midterm02solutions-3072-f15 - ECE 3072 2nd Midterm Exam...

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Unformatted text preview: ECE 3072 2nd Midterm Exam November 17, 2015 Closed Book (One sheets of notes allowed) Note: Use opposite side of page if necessaly. Name 7337; SCORE 1. (20) 2. (20) 3. (20) Total .(50) Lr~7 , fl 4‘ A V W A: \lJ fl lgx: 15) ‘PéRAtVR - $13 K {AmSW‘J (D (Nab 1 . iVoh . A PV module IS made up of 48 identical cells, all wired in series. With full sun insolation of 1 kW/m2 (“full sun” means 1 kW/m2 of solar power), each cell has short- circuit current per cm2, [so = 25 mA/cm, and its reverse saturation current, lo, per cm is 2 x 10'12 A/cmz. The parallel resistance, Flp = 5.09 and the series resistance, Fls = O Q. The area of each cell is 180 cm2. Assume that the temperature is 25°C. a) Calculate the short circuit current and open—circuit voltage at 75% sun insolation. q 4&3] ,qfi Li, A V ,. ,;,, PW :7 0\ {3‘2 3 \j 9 ' .\. ~ PU ‘ » “3'2" \ 13C : Islwtlmfi ME '~ $375 A, (U J - mitotic N 7, «a , M» «e: H . L5? “‘9UCC 5 V1 “ Li“, ( """"" AV {)13 (M, ZWOL J “3 @b) At full sun, find the voltage, current and power delivered by the module when the diode voltage in the equivalent circuit for each cell is 0.57 V. 14: IO ( {It/xii? A} :;_. ZNO\ (as \ [CONN “k 3 ’3 llllA s) 1 It 1 e {‘1‘ A V“; M> 1: I“: fit“ it“ {Q : lswitjm“<143,g ,— lewi sowi- l \J} (9 In full sun, if one cell is shaded, find the following if somehow the same current found in b) is forced to flow through the module: @c) the module output voltage. 11'): 0 ® - i All 7' — ? WA l I "‘>1:mlc,\/>°"1Y§ _ w ’ r \llé Utensil, :,l© ( 6W: #6 ~— MPW ‘ V \ “1 5153“ ~ m : g "‘ ‘ \ ( / fl awn “\r "“"‘ WW '5 mmj U45 4er \1 \i HQ Vw-AQ +\/ O 15’— ~ EX 0\ \ Q 0 @ \lxglgwfi 5’ Q \1: host @ \ : mm / le. \119‘1 VA in other ceLL we ShW om \/ ‘: evil“ single add some WVYefltd) the power dissipated in the shaded cell. (pkg f Liana? Ptl :gkX/ijblme :“ \‘Ll‘m‘ “LAM?” // 40‘ mi 2. Consider a wind turbine with a blade length of 20 m at an elevation of 200 m. Suppose the wind speed is 20 m/sec. Furthermore, at the fixed angle of attack, the lift force on the blades is 15 kN. a) At a temperature of 15°C, find the wind power. —920 353 329.3(1s+273) = 1_0991kg/m3 15+273 Solution: 6 = 1 kW p = ~6v3 — .1.O991.(20)3 = 4.396;; " E Pwmd = Ap = nr2.p = 3.14 . (20)2 .4396 = 5.5 MW 1) Total Points for this part of the question is 10. 2) If only formula with no values substituted: 4 points. 3) if only 6 and p is calculated and no Pwind formula: 6 points. 4) Mathematical calculation mistake: 9 points. b) Compute the power coefficient. Solution: d = gravity point = 10 m (given during exam) No. of blades = 3 (given during exam) wmade — speed of the blade T= 3FLd= 3.15x103.10=450kNm PBlade = T-wBlade = 450 X 103 -wBiade P 450 x103 .w CP = Blade = Blade = 0.081w31ade Pwl'nd 5.5 X 106 1) Total Points for this part of the question is 10‘ 2) If only formula with no values substituted: 4 points. 3) If onlyPBlade is calculated and no 6,: formula: 5 points. 4) Mathematical calculation mistake: 9 points. 3. a) Once the blade power in a wind turbine hits the maximum allowable value, the power must be reduced to zero. How is this done? What is this process called? By moving to angle of attack to 90° or a negative value and this causes there to be only drag force. (3) This is called “feathering.” (3) b) What is the capacity factor and why is it so important in solar and wind energy? The capacity factor is the ratio of the installed power to the generated power. (3) Both solar and wind have poor capacity factor (1) since the sun does not always shine, and the wind does not always blow. (2) 0) Draw a sketch of an airfoil, and in doing so define the cord and camber lines. Be very clear and VERY specific. (6) Sketch (3) Three labeling (3) Center of gravity Mean camber lint-z Cord tine d) What is the part to which the blades are attached called in a wind turbine? (2) The hub. ...
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