Add assignment 2 and 3

elements_of_mech/assignment-work-and-heat/assign2.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- First Law of Thermodynamics}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{}
+\maketitle
+\section*{Question 1}
+First law of thermodynamics suggests that $\sum Q = \sum W$.
+\begin{align*}
+	Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\
+	75 - 40 + Q_3 &= -15 + 44 \\
+	Q_3 &= -6 kJ
+\end{align*}
+i.e.\ from the system to the surroundings.
+\section*{Question 2}
+We know that,
+\begin{align*}
+	Q &= \Delta E + W \\
+	Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\
+	Q &= -8600 kJ \\
+	Q &= -8.6 MJ
+\end{align*}
+\section*{Question 3}
+There is no heat transfer, therefore
+\begin{align*}
+	Q &=  \Delta E + W \\
+	W &= - \Delta E - \Delta V \\
+	W &= \int_{1}^{0} C_v \cdot dT \\
+	W &= -0.718 (T_2 - T_1) \\
+	W &= -50.26 \frac{kJ}{kg} \\
+	\text{Total Work} &= 2 \times (-50.26) = -100 kJ
+\end{align*}
+\section*{Question 4}
+\begin{align*}
+	1000 &= a + b \times 0.2 \\
+	200 &= a + b \times 1.2 \\
+	b &= -800 \\
+	a &= 1000 + 2 \times 800 = 1160 \\
+	\therefore P &= 1160 - 800 V \\
+	W &= \int_{V_1}^{V_2} P \cdot dV \\
+	  &= \int_{0.2}^{1.2} (1160 - 800V)dV \\
+	  &= 6000 kJ \\
+	V_1 &= 1.5 \times  1000 \times \frac{0.2}{1.5} - 85 \\
+		&= 215 \frac{kJ}{kg} \\
+	V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\
+		&= 155 \frac{kJ}{kg} \\
+	\Delta V &= V_2 - V_1 \\
+			 &= 40 \frac{kJ}{kg} \\
+	\Delta U &= m \Delta V  = 60 kJ \\
+	Q &= \Delta U + W \\
+	  &= 60 + 600 \\
+	  &= 660 kJ \\
+	U &= 1.5 PV - 85 \frac{kJ}{kg} \\
+	U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\
+	  &= 800 V^2 - 85 \frac{kJ}{kg} \\
+	\frac{\delta U}{\delta V} &= 1160 - 1600 V \\
+	\text{For maximum V,} \\
+	V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\
+	V &= 0.725 \\
+	u_{\max} = 335.5 \frac{kJ}{kg} \\
+\end{align*}
+\begin{align*}
+	U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\
+\end{align*}
+\section*{Question 5}
+\subsection*{Part a}
+\begin{align*}
+	Q &= \int_{273}^{373} C_p \cdot dT \\
+	t &= T - 273 K \\
+	\therefore t + 100 &= T - 173 \\
+	Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\
+	Q &= 238.32 J
+\end{align*}
+\subsection*{Part b}
+\begin{align*}
+	Q &= \Delta E + \int p\cdot dV \\
+	\Delta E &= Q - p(V_2 - V_1) \\
+	\Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\
+	\Delta E &= 197.79J
+\end{align*}
+\end{document}

elements_of_mech/assignment-work-and-heat/assign3.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\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}