Key to Biomorph 4: Water, Air, and Energy
The Biomorphs keep trying:
"Our advanced Biomorph technology will require only 5 x 10^{13} kilowatts of power. We'll provide this on Earth with solar energy." On Earth, 0.3 kilowatts (kW) of solar energy is available per square meter. So 5 x 10^{13} kW divided by 0.3 kW = 1.7 x 10^{14 }square meters, or 1.7 x 10^{8 }square kilometers. This is about a third the surface area of Earthabout the amount of our dry land mass. So the Biomorphs will take up all of our dry land! 
Facts about WaterWater accounts for 60% of our total weight Water consumption Water use Useful conversions: 1 kg = 2.2 pounds 
Written assignment  turn in to professor
100 L/x = 30%/100% 100/x = 0.3 x = 100/0.3 x = 333 liters of camel body fluid Since the density of water is 1.0 kg/L: 333 L x (1000 ml/L) x (1g/ml) x (1kg/1000g) = 333 L = 333 kg 333 kg water/700 kg total body mass = 47.6% of the camel’s mass is water
2. Jamis’ body produced a little over 33 liters of water. How many pounds did Jamis weigh?
If 60% of the human body mass is water, then 60% of Jamis's body mass X = 33 L. So what is 100% of X? 60%/100% = 33 L/x 0.6 = 33/x x = 33/0.6 x = 55 L 55 L = 55 kg (see #1) 55 kg x (2.2 lbs/kg) = 121 lbs
3. They had 38 million dekaliters (380 million liters) of water stored in the sietch where Jamis was killed. If there were 1000 persons in the sietch, how many years would the hoard of water last if they did not use their stillsuits or reclaim water in any way?
38 x 10^{6} deciliters x (10 L/deciliter) = 3.8 x 10^{8} L of water Summing up water disposal figures = 0.5 + 0.9 + 1.5 + 0.1 = 3 L water needed per person each day You could make a case for needing only 1.5 L (drinking water) per day
1000 persons x (3.0 L/person x day) x (365 days/year) = 1095000 L used per year 3.8 x 10^{8} L / (1095000 L/day) = 347 years
4. The atmosphere of Dune is 23% oxygen, 75.4% nitrogen, 0.023% carbon dioxide plus trace gases. The atmosphere of Earth is 20.9% oxygen, 78.1% nitrogen, 0.03% carbon dioxide plus trace gases (including water vapor.) Discuss the difference in the atmospheres and possible reasons for these differences. Do these numbers seem reasonable for the atmosphere of Dune? Explain your reasoning.
The content of oxygen (O_{2}), nitrogen (N_{2}), and carbon dioxide (CO_{2}) gases are similar to those of Earth. (If they were not, humans could not breathe on Dune as they do on Earth.) But where would all the oxygen come from on Dune, if there are no plants or photosynthetic microbes to produce it? Also, where would all the water come from to make the dew for dew collectors, if there is no ocean?
5. The source of energy for life on Dune is not explained. One could easily imagine solar energy being used on this planet. On Earth, 0.3 kilowatts (kW) of solar energy is available per square meter. Presently, photovoltaic solar panels are able to convert 10% of the available energy to electricity. Assume a glowglobe uses as much electricity as a 25 W light bulb. How large an area of solar panels would be needed to light one glowglobe for each of the 1000 persons living in the sietch? a. Calculate with typical Earth values. b. Calculate using values you propose for Dune. Support/explain the numbers you use.
0.3 kW/m^{2} energy is typically available on Earth. 10% of that energy can be utilized by the solar panels. So, (0.3 kW/m^{2}) x 10% efficiency = (0.3 kW/m^{2}) x 0.10 = 0.030 kW/m^{2} usable 1000 persons x 25 W glowglobe = 25 x 10^{3} W = 25 kW 25 kW / (.030 kW/m^{2} ) = 833 m^{2} (equals an area 29 m x 29 m)
Calculation is repeated using your estimation of power available on Dune. Must include discussion of why the estimated value was chosen.
6. In a previous assignment, you calculated the energy of one photon of blue light. Repeat that calculation here and further calculate how many photons per second of that wavelength of light you would need to provide 100 W of energy.
Blue light – pick a wavelength in the blue range (I’ve chosen to do the high and low values)
450 nm: E = (6.62 x 10^{34} J sec) ( 3.00 x 10^{8} m/sec) / 450 x 10^{9} m E = 4.41 x 10^{19} J in a photon
490 nm: E = (6.62 x 10^{34} J sec) ( 3.00 x 10^{8} m/sec) / 490 x 10^{9} m E = 4.05 x 10^{19} J in a photon
At 450 nm 100 W x (J sec^{1}/W) = 100 J/sec (100 J/sec)/(4.41 x 10^{19} J/photon) = 2.26 x 10^{20} photons/sec
At 490 nm = 2.46 x 10^{20} photons/sec
7. Describe how dew collectors can be used to water plants.
A piece of plastic can be set out on the ground, in the form of a slanted depression, to collect water that condenses in the cool of the night (dew). The dew will collect in the center of the plastic. It must be harvested early in the morning, before the sun rises and the water evaporates.
8. Discuss the purpose of the stillsuit and how it functions. Explain why the stillsuits won’t work in real life.
The stillsuit distills water out of the wearer's breath, presumably by chilling it until the exhaled air reaches saturation and the vapor condenses. The stillsuit also filters water out of bodily fluids and secretions, including saliva, sweat, urine, and feces. Filtration requires power, as does chilling the breath.
