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\begin{document}
\title{Assignment --- Second Law of Thermodynamics}
\author{Ahmad Saalim Lone, 2019BCSE017}
\date{}
\maketitle
\section*{Question 1}
We know that the net power output is the difference between the heat input and the heat rejected (cyclic device).
\begin{align*}
	W_{net,out} &= Q_H + Q_L \\
	W_{out} &= 90 - 55 = 35 MW
\end{align*}
The thermal efficiency is the ratio of net work output and the heat output.
\[
	\eta_{thermal} = W_{out}\frac{W_{out}}{Q_H} =  \frac{35}{90} = 0.3889
\]
\section*{Question 2}
\begin{align*}
	\text{efficiency} = 1 - \frac{T_1}{T_2}  &= \frac{\text{work input}}{\text{work output}} \;\;\text{(must)}\\
	1 - \frac{300}{1000} &\implies \frac{6}{1} \;\;\text{(claimed)} \\
	0.7 &> 0.6 \\
	\text{output} &> \text{claimed $\implies$ good}
\end{align*}
$\therefore$ we can agree to this claim.
\section*{Question 3}
\[
	\eta = \frac{\text{output}}{\text{input}} = \frac{0.65}{0.65 + 0.4} = 0.619 = 61.9%
\]
now,
\begin{align*}
	\eta &= 1 - \frac{T_2}{T_1} \\
	T_1 &= \frac{T_2}{1 - \eta} \\
	T_1 &= 787.5 K
\end{align*}
$\therefore$, the temperature at which energy is absorbed is 787.5 K
\section*{Question 4}
\begin{align*}
	\eta &= 1 - \frac{T_1}{T_2} \\
\end{align*}
where,
\begin{align*}
	\eta &= 0.6 \\
	T_2 &= 800 K \\
	T_1 &= T_{sink} \\
	T_{sink} &= (1 - \eta)T_2 \\
			  &= 324K
\end{align*}
Also,
\begin{align*}
	\eta &=  \frac{Q_1 - Q_{\text{rejected}}}{Q_1} \\
	Q_{\text{rejected}} &= Q_1 (1 - \eta) \\
							 &= 400(1 - 0.6) \\
							 &= 160 kJ
\end{align*}
\section*{Question 5}
Max COP will be achieved only in a Carnot refrigerator.
\[
	{(COP_{\max})}_{R} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2} = \frac{-20 + 273}{57} = 4.438
\]
Minimum Power $\to$ Max COP\@. We know that $P = \frac{Q}{\eta}$.
\[
	P = \frac{3 \times 57}{253} = 0.657 W
\]
\section*{Question 6}
Max COP will be achieved only in a Carnot refrigerator.
\begin{align*}
	{(COP)}_{\text{Carnot refrigerator}} &= \frac{T_{low}}{T_{high} - T_{low}} = 1.37 \times 10^{-2} \\
	{(COP)}_{\text{refrigerator}} &= \frac{Q_L}{W} = 1.37 \times 10^{-2} \\
	Q_L &= 83.3 \\
\end{align*}
\section*{Question 7}
For process $1-2$
\begin{align*}
	Q_{1-2} &= U_2 - U_1 + W_{1-2} \\
	-732 &= U_2 - U_1 - 2.8 \times 3600 \\
	U_2 - U_1 &= 9348kJ
\end{align*}
For process $2-1$
\begin{align*}
	Q_{2-1} &= U_1 - U_2 + W_{2-1} \\
	-732 &= -9348 - 2.8 \times 3600 \\
	Q_{2-1} &= -708 kJ
\end{align*}
Now,
\[
	Q_{2-1} = U_1 - U_2 + W_{2-1}
\]
For maximum work, $Q_{2-1} = 0$.
\[
	\therefore {(W_{2-1})}_{\max} = U_2 - U_1 = 9348kJ
\]
\section*{Question 8}
\begin{align*}
	{(COP)}_R &= \frac{268}{298 - 268} = 8.933 \\
	{(COP)}_R &= \frac{5}{W} \\
	W &= 0.56KW
\end{align*}
\section*{Question 9}
\begin{align*}
	\eta &= 1 - \frac{273+60}{273+671} = 0.64725 \\
	\eta &= 0.3236 \\
	\implies 1 - 0.3236 &= 0.6764 \\
	\text{Ideal COP} &= \frac{305.2}{305.2 - 266.4} = 7.866 \\
	\text{Actual COP} &= 3.923 = \frac{Q_3}{W} \\
	if \;Q_3 &= 1kJ \\
	\therefore W &= \frac{Q_3}{3.923} = 0.2549kJ \\
	W &= Q_{\text{output}} - Q_{\text{input}} = \frac{1}{3.923} Q_1 - 0.6764Q_1 = 0.2549 \\
	Q_{\text{out}} &= \frac{0.2549}{1 - 0.6764} = 0.7877 kJ \\
\end{align*}
\section*{Question 10}
\begin{align*}
	\frac{10}{W} &= \frac{293}{293-273} \\
	or\;W &= 683 W
\end{align*}
\end{document}