Solutions_1
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Solutions_1

Course Number: ENGR 120, Spring 2011

College/University: Washington State

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CE 341 Homework Assignment #1 Due Friday, Sept 3, 2009 [10] 1. Mars radiates energy with a peak wavelength of 13.2 um a) Treating Mars as a blackbody, what would its temperature be? b) What would be the frequency and energy content of a photon at that wavelength? [10] 2. What is the O3 concentration (ug/m3) in the stratosphere at 30 km if its mixing ratio is 10 ppm? (use the US Standard Atmosphere to get P, T...

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341 CE Homework Assignment #1 Due Friday, Sept 3, 2009 [10] 1. Mars radiates energy with a peak wavelength of 13.2 um a) Treating Mars as a blackbody, what would its temperature be? b) What would be the frequency and energy content of a photon at that wavelength? [10] 2. What is the O3 concentration (ug/m3) in the stratosphere at 30 km if its mixing ratio is 10 ppm? (use the US Standard Atmosphere to get P, T values, http://www.digitaldutch.com/atmoscalc/index.htm) [30] 3. Simple Greenhouse Model. Derive the dependence of the Earth's surface temperature (To) on the fraction f of terrestrial radiation absorbed by the atmosphere. Assume the atmosphere is an isothermal layer some distance above the earths surface. It has temperature T1 and is transparent to solar radiation. Below is a sketch to get you started. Hint what is the energy balance of the atmospheric layer? [30] 4. What is the equilibrium temperature of a cube orbiting the sun at the same orbital distance as the planet Mercury (46 x 106 km)? Assume only one face of the cube is illuminated by the sun. (Hint you will first need to calculate the Solar constant for this orbital distance). [20] 5. To offset global warming it has been proposed that giant solar reflectors should be put into orbit around the earth to reflect sunlight. Placing reflectors in space will reduce the solar constant. What percentage reduction in the Earths solar constant S is necessary to reduce the effective temperate (Te) by 2 C from its current 288 K average temperature. CE 341 Homework Assignment 1 Solutions [10] 1. Mars. a) use Wiens law : max (um) = 2898 / T 13.2 um = 2898 / T T = 219 K = - 53 C b) energy of a photon E = hv E = h c/ E = 6.626 x 10-34 J s-1 * ( 2.998 x 108 m/s / 13.2 x 10-6 m) E = 1.50 x 10-20 J [10] 2. Ozone concentration in stratosphere. O3 mixing ratio = 10 ppmV Pressure @ 30 Km = 0.011565 atm Temperature @ 30 km = 226.65 K ug/m3 = fractional abundance * molar concentration of air * molecular weight * unit conversion factors fractional abundance : 10 ppmv = 10 x 10-6 molar concentration air : P / RT molecular weight : 48 g/mol (see http://webbook.nist.gov/chemistry/ for data) ug/m3 = 10 x 10-6 * (0.01156 atm/ 0.08206 atm L mol-1 K-1 * 226.7 K) * 48 g/mol * 103 L / m3 * 106 ug / g ug/m3 = 305 [30] 3. Greenhouse Model. The energy balance of the Earth + atmosphere is system achieved by balancing the energy fluxes in and out of the system. Energy absorbed from sun = energy radiated to space by Earth and atmosphere S 1 1 f T04 fT14 (1) 4 The atmospheric layer is not a black body (since it absorbs only a fraction of the energy emitted by the Earth). Because it is not a black body the amount of energy it radiates is given by a modified version of the StefanBoltzmann law: Erad = T4 where (0 < < 1) is the emissivity of the object. The emissivity is simply the ratio of the energy radiated by an object to that emitted by a black body of the same temperature. The emissivity of the atmospheric layer is given by the fraction f. We know this from Kirchhoff's law that states "at thermal equilibrium the emissivity of an object is equal to its absorptivity". Since the object absorbs a fraction f of the energy radiated by the Earth, it follows that = f. The energy balance for the atmospheric layer: energy absorbed = energy emitted (2) fT04 2 fT14 The atmospheric layer emits from both sides, hence factor of 2. This yields a relationship for T1 that we can substitute into equation (1). T14 To4 2 (3) Substituting, S 1 1 f T04 fT14 4 fTo4 S 1 f 4 1 f T0 1 T04 4 2 2 4 S 1 T0 4 1 f 2 1 [30] 4. Equilibrium temperature of a cube. First determine solar constant for Mercury's orbital distance. This is a simple geometry problem (think illuminated orbital spheres) Smercury = Searth * (R2earth / R2mercury) Smercury = 1370 W/m2 * ((150 x 106 km)2 / (46 x 106 km)2) Smercury =14,567 W/m2 Treating the cube with dimension x as an isothermal black body with albedo = 0: Energy absorbed = energy radiated Smercury * illuminated surface area = T4 * (total surface area) Smercury x2 = T4 6x2 1 4 14,567 Wm -2 S T0 8 2 4 6 6 * 5.67x 10 Wm K 1 4 To = 454 K [20] 5. Solar reflectors. What change in S leads to a 2 C temperature change? For Earth : S = 1370 W/m2 = 0.31 T0 = 288 K (current average surface temperature) T1 = 286 K 4T 4 S 1 4 T 1 S 4 S 4 4 4 4 2864 2884 T1 T0 1 1 0.31 S = -62 W/m2 So about a 4.5% reduction in the solar constant would be estimated to yield a 2 C temperature decrease. This implies reflectors that would cover 4.5% of the cross sectional area of the Earth, an imposing figure.

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