MIT OpenCourseWare http://ocw.mit.edu 1.020 Ecology II: Engineering for Sustainability Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Civil and Environmental Engineering 1.020 Ecology II: Engineering for Sustainability Practice Questions and Solutions for Exam 1 on March 17, 2008 Exam Review Friday March 14th, 2008 9-11am 1. Mass balance A deep reservoir with vertical sides has a surface area of Ar m2. Water (density ρw) enters the reservoir at a rate of qin. The depth of water at a given time is h. An outlet of cross sectional area Apipe m2 is situated at the bottom of the reservoir, and the flow out of the pipe can be expressed as 2ghApipeqout = a) Write a conservation equation for the mass of the water in the reservoir: (i) as a difference equation Mt+Δt − Mt =ρwqin −ρwqout =ρwqin Δt −ρw 2ght Apipe Δt (ii) as a differential equation dM =ρwqin −ρw 2ghApipe dt Leave your answer in terms of the water mass, ρw, qin, Ar and Apipep. b) Using the relations given above rewrite the differential conservation equation to give the change of water depth with time. M =ρ A hw r dM dh =ρw Ar =ρwqin −ρw 2ghApipe since ρw and Ar are constant. dt dt Apipedh = qin − 2ghdt A Ar r2. Chemical kinetics In the absence of oxygen, certain bacteria are able to respire anaerobically, using acetic acid as a substrate. In addition to CO2, the bacteria give off methane (CH4) in the process, and so the respiration process is also known as methanogenesis. In a closed volume V Liters (e.g. an anoxic region fairly deep under a lake’s bottom) the following reaction occurs: CH3COOH → CH4 + CO2 Acetic acid methane The rate (per second) at which the number of moles of acetic acid is used is = k1VCCH3COOH, with CCH3COOH in mol/L. The molecular masses of the three species are: CH3COOH 60 g/mol, CH4 16 g/mol, CO2 44 g/mol. The system starts off with a concentration of 1 mole/L of CH3COOH. a) Using the mass balance (stoichiometry of the reaction) write differential equations for the rate change in mass concentration (g/L) of acetic acid, methane and CO2. Leave your answer in terms of k1. b) Draw a rough plot of the changes in masses of the three species with time. VdCCH3COOH = −k1VCCH3COOHdt dCCH COOH ⎡mol ⎤Acetic acid: dt 3 ⎢⎣ Ls ⎥⎦ = −k1CCH3COOH dCCH COOH ⎡ g ⎤ g3 dt ⎢⎣ Ls ⎥⎦ = −k1CCH3COOH 60 mol dCCH4 ⎡ g ⎤ g dCCO2 ⎡ g ⎤ gMethane: dt ⎢⎣ Ls ⎥⎦ = k1CCH COOH 16 mol CO2: dt ⎢⎣ Ls ⎥⎦ = k1CCH COOH 44 mol3 3 Plot of concentration evolution with time: t(s) c (g/L) 60 44 16 CH4 CO2 CH3COOH03. Closed system, nutrient cycle: benthic layer nitrogen The nutrient modeling example given in the Lecture08_5 outline turns out to far underestimate the measured levels of nitrogen on the bottom of the ecosystem (benthic layer). It is found that nitrogen fixing bacteria are present on particles in suspension. They uptake the dissolved nitrogen in a first order process (rate of uptake directly proportional to dissolved nitrogen mass), and then they settle to the bottom. In addition, a small fraction of the dead phytoplankton and zooplankton (denoted by f), instead of becoming dissolved nitrogen, directly enters the sediment. The original model is shown below. Subscripts 1=dissolved N mass, 2 = N mass in phytoplankton, 3 = N mass in zooplankton. If we introduce a fourth compartment 4 = N mass in sediment, write down the rate equation for the sediment nitrogen mass. Also, update the above three equations of this closed system model. Solution: Let the first order removal of dissolved N be equal sM1. Since we have a closed system, the sink terms in one compartment must enter as source terms in the other compartments. The four new equations, with the added terms in bold, are:Link 1 Link 2 xin = 66 vehicles hr-1 xout = 66 vehicles hr-1 Node 1 Node 2 x1, t1 x2, t2 Link 3 Link 4 x3, t3 x4, t4Node 3 Node 4 1 1 2dM = m& + (1− f)(d M +d M )− g ⎜⎜⎛ M ⎟⎟⎞M + (1−ε )g ⎜⎜⎛ M ⎟⎟⎞M − sM dt boundary,1 2 2 3 3 2 M + M 2 3 M + M 3 1 ⎝ 1h 1 ⎠ ⎝ 2h 2 ⎠ dM 2 ⎛ M1 ⎞ ⎛ M 2 ⎞ = g2 ⎜⎜ ⎟⎟M 2 −d 2M 2 − g3 ⎜⎜ ⎟⎟M 3 dt ⎝ M1h + M1 ⎠ ⎝ M 2h + M 2 ⎠ dM 3 ⎛ M 2 ⎞ =εg3 ⎜⎜ ⎟⎟M 3 −d 3M 3 dt ⎝ M 2h + M 2 ⎠ dM4 = f(d2M2 +d3M3 ) + sM1dt 4. Transportation The traffic network shown below consists of four links (1 to 4) and four nodes (1 to 4) You are given the following information: i) The travel time/flow equations for each link must be satisfied: Link 1: t1 = 15 ; t1 = Link 1 travel time (hrs) , x1 = Link 1 traffic flow (vehicles hr-1) Link 2: t2 = 1 + 0.2x2 ; t2 = Link 2 travel time (hrs) , x2 = Link 2 traffic flow (vehicles hr-1) Link 2: t3 = ? ; t3 = Link 3 travel time (hrs) , x3 = Link 3 traffic flow (vehicles hr-1) Link 2: t4 = 2 + 0.1x4 ; t4 = Link 4 travel time (hrs) , x4 = Link 4 traffic flow (vehicles hr-1) The equilibrium time taken for both routes = 24.4 hours. a) Find t3, the travel time for Link 3. From the equilibrium condition, 1. t1+t2 = t3+t4 Æ 15 + (1+0.2x2) = t3+(2+0.1x4) = 24.4 hours This gives 16+0.2x2 = 24.4 Æ x2 = 42 vehicles hr-1 = x1. For nodes 2. Node 3: x3=x4;3. Node 2: x1=x2 4. Node 4: x2+x4 = 66 Æ x4 = 66-42 = 24 = x3 Therefore from (1), t3 = 24.4 – (2+0.1*24) = 20 hours b) If the link lengths are equal and all vehicles are identical, which link (1-2 or 3-4) do you think has the greater CO emissions rate (in gm CO km-1 hr-1)? Since x2>x4, everything else being equal there is more traffic volume in the top route and therefore has greater CO emissions. 5. Heat Transfer A 1 kg aluminum pan (specific heat cp = 900 J kg-1K-1), of surface area 0.01 m2, is being heated on a butane burner in a room of fixed temperature Ta=20 degrees C. a) Write a differential equation describing the time change in temperature of the aluminum pan. Assume that the pan’s temperature is uniform. We are taking the pan as the control volume, and assume that the temperature is uniform throughout. This means that we can neglect any internal conduction effects of the pan. Energy balance: dU = Q& radin − Q& radout − Q& conv + Q& burnerdt dU dT dT dTThe change of internal energy can be expressed as = C = C = mcdt vdt pdt pdt Cv and Cp …