Add assignment 2 and 3
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elements_of_mech/assignment-work-and-heat/assign2.tex
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elements_of_mech/assignment-work-and-heat/assign2.tex
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\begin{document}
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\title{Assignment --- First Law of Thermodynamics}
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\author{Ahmad Saalim Lone, 2019BCSE017}
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\date{}
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\maketitle
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\section*{Question 1}
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First law of thermodynamics suggests that $\sum Q = \sum W$.
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\begin{align*}
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Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\
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75 - 40 + Q_3 &= -15 + 44 \\
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Q_3 &= -6 kJ
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\end{align*}
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i.e.\ from the system to the surroundings.
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\section*{Question 2}
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We know that,
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\begin{align*}
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Q &= \Delta E + W \\
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Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\
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Q &= -8600 kJ \\
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Q &= -8.6 MJ
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\end{align*}
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\section*{Question 3}
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There is no heat transfer, therefore
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\begin{align*}
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Q &= \Delta E + W \\
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W &= - \Delta E - \Delta V \\
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W &= \int_{1}^{0} C_v \cdot dT \\
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W &= -0.718 (T_2 - T_1) \\
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W &= -50.26 \frac{kJ}{kg} \\
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\text{Total Work} &= 2 \times (-50.26) = -100 kJ
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\end{align*}
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\section*{Question 4}
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\begin{align*}
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1000 &= a + b \times 0.2 \\
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200 &= a + b \times 1.2 \\
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b &= -800 \\
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a &= 1000 + 2 \times 800 = 1160 \\
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\therefore P &= 1160 - 800 V \\
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W &= \int_{V_1}^{V_2} P \cdot dV \\
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&= \int_{0.2}^{1.2} (1160 - 800V)dV \\
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&= 6000 kJ \\
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V_1 &= 1.5 \times 1000 \times \frac{0.2}{1.5} - 85 \\
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&= 215 \frac{kJ}{kg} \\
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V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\
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&= 155 \frac{kJ}{kg} \\
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\Delta V &= V_2 - V_1 \\
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&= 40 \frac{kJ}{kg} \\
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\Delta U &= m \Delta V = 60 kJ \\
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Q &= \Delta U + W \\
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&= 60 + 600 \\
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&= 660 kJ \\
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U &= 1.5 PV - 85 \frac{kJ}{kg} \\
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U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\
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&= 800 V^2 - 85 \frac{kJ}{kg} \\
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\frac{\delta U}{\delta V} &= 1160 - 1600 V \\
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\text{For maximum V,} \\
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V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\
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V &= 0.725 \\
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u_{\max} = 335.5 \frac{kJ}{kg} \\
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\end{align*}
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\begin{align*}
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U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\
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\end{align*}
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\section*{Question 5}
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\subsection*{Part a}
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\begin{align*}
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Q &= \int_{273}^{373} C_p \cdot dT \\
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t &= T - 273 K \\
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\therefore t + 100 &= T - 173 \\
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Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\
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Q &= 238.32 J
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\end{align*}
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\subsection*{Part b}
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\begin{align*}
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Q &= \Delta E + \int p\cdot dV \\
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\Delta E &= Q - p(V_2 - V_1) \\
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\Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\
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\Delta E &= 197.79J
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\end{align*}
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\end{document}
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elements_of_mech/assignment-work-and-heat/assign3.tex
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elements_of_mech/assignment-work-and-heat/assign3.tex
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\documentclass{article}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\begin{document}
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\title{Assignment --- Second Law of Thermodynamics}
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\author{Ahmad Saalim Lone, 2019BCSE017}
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\date{}
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\maketitle
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\section*{Question 1}
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We know that the net power output is the difference between the heat input and the heat rejected (cyclic device).
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\begin{align*}
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W_{net,out} &= Q_H + Q_L \\
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W_{out} &= 90 - 55 = 35 MW
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\end{align*}
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The thermal efficiency is the ratio of net work output and the heat output.
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\[
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\eta_{thermal} = W_{out}\frac{W_{out}}{Q_H} = \frac{35}{90} = 0.3889
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\]
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\section*{Question 2}
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\begin{align*}
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\text{efficiency} = 1 - \frac{T_1}{T_2} &= \frac{\text{work input}}{\text{work output}} \;\;\text{(must)}\\
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1 - \frac{300}{1000} &\implies \frac{6}{1} \;\;\text{(claimed)} \\
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0.7 &> 0.6 \\
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\text{output} &> \text{claimed $\implies$ good}
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\end{align*}
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$\therefore$ we can agree to this claim.
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\section*{Question 3}
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\[
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\eta = \frac{\text{output}}{\text{input}} = \frac{0.65}{0.65 + 0.4} = 0.619 = 61.9%
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\]
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now,
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\begin{align*}
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\eta &= 1 - \frac{T_2}{T_1} \\
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T_1 &= \frac{T_2}{1 - \eta} \\
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T_1 &= 787.5 K
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\end{align*}
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$\therefore$, the temperature at which energy is absorbed is 787.5 K
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\section*{Question 4}
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\begin{align*}
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\eta &= 1 - \frac{T_1}{T_2} \\
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\end{align*}
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where,
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\begin{align*}
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\eta &= 0.6 \\
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T_2 &= 800 K \\
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T_1 &= T_{sink} \\
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T_{sink} &= (1 - \eta)T_2 \\
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&= 324K
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\end{align*}
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Also,
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\begin{align*}
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\eta &= \frac{Q_1 - Q_{\text{rejected}}}{Q_1} \\
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Q_{\text{rejected}} &= Q_1 (1 - \eta) \\
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&= 400(1 - 0.6) \\
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&= 160 kJ
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\end{align*}
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\section*{Question 5}
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Max COP will be achieved only in a Carnot refrigerator.
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\[
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{(COP_{\max})}_{R} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2} = \frac{-20 + 273}{57} = 4.438
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\]
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Minimum Power $\to$ Max COP\@. We know that $P = \frac{Q}{\eta}$.
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\[
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P = \frac{3 \times 57}{253} = 0.657 W
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\]
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\section*{Question 6}
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Max COP will be achieved only in a Carnot refrigerator.
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\begin{align*}
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{(COP)}_{\text{Carnot refrigerator}} &= \frac{T_{low}}{T_{high} - T_{low}} = 1.37 \times 10^{-2} \\
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{(COP)}_{\text{refrigerator}} &= \frac{Q_L}{W} = 1.37 \times 10^{-2} \\
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Q_L &= 83.3 \\
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\end{align*}
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\section*{Question 7}
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For process $1-2$
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\begin{align*}
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Q_{1-2} &= U_2 - U_1 + W_{1-2} \\
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-732 &= U_2 - U_1 - 2.8 \times 3600 \\
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U_2 - U_1 &= 9348kJ
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\end{align*}
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For process $2-1$
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\begin{align*}
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Q_{2-1} &= U_1 - U_2 + W_{2-1} \\
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-732 &= -9348 - 2.8 \times 3600 \\
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Q_{2-1} &= -708 kJ
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\end{align*}
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Now,
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\[
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Q_{2-1} = U_1 - U_2 + W_{2-1}
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\]
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For maximum work, $Q_{2-1} = 0$.
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\[
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\therefore {(W_{2-1})}_{\max} = U_2 - U_1 = 9348kJ
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\]
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\section*{Question 8}
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\begin{align*}
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{(COP)}_R &= \frac{268}{298 - 268} = 8.933 \\
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{(COP)}_R &= \frac{5}{W} \\
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W &= 0.56KW
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\end{align*}
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\section*{Question 9}
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\begin{align*}
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\eta &= 1 - \frac{273+60}{273+671} = 0.64725 \\
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\eta &= 0.3236 \\
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\implies 1 - 0.3236 &= 0.6764 \\
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\text{Ideal COP} &= \frac{305.2}{305.2 - 266.4} = 7.866 \\
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\text{Actual COP} &= 3.923 = \frac{Q_3}{W} \\
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if \;Q_3 &= 1kJ \\
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\therefore W &= \frac{Q_3}{3.923} = 0.2549kJ \\
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W &= Q_{\text{output}} - Q_{\text{input}} = \frac{1}{3.923} Q_1 - 0.6764Q_1 = 0.2549 \\
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Q_{\text{out}} &= \frac{0.2549}{1 - 0.6764} = 0.7877 kJ \\
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\end{align*}
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\section*{Question 10}
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\begin{align*}
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\frac{10}{W} &= \frac{293}{293-273} \\
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or\;W &= 683 W
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\end{align*}
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\end{document}
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