Modern Engineering Thermodynamics

17.4 Energy Conversion Efficiency of Biological Systems 701
On the other hand, in the case of most plants and some animals, there is either direct energy conversion of
incoming solar radiation or a metabolic reduction resulting from direct body warming from incoming solar
radiation. Since radiation is one of the classical heat transfer mechanisms, solar radiation belongs to the _Q term.
In this case, part of the system’s _Q is actually incoming and used within the system and must be considered as
part of the total energy input.
The energy conversion efficiency of plant photosynthesis can be defined as
ðη E Þ photosynthesis =
Energy converted to organic molecules by photosynthesis ðper unit areaÞ
Solar energy input to earth ðper unit areaÞ
(17.14)
This is quite low, typically ranging between 0.01 and 1.0%. Part of the reason for the low efficiency is that not
all the solar energy incident on a unit area of the Earth is intercepted by a plant. As plants become smaller and
more uniformly cover the Earth, this efficiency rises somewhat. For example, in the case of algae (a microscopic
one-celled plant), the photosynthetic energy conversion efficiency at a small densely packed test site can be as
high as 10%.
The energy conversion efficiency of animals can be defined as
ðη E Þ animal =
Rate of food energy stored in the body as complex organic molecules
Rate of energy taken into the body as food
(17.15)
EXAMPLE 17.2
Everyone has heard about the food chain, but few realize how inefficient it is in nature. The energy conversion efficiency from
sunlight to plant growth is only about 1.00%, the energy conversion efficiency of the plants eaten by grazing herbivores is about
20.0%, and the energy conversion efficiency of the carnivores who hunt and eat the herbivores is only about 5.00%. So the overall
energy conversion efficiency from sunlight to carnivore is about (0.0100)(0.200)(0.0500) = 1.00 × 10 −4 = 0.0100%. If the
average daily solar energy reaching the surface of the Earth is 15.3 MJ/d · m 2 , then how much land is required to grow the plants
needed to feed the herbivores eaten by a large carnivore that requires 10.0 MJ/d to stay alive?
Solution
Since our hunting carnivore requires 10.0 MJ of food per day, at a 5.00% food energy conversion rate, it must consume
10:0 MJ/d
= 200: MJ/d
0:0500
of herbivore meat. The food energy conversion rate of the grazing herbivores is 20.0%, so they must consume
200: MJ/d
= 1000 MJ/d
0:200
in plant food. At a 1.0% energy conversion rate, the plants consumed by the herbivore require
1000 MJ/d
= 1:00 × 10 5 MJ/d
0:0100
of solar energy. Since the average solar energy intensity on the surface of the Earth is 15.3 MJ/d · m 2 , 100,000 MJ/d of solar
energy require an area of
and since 1 acre = 4047 m 2 , then
100,000 MJ/d
= 6540 m2
15:3 MJ/d.m2
6540 m 2 1 acre
¼ 1:62 acres
4047 m 2
of plant food is required to supply the food chain energy required to meet the 10 MJ/d needs on our carnivore.
Exercises
4. Using the results of Example 17.2, determine the number of carnivores that can be supported by herbivores living off
1500 acres of plants. Answer: 926 carnivores.
5. If the number of available herbivores in Example 17.2 increases dramatically and the carnivores’ hunting energy
expenditure is reduced to the point where their food energy conversion efficiency increases from 5.00% to 12.0%,
determine the amount of land required to support one carnivore. Answer: 0.675 acres/carnivore.
6. If the carnivore in Example 17.2 moves to a tropical climate where the solar intensity and the photosynthetic energy
conversion efficiency double, how much land would be required to support its food chain? Answer: 0.81 acres.