SME
Ecology
Ecosystem energetics
Use this for “GPP is 20,000 kJ/(m²·yr) and plants respire 11,000. Herbivores eat 1800, assimilate 720 and make 72. Find NPP and the efficiencies.”
Chart
U = GPP − Rₐ − F − Rₕ − P
20,000 − 11,000 − 1,080 − 648 − 72 = 7,200
The flows in add and the flows out take away: U = 7,200.
This is an example. Type your own values to start a new problem.
GPP
Gross primary production
example · kJ/(m²·yr)
Rₐ
Plant respiration
example · kJ/(m²·yr)
NPP
Net primary production
calculated · kJ/(m²·yr)
I
Ingested by herbivores
example · kJ/(m²·yr)
A
Assimilated
example · kJ/(m²·yr)
P
Herbivore production
example · kJ/(m²·yr)
AE
Assimilation efficiency
calculated · %
PE
Production efficiency
calculated · %
TE
Trophic efficiency
calculated · %
U
Not eaten
worked out (not typed) · kJ/(m²·yr)
F
Egested
worked out (not typed) · kJ/(m²·yr)
Rₕ
Herbivore respiration
worked out (not typed) · kJ/(m²·yr)
Formulas
NPP = GPP − Rₐ
9,000 = 20,000 − 11,000
AE
=
100
×
A
I
40
=
100
×
720
1,800
PE
=
100
×
P
A
10
=
100
×
72
720
TE
=
100
×
P
NPP
0.8
=
100
×
72
9,000
U
=
NPP
−
I
7,200 = 9,000 − 1,800
F
=
I
−
A
1,080 = 1,800 − 720
Rₕ
=
A
−
P
648 = 720 − 72
Assumptions
•
Plants respire Rₐ of what they fix (GPP); the rest, NPP, is new plant matter herbivores can eat.
•
Herbivores egest F of what they ingest (I); they assimilate A, respire Rₕ of it and keep P as new growth.
•
Trophic efficiency compares herbivore production with plant production, usually about 10% or less.
Step-by-step
WE KNOW
GPP = 20,000 kJ/(m²·yr), Rₐ = 11,000 kJ/(m²·yr), I = 1,800 kJ/(m²·yr), A = 720 kJ/(m²·yr), P = 72 kJ/(m²·yr)
FIND
net primary production (NPP), assimilation efficiency (AE), production efficiency (PE), trophic efficiency (TE), not eaten (U), egested (F), herbivore respiration (Rₕ)
Step 1 · Find net primary production (NPP)
Use NPP = GPP − Rₐ
What plants fix, less what they burn in respiration, is left as new plant matter.
NPP = 20,000 − 11,000
NPP
=
9,000
kJ/(
m
2
·yr)
Step 2 · Find assimilation efficiency (AE)
Use AE = 100 × A ÷ I
The share of the food eaten that is absorbed, as a percent.
AE
=
100
×
720
1,800
AE
=
72,000
1,800
AE = 40%
Step 3 · Find production efficiency (PE)
Use PE = 100 × P ÷ A
The share of the energy absorbed that becomes new growth, as a percent.
PE
=
100
×
72
720
PE
=
7,200
720
PE = 10%
Step 4 · Find trophic efficiency (TE)
Use TE = 100 × P ÷ NPP
Herbivore production as a percent of plant production.
TE
=
100
×
72
9,000
TE
=
7,200
9,000
TE = 0.8%
Step 5 · Find not eaten (U)
Use U = NPP − I
Plant matter herbivores do not eat goes to the decomposers.
U
=
9,000
−
1,800
U
=
7,200
kJ/(
m
2
·yr)
Step 6 · Find egested (F)
Use F = I − A
What is eaten but not absorbed leaves as feces.
F
=
1,800
−
720
F
=
1,080
kJ/(
m
2
·yr)
Step 7 · Find herbivore respiration (Rₕ)
Use Rₕ = A − P
What is absorbed but not kept as growth is burned in respiration.
Rₕ = 720 − 72
Rₕ
=
648
kJ/(
m
2
·yr)
Check
Put the numbers back in. Both sides should match.
9,000 = 20,000 − 11,000 ✓
40
=
100
×
720
1,800
✓
10
=
100
×
72
720
✓
0.8
=
100
×
72
9,000
✓
7,200 = 9,000 − 1,800 ✓
1,080 = 1,800 − 720 ✓
648 = 720 − 72 ✓
Topic check
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