84 lines
2.3 KiB
TeX
84 lines
2.3 KiB
TeX
\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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