioxide and perform photosynthesis, they have to find a way to get the carbon dioxide into the leaf. But diffusion of this gas is prevented by the cuticle. In response, plants have evolved small pores in the leaf, called stomata, through which gas exchange can occur. Stomata are made up of two guard cells, which form the actual pore opening, and several modified epidermal cells that assist in the functioning of the stomata.
Plants can open and close the pores by changing the water status of the guard cells. When they take up water from the surrounding epidermal cells, they swell, and their inner surfaces pull away from each other, opening the pore. When they lose water, they come back together, and the pore closes. So, changes in water potential of the guard and epidermal cells are responsible for regulating the size of the stomatal pore.
This means, of course, that in order to carry out photosynthesis, plants must open their pores, which means that they will lose large amounts of water. This is the "cost" of doing photosynthesis on land - in order to take up carbon dioxide, plants have to lose water. But, as we will see, by regulating the stomatal opening, they can control the amount of water lost and keep it at a reasonable level, while still taking in adequate amounts of carbon dioxide.
Forces Governing Water Loss from Leaves
Water moves down the
water potential gradient, and the same holds true for water evaporating
from leaves. In this case, the atmosphere has a very low water potential
(sometimes -300 bars, -30 MPa) so the gradient is almost always OUT of
the leaf. But what about evaporation from the inside of a leaf?
Water is thought to evaporate from the inner walls of mesophyll cells, beneath the stomatal pore. The space beneath the pore is often called the "substomatal cavity". Most physiologists assume the air space is saturated with water vapor, so the RH is 100%. Thus, the water potential is near 0 MPa. This also means that we can easily know the vapor pressure of the leaf interior, since at 100% RH, at a particular temperature, the air would be saturated. Thus, if we know leaf temperature, we know internal leaf vapor pressure!! A leaf at 25oC would have a vapor pressure of 3.167 KPa (see your vapor pressure chart).
If the air was also at 25 C, but only 50% RH, the vapor pressure (VP) of the air would be:
0.50 * 3.167 = 1.584 KPa
The gradient in VP between the leaf and air represents the true driving force for diffusion of water vapor out of the leaf, and is called the Vapor Pressure Deficit (VPD). It is:
VPleaf - VPair = 3.167 - 1.584 = 1.584 KPa
To relate transpiration (E) to VPD, we use the following equation:
E = [VPD/BP]*gsw
where:
E is the transpiration rate in mmol H2O m-2 s-1
VPD is the vapor pressure deficit in KPa
BP is the barometric pressure in KPa (101.3 at sealevel)
gsw is the conductance to water vapor, usually the total or
stomatal
conductance
As an example:
Suppose you had the following:
Air Temperature is 20oC
Leaf Temperature is 16oC (lower than air temperature because
of
evaporative cooling)
Relative Humidity is 60%
Conductance is 400 mmol H2O m-2 s-1
What is the transpiration rate?
1. Solve to get VPair first. This is RH*VPsat
where VPsat is the saturation
vapor pressure of the air at air temperature (refer to your VP chart).
(0.60)*(2.338) = 1.4028 KPa
2. The leaf VP is simply the saturation VP at leaf temperature, which is
1.818 KPa.
3. The VPD is VPleaf - VPair, which is 1.818 - 1.403
= 0.415 KPa
4. Now, insert known terms into the equation to solve for E:
E = (0.415*400)/101.3 = 1.64 mmol H2O m-2 s-1
Now, assume that the air temperature rises by 10oC, and that leaf temperature rises by 5oC. What is the new transpiration rate?
1. Remember, the VPair is still the same!!! Just because
the air heated up
does not mean that the amount of water in the air rose also!
So VPair = 1.403 KPa still.
2. However, the VPleaf has increased, because the leaf is hotter.
The new VPleaf = 2.486 KPa
3. The new VPD is now 2.486 - 1.403 = 1.083 KPa
4. Our new E is:
E = (1.083*400)/101.3 = 4.28 mmol H2O m-2 s-1
Note that the increase in air and leaf temperature result in a 2.6X increase in E. Of course, we have assumed that conductance did not change. This is not always the case, but we kept it the same for simplicity's sake.
You should work out other problems to get familiar with these concepts.
They are posted on the web also.