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