add uncommitted year 1 content

uni/mmme/1xxx/1026_maths_for_engineering/eigenvalues.md

@@ -0,0 +1,64 @@
+---
+author: Akbar Rahman
+date: \today
+title: Eigenvalues
+tags: [ mmme1026, maths, eigenvalues, uni ]
+uuid: f2220395-bc97-432e-a1d2-74085f16991d
+---
+
+An eigenvalue problem takes the form:
+
+Find all the values of $\lambda$ for which the equation
+
+$$A\pmb{x} = \lambda \pmb{x}$$
+
+has a nonzero solution $\pmb x$, where $A$ is an $n\times n$ matrix and
+$\pmb x$ is a column vector.
+
+The equation may be written as
+
+\begin{align*}
+A\pmb x &= \lambda I \pmb x \\
+\Leftrightarrow A \pmb x - \lambda I \pmb x & = 0 \\
+\Leftrightarrow (A-\lambda I)\pmb x &= 0
+\end{align*}
+
+($\Leftrightarrow$ means "if and only if")
+
+Non-zero solutions will exist if 
+
+$\det(A-\lambda I) = 0$
+
+There are infinitely many eigenvectors for a given eigenvalue.
+This is because if $\pmb x$ is an eigenvector of $A$ corresponding to the
+eigenvalue $\lambda$ and $c$ is a non-zero scalar, then $c\pmb x$ is also
+an eigenvector of $A$:
+
+$$A(c\pmb x) = cA\pmb x = c\lambda \pmb x = \lambda(c\pmb x)$$
+
+In general, if $A$ is an $n\times n$ matrix, then $|A-\lambda I|$ is a
+polynomial of degree $n$ in $\lambda$, called the characteristic polynomial.
+The characteristic equation is:
+
+$$\lambda^n + c_{n-1}\lambda^{n-1} + c_{n-2}\lambda^{n-2} + \cdots + c_0 = 0$$
+
+<details>
+<summary>
+
+#### Example 1 ($2\times2$ example)
+
+</summary>
+
+If $A$ is the matrix
+
+$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
+
+then
+
+$$|A - \lambda I| = \lambda^2 - (a+d)\lambda + (ad-bc)$$
+
+And the standard method for solving a quadratic can be used to find $\lambda$.
+
+</details>
+
+

uni/mmme/1xxx/1028_statics_and_dynamics/dynamics.md

@@ -11,3 +11,62 @@ uuid: e6d3a307-b2e6-40e3-83bb-ef73512d69ad
 $$a_c = r\omega^2$$
 
 $$a = r\alpha \hat{e}_\theta - r\omega^2\hat{e}_r$$
+
+## Moment of Inertia
+
+$$J = mr^2 = \frac{M}{\ddot\theta}$$
+
+The unit of $J$ is kgm$^2$.
+
+Consider a particle of mass $m$ attached to one end of a rigid rod of length $r$.
+The rod is pivoting at its other end about point $O$, and experiences a torque $M$.
+This torque will cause the mass and the rod to rotate about $O$ with angular velocity
+$\dot{\theta}$ an angular acceleration $\ddot{\theta}$.
+
+![](./images/vimscrot-2022-03-10T14:40:59,716300890+00:00.png)
+
+What is the expression for $M$?
+
+Well if break down the moment $M$ into a force, $F$, acting on the mass, we know that the
+moment $M = Fr$.  
+We know $F = ma$, and this case $a = r\ddot{\theta}$ so $M = mr^2\ddot\theta$.
+
+The moment of inertia is $J = mr^2$ so $M = J\ddot\theta$.
+
+If multiple torques are applied to a body the *rotational equation* of the motion is
+
+$$\overrightarrow{M} = \sum_i \overrightarrow{M}_i = J \overrightarrow{\ddot\theta} = J \overrightarrow{\alpha}$$
+
+The moment of inertia of any object is found by considering the object to be made up of lots of
+small particles and adding the moments of inertia for each small particle.
+The moments of inertia for a body depends on the mass and its distribution about the axis in
+consideration.
+
+$$J = \sum_i m_ir^2_i \rightarrow \int\! r^2 \mathrm{d}m$$
+
+### Perpendicular Axis Rule
+
+The perpendicular axis rule states that, for lamina object:
+
+$$J_z = J_x + J_y$$
+
+where $J_x$, $J_y$, and $J_z$ are the moments of inertia along their respective axes.
+
+### Parallel Axes Rule (Huygens-Steiner Theorem)
+
+The parallel axes rule states that:
+
+$$J_A = J_G = md^2$$
+
+where $d$ is the perpendicular distance between the two axes.
+
+![](./images/vimscrot-2022-03-10T15:06:48,355133323+00:00.png)
+
+### Moment of a Compound Object
+
+The moment of inertia for any compound object can be calculated by adding and subtracting the
+moments of inertia for its 'standard' components.
+
+### Moment of Inertia of Standard Objects
+
+// TODO

uni/mmme/1xxx/1048_thermodynamics_and_fluid_mechanics/thermodynamics.md

@@ -547,3 +547,90 @@ $$Q = m (c_v-c_v)(T_2-T_1) = 0 $$
 
 This proves that the isentropic version of the process adiabatic (no heat is transferred across the
 boundary).
+
+# 2nd Law of Thermodynamics
+
+The 2nd Law recognises that processes happen in a certain direction.
+It was discovered through the study of heat engines (ones that produce mechanical work from heat).
+
+> Heat does not spontaneously flow from a cooler to a hotter body.
+
+~ Clausius' Statement on the 2nd Law of Thermodynamics
+
+> It is impossible to construct a heat engine that will operate in a cycle and take heat from a
+> reservoir and produce an equivalent amount of work.
+
+~ Kelvin-Planck Statement of 2nd Law of Thermodynamics
+
+
+## Heat Engines
+
+A heat engine must have:
+
+- Thermal energy reservoir --- a large body of heat that does not change in temperature
+- Heat source --- a reservoir that supplies heat to the engine
+- Heat sink --- a reservoir that absorbs heat rejected from a heat engine (this is usually
+  surrounding environment)
+
+![](./images/vimscrot-2022-03-22T09:17:36,214723827+00:00.png)
+
+#### Steam Power Plant
+
+![](./images/vimscrot-2022-03-22T09:19:07,697440371+00:00.png)
+
+## Thermal Efficiency
+
+For heat engines, $Q_{out} > 0$ so $W_{out} < Q_{in}$ as $W_{out} = Q_{in} - Q_{out}$
+
+$$\eta = \frac{W_{out}}{Q_{in}} = 1 - \frac{Q_{out}}{Q_{in}}$$
+
+Early steam engines had efficiency around 10% but large diesel engines nowadays have efficiencies
+up to around 50%, with petrol engines around 30%.
+The most efficient heat engines we have are large gas-steam power plants, at around 60%.
+
+## Carnot Efficiency
+
+The maximum efficiency for a heat engine that operates reversibly between the heat source and heat
+sink is known as the *Carnot Efficiency*:
+
+$$\eta_{carnot} = 1 - \frac{T_2}{T_1}$$
+
+where $T$ is in Kelvin (or any unit of absolute temperature, I suppose)
+
+Therefore to maximise potential efficiency, you want to maximise input heat temperature, and
+minimise output heat temperature.
+
+The efficiency of any heat engine will be less than $\eta_{carnot}$ if it operates between more than
+two reservoirs.
+
+## Reversible and Irreversible Processes
+
+### Reversible Processes
+
+A reversible process operate at thermal and physical equilibrium.
+There is no degradation in the quality of energy.
+
+There must be no mechanical friction, fluid friction, or electrical resistance.
+
+Heat transfers must be across a very small temperature difference.
+
+All expansions must be controlled.
+
+### Irreversible Processes
+
+In irreversible processes, the quality of the energy degrades.
+For example, mechanical energy degrades into heat by friction and heat energy degrades into lower
+quality heat (a lower temperature), including by mixing of fluids.
+
+Thermal resistance at both hot sources and cold sinks are an irreversibility and reduce efficiency.
+
+There may also be uncontrolled expansions or sudden changes in pressure.
+
+# Energy Quality
+
+## Quantifying Disorder (Entropy)
+
+$$S = k\log_eW$$
+
+where $S$ is entropy, $k = 1.38\times10^{-23}$ J/K is Boltzmann's constant, and $W$ is the number of
+ways of reorganising energy.s