Add elements of mechanical engineering assignment.

mech_engg/assignment-work-and-heat/assign.tex

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+\documentclass{article}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\begin{document}
+\title{Assignment --- Work \& Heat}
+\author{Ahmad Saalim Lone, 2019BCSE017}
+\date{07 May, 2020}
+\maketitle
+\section*{Question 1}
+The piston of an oil engine of area $0.0045 m^{3}$ moves downward $75 mm$,
+drawing in $0.00028 m^3$ of fresh air from the atmosphere. The pressure in the
+cylinder is uniform during the process at $80 kPa$, while atmospheric pressure
+is $101.325 kPa$, the difference being due to the flow resistance in the
+induction pipe and the inlet valve. Estimate the displacement work done by the
+air finally in the cylinder.
+
+\subsection*{Solution}
+\begin{align*}
+	\text{Area of the Piston} &= 0.0045m^2 \\
+	\text{displacement} &= 75 mm = 0.075 m \\
+	\text{Volume covered} = \Delta V &= 0.075 \times 0.0045 m^3 \\
+	\Delta V &= 0.0003375 m^3 \\
+	\text{pressure in the cylinder is constant} &= 80 kPa \\
+	\text{Work done} = p \Delta V &= 80 kPa \times 0.0003375 m^3 = 27 J
+\end{align*}
+
+\section*{Question 2}
+
+A mass of gas is compressed in a quasi-static process from $80 kPa, 0.1m^3$ to
+$0.4 MPa, 0.03 m^3$. Assuming that the pressure and volume are related by
+$pv^n = constant$, find the work done by the gas system.
+
+\subsection*{Solution}
+
+Quasi-static process is a thermodynamic process that happens slowly enough for
+the system to remain in internal equilibrium.
+
+Given
+\begin{align*}
+	P_1 &= 80 kPa \\
+	P_2 &= 0.4 MPa = 400 kPa \\
+	V_1 &= 0.1 m^3 \\
+	V_2 &= 0.03 m^3 \\
+\end{align*}
+
+Since, $PV^n = $ constant, the compression remains constant. So,
+
+\begin{align*}
+	P_1 V_1^n &= P_2 V_2^n \\
+	n &= \frac{\ln{\left(\frac{P_2}{P_1}\right)}}{\ln{\left(\frac{V_1}{V_2}\right)}} \\
+	n &= \frac{\ln{\left(\frac{400}{80}\right)}}{\ln{\left(\frac{0.1}{0.03}\right)}} \\
+	n &= \frac{\ln{5}}{\ln{3.\overline{3}}} \\
+	n &= \frac{\ln{5}}{\ln{3.\overline{3}}} \\
+	n &= 1.34
+\end{align*}
+
+\begin{align*}
+	\text{Work Done} &= \frac{P_1 V_1 - P_2 V_2}{n - 1} \\
+	&= \frac{80 \times 0.1 - 400 \times 0.03}{1.34 - 1} \\
+	&= - 11.76 J
+\end{align*}
+
+\section*{Question 3}
+
+A system of volume $V$ contains a mass $m$ of gas at pressure $p$ and
+temperature $T$. The macroscopic properties of the system obey the following
+relationship
+
+\[\left(P + \frac{a}{V^2}\right)\left(V - b \right) = mRT\]
+
+Where $a$, $b$ and $R$ are constants. Obtain an expression for the displacement
+work done by the system during a constant temperature expansion from volume
+$V_1$ to volume $V_2$. Calculate the work done by a system which contains $10
+kg$ of this gas expanding from $1m^3$ to $10 m^3$ at a temperature of $293 K$.
+Use the values $a= 15.7 \times 10 Nm^4$, $b= 1.07 \times 10-2 m^3$ and $R= 0.278 KkJ/kg$.
+
+\subsection*{Solution}
+
+\[\text{Macroscopic Properties} = \left(P + \frac{a}{V^2}\right)\left(V - b \right) = mRT\]
+
+For constant temperature expansion, as $m$, $R$ \& $T$ are constants, we can
+deduce,
+
+\[ mRT = \left(P_1 + \frac{a}{V_1^2}\right)\left(V_1 - b\right) = \left(P_2 + \frac{a}{V_2^2}\right)\left(V_2 - b\right)  = k\]
+
+\[ \therefore k = mRT = 10 \times 0.278 \times 293 KJ = 814.54 KJ\]
+
+We can find out $P_1$ and $P_2$ using the above equation
+
+\begin{gather*}
+	\left(P_1 + \frac{a}{V_1^2}\right)\left(V_1 - b\right) = k \\
+	P_1 + \frac{a}{V_1^2} = \frac{k}{V_1 - b} \\
+	P_1  = \frac{k}{V_1 - b} -  \frac{a}{V_1^2}\\
+	P_1  = 666.35 KPa
+\end{gather*}
+
+Similarly, $P_2 = 80 KPa$
+
+We know, the equation for work is
+\begin{align*}
+	W &= \int_{V_1}^{V_2}P dV \\
+	W &= \int_{V_1}^{V_2}\left(\frac{k}{V -b} - \frac{a}{V^2}\right) dV \\
+	W &= \left(P_1 + \frac{a}{V_1^2}\right)(V_1 - b) \ln\left(\frac{V_2 - b}{V_1 -b}\right) + a\left(\frac{1}{V_2} - \frac{1}{V_1}\right) \\
+	\text{Substituting values, we get} \\
+	W &= 1742 KJ
+\end{align*}
+
+
+\section*{Question 4}
+
+If a gas of volume $6000 cm^3$ and at pressure of $100 kPa$ is compressed
+quasi-statically according to $pV^2 = constant$ until the volume becomes $2000
+cm^3$, determine the final pressure and the work transfer.
+
+\subsection*{Solution}
+
+\begin{align*}
+	V_1 &= 6000 cm^3 = 0.006 m^3 \\
+	V_2 &= 2000 cm^3 = 0.002 m^3 \\
+	P_1 &= 100 KPa
+\end{align*}
+
+We know that
+\begin{align*}
+	P_1 V_1^2 &= P_2 V_2^2  \\
+	\therefore P_2 &= 900 KPa
+\end{align*}
+
+\begin{align*}
+	\text{Work Done} = W &= \frac{P_2 V_2 - P_1 V_1}{n - 1} \\
+						 &= \frac{900 \times 0.002 - 100 \times 0.006}{2 - 1} \\
+						 &= 1.2 KJ
+\end{align*}
+
+\section*{Question 5}
+
+A mass of $1.5kg$ of air is compressed in a quasi-static process from $0.1 MPa$
+to $0.7 MPa$ for which $pv= Constant$. The initial density of air is $1.16
+kg/m3$. Find the work done by the piston to compress the air.
+
+\subsection*{Solution}
+
+\begin{align*}
+	PV &= \text{constant} \\
+	\therefore P_1 V_1 &= P_2 V_2 \\
+	0.1 MPa \times V_1 &= 0.7 MPa \times \frac{d_1}{m_1} \\
+	0.1 MPa \times V_1 &= 0.7 MPa \times \frac{1.16 kg/m^3}{1.5 kg} \\
+	V_1 &= 1.293 m^3
+\end{align*}
+
+\begin{align*}
+	\text{Work Done} &= \int P dV \\
+					 &= \int_{V_1}^{V_2} \frac{P_1 V_1}{V} dV \\
+					 &= P_1 V_1 \int_{V_1}^{V_2} \frac{dV}{V} \\
+					 &= 251.63 KJ
+\end{align*}
+
+\end{document}