A latex template based on the "Exam" class for creating paper-based exams, with lots of nice features. See the README for details.
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1 year ago
  1. \question[10] These are some ``long form" questions. A PV Panel is found to have a maximum power point of 34.1V and 9.83A when tested at STC (1kW/m\textsuperscript{2}), and has a stated efficiency of 19.6\%.
  2. Estimate the active area of this panel (i.e. the area of semiconductor that light falls on) in \si{\meter\squared}.
  3. \begin{solutionorbox}[7.75in]
  4. At Standard test conditions, the incident radiant energy is 1kW/m\textsuperscript{2}.
  5. Assume that the area of the panel is $A$. The radiant power falling on this panel, i.e. incident power, is therefore:
  6. \begin{equation*}
  7. P_\text{incident}= 1\text{kW}/m^2 * A
  8. \end{equation*}
  9. The output power is just the incident power times the efficiency:
  10. \begin{equation*}
  11. P_\text{output}= 1\text{kW}/m^2 * A * \eta
  12. \end{equation*}
  13. Where efficiency $\eta=(19.6/100)$.
  14. The maximum output power can be determined from the maximum power point details:
  15. Pout
  16. \begin{equation*}
  17. P_\text{out}= V_{oc}*I_{sc}.
  18. \end{equation*}
  19. Thus, equating the two values of P\textsubscript{out}, we can calculate $A$:
  20. \begin{equation*}
  21. A= (V_{oc}*I_{sc})/(\eta * 1kW/m^2) = 1.71 m^2 = 18\text{\ square feet}.
  22. \end{equation*}
  23. \end{solutionorbox}
  24. \clearpage
  25. \question[10] This is a question that requres a graph / plot as the response. The graph can be generated in \TeX. A PV Panel has a maximum power point of 34.1V and 9.83A when tested at STC (standard testing conditions) and a fill factor of 83.8\%. The open circuit voltage is found to be 40\si{\volt}. Compute the short circuit current for this panel and then sketch the VI curve, and label the maximum power point.
  26. \fillwithgrid{8in}
  27. \clearpage
  28. \question This is an example of a complex, multi-part question with multiple types of sub-parts. A user wants to connect an inductive load (Z) with a rating of 10kW and a power factor of 0.5 to the utility supply, as shown in the figure below. The supply voltage is $v_g(t) = 170\sin({\omega t + 0^\circ})$ , with a frequency of 60Hz.
  29. \begin{figure}[h]
  30. \centering
  31. \begin{circuitikz}[scale=0.7]
  32. \draw
  33. (0,0) to[sinusoidal voltage source, , v_<=$v_g(t)$] (0,4)
  34. to[short,i=${i_g(t)}$] (5,4)
  35. %(5,0) to[I, color=blue, *-*, l=$i_c(t)$] (5,4)
  36. (5,4) -- (7,4)
  37. to[european resistor, l=$Z$] (7,0) -- (0,0);
  38. \end{circuitikz}
  39. %\caption*{Problem 1}
  40. \label{fig:prob1}
  41. \end{figure}
  42. \begin{parts}
  43. \part[5] Calculate values of P, Q and S (with the appropriate units):
  44. \begin{solutionorbox}[4in]
  45. Given that P is 10kW and $cos\theta=0.5$ therefore current lags voltage by $60^\circ$.
  46. Magnitude of apparent power is therefore P/$\cos60^\circ$=20kVA.
  47. Therefore $Q=S\sin\theta=17.32\text{kVAR}$.
  48. Since the load is inductive, Q has a positive value.
  49. \end{solutionorbox}
  50. \vspace{5mm}
  51. P = \fillin[10kW][2in]
  52. \vspace{2mm}
  53. Q = \fillin[17.32kVAr][2in]
  54. \vspace{2mm}
  55. S = \fillin[20kVA][2in]
  56. \part[5]Draw the power triangle for this load. You can change the spacing of the grid too:
  57. \vspace{5mm}
  58. \setlength{\gridsize}{\dimexpr.025\linewidth-41\gridlinewidth}
  59. \fillwithgrid{3in}
  60. \clearpage
  61. \part[5] Calculate the net impedance now.
  62. \begin{solutionorbox}[3.5in]
  63. The net impedance is the parallel combination of the capacitor's impedance and the existing load.
  64. We found that $Z=0.181 +0.313j$. The impedance of the newly added capacitor is:
  65. \[X_C = \cfrac{1}{j\omega C} = -0.4171j \Omega \]
  66. Therefore net impedance is:
  67. \[ X_C || Z = \cfrac{ZX_C}{Z+X_C} = 0.7222 - 0.0027j\Omega \approx 0.7222\Omega \]
  68. \end{solutionorbox}
  69. \part[5] What is the power factor seen by the grid after the capacitor is installed?
  70. \begin{solutionorbox}[2in]
  71. The power factor is now nearly unity.
  72. Phase angle is about 0.2 degrees which for all practical purposes is almost zero.
  73. \end{solutionorbox}
  74. \end{parts}