Solved thermodynamics exercise

Evapotranspiration from a large tree: a natural air conditioner?

Exercise 2 · Lesson 2 — A History of Thermodynamics and Calorimetry

  • order of magnitude
  • evapotranspiration
  • tree
  • latent heat
  • cooling power
  • air conditioner
  • COP

Statement

On a hot, dry day, a large tree with an adequate water supply may evapotranspire about 500 L=5,0×10−1 m3500\,\text{L}=5{,}0\times10^{-1}\,\text{m}^3 of water. Since transpiration occurs mainly in daylight, assume that it is spread over twelve hours. The tree canopy covers a ground area Atree=50 m2A_{\text{tree}}=50\,\text{m}^2. The density of water and its latent heat of vaporization at atmospheric pressure are

ρwater=1,0×103 kg ⁣⋅ ⁣m−3,Lv=2,45×106 J ⁣⋅ ⁣kg−1.\rho_{\text{water}}=1{,}0\times10^3\,\text{kg}\!\cdot\!\text{m}^{-3}, \qquad L_v=2{,}45\times10^6\,\text{J}\!\cdot\!\text{kg}^{-1}.
  1. Calculate the mass of water evapotranspired during the day and the energy required to vaporize it.
  2. Explain why this produces a cooling effect.
  3. Calculate the tree's average cooling power per square metre during the twelve hours of active transpiration.
  4. For comparison, consider an air conditioner supplied with electrical power Pelec=2,0×103 WP_{\mathrm{elec}}=2{,}0\times10^3\,\text{W}, with cooling coefficient of performance COP=3,0\mathrm{COP}=3{,}0, cooling a room of area Aroom=20 m2A_{\mathrm{room}}=20\,\text{m}^2. By definition, COP=Pcool/Pelec\mathrm{COP}=P_{\mathrm{cool}}/P_{\mathrm{elec}}. Calculate its cooling power and cooling power per square metre, then compare them with the tree's values.
  5. Why can we not conclude that the tree and the air conditioner provide exactly the same perceived cooling, even if their powers per unit area have the same order of magnitude?
  6. Could many houseplants cool a home? (General knowledge: the answer depends on biological processes.)

Hint
Use m=ρVm=\rho V, Qevap=mLvQ_{\mathrm{evap}}=mL_v, and P=Q/τP=Q/\tau. For the air conditioner, Pcool=COP PelecP_{\mathrm{cool}}=\mathrm{COP}\,P_{\mathrm{elec}}.

Detailed solution

Solution
1. The corresponding mass is

m=ρwaterV=1,0×103×5,0×10−1=5,0×102 kg.m=\rho_{\text{water}}V =1{,}0\times10^3\times5{,}0\times10^{-1} =5{,}0\times10^2\,\text{kg}.

The vaporization energy is

Qevap=mLv=5,0×102×2,45×106=1,225×109 J,Q_{\mathrm{evap}}=mL_v =5{,}0\times10^2\times2{,}45\times10^6 =1{,}225\times10^9\,\text{J},

whose order of magnitude is one gigajoule.

2. Vaporization is endothermic. It draws this energy from the leaves, soil, and surrounding air, thereby lowering the temperature of the leaves and their immediate environment.

3. Twelve hours correspond to τ=12×3600=4,32×104 s\tau=12\times3600=4{,}32\times10^4\,\text{s}. Hence

Ptree=Qevapτ=1,225×1094,32×104≈2,84×104 W,P_{\text{tree}}=\frac{Q_{\mathrm{evap}}}{\tau} =\frac{1{,}225\times10^9}{4{,}32\times10^4} \approx2{,}84\times10^4\,\text{W},

and

PtreeAtree=2,84×10450≈5,7×102 W ⁣⋅ ⁣m−2.\frac{P_{\text{tree}}}{A_{\text{tree}}} =\frac{2{,}84\times10^4}{50} \approx5{,}7\times10^2\,\text{W}\!\cdot\!\text{m}^{-2}.

4. The useful cooling power is

Pcool=COP Pelec=3,0×2,0×103=6,0×103 W.P_{\text{cool}}=\mathrm{COP}\,P_{\mathrm{elec}} =3{,}0\times2{,}0\times10^3 =6{,}0\times10^3\,\text{W}.

Per unit floor area,

PcoolAroom=6,0×10320=3,0×102 W ⁣⋅ ⁣m−2.\frac{P_{\text{cool}}}{A_{\text{room}}} =\frac{6{,}0\times10^3}{20} =3{,}0\times10^2\,\text{W}\!\cdot\!\text{m}^{-2}.

In this model, the tree's evapotranspiration power per unit area is almost twice that of the air conditioner.

5. This comparison concerns energy fluxes only. An air conditioner cools a controlled enclosed volume, whereas the vapour from a tree disperses into the atmosphere and wind brings in warm air. Perceived cooling depends on ventilation, humidity, ambient temperature, and distance from the tree. Transpiration also varies with sunlight, wind, air humidity, species, and available water, and can decrease sharply during drought. The tree additionally provides shade, which directly reduces the solar radiation received by the ground and by people. The calculation therefore compares orders of magnitude, not equal thermal comfort.

6. In practice, houseplants provide very little cooling. In most plants, light encourages stomata to open, allowing water vapour to escape. Indoor light is generally weak, so stomata open less and transpiration decreases. In a poorly ventilated room, rising humidity also progressively slows evaporation.

Many plants may therefore cause slight local evaporative cooling, but cannot replace an air conditioner. Removing heat durably would require moist air to be vented outdoors. Intense artificial lighting might stimulate transpiration, but its electrical energy would itself end up almost entirely as heat in the room. CAM plants adapted to arid environments, including cacti, are an exception: their stomata open mainly at night.