Microbes to our rescue
Plan B
Edward Chesky
Major (Retd) US Army, USA
Mohideen Ibramsha
1968 Alumni of Thiagarajar College of Engineering, Madurai, TN, India
1974 intellectual son of PhD guide Prof. V.Rajaraman & Mrs. Dharma Rajaramn, CS, EE, IIT, Kanpur, UP, India
1991 First HOD of CSE, CEC [now BSAU] Chennai, TN, India
Associate Professor (Retd), Computer Science, Framingham, MA, USA
Consultant R&D, M A M College of Engineering, Trichy, TN, India
Advisor, HyDIGIT Pte Ltd, Singapore
Email: ibramsha7@yahoo.com
Introduction: In https://mohideenibramsha7.wixsite.com/website/single-post/2018/09/16/Hydrogen-Economy-Now-11 we considered the CO2 consuming microbes to produce CH4 from CO2 and H2. Here we consider the use of microbes that grow consuming CO and CH3COOH to produce CH4 and Hydrogen.
Microbe details: The PDF file of a review on “Acetate Metabolism in Anaerobes from the Domain Archaea” gets downloaded through the URL www.mdpi.com/2075-1729/5/2/1454/pdf In the PDF file there are two figures of interest to us. Figure 3 gives the actions by the microbe Methanosarcina acetivorans in producing acetate from CO. Figure 4 illustrates the actions by the microbe while consuming the acetate.
Acetate production: By considering the inputs into the membrane we find that 8 CO enters the membrane. From the membrane we find CH3COOH and CH4 leave. Combining the input and the output, we might write the following equation.
8CO → CH3COOH + CH4 [1]
Equation 1 does not have the same number of Carbon atoms, Oxygen atoms and Hydrogen atoms in its LHS and RHS. The missing atoms are supplied by 4 molecules of water, and producing 5 molecules of CO2. Now we get:
8CO + 4H2O → CH3COOH + CH4 + 5CO2 [2]
Equation 2 is balanced.
The CO2 molecules exit the membrane.To ensure that the CO2 molecules do not enter the membrane, the enzyme Cam shown in Figure 4 combines one H2O molecule with one CO2 molecule and produces HCO3 which cannot enter the membrane. Considering the molecules that leave the membrane to be converted, we get
5CO2 + 5H2O → 5HCO3(-) + 5H(+) [3]
The full reaction involves equations 2 and 3. Adding all the elements of the LHS of equations 2 and 3, we get 8CO + 4H2O + 5CO2 + 5H2O. Likewise adding the elements of the RHS we get CH3COOH + CH4 + 5CO2 + 5HCO3(-) + 5H(+). Cancelling the CO2 molecules that appear on both sides, we get:
8CO + 9H2O → CH3COOH + CH4 + 5HCO3(-) + 5H(+) [4]
Equation 4 is balanced. As the microbe consumes more CO and grows, the concentration of HCO3(-) continues to increase until it reaches saturation. Once saturation is reached, HCO3 precipitates. The detail about the precipitation of HCO3 is given below.
From http://weppi.gtk.fi/publ/foregsatlas/text/HCO3_.pdf we find that the measurements done by the Forum of European Geological Surveys found an average of 126 milli gram of HCO3 per liter of water, while in the waters of the Barents region it was 28 milli grams per liter of water. From http://employee.heartland.edu/rmuench/chem161/solubilityrules.pdf we find that HCO3 it is seen that HCO3 exists in the aqueous form or as solid. We understand that the HCO3 produced by the microbes remain dissolved in water up to a concentration and then precipitate.
As HCO3 precipitates two H(+) would join and become Hydrogen. The Hydrogen escapes as gas. The CH4 escapes as gas. The acetate remains in the water.
Acetic acid is a weak acid. From https://chemistry.stackexchange.com/questions/22036/how-is-the-dissolution-of-acetic-acid-that-makes-its-aqueous-solution-a-poor-ele we find:
When acetic acid is dissolved in water there is an equilibrium reaction:
CH3COOH + H2O ← → CH3COO− + H3O+
Since acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.
The rest of CH3COOH remains surrounded by water molecules as explained in https://www.quora.com/Why-is-acetic-acid-soluble-in-water by Yashit Maheshwary.Quoting her, we get:
Acetic Acid (CH3COOH), has only one carbon atom (Other than the carbonyl carbon), so steric hindrance provided by the methyl group is substantially low. This allows water molecule to easily form bonds with the acetic acid molecules (Hydrogen Bonds).
Thus acetic acid remains in water. As the concentration of Acetic Acid increases, we expect that the microbe would grow consuming the CH3COOH as well. We consider the consumption of CH3COOH now.
Acetate consumption: Figure 4 in the PDF is about acetate consumption. Inside the cell, CH3COOH becomes CH4 + CO2.Both CH4 and CO2 leave the cell membrane. To avoid the CO2 re-entering the cell, the enzyme Cam converts CO2 to HCO3- using one H2O molecule.
The reactions are:
CH3COOH → CH4 + CO2 [5]
CO2 + H2O → HCO3(-) + H(+) [6]
Combining the two equations, we get
CH3COOH + H2O → CH4 + HCO3(-) + H(+) [7]
One bioreactor or two bioreactors?: We feed CO to the microbe. As the microbe produces CH3COOH, after sometime the microbe has both CO and CH3COOH available for consumption. As CH3COOH accumulates it is possible that some microbes would consume CH3COOH while other microbes might consume CO. As CO is a gas chances are that CO would leave the bioreactor without being consumed by the microbes.
To avoid partial consumption of CO, it is preferable to give CO to the microbe when CH3COOH is not available in plenty.
When the microbe consumes CO, we get 5 Hydrogen atoms for every CH4 molecule. In contrast when the microbe consumes CH3COOH we get just one Hydrogen atom for every CH4 molecule.
Let us assume that X fraction of microbes consume CO and the remaining (1 - X) fraction of microbes consume CH3COOH. Then we get 5 Hydrogen atoms for every CH4 molecule from X fraction and 1 Hydrogen atom for every CH4 molecule from (1 - X) fraction. Thus we get [5X + (1 - X)] Hydrogen atoms for [X + (1 - X)] CH4 molecules. The ratio between the Hydrogen atoms and the CH4 molecules is (1 + 4X) Hydrogen atoms for every CH4 molecule. So X = (R - 1)/4 where R is the ratio. We decide on a value of X. When we start with no Acetate in the bioreactor R would be 5 and would continue decreasing as Acetate accumulates. When R reaches the value corresponding to the decided value of X, we stop feeding CO and let the microbes consume the dissolved Acetate only.
We start feeding the microbes in another bioreactor with CO. When the second bioreactor reaches the desired value of X, we stop feeding CO in the second reactor and start feeding CO in the first reactor. As the microbes in the first reactor grew consuming CH3COOH, when we restart feeding CO to the first reactor, we assume that the CH3COOH concentration in the first reactor is sufficiently low so that on feeding CO most of the microbes would consume CO.
By experiments decide the best value for X such that the production of CH4 is maximized. Since the H2 produced is proportional to the CH4, when we maximize CH4 we maximize H2 as well.
CO for desired CH4: When the microbe consumes CO, it produces one molecule of CH4 and one molecule of CH3COOH. When the CH3COOH is consumed by the microbe it gives one molecule of CH4. Thus we get 2 molecules of CH4 when 8 molecules of CO are consumed by the microbe. We are not considering the production of H2, as H2 is not required to generate CH4. We need the CO molecules to generate CH4.
By performing the high temperature reaction of the SMR we get CH4 + H2O → CO + 3H2. Since we produce 2 CH4 molecules from 8 CO molecules, from the SMR we get 2 CO molecules and we need to supply 6 more CO molecules to sustain the process of using the microbes to produce CH4 and H2. For these additional 6 molecules of CO we perform reaction of Carbon with steam following the reaction C + H2O → CO + H2.
We quote to substantiate the approach.
https://www.quora.com/When-solid-carbon-reacts-with-steam-carbon-monoxide-and-hydrogen-gases-are-formed-This-is-a-reversible-reaction-If-pressure-is-increased-which-side-will-the-equilibrium-shift-to-Does-the-solid-carbon-count-as-a-mole-when-answering-this
===
Michael Mombourquette, Chemistry Professor at Queen's University at Kingston (1996-present)
Answered Jan 9, 2018
The balanced equation is
C(s) + H2O(g) → CO(g) and H2(g)
LeChatelier’s principle says if you apply a stress to a system in equilibrium, the equilibrium will change to reduce the stress.
If you add more steam, H2O(g), the system will shift to products to use up the extra pressure of H2O.
===
The temperature of the above process could be inferred from the following abstract.
https://pubs.acs.org/doi/abs/10.1021/ie50519a054
===
Reaction of Carbon with Steam at Elevated Temperatures
The present discussion is concerned primarily with the reaction at temperatures above 2000 F, the region of greatest interest to the manufactured gas industry. At temperatures above 1700 F, it has been shown that the equilibrium products of the reaction are almost entirely carbon monoxide and hydrogen. Experimentally, however, some carbon dioxide and water are usually present, even at very high temperatures, indicating the incomplete nature of the reaction.
===
We suggest that the heat of the reaction for the SMR and the generation of CO from Carbon be supplied by the biomass. We extract Syngas from the biomass to generate steam for both processes. The carbon residue of the biomass after extracting Syngas is used to supply the Carbon for the production of CO.
As in plan A discussed in https://mohideenibramsha7.wixsite.com/website/single-post/2018/09/16/Hydrogen-Economy-Now-11 we consider the regeneration of CH4.
Regeneration of CH4: The bioreactor consuming CO produces CH4 following
8CO + 9H2O → CH3COOH + CH4 + 5HCO3(-) + 5H(+) [8].
The bioreactor consuming CH3COOH produces CH4 following
CH3COOH + H2O → CH4 + HCO3(-) + H(+) [9].
Both bioreactors produce 2CH4, 6HCO3(-) and 6H(+) using 8CO and 10H2O.
In the high temperature part of the SMR, we use 2CH4 as follows.
2CH4 + 2H2O → 2CO + 6H2 [10].
Using steam on Carbon we produce
6C + 6H2O --: 6CO + 6H2 [11].
Having produced 8CO, the process could be repeated.
In one cycle we produce 5H(+) in CO consuming bioreactor, 1H(+) in CH3COOH consuming bioreactor, 6H2 in the high temperature reaction of SMR, and 6H2 when producing CO from Carbon and steam. Thus we get a total of 5H(+) + 1H(+) + 6H2 + 6H2 = 15H2.
To identify the inputs we rewrite the equations and simplify the LHS and RHS sides of the equations cancelling out common terms.
8CO + 9H2O → CH3COOH + CH4 + 5HCO3(-) + 5H(+) [8].
CH3COOH + H2O → CH4 + HCO3(-) + H(+) [9].
2CH4 + 2H2O → 2CO + 6H2 [10].
6C + 6H2O --: 6CO + 6H2 [11].
8CO appearing in the LHS of [8] is cancelled by the 2CO in the RHS of [10] and 6CO in the RHS of [11]. CH3COOH appearing in the LHS of [9] is cancelled by the CH3COOH appearing in the RHS of [8]. The 2CH4 appearing in the LHS of [10] are cancelled by the CH4 appearing in the RHS of [8] and the CH4 appearing in the RHS of [9].
Indicating the cancelled terms by underlining them, we rewrite the equations below:
8CO + 9H2O → CH3COOH + CH4 + 5HCO3(-) + 5H(+) [8].
CH3COOH + H2O → CH4 + HCO3(-) + H(+) [9].
2CH4 + 2H2O → 2CO + 6H2 [10].
6C + 6H2O --: 6CO + 6H2 [11].
Adding the remaining terms in the LHS and RHS, we get:
9H2O + H2O + 2H2O + 6C + 6H2O → 5HCO3(-) + 5H(+) + HCO3(-) + H(+) + 6H2 + 6H2.
Consolidating the terms, we get:
6C + 18H2O → 6HCO3 + 15H2
As HCO3(-) precipitates as solid, the H(+) leaves and becomes Hydrogen gas.
The inputs are Carbon supplied by the biomass and water. The output is Hydrogen that would be collected for the Hydrogen Economy. The solid HCO3 is collected and stored for possible use later or simply stored to protect it from future release of CO2 by appropriate microbes or by heat..
Weather security: We have the processes of SMR, bioreactors, and generators of CO from Carbon. All these processes could be housed in weather-secure enclosures as in Plan A. The Carbon needs to be supplied by biomass. Normally the trees are grown in the open and are subject to harm by hostile weather.
We could make greenhouses weather secure. We propose to grow fastest growing plants inside the greenhouses to supply the biomass for the Hydrogen Economy.
The Guinness Record for the fastest growing tree is held by the Empress Tree.
http://www.guinnessworldrecords.com/world-records/fastest-growing-tree-
===
The world´s fastest-growing tree is the empress or foxglove tree (Paulownia tomentosa), named after its purple foxglove-like flowers. It can grow 6 m in its first year, and as much as 30 cm in three weeks.
===
In the UK climate this tree could grow to a height of 12 meters.
http://www.deepdale-trees.co.uk/trees/2016/07-Paulownia-tomentosa.html
===
The most common grown variety in the UK is the Paulownia tomentosa Award of Garden Merit (AGM) usually reaching up to 12m tall in our climate.
===
We believe the growth rate decreases after the first year. We would like to reap as much carbon as possible in a given period. More details about this tree are found in an article describing the attempts to introduce this in Albania.
https://www.researchgate.net/publication/309656225_Paulownia_Tomentosa_a_Fast_Growing_Timber
===
Paulownias plant has a tendency to form many branches if it is grown in open space, whereas in the forest it tends to form a straight trunk. … Paulownia tree aged 5 – 7 years old can generate 1 m3 timber ... Paulownia trunk is light, … with a specific mass of 0,35 g/cm3. … Paulownia is a very fast growing plant, trees can reach 4 – 6 m only during the first year (after cutting) growing further 2 – 3 m during the second year … Calorific value of Paulownia biomass is higher than that of coal … in Bulgaria are planted plantations for biomass with density of 3000 - 3300 plants/ha in distances 3,3 x 1 m and 2 x 1,5 m respectively. … the most interesting indicators are the average plant height of 4,25 m and the trunk diameter on chest height by 5,22 cm … Paulownia planted since 13 years ago in the garden of a house … a height over 10 m and a diameter of 38 – 40 cm on chest height,
===
From the above quote we find that in one year the trunk had a height of 4.25 meters with a diameter of 5.22 cm. The cross sectional area of the trunk is 21.40 sq cm. The volume of the trunk is 9,095.36 cubic cm. A tree after 13 years of growth has a height of 10 meters and a trunk with a diameter of 38 to 40 cm. Using 40 cm as the diameter, the cross sectional area of the 13 year old trunk is 5,026.55 sq cm and the volume of the trunk is 5026548.25 cubic cm.
If we cut the tree every year, in 13 years we would have 9095.36 x 13 = 118239.68 cubic cm. Cutting the tree when it is fully grown gives 42.51 times timber. Thus it is advantageous to cut the tree when it is fully grown.
The tree grows to a height of 12 meters in UK. Let us make a geodesic dome of 12 meter height. The radius of the dome is 12 meters. The land covered by the dome is 452.39 sq meters. With one tree every 3 sq meter, we could plant 150 trees. Let us start planting at a distance of 1 meter from the edge of the dome. The radius is 11 meters, with a circumference of 69 meters. Planting at a distance of 3 meters between adjacent trees, we plant 23 trees. The height of the dome at these points is 4.80 meters. Thus these 23 trees could be cut every year. At a distance of 2 meters from the edge, we could plant 20 trees. These trees could grow up to 6.63 meters. Thus these 20 trees could be cut every two years. At a distance of 3 meters from the edge, the circumference is 56.55 meters and we could plant 18 trees. These trees could grow to a height of 7.94 meters and could be cut every three years or so. The circumference at a distance of 4 meters from the edge is 50.27 meters permitting 16 trees to be planted. These 16 trees could grow 8.94 meters tall and could be cut possibly every 5 years. The trees planted at a distance of 5 meters from the edge could grow 9.75 meters tall. We could plant 14 trees at this distance, and these trees could be cut every 8 years. From a distance of 6 meters and more the trees could grow more than 10 meters and these could be cut every 10 years. The number of trees in this 113.1 sq meters is 37 giving an area of 3 sq meters for every tree. The total number of trees within the dome is 37 + 14 + 16 + 18 + 20 + 23 = 128 trees with different periods of growth.
Let us assume that the diameter of the trunk increases linearly from 5.22 cm after one year to 40 cm after 10 years. On average the diameter increases by (40 - 5.22)/9 = 34.78/9 = 3.86 cm every year. We calculate the average volume of wood available after ten years when the trees have stabilised.
No Distance Height cm Width cm Volume Cubic cm Number Period Annual yield
1 1 m 480 5.22 10272 23 1 year 236265
2 2 m 663 9.08 42931 20 2 years 429314
3 3 m 794 12.94 104418 18 3 years 626513
4 4 m 894 20.66 299700 16 5 years 959042
5 5 m 975 32.24 795947 14 8 years 1392908
6 6 m+ 1000 40.00 1256637 37 10 years 4649557
Total 8293599
The volume of timber harvested every year is 8,293,599 cubic cm. The mass of the timber is 0.35 gm per cubic cm. Hence weight of timber harvested every year is 8,293,599 x 0.35 = 2902759 gm = 2902.759 Kg. Let us assume that the carbon is half the mass of the timber. The other half is moisture and Syngas. Their actual amounts could be decided during the harvesting of the trees. We are interested in the amount of carbon only. Thus the carbon from one plantation under the geodesic dome is 1451.379 Kg.
The final equation for our process to generate Hydrogen is 6C + 18H2O → 6HCO3 + 15H2. From the equation we find that 72 Kg of Carbon yields 30 Kg of Hydrogen. Thus 1451.379 Kg of Carbon would give us 604.74 Kg of Hydrogen per year.
With 120 MJ of energy per Kg of Hydrogen, the energy yield per dome is 72,568.80 MJ per year or 0.002301 MJ per second. If this Hydrogen is used to produce electricity in a CAM ALLAM power plant, with 50% efficiency, we get 1150 J per second or 1150 watts. We could plan for one forest under one geodesic dome for 1 KW of power. We might assume that the 150 watts out of 1,150 watts is enough for the energy required to cut the trees, transport the cut pieces, and to produce the Carbon for the generation of Hydrogen. Now we ignore the energy in the Syngas produced from the wood.
We need to plant as many plantations as required depending on the amount of Hydrogen we need every year. The area of one plantation is 452.39 sq meter. The geodesic dome protects the plantation from hurricanes and tornadoes only. It does not protect the plants from earthquake.
In forests one could simply cut a tree close to the ground and after the tree falls to the ground, the fallen tree could be cut to small pieces for transportation. We cannot cut the trees inside the geodesic dome close to the surface and push the cut tree to the ground to avoid damaging the nearby closely planted trees. We need to first trim the leaves and the branches leaving the trunk. The trunk should be cut from the top in pieces small enough to be taken down and transported. Equipment to cut the trunk starting with the top and moving down could be built.
As the geodesic dome protects the trees from hurricanes and tornadoes, we need to protect the trees from fire and earthquakes. We first consider the protection from earthquakes.
Earthquake protection: Our geodesic dome protected carbon source has trees planted in horizontal areas. First we consider some tree losses from earthquakes.
One recent earthquake occured in China. The details are found at www.mdpi.com/2072-4292/8/3/252/pdf
The above URL downloads the PDF file of the article on “Assessing Earthquake-Induced Tree Mortality in Temperate Forest Ecosystems: A Case Study from Wenchuan, China”, published on 17th March, 2016. We quote from the PDF.
Earthquakes can produce significant tree mortality … Through surface faulting and ground shaking, earthquakes induce extensive forest loss … The Wenchuan earthquake occurred on 12 May 2008 … MODIS Terra images with a spatial resolution of 250 m were selected from two dates, 9 May 2007 and 24 May 2008, to represent the pre- and post-earthquake conditions, respectively … The earthquake impact boundary was estimated by comparing the ∆GV map with surface ground shaking experienced by the Wenchuan Earthquake (seismic intensity field). … all trees with a diameter at breast height (DBH) >= 5 cm were measured with variables including tree species, DBH, tree height and tree status (living or dead). … The boundary line of 75 km away from the seismic intensity isoline of 10 was set as the earthquake impact area … The turning point, i.e., 75 km away from seismic isoline 10, was therefore set as the average maximum distance (boundary) for forests affected by the Wenchuan earthquake … The larger trees sustained more damages from the earthquake … The aboveground biomass loss (117 Mg¨ ha´1 ) in the Wenchuan fault zone was even higher than the biomass loss following hurricane Katrina (87 Mg¨ ha´1 ) … The prime impact factor on earthquake-induced tree mortality is topography. Trees located on hills with steep slopes experienced higher mortality rates (Figure 9a). This is consistent with the findings in other earthquake-related studies [31,32] as well as the field investigation in our work, i.e., the landslides and mountain collapses were more often found in topographically steep regions and consequently buried more trees. … Since trees are usually buried following earthquake events, pre-earthquake forest conditions were approximated by measuring the adjacent undisturbed forest … the study area is a mountainous area with steep valleys. The land use and vegetation are fragmented. This is significantly different from plains … the Wenchuan earthquake had a strong and measurable impact on tree mortality with a total biomass carbon loss of 10.9 Tg¨C.
In Figure 9a of the above PDF, the smallest inclination of the forest area is about 28 degrees. Most of the loss was due to the trees falling down and subsequently covered by the sliding land. We would like to record that the buried wood would eventually become compost after releasing substantial amounts of Methane. We recommend, if possible, to dig out the fallen trees and use them as source of carbon avoiding Methane production.
What causes the tree to fall during an earthquake? The waves produced by an earthquake are classified into four types. We quote from http://eqseis.geosc.psu.edu/~cammon/HTML/Classes/IntroQuakes/Notes/waves_and_interior.html
There are many different seismic waves, but all of basically of four types:
1. Compressional or P (for primary)
2. Transverse or S (for secondary)
3. Love
4. Rayleigh
An earthquake radiates P and S waves in all directions and the interaction of the P and S waves with Earth's surface and shallow structure produces surface waves. … Near an earthquake the shaking is large and dominated by shear-waves and short-period surface waves. These are the waves that do the most damage to our buildings, highways, etc. … The first two wave types, P and S , are called body waves because they travel or propagate through the body of Earth. The latter two are called surface waves they the travel along Earth's surface and their amplitude decreases with depth into Earth. … Secondary , or S waves, travel slower than P waves and are also called "shear" waves because they don't change the volume of the material through which they propagate, they shear it. S-waves are transverse waves because they vibrate the ground in a the direction "transverse", or perpendicular, to the direction that the wave is traveling. … An important distinguishing characteristic of an S-wave is its inability to propagate through a fluid or a gas because a fluids and gasses cannot transmit a shear stress and S-waves are waves that shear the material. … Love waves are transverse waves that vibrate the ground in the horizontal direction perpendicular to the direction that the waves are traveling. They are formed by the interaction of S waves with Earth's surface and shallow structure and are dispersive waves. … Another important characteristic of Love waves is that the amplitude of ground vibration caused by a Love wave decreases with depth - they're surface waves. … Rayleigh waves are … dispersive … and they also decrease in amplitude with depth.
If we could avoid Love waves and Rayleigh waves from the trees inside the geodesic dome, we could save the trees from earthquakes.
The Love and Rayleigh waves are generated by the interaction of the S wave with the surface. Let us dig a trench around the geodesic dome to a depth of t meters. The slope connecting the centre with the bottom is t/12 meter per meter. However, the slope of a line connecting a point on the geodesic dome at the earth’s surface and the bottom of the trench diagonally opposite to the point is t/24 meter per meter.
So far it is assumed that the earthquakes occur at the fault lines. We decide the depth t such that the nearest fault line at the expected depth of the future earthquake is above the cone with the slope t/24 meter per meter from the center of the geodesic dome. Since S wave cannot jump the trench, we need to consider only those earthquakes that might occur within the cone. We have selected t such that the chance of an earthquake within the cone is negligibly small.
The S waves from such rare earthquakes could reach the trees. To protect the trees from the potential Love and Rayleigh waves that could be generated within the geodesic dome we take the following two actions.
Avoid areas that amplify the earthquake waves and might lead to liquefaction of the surface.
Bind every tree, above a height of 2 meters such that a triangular mesh connects all the trees. The triangular mesh ensures that no tree could be shook in isolation. As all trees need to shake together, the combined moment of inertia of all the trees is expected to be very high. This moment of inertia should overcome the shaking force due to the Love and Rayleigh waves of earthquakes of a certain magnitude. Decide the strength of the rods connecting the trees based on the strength of the earthquake to be protected against as we expect the trees might shake due to the failure of the rods..
To avoid risky surfaces, we perform the Standard Penetration Test quoted from https://oregonstate.edu/instruct/oer/earthquake/10%20chapter%208_color.html
The liquefaction susceptibility of sand can be determined by standard geotechnical engineering tests such as the Standard Penetration Test. During this test, a sampling tube is driven into the ground by dropping a 140-pound weight from a height of thirty inches (okay, it isn’t rocket science, but it works because every foundation engineer does it exactly the same way). The penetration resistance is the number of blows (number of times the weight is dropped) it takes to drive the sampler one foot into the soil. A low penetration resistance would be fewer than ten blows per foot; a high resistance would be greater than thirty blows per foot. Liquefiable sands have a very low penetration resistance; it’s very easy to drive the sampling tube into the sand.
We hope that the above precautions would protect the geodesic dome and the trees inside the dome from earthquake. The circular trench must be dug around each dome. We have indicated the process to decide the depth of the trench. The width of the trench could depend on the slope formed by a freestanding cone of the soil. As this is linearly dependent on the depth of the trench, it could become quite large. To reduce the area of the dome and the trench, we use techniques to keep the soil from sliding into the trench. One way would be to erect circular concrete walls on both sides of the trench providing struts between the walls to overcome the slipping force of the soil. We assume just 1 meter wide trench around every dome. Now let us consider protection from forest fires.
Fire protection: The risk of forest fires are reduced by fuel breaks. From https://efotg.sc.egov.usda.gov/references/public/CO/CO383_Spec.pdf we find:
5. The dimensions of the fuel break (width and length) shall be sufficient to reduce fire spread and intensity. Width on level ground shall be a minimum of 150 feet for cropland, rangeland, and other non-forestland sites and a minimum of 300 feet on forest land sites. Add 10 feet to the width for every 10 percent increase in slope (e.g., a width of 360 feet would be used on a 60 percent slope). Length shall match the length of the ignition source to the extent feasible.
We cannot leave 300’ around every geodesic dome of 80’ diameter. We expect that more than one geodesic domes would be required to supply the carbon required. Place the geodesic domes with trenches as close as possible providing access roads between the domes so that the trees could be maintained and harvested easily. Surrounding all the domes together dig a trench of desired depth. Keep the width of the trench at least equal to two times the breadth to which the soil would incline naturally without holding walls. The cross section of the trench would be a trapezoid with the minimum side at the bottom being zero feet. Providing more width is better. Fill the trench with water and use it for aqua farming. This water is available for spraying on the sides of the geodesic domes exposed to the surrounding forest during a forest fire. Include the top width of the trench towards the width of the fuel break. Make the fuel break a closed curve surrounding the trench.
Comparison with a solar power plant: https://www.dailymail.co.uk/sciencetech/article-1393879/Gemasolar-Power-Plant-The-worlds-solar-power-station-generates-electricity-NIGHT.html?ITO=1490 describes the first solar power plant that generates power during darkness. Quoting from the article, we have:
The Gemasolar Power Plant near Seville in southern Spain consists of an incredible 2,650 panels spread across 185 hectares of rural land. … The regular sunshine in southern Spain means the facility can therefore operate through most nights, guaranteeing electrical production for a minimum of 270 days per year, up to three times more than other renewable energies. … It is expected to produce 110 GWh/year
The energy 110 GWh/year is equivalent to 301.37 MWh/day, which is equal to 12.56 MWh/hour or 12.56 MW.
We have found that one geodesic dome of 12 meter radius is enough to produce 1 KW. We provide a 1 meter wide trench to protect against earthquake. For access to the domes, we provide one way roads of 4 meter width. This 4 meter width contributes additional 2 meters to the radius of the circle for each dome. Thus the radius of the dome is 12 + 1 + 2 = 15 meters. To produce 12.56 MW we need 12,560 domes. The surface area of the circle for each dome is 706.85 sq m. The area for 12,560 domes is 8878141 sq m. At 10,000 sq m per ha, the area is 887.81 ha.
Let us assume that the 12,560 domes could be located inside a circle of area 8878141 sq m. The radius of this circle is 1681.06 m. We add 300.00 m for the fuel break, giving 1981.06 m as the radius of the circle to supply the required Carbon to produce same power as the Gemasolar power plant. The total area is 12329504.98 sq m or approximately 1,233 ha.
The Gemasolar power plant uses 185 ha. Thus purely on the question of area alone our Plan B needs 6.66 times the area of the Gemasolar power plant. The Gemasolar power plant avoids the emission of GreenHouse Gases. In contrast, Plan B extracts 75 Kg of Carbon per year per dome. A total of 12,560 domes remove 904,320 Kg of Carbon every year from the atmosphere. As 12 Kg of Carbon produces 44 Kg of CO2, the domes remove 3315,840 Kg or 3,316 tons of CO2 from the atmosphere every year.
Conclusion: In comparison to Plan A described in https://mohideenibramsha7.wixsite.com/website/single-post/2018/09/16/Hydrogen-Economy-Now-11 Plan B requires substantially more land and resources to generate Hydrogen. It is important to notice that even if Company-X does not appear in the near future, Plan B permits the evolution of the Hydrogen Economy from day one. Like the Gemasolar power plant Plan A does not remove any CO2 from the atmosphere, while Plan B does remove CO2 from the atmosphere.
Previous posts in this series:
https://mohideenibramsha7.wixsite.com/website/single-post/2018/07/25/Hydrogen-Economy-Now-1
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/02/Hydrogen-Economy-2
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/06/Hydrogen-Economy-Now-3
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/07/Hydrogen-Economy-Now-4
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/07/Hydrogen-Economy-Now-5
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/08/Hydrogen-Economy-Now-6
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/09/Hydrogen-Economy-Now-7
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/09/Hydrogen-Economy-Now-8
https://mohideenibramsha7.wixsite.com/website/single-post/2018/08/24/Hydrogen-Economy-Now-9
https://mohideenibramsha7.wixsite.com/website/single-post/2018/09/10/Hydrogen-Economy-Now-10
https://mohideenibramsha7.wixsite.com/website/single-post/2018/09/16/Hydrogen-Economy-Now-11