Food and Energy

Estimate the maximum plant production rate for a mid-latitude farm by considering the amount of solar energy that gets stored in sugar bonds.

• About 50% of the 1400 W/m² solar energy reaches Earth's surface,
• only 25% of that is in the right frequency range for photosynthesis,
• 60% of that is absorbed by plants if foliage is dense,
• 35% of that is actually converted to bond energy,
• cloud coverage averages ~50%

• 1400*0.5*0.25*0.60*0.35*0.5 = 18W = 18 J/s


• The sun is only "up" and producing plant material ~120 day/yr *11 hr/dy *3600 s/hr = 4.75 Ms per year. Multiplying this by the bond energy production rate computed above gives 4.75 Ms * 18 J/s = 86MJ per year per square meter.
• The heat of combustion of sugar (roughly the same for cellulose) is 3744 cal/g or (3744000cal/1kg) *4.186J/cal = 15.67 MJ/kg. This is approximately the energy of formation.
• The theoretical plant-production-rate is thus about (86MJ/yr/m²) /(15.67 MJ/kg) = 5.6 kg/yr/m²
  Q1: This is an overestimate; list as many reasonable systematic errors that may account for this as you can, explaining each.
• For the sake of comparison:
  • US canola production rate ~0.16 kg/m²/yr (1400 lb/acre/yr).
  • A 5ton/acre hay rate = 1.2 kg/m²
  • Sweet corn yields, per year, about ~(12000 ears sweet corn)/acre at ~(3oz edible corn)/ear = (3oz/ear *(.0284kg/oz) *12000ears) / (4045m²/acre) =0.25 kg/m².
  • Feed corn yields 140 bushels/acre at ~55 lb/bushel = 7700 lb/acre == 0.86 kg/m².
  • Q2: What, do you suppose, accounts for the vastly different yield values among canola oil (0.16 kg/m²), sweet corn, feed corn, and hay (1.2 kg/m²)?

Compute the efficiency (food-energy)/(fuel-energy) of the US farming system.

For this calculation, we will use a 250 acre crop (1km by 1km) patch and an average production rate of 0.3 kg/m²/yr. This gives 300,000 kg/yr = 300 metric tons/yr. For each of the following six energy uses, compute the amount of energy in mega-Joules and then divide the total by the total yield of 3E5 kg to get MJ/kg of grain:

3. To disk, plant, fertilize, and harvest takes 4 operations. Assume a 400 hp tractor with 15% efficiency pulling a 5m wide tool at 5km/hr.
4. Estimate the energy required to produce the fertilizer for the field. The heat of formation of ammonium nitrate is ~ 40000 BTU/lb =93 MJ/kg and is applied at ~ 10-30 kg/acre.
5. Estimate the fuel energy requirements to transport this corn to food processing locations. Hint: First show that a fuel efficiency of ~40 ton*miles/gallon makes sense for a fully loaded semi.
6. Estimate the packaging energy using the data below and the measured values from a box of microwave popcorn.
7. Estimate the cooking energy per kg. Base this on the energy required to cook microwave popcorn.
8. Look up the caloric content of popcorn and compute the food-energy value, in MJ/kg. .
   
9. Summarize the above information by tabulating all of the main energy terms, each in MJ/kg.
10. Compute the sum of all the fossil fuel components. This "energy per kg-of-food" is called a food energy subsidy, why?
11. Compute the ratio of food-energy-per-kg divided by the fuel-energy-per-kg.
12. Sketch a full-sheet-size diagram of energy flows into and out of the US agriculture sector. Label each of these energy flows with source type - solar, petroleum, nuclear, coal, wood, food, ... - and indicate the relative magnitude by the width of the flow arrow.
13. Consider the fact that producing meat from grains uses roughly 10kcal feed for each 1kcal of meat. What impact would that have on the overall efficiency of the US agricultural "engine?" This should be a numerical estimate with discussion.
14. What simple practices could we use to reduce this food subsidy? Comment on the effectiveness of each practice.

Some constants:
1 acre = 4045m2
1 lb = 16 oz = 454 gm
1 oz = 28.35 gm
1 ton = 2000 lb while 1 metric ton = 1000kg = 2200 lb
1 gal gasoline -> 130MJ