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