GE 111 2013Midterm Answers Eam Version 1 2 (μg/m 3 ) 3 4a ω rot (rad/s) 4a ω rev (rad/s) 4b V rot (m/s) 4c V rev (m/s) 1-14 109.99 Y=200X 1.5 7.2710-5 1.9910-7 231.59 2.9710 4 2-11 218.93 231.84 3-10 146.07 231.08 Eam Version 5a (km) 5b (km) 5c (km) 6b (%) 6c tanks 6c (%) 1 365.2 411.7 83.9 39.9 11 55.4 2 383.1 436.6 79.1 42.5 3 60.4 3 347.7 386.8 90.2 36.9 7 60.4
Question 1 When log a and log b are two solutions of 2-6 3 = 0, find log a b + log b a. Don t use your calculator. If your work reflects the use of a calculator (i.e. a quadratic formula), you will receive no marks. 2 6-3 = ( 1 )( 2 ) = 0 1 = log a, 2 = log b 1 + 2 = 6, 1 2 = -3
Q2: Formaldehyde A fiberboard factory uses activated charcoal screens to keep formaldehyde in the air to less than 300 μg/m 3. The relationship between age of the screen (t, days it has been in use) and the concentration removed from the airflow (C r, μg/m 3 ) is an eponential relationship: C r = C i e kt, where C i = concentration of formaldehyde in the inflow airstream (μg/m 3 ). The concentration of formaldehyde in the outflow airstream is thus, C o = C i -C r. From a newly installed screen the outflow concentration was measured at 164.8/91/130 μg/m 3 for a time of 2.00 days, given a C i of 500 μg/m 3. Given a new screen how many hours (to nearest 0.01 hr) will it take for the outflow airstream (Co) to reach a formaldehyde concentration of 300 μg/m 3? Solution: First find k using known values for Cr, Ci, and t k = -0.1999/-0.1004/-0.1506 Then solve for t given a Co of 300 μg/m 3 and a Ci of 500 μg/m 3 t = ln(cr/ci)/k = 4.58/9.12/6.09 days = 109.99/218.93/146.07 hrs
Q3: Graphing Imagine you have been hired as a summer student. The professor who is supervising you is conducting research on the acoustic attenuation of comple media. She gives you a set of data and asks you to find the empirical equation that fits the data. She suggests you use the simple variables, y and, where y is the dependent variable. Plot the data from Table 1 on the provided graph. Figure 1. Acoustic attenuation of comple media Use the rules taught in GE 111 to construct 10,000,000 the figure that you would present to your Pick three y points from 1040 the line: 3 professor. Draw a line of best fit and use the (3,1000), 23500 24 method of selected points to find the 1000,000 (400,2000000), and 200200 (100, 200,000) 100 equation. Clearly show the resultant 565600 200 equation, along with any required calculations. Solution: Because we used a log-log plot on the previous slide and got a line, we know this data is represented by a power function: As data plots linear on a log-log graph it is represented by a power function: y = b m Answer => y = 200 1.5 y 100,000 10,000 1000 0.1 1 10 100 1000
Q4: Rotation & Revolution The Earth's rotation is the spinning of the earth on its ais. This therefore eplains the change from daylight to night-time. A revolution is the Earth's journey around the Sun, as it starts from one point and returns to it. This corresponds with a year, and also eplains the seasons. Assume the Earth is a complete sphere with a radius of 6371/6378/6357 km and the distance between the Sun and the Earth is 150 million km. You don t have to know if this distance is core-to-core or surface-to-surface because the size of the Earth is ignorable to this distance. a) Find the angular rotation velocity, ω rot and the angular revolution velocity, ω rev in rad/s. b) Find the linear rotation velocity, v rot at Saskatoon in m/s. Angle between the rotation ais and direct line from the core of the Earth and the city of Saskatoon makes 30. In other words, θ in the picture is 60. v rot = r ω rot = Rcosθ ω rot = 231.59 / 231.84 / 231.08 m/s c) Find the linear revolution velocity v rev in m/s. v rev = R ES ω rev = 2.97 10 4 m/s
30 km/h Q5: Maple Creek Airplane A small airplane takes off from Maple Creek at night. The pilot sets the rotational velocity of the single propeller so that the speed of the airplane is 138/148/128 km/h. To maintain a correct course, the pilot is depending upon the cockpit instrumentation for correct information. The cockpit instrumentation tells the pilot that the weather is cloudy and there is a small constant breeze from the West to the East at 15/10/20 km/h; however, all of the cockpit instrumentation, ecept for the compass, is faulty. The pilot sets a heading of 30 (east of North) and intends on flying for 2.5 hours before landing at Saskatoon. N a) Although the cockpit instrumentation is faulty, assume that the pilot s 15 km/h calculations are correct. What is the distance that the airplane should fly, 120 Saskatoon between Maple Creek and Saskatoon? (give answer to nearest 0.1 km) S 2 = 138 2 + 15 2 2 138 15 cos 120, S = speed 30 Distance = Speed time = 365.2 km / 383.1 km / 347.7 km Maple Creek b) The real weather states that there is actually a breeze from South to North at 30 km/h. How far from Maple Creek will the airplane be after 2.5 hours? V 2 = 138 2 + 30 2 2 138 30 cos 150 Distance = Speed time = 411.7 km / 436.6 km / 386.8 km c) For the case where the breeze is actually blowing from South to North at 30 km/h. How far from Saskatoon will the airplane be after 2.5 hours? From a) Use Sin Law for Saskatoon angle relative to wind from Maple Creek: b = 5.101 / 3.239 / 7.154 then add it to heading from Maple Creek = 30 + 5.101 = 35.101 / 33.239 / 37.154. From b) Use Sine Law to find angle of actual Course taken: c = 5.227 / 4.928 / 5.564. Therefore, angle of Course from Maple Creek = 30-5.227 = 24.773. Angle between bearing and course: 35.101 24.773 = 10.328 / 8.167 / 12.718 D 2 = 411.663 2 + 365.196 2 2 411.663 365.196 cos 10.328 = 83.850 km ANSWER c) = 83.9 km / 79.1 km / 90.2 km from Saskatoon V Maple Creek D D Saskatoon Saskatoon
Q6 Oil Refinery An oil refinery receives two different streams of crude oil for processing, 20/15/35 rail tank cars a day of Dakota Bakken and 35/40/40 rail tank cars a day of Great West. Each rail tank car holds 130 m 3. The refinery receives both streams at the same time, mies them, and then fractionates the mied stream into three outflow streams; gas, light oil, and heavy oil. Dakota Bakken (with a density of 0.80 tonnes/m 3 ) fractionates into 15% gas, 65% light oil, and the remainder as heavy oil. Great West (0.90 tonnes/m 3 ) fractionates into 10% gas, 40% light oil, and the remainder as heavy oil. All percentages are on a mass basis. b) For a one day period, what is the mass percentage (to 0.1%) of the heavy oil outflow stream of the three outflow streams Heavy Oil = 2463.5/2652/3068(t/d) Total Crude = 6175/6240/8320(t/d) Percentage Heavy Oil = 2463.5/6175 = 39.9/42.5/36.9% a) Draw and label a material flow diagram using known values. c) The refinery must produce the outflow streams such that light oil forms at least 55 60 60% of their three outflow streams and 20 15 35 tank cars/d of Dakota Bakken are processed. How many whole tank cars of Great West are used per day and what is the final percentage light oil composition (to nearest 0.1%) of the three outflow streams? Light oil amount = 0.65 DB + 0.40 GW = 0.55 (DB + GW), where DB is 2080 t/d GW (t/d) = DB (0.65 0.55)/(0.55 0.40) = 1386.67 t/d No of tank cars = 1386.67/117 = 11.85 tank cars/d it must be whole tank cars and 55%, therefore it must be 11/3/7 tank cars/d Check: (0.65*2080 + 0.40*1287)/(2080 + 1287) = 55.4% / 60.4% / 60.4% DB (20 tanks/d 130 m 3 /tank 0.80 t/m 3 ) G=0.15 DB L=0.65 DB H=0.20 DB GW (35 tanks/d 130 m 3 /tank 0.90 t/m 3 ) G=0.10 GW L=0.40 GW H=0.50 GW Mi and fractionates Gas (t/d) Light Oil (t/d) Heavy Oil (t/d)