From 10f57a032386095b653a76cd1f6586a45229298b Mon Sep 17 00:00:00 2001 From: Ceda EI Date: Thu, 7 May 2020 19:59:02 +0530 Subject: [PATCH] Add elements of mechanical engineering assignment. --- mech_engg/assignment-work-and-heat/assign.tex | 158 ++++++++++++++++++ 1 file changed, 158 insertions(+) create mode 100644 mech_engg/assignment-work-and-heat/assign.tex diff --git a/mech_engg/assignment-work-and-heat/assign.tex b/mech_engg/assignment-work-and-heat/assign.tex new file mode 100644 index 0000000..dd31388 --- /dev/null +++ b/mech_engg/assignment-work-and-heat/assign.tex @@ -0,0 +1,158 @@ +\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}