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% Most of the following are not required for the majority
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% Document Info
\author{Bama}
\pdfinfo{
  /Title (thermodynamics-final.pdf)
  /Creator (Cheatography)
  /Author (Bama)
  /Subject (Thermodynamics final Cheat Sheet)
}

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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Thermodynamics final Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Bama} via \textcolor{DarkBackground}{\uline{cheatography.com/66428/cs/16580/}}}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Bama \\
  \uline{cheatography.com/bama} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 7th August, 2018.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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% Set font size to small. Switch to any value
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% www.emerson.emory.edu/services/latex/latex_169.html
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Air Standard Otto Cycle}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Process 1–2: An isentropic compression of the air as the piston moves from bottom dead center to top dead center. \newline % Row Count 3 (+ 3)
Process 2–3: A constant-volume heat transfer to the air from an external source while the piston is at top dead center. This process is intended to represent the ignition of the fuel–air mixture and the subsequent rapid burning. \newline % Row Count 8 (+ 5)
Process 3–4: An isentropic expansion (power stroke). \newline % Row Count 10 (+ 2)
Process 4–1: Completes the cycle by a constant-volume process in which heat is rejected from the air while the piston is at bottom dead center.% Row Count 13 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Air Standard Otto Cycle Diagram and Schematic}}  \tn
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\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/bama_1533600852_otto cycle.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Second Law}}  \tn
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\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{3.6 cm} x{4.4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Efficiencies}}  \tn
% Row 0
\SetRowColor{LightBackground}
Power = 1 - Qᴄ / Qʜ & Turbine = Wᴀᴄᴛᴜᴀʟ / Wɪᴅᴇᴀʟ \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
Refrigeration = Qᴄ / Qʜ - Qᴄ & Comp or Pump = Wɪᴅᴇᴀʟ / Wᴀᴄᴛᴜᴀʟ \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Heat = Qʜ / Qʜ - Qᴄ & Regenerator = Qᴀᴄᴛᴜᴀʟ /  Qɪᴅᴇᴀʟ \tn 
% Row Count 9 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{1.8 cm} x{1.44 cm} x{1.944 cm} x{2.016 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{4}{x{8.4cm}}{\bf\textcolor{white}{Ideal Gases}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\emph{pv}} = RT & {\emph{p}}V = {\emph{m}}RT & C{\emph{ᴘ}} = C{\emph{ᴠ}} + R & Polytropic Process: \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
k = C{\emph{ᴘ}} / C{\emph{v}} & Pʀ = P / Pᴄ & Tʀ = T / Tᴄ & {\emph{pv}}\textasciicircum{}n\textasciicircum{} = constant \tn 
% Row Count 4 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}----}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{2.128 cm} x{3.04 cm} x{2.432 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{8.4cm}}{\bf\textcolor{white}{Exact Analysis of Isentropic Process for Ideal Air}}  \tn
% Row 0
\SetRowColor{LightBackground}
Pʀ2 / Pr1 = P1 / P2 & {\emph{V}}ʀ2 / {\emph{V}}ʀ1 = {\emph{V}}2 / {\emph{V}}1 & Pressure ratio = P2/ P1 \tn 
% Row Count 2 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Exact Analysis of Entropy Change in Ideal Gasses}}  \tn
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Air Standard Diesel Cycle}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{The air-standard Diesel cycle is an ideal cycle that assumes heat addition occurs during a constant-pressure process that starts with the piston at top dead center. The cycle consists of four internally reversible processes in series. \newline % Row Count 5 (+ 5)
{\bf{Process 1-2}} is the same as in the Otto cycle: an isentropic compression. Heat is not transferred to the working fluid at constant volume as in the Otto cycle, however. {\bf{In the Diesel cycle, heat is transferred to the working fluid at constant pressure}}. \newline % Row Count 11 (+ 6)
{\bf{Process 2–3}} also makes up the first part of the power stroke. \newline % Row Count 13 (+ 2)
{\bf{Process 3-4}} is an isentropic expansion and is the remainder of the power stroke. \newline % Row Count 15 (+ 2)
{\bf{Process 4-1}} As in the Otto cycle, the cycle is completed by a constant-volume process in which heat is rejected from the air while the piston is at bottom dead center. This process replaces the exhaust and intake processes of the actual engine.% Row Count 20 (+ 5)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Air Standard Diesel Cycle Diagram and Schematic}}  \tn
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\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/bama_1533602008_imgonline-com-ua-twotoone-jILcjdt3CA7.jpg}}} \tn 
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Cycles}}  \tn
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Work}}  \tn
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\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/bama_1533594282_Work.jpg}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{Miscellaneous}}  \tn
% Row 0
\SetRowColor{LightBackground}
Tds = du + pdv & Tds = du + vdp \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
x = {\emph{M}}ᴠᴀᴘᴏʀ / {\emph{M}}ᴛᴏᴛᴀʟ & S = Sғ + (x)Sғɢ = Sғ(1 -x) + Sɢ \tn 
% Row Count 4 (+ 3)
% Row 2
\SetRowColor{LightBackground}
Expansion valve: Δh = 0 & Condensor: Qᴏᴜᴛ \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
Evaporator: Qɪɴ & First law: Eɪɴ - Eᴏᴜᴛ = ΔEsʏsᴛᴇᴍ \tn 
% Row Count 9 (+ 3)
% Row 4
\SetRowColor{LightBackground}
h = u +pv & W = VI (electrical) \tn 
% Row Count 10 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Specific Heat}}  \tn
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\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}