From 7f7a385591e37bd95329f5e8a5e2417dd8b5da9c Mon Sep 17 00:00:00 2001 From: Ceda EI Date: Tue, 18 Aug 2020 17:57:04 +0530 Subject: [PATCH] Add assignment 2 and 3 --- .../assignment-work-and-heat/assign2.tex | 83 +++++++++++++ .../assignment-work-and-heat/assign3.tex | 117 ++++++++++++++++++ 2 files changed, 200 insertions(+) create mode 100644 elements_of_mech/assignment-work-and-heat/assign2.tex create mode 100644 elements_of_mech/assignment-work-and-heat/assign3.tex diff --git a/elements_of_mech/assignment-work-and-heat/assign2.tex b/elements_of_mech/assignment-work-and-heat/assign2.tex new file mode 100644 index 0000000..2412aff --- /dev/null +++ b/elements_of_mech/assignment-work-and-heat/assign2.tex @@ -0,0 +1,83 @@ +\documentclass{article} +\usepackage{amsmath} +\usepackage{amssymb} +\begin{document} +\title{Assignment --- First Law of Thermodynamics} +\author{Ahmad Saalim Lone, 2019BCSE017} +\date{} +\maketitle +\section*{Question 1} +First law of thermodynamics suggests that $\sum Q = \sum W$. +\begin{align*} + Q_1 + Q_2 + Q_3 &= W_1 + W_2 \\ + 75 - 40 + Q_3 &= -15 + 44 \\ + Q_3 &= -6 kJ +\end{align*} +i.e.\ from the system to the surroundings. +\section*{Question 2} +We know that, +\begin{align*} + Q &= \Delta E + W \\ + Q &= -50000 + \frac{-1000 \times 3600}{1000} kJ \\ + Q &= -8600 kJ \\ + Q &= -8.6 MJ +\end{align*} +\section*{Question 3} +There is no heat transfer, therefore +\begin{align*} + Q &= \Delta E + W \\ + W &= - \Delta E - \Delta V \\ + W &= \int_{1}^{0} C_v \cdot dT \\ + W &= -0.718 (T_2 - T_1) \\ + W &= -50.26 \frac{kJ}{kg} \\ + \text{Total Work} &= 2 \times (-50.26) = -100 kJ +\end{align*} +\section*{Question 4} +\begin{align*} + 1000 &= a + b \times 0.2 \\ + 200 &= a + b \times 1.2 \\ + b &= -800 \\ + a &= 1000 + 2 \times 800 = 1160 \\ + \therefore P &= 1160 - 800 V \\ + W &= \int_{V_1}^{V_2} P \cdot dV \\ + &= \int_{0.2}^{1.2} (1160 - 800V)dV \\ + &= 6000 kJ \\ + V_1 &= 1.5 \times 1000 \times \frac{0.2}{1.5} - 85 \\ + &= 215 \frac{kJ}{kg} \\ + V_2 &= 1.5 \times 200 \times \frac{1.2}{1.5} - 85 \\ + &= 155 \frac{kJ}{kg} \\ + \Delta V &= V_2 - V_1 \\ + &= 40 \frac{kJ}{kg} \\ + \Delta U &= m \Delta V = 60 kJ \\ + Q &= \Delta U + W \\ + &= 60 + 600 \\ + &= 660 kJ \\ + U &= 1.5 PV - 85 \frac{kJ}{kg} \\ + U &= 1.5 \left(\frac{1160 - 800 V}{1.5}\right)V - 85 \frac{kJ}{kg} \\ + &= 800 V^2 - 85 \frac{kJ}{kg} \\ + \frac{\delta U}{\delta V} &= 1160 - 1600 V \\ + \text{For maximum V,} \\ + V_1 \rightarrow \frac{\delta U}{\delta V} &= 0 \\ + V &= 0.725 \\ + u_{\max} = 335.5 \frac{kJ}{kg} \\ +\end{align*} +\begin{align*} + U_{\max} &= 1.5 \times u_{\max} = 503.25 kJ \\ +\end{align*} +\section*{Question 5} +\subsection*{Part a} +\begin{align*} + Q &= \int_{273}^{373} C_p \cdot dT \\ + t &= T - 273 K \\ + \therefore t + 100 &= T - 173 \\ + Q &= \int_{273}^{373} \left(2.093 + \frac{41.87}{T - 173}\right) \cdot dT \\ + Q &= 238.32 J +\end{align*} +\subsection*{Part b} +\begin{align*} + Q &= \Delta E + \int p\cdot dV \\ + \Delta E &= Q - p(V_2 - V_1) \\ + \Delta E &= 238.32 - 101.325 (0.0024 - 0.002) \times 1000 J \\ + \Delta E &= 197.79J +\end{align*} +\end{document} diff --git a/elements_of_mech/assignment-work-and-heat/assign3.tex b/elements_of_mech/assignment-work-and-heat/assign3.tex new file mode 100644 index 0000000..1d6a904 --- /dev/null +++ b/elements_of_mech/assignment-work-and-heat/assign3.tex @@ -0,0 +1,117 @@ +\documentclass{article} +\usepackage{amsmath} +\usepackage{amssymb} +\begin{document} +\title{Assignment --- Second Law of Thermodynamics} +\author{Ahmad Saalim Lone, 2019BCSE017} +\date{} +\maketitle +\section*{Question 1} +We know that the net power output is the difference between the heat input and the heat rejected (cyclic device). +\begin{align*} + W_{net,out} &= Q_H + Q_L \\ + W_{out} &= 90 - 55 = 35 MW +\end{align*} +The thermal efficiency is the ratio of net work output and the heat output. +\[ + \eta_{thermal} = W_{out}\frac{W_{out}}{Q_H} = \frac{35}{90} = 0.3889 +\] +\section*{Question 2} +\begin{align*} + \text{efficiency} = 1 - \frac{T_1}{T_2} &= \frac{\text{work input}}{\text{work output}} \;\;\text{(must)}\\ + 1 - \frac{300}{1000} &\implies \frac{6}{1} \;\;\text{(claimed)} \\ + 0.7 &> 0.6 \\ + \text{output} &> \text{claimed $\implies$ good} +\end{align*} +$\therefore$ we can agree to this claim. +\section*{Question 3} +\[ + \eta = \frac{\text{output}}{\text{input}} = \frac{0.65}{0.65 + 0.4} = 0.619 = 61.9% +\] +now, +\begin{align*} + \eta &= 1 - \frac{T_2}{T_1} \\ + T_1 &= \frac{T_2}{1 - \eta} \\ + T_1 &= 787.5 K +\end{align*} +$\therefore$, the temperature at which energy is absorbed is 787.5 K +\section*{Question 4} +\begin{align*} + \eta &= 1 - \frac{T_1}{T_2} \\ +\end{align*} +where, +\begin{align*} + \eta &= 0.6 \\ + T_2 &= 800 K \\ + T_1 &= T_{sink} \\ + T_{sink} &= (1 - \eta)T_2 \\ + &= 324K +\end{align*} +Also, +\begin{align*} + \eta &= \frac{Q_1 - Q_{\text{rejected}}}{Q_1} \\ + Q_{\text{rejected}} &= Q_1 (1 - \eta) \\ + &= 400(1 - 0.6) \\ + &= 160 kJ +\end{align*} +\section*{Question 5} +Max COP will be achieved only in a Carnot refrigerator. +\[ + {(COP_{\max})}_{R} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2} = \frac{-20 + 273}{57} = 4.438 +\] +Minimum Power $\to$ Max COP\@. We know that $P = \frac{Q}{\eta}$. +\[ + P = \frac{3 \times 57}{253} = 0.657 W +\] +\section*{Question 6} +Max COP will be achieved only in a Carnot refrigerator. +\begin{align*} + {(COP)}_{\text{Carnot refrigerator}} &= \frac{T_{low}}{T_{high} - T_{low}} = 1.37 \times 10^{-2} \\ + {(COP)}_{\text{refrigerator}} &= \frac{Q_L}{W} = 1.37 \times 10^{-2} \\ + Q_L &= 83.3 \\ +\end{align*} +\section*{Question 7} +For process $1-2$ +\begin{align*} + Q_{1-2} &= U_2 - U_1 + W_{1-2} \\ + -732 &= U_2 - U_1 - 2.8 \times 3600 \\ + U_2 - U_1 &= 9348kJ +\end{align*} +For process $2-1$ +\begin{align*} + Q_{2-1} &= U_1 - U_2 + W_{2-1} \\ + -732 &= -9348 - 2.8 \times 3600 \\ + Q_{2-1} &= -708 kJ +\end{align*} +Now, +\[ + Q_{2-1} = U_1 - U_2 + W_{2-1} +\] +For maximum work, $Q_{2-1} = 0$. +\[ + \therefore {(W_{2-1})}_{\max} = U_2 - U_1 = 9348kJ +\] +\section*{Question 8} +\begin{align*} + {(COP)}_R &= \frac{268}{298 - 268} = 8.933 \\ + {(COP)}_R &= \frac{5}{W} \\ + W &= 0.56KW +\end{align*} +\section*{Question 9} +\begin{align*} + \eta &= 1 - \frac{273+60}{273+671} = 0.64725 \\ + \eta &= 0.3236 \\ + \implies 1 - 0.3236 &= 0.6764 \\ + \text{Ideal COP} &= \frac{305.2}{305.2 - 266.4} = 7.866 \\ + \text{Actual COP} &= 3.923 = \frac{Q_3}{W} \\ + if \;Q_3 &= 1kJ \\ + \therefore W &= \frac{Q_3}{3.923} = 0.2549kJ \\ + W &= Q_{\text{output}} - Q_{\text{input}} = \frac{1}{3.923} Q_1 - 0.6764Q_1 = 0.2549 \\ + Q_{\text{out}} &= \frac{0.2549}{1 - 0.6764} = 0.7877 kJ \\ +\end{align*} +\section*{Question 10} +\begin{align*} + \frac{10}{W} &= \frac{293}{293-273} \\ + or\;W &= 683 W +\end{align*} +\end{document}