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89 lines
3.6 KiB
89 lines
3.6 KiB
\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\%.
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Estimate the active area of this panel (i.e. the area of semiconductor that light falls on) in \si{\meter\squared}.
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\begin{solutionorbox}[7.75in]
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At Standard test conditions, the incident radiant energy is 1kW/m\textsuperscript{2}.
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Assume that the area of the panel is $A$. The radiant power falling on this panel, i.e. incident power, is therefore:
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\begin{equation*}
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P_\text{incident}= 1\text{kW}/m^2 * A
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\end{equation*}
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The output power is just the incident power times the efficiency:
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\begin{equation*}
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P_\text{output}= 1\text{kW}/m^2 * A * \eta
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\end{equation*}
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Where efficiency $\eta=(19.6/100)$.
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The maximum output power can be determined from the maximum power point details:
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Pout
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\begin{equation*}
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P_\text{out}= V_{oc}*I_{sc}.
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\end{equation*}
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Thus, equating the two values of P\textsubscript{out}, we can calculate $A$:
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\begin{equation*}
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A= (V_{oc}*I_{sc})/(\eta * 1kW/m^2) = 1.71 m^2 = 18\text{\ square feet}.
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\end{equation*}
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\end{solutionorbox}
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\clearpage
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\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.
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\fillwithgrid{8in}
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\clearpage
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\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.
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\begin{figure}[h]
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\centering
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\begin{circuitikz}[scale=0.7]
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\draw
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(0,0) to[sinusoidal voltage source, , v_<=$v_g(t)$] (0,4)
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to[short,i=${i_g(t)}$] (5,4)
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%(5,0) to[I, color=blue, *-*, l=$i_c(t)$] (5,4)
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(5,4) -- (7,4)
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to[european resistor, l=$Z$] (7,0) -- (0,0);
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\end{circuitikz}
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%\caption*{Problem 1}
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\label{fig:prob1}
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\end{figure}
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\begin{parts}
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\part[5] Calculate values of P, Q and S (with the appropriate units):
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\begin{solutionorbox}[4in]
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Given that P is 10kW and $cos\theta=0.5$ therefore current lags voltage by $60^\circ$.
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Magnitude of apparent power is therefore P/$\cos60^\circ$=20kVA.
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Therefore $Q=S\sin\theta=17.32\text{kVAR}$.
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Since the load is inductive, Q has a positive value.
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\end{solutionorbox}
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\vspace{5mm}
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P = \fillin[10kW][2in]
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\vspace{2mm}
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Q = \fillin[17.32kVAr][2in]
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\vspace{2mm}
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S = \fillin[20kVA][2in]
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\part[5]Draw the power triangle for this load. You can change the spacing of the grid too:
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\vspace{5mm}
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\setlength{\gridsize}{\dimexpr.025\linewidth-41\gridlinewidth}
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\fillwithgrid{3in}
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\clearpage
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\part[5] Calculate the net impedance now.
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\begin{solutionorbox}[3.5in]
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The net impedance is the parallel combination of the capacitor's impedance and the existing load.
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We found that $Z=0.181 +0.313j$. The impedance of the newly added capacitor is:
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\[X_C = \cfrac{1}{j\omega C} = -0.4171j \Omega \]
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Therefore net impedance is:
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\[ X_C || Z = \cfrac{ZX_C}{Z+X_C} = 0.7222 - 0.0027j\Omega \approx 0.7222\Omega \]
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\end{solutionorbox}
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\part[5] What is the power factor seen by the grid after the capacitor is installed?
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\begin{solutionorbox}[2in]
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The power factor is now nearly unity.
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Phase angle is about 0.2 degrees which for all practical purposes is almost zero.
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\end{solutionorbox}
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\end{parts}
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